# Articles - Principal Component Methods in R: Practical Guide

## Correspondence Analysis: Theory and Practice

This article presents the theory and the mathematical procedures behind correspondence Analysis. We write all the formula in a very simple format so that beginners can understand the methods.

Contents:

## Required packages

• FactoMineR for computing CA
• factoextra for visualizing the results

Install:

install.packages("FactoMineR")
install.packages("factoextra")

Load:

library("FactoMineR")
library("factoextra")

## Data format

• Demo data: housetasks [in factoextra package]
• Data contents: housetasks repartition in the couple.
• rows are the different tasks
• values are the frequencies of the tasks done by the i) wife only, ii) alternatively, iii) husband only, iv) jointly.
• Data illustration:

Load the data:

library(factoextra)
data(housetasks)
# head(housetasks)

As the above contingency table is not very large, with a quick visual examination it can be seen that:

• The house tasks Laundry, Main_Meal and Dinner are dominant in the column Wife
• Repairs are dominant in the column Husband
• Holidays are dominant in the column Jointly

## Visualize a contingency table

To easily interpret the contingency table, a graphical matrix can be drawn using the function balloonplot() [gplots package]. In this graph, each cell contains a dot whose size reflects the relative magnitude of the value it contains.

library("gplots")
# 1. convert the data as a table
dt <- as.table(as.matrix(housetasks))
# 2. Graph
balloonplot(t(dt), main ="housetasks", xlab ="", ylab="",
label = FALSE, show.margins = FALSE)

For a very large contingency table, the visual interpretation would be very hard. Other methods are required such as correspondence analysis.

## Key terms

• Row margins: Row sums (row.sum)
• Column margins: Column sums (col.sum)
• grand total: Total sum of all values in the contingency table.
# Row margins
row.sum <- apply(housetasks, 1, sum)
head(row.sum)
# Column margins
col.sum <- apply(housetasks, 2, sum)
head(col.sum)
# grand total
n <- sum(housetasks)

The contingency table with row and column margins are shown below:

 Wife Alternating Husband Jointly TOTAL Laundry 156 14 2 4 176 Main_meal 124 20 5 4 153 Dinner 77 11 7 13 108 Breakfeast 82 36 15 7 140 Tidying 53 11 1 57 122 Dishes 32 24 4 53 113 Shopping 33 23 9 55 120 Official 12 46 23 15 96 Driving 10 51 75 3 139 Finances 13 13 21 66 113 Insurance 8 1 53 77 139 Repairs 0 3 160 2 165 Holidays 0 1 6 153 160 TOTAL 600 254 381 509 1744
• Row margins: light gray
• Column margins: light blue
• The grand total (the total of all values in the table): pink

## Row variables

To compare rows, we can analyse their profiles in order to identify similar row variables.

### Row profiles

The profile of a given row is calculated by taking each row point and dividing by its margin (i.e, the sum of all row points). The formula is:

$row.profile = \frac{row}{row.sum}$

For example the profile of the row point Laundry/wife is P = 156/176 = 88.6%.

The R code below can be used to compute row profiles:

row.profile <- housetasks/row.sum
# head(row.profile)
 Wife Alternating Husband Jointly TOTAL Laundry 0.8864 0.07955 0.0114 0.0227 1 Main_meal 0.8105 0.13072 0.0327 0.0261 1 Dinner 0.7130 0.10185 0.0648 0.1204 1 Breakfeast 0.5857 0.25714 0.1071 0.0500 1 Tidying 0.4344 0.09016 0.0082 0.4672 1 Dishes 0.2832 0.21239 0.0354 0.4690 1 Shopping 0.2750 0.19167 0.0750 0.4583 1 Official 0.1250 0.47917 0.2396 0.1562 1 Driving 0.0719 0.36691 0.5396 0.0216 1 Finances 0.1150 0.11504 0.1858 0.5841 1 Insurance 0.0576 0.00719 0.3813 0.5540 1 Repairs 0.0000 0.01818 0.9697 0.0121 1 Holidays 0.0000 0.00625 0.0375 0.9563 1 TOTAL 0.3440 0.14564 0.2185 0.2919 1

In the table above, the row TOTAL (in light blue) is called the average row profile (or marginal profile of columns or column margin)

The average row profile is computed as follow:

$average.rp = \frac{column.sum}{grand.total}$

For example, the average row profile is : (600/1744, 254/1744, 381/1744, 509/1744). It can be computed in R as follow:

# Column sums
col.sum <- apply(housetasks, 2, sum)
# average row profile = Column sums / grand total
average.rp <- col.sum/n
average.rp
##        Wife Alternating     Husband     Jointly
##       0.344       0.146       0.218       0.292

### Distance (or similarity) between row profiles

If we want to compare 2 rows (row1 and row2), we need to compute the squared distance between their profiles as follow:

$d^2(row_1, row_2) = \sum{\frac{(row.profile_1 - row.profile_2)^2}{average.profile}}$

This distance is called Chi-square distance

For example the distance between the rows Laundry and Main_meal are:

$d^2(Laundry, Main\_meal) = \frac{(0.886-0.810)^2}{0.344} + \frac{(0.0795-0.131)^2}{0.146} + ... = 0.036$

The distance between Laundry and Main_meal can be calculated as follow in R:

# Laundry and Main_meal profiles
laundry.p <- row.profile["Laundry",]
main_meal.p <- row.profile["Main_meal",]
# Distance between Laundry and Main_meal
d2 <- sum(((laundry.p - main_meal.p)^2) / average.rp)
d2
## [1] 0.0368

The distance between Laundry and Driving is:

# Driving profile
driving.p <- row.profile["Driving",]
# Distance between Laundry and Driving
d2 <- sum(((laundry.p - driving.p)^2) / average.rp)
d2
## [1] 3.77

Note that, the rows Laundry and Main_meal are very close (d2 ~ 0.036, similar profiles) compared to the rows Laundry and Driving (d2 ~ 3.77)

You can also compute the squared distance between each row profile and the average row profile in order to view rows that are the most similar or different to the average row.

### Squared distance between each row profile and the average row profile

$d^2(row_i, average.profile) = \sum{\frac{(row.profile_i - average.profile)^2}{average.profile}}$

The R code below computes the distance from the average profile for all the row variables:

d2.row <- apply(row.profile, 1,
function(row.p, av.p){sum(((row.p - av.p)^2)/av.p)},
average.rp)
as.matrix(round(d2.row,3))
##             [,1]
## Laundry    1.329
## Main_meal  1.034
## Dinner     0.618
## Breakfeast 0.512
## Tidying    0.353
## Dishes     0.302
## Shopping   0.218
## Official   0.968
## Driving    1.274
## Finances   0.456
## Insurance  0.727
## Repairs    3.307
## Holidays   2.140

The rows Repairs, Holidays, Laundry and Driving have the most different profiles from the average profile.

### Distance matrix

In this section the squared distance is computed between each row profile and the other rows in the contingency table.

The result is a distance matrix (a kind of correlation or dissimilarity matrix).

The custom R function below is used to compute the distance matrix:

## data: a data frame or matrix;
## average.profile: average profile
dist.matrix <- function(data, average.profile){
mat <- as.matrix(t(data))
n <- ncol(mat)
dist.mat<- matrix(NA, n, n)
diag(dist.mat) <- 0
for (i in 1:(n - 1)) {
for (j in (i + 1):n) {
d2 <- sum(((mat[, i] - mat[, j])^2) / average.profile)
dist.mat[i, j] <- dist.mat[j, i] <- d2
}
}
colnames(dist.mat) <- rownames(dist.mat) <- colnames(mat)
dist.mat
}

Compute and visualize the distance between row profiles. The package corrplot is required for the visualization. It can be installed as follow: install.packages("corrplot").

# Distance matrix
dist.mat <- dist.matrix(row.profile, average.rp)
dist.mat <-round(dist.mat, 2)
# Visualize the matrix
library("corrplot")
corrplot(dist.mat, type="upper",  is.corr = FALSE)

The size of the circle is proportional to the magnitude of the distance between row profiles.

When the data contains many categories, correspondence analysis is very useful to visualize the similarity between items.

### Row mass and inertia

The Row mass (or row weight) is the total frequency of a given row. It’s calculated as follow:

$row.mass = \frac{row.sum}{grand.total}$

row.sum <- apply(housetasks, 1, sum)
grand.total <- sum(housetasks)
row.mass <- row.sum/grand.total
head(row.mass)
##    Laundry  Main_meal     Dinner Breakfeast    Tidying     Dishes
##     0.1009     0.0877     0.0619     0.0803     0.0700     0.0648

The row inertia is calculated as the row mass multiplied by the squared distance between the row and the average row profile:

$row.inertia = row.mass * d^2(row)$

• The inertia of a row (or a column) is the amount of information it contains.
• The total inertia is the total information contained in the data table. It’s computed as the sum of rows inertia (or equivalently, as the sum of columns inertia)
# Row inertia
row.inertia <- row.mass * d2.row
head(row.inertia)
##    Laundry  Main_meal     Dinner Breakfeast    Tidying     Dishes
##     0.1342     0.0907     0.0382     0.0411     0.0247     0.0196
# Total inertia
sum(row.inertia)
## [1] 1.11

The total inertia corresponds to the amount of the information the data contains.

### Row summary

The result for rows can be summarized as follow:

row <- cbind.data.frame(d2 = d2.row, mass = row.mass, inertia = row.inertia)
round(row,3)
##               d2  mass inertia
## Laundry    1.329 0.101   0.134
## Main_meal  1.034 0.088   0.091
## Dinner     0.618 0.062   0.038
## Breakfeast 0.512 0.080   0.041
## Tidying    0.353 0.070   0.025
## Dishes     0.302 0.065   0.020
## Shopping   0.218 0.069   0.015
## Official   0.968 0.055   0.053
## Driving    1.274 0.080   0.102
## Finances   0.456 0.065   0.030
## Insurance  0.727 0.080   0.058
## Repairs    3.307 0.095   0.313
## Holidays   2.140 0.092   0.196

## Column variables

### Column profiles

These are calculated in the same way as the row profiles table.

The profile of a given column is computed as follow:

$col.profile = \frac{col}{col.sum}$

The R code below can be used to compute column profile:

col.profile <- t(housetasks)/col.sum
col.profile <- as.data.frame(t(col.profile))
# head(col.profile)
 Wife Alternating Husband Jointly TOTAL Laundry 0.2600 0.05512 0.00525 0.00786 0.1009 Main_meal 0.2067 0.07874 0.01312 0.00786 0.0877 Dinner 0.1283 0.04331 0.01837 0.02554 0.0619 Breakfeast 0.1367 0.14173 0.03937 0.01375 0.0803 Tidying 0.0883 0.04331 0.00262 0.11198 0.0700 Dishes 0.0533 0.09449 0.01050 0.10413 0.0648 Shopping 0.0550 0.09055 0.02362 0.10806 0.0688 Official 0.0200 0.18110 0.06037 0.02947 0.0550 Driving 0.0167 0.20079 0.19685 0.00589 0.0797 Finances 0.0217 0.05118 0.05512 0.12967 0.0648 Insurance 0.0133 0.00394 0.13911 0.15128 0.0797 Repairs 0.0000 0.01181 0.41995 0.00393 0.0946 Holidays 0.0000 0.00394 0.01575 0.30059 0.0917 TOTAL 1.0000 1.00000 1.00000 1.00000 1.0000

In the table above, the column TOTAL is called the average column profile (or marginale profile of rows)

The average column profile is calculated as follow:

$average.cp = row.sum/grand.total$

For example, the average column profile is : (176/1744, 153/1744, 108/1744, 140/1744, …). It can be computed in R as follow:

# Row sums
row.sum <- apply(housetasks, 1, sum)
# average column profile= row sums/grand total
average.cp <- row.sum/n
head(average.cp)
##    Laundry  Main_meal     Dinner Breakfeast    Tidying     Dishes
##     0.1009     0.0877     0.0619     0.0803     0.0700     0.0648

### Distance (similarity) between column profiles

If we want to compare columns, we need to compute the squared distance between their profiles as follow:

$d^2(col_1, col_2) = \sum{\frac{(col.profile_1 - col.profile_2)^2}{average.profile}}$

For example the distance between the columns Wife and Husband are:

$d^2(Wife, Husband) = \frac{(0.26-0.005)^2}{0.10} + \frac{(0.21-0.013)^2}{0.09} + ... + ... = 4.05$

The distance between Wife and Husband can be calculated as follow in R:

# Wife and Husband profiles
wife.p <- col.profile[, "Wife"]
husband.p <- col.profile[, "Husband"]
# Distance between Wife and Husband
d2 <- sum(((wife.p - husband.p)^2) / average.cp)
d2
## [1] 4.05

You can also compute the squared distance between each column profile and the average column profile

### Squared distance between each column profile and the average column profile

$d^2(col_i, average.profile) = \sum{\frac{(col.profile_i - average.profile)^2}{average.profile}}$

The R code below computes the distance from the average profile for all the column variables

d2.col <- apply(col.profile, 2,
function(col.p, av.p){sum(((col.p - av.p)^2)/av.p)},
average.cp)
round(d2.col,3)
##        Wife Alternating     Husband     Jointly
##       0.875       0.809       1.746       1.078

### Distance matrix

# Distance matrix
dist.mat <- dist.matrix(t(col.profile), average.cp)
dist.mat <-round(dist.mat, 2)
dist.mat
##             Wife Alternating Husband Jointly
## Wife        0.00        1.71    4.05    2.93
## Alternating 1.71        0.00    2.67    2.58
## Husband     4.05        2.67    0.00    3.70
## Jointly     2.93        2.58    3.70    0.00
# Visualize the matrix
library("corrplot")
corrplot(dist.mat, type="upper", order="hclust", is.corr = FALSE)

### column mass and inertia

The column mass(or column weight) is the total frequency of each column. It’s calculated as follow:

$col.mass = \frac{col.sum}{grand.total}$

col.sum <- apply(housetasks, 2, sum)
grand.total <- sum(housetasks)
col.mass <- col.sum/grand.total
head(col.mass)
##        Wife Alternating     Husband     Jointly
##       0.344       0.146       0.218       0.292

The column inertia is calculated as the column mass multiplied by the squared distance between the column and the average column profile:

$col.inertia = col.mass * d^2(col)$

col.inertia <- col.mass * d2.col
head(col.inertia)
##        Wife Alternating     Husband     Jointly
##       0.301       0.118       0.381       0.315
# total inertia
sum(col.inertia)
## [1] 1.11

Recall that the total inertia corresponds to the amount of the information the data contains. Note that, the total inertia obtained using column profile is the same as the one obtained when analyzing row profile. That’s normal, because we are analyzing the same data with just a different angle of view.

### Column summary

The result for rows can be summarized as follow:

col <- cbind.data.frame(d2 = d2.col, mass = col.mass,
inertia = col.inertia)
round(col,3)
##                d2  mass inertia
## Wife        0.875 0.344   0.301
## Alternating 0.809 0.146   0.118
## Husband     1.746 0.218   0.381
## Jointly     1.078 0.292   0.315

## Association between row and column variables

When the contingency table is not very large (as above), it’s easy to visually inspect and interpret row and column profiles:

• It’s evident that, the housetasks - Laundry, Main_Meal and Dinner - are more frequently done by the “Wife”.
• Repairs and driving are dominantly done by the husband
• Holidays are more frequently taken jointly

Larger contingency table is complex to interpret visually and several methods are required to help to this process.

Another statistical method that can be applied to contingency table is the Chi-square test of independence.

### Chi-square test

Chi-square test issued to examine whether rows and columns of a contingency table are statistically significantly associated.

• Null hypothesis (H0): the row and the column variables of the contingency table are independent.
• Alternative hypothesis (H1): row and column variables are dependent

For each cell of the table, we have to calculate the expected value under null hypothesis.

For a given cell, the expected value is calculated as follow:

$e = \frac{row.sum * col.sum}{grand.total}$

The Chi-square statistic is calculated as follow:

$\chi^2 = \sum{\frac{(o - e)^2}{e}}$

• o is the observed value
• e is the expected value

This calculated Chi-square statistic is compared to the critical value (obtained from statistical tables) with $$df = (r - 1)(c - 1)$$ degrees of freedom and p = 0.05.

• r is the number of rows in the contingency table
• c is the number of column in the contingency table

If the calculated Chi-square statistic is greater than the critical value, then we must conclude that the row and the column variables are not independent of each other. This implies that they are significantly associated.

Note that, Chi-square test should only be applied when the expected frequency of any cell is at least 5.

Chi-square statistic can be easily computed using the function chisq.test() as follow:

chisq <- chisq.test(housetasks)
chisq
##
##  Pearson's Chi-squared test
##
## data:  housetasks
## X-squared = 2000, df = 40, p-value <2e-16

In our example, the row and the column variables are statistically significantly associated (p-value = 0)

Note that, while Chi-square test can help to establish dependence between rows and the columns, the nature of the dependency is unknown.

The observed and the expected counts can be extracted from the result of the test as follow:

# Observed counts
chisq$observed ## Wife Alternating Husband Jointly ## Laundry 156 14 2 4 ## Main_meal 124 20 5 4 ## Dinner 77 11 7 13 ## Breakfeast 82 36 15 7 ## Tidying 53 11 1 57 ## Dishes 32 24 4 53 ## Shopping 33 23 9 55 ## Official 12 46 23 15 ## Driving 10 51 75 3 ## Finances 13 13 21 66 ## Insurance 8 1 53 77 ## Repairs 0 3 160 2 ## Holidays 0 1 6 153 # Expected counts round(chisq$expected,2)
##            Wife Alternating Husband Jointly
## Laundry    60.5        25.6    38.5    51.4
## Main_meal  52.6        22.3    33.4    44.6
## Dinner     37.2        15.7    23.6    31.5
## Breakfeast 48.2        20.4    30.6    40.9
## Tidying    42.0        17.8    26.6    35.6
## Dishes     38.9        16.5    24.7    33.0
## Shopping   41.3        17.5    26.2    35.0
## Official   33.0        14.0    21.0    28.0
## Driving    47.8        20.2    30.4    40.6
## Finances   38.9        16.5    24.7    33.0
## Insurance  47.8        20.2    30.4    40.6
## Repairs    56.8        24.0    36.0    48.2
## Holidays   55.0        23.3    35.0    46.7

As mentioned above the Chi-square statistic is 1944.456.

Which are the most contributing cells to the definition of the total Chi-square statistic?

If you want to know the most contributing cells to the total Chi-square score, you just have to calculate the Chi-square statistic for each cell:

$r = \frac{o - e}{\sqrt{e}}$

The above formula returns the so-called Pearson residuals (r) for each cell (or standardized residuals). Cells with the highest absolute standardized residuals contribute the most to the total Chi-square score.

Pearson residuals can be easily extracted from the output of the function chisq.test():

round(chisq$residuals, 3) ## Wife Alternating Husband Jointly ## Laundry 12.27 -2.298 -5.878 -6.61 ## Main_meal 9.84 -0.484 -4.917 -6.08 ## Dinner 6.54 -1.192 -3.416 -3.30 ## Breakfeast 4.88 3.457 -2.818 -5.30 ## Tidying 1.70 -1.606 -4.969 3.58 ## Dishes -1.10 1.859 -4.163 3.49 ## Shopping -1.29 1.321 -3.362 3.38 ## Official -3.66 8.563 0.443 -2.46 ## Driving -5.47 6.836 8.100 -5.90 ## Finances -4.15 -0.852 -0.742 5.75 ## Insurance -5.76 -4.277 4.107 5.72 ## Repairs -7.53 -4.290 20.646 -6.65 ## Holidays -7.42 -4.620 -4.897 15.56 Let’s visualize Pearson residuals using the package corrplot: library(corrplot) corrplot(chisq$residuals, is.cor = FALSE)

For a given cell, the size of the circle is proportional to the amount of the cell contribution.

The sign of the standardized residuals is also very important to interpret the association between rows and columns as explained in the block below:

1. Positive residuals are in blue. Positive values in cells specify an attraction (positive association) between the corresponding row and column variables.
• In the image above, it’s evident that there are an association between the column Wife and, the rows Laundry and Main_meal.
• There is a strong positive association between the column Husband and the row Repair
2. Negative residuals are in red. This implies a repulsion (negative association) between the corresponding row and column variables. For example the column Wife are negatively associated (~ “not associated”) with the row Repairs. There is a repulsion between the column Husband and, the rows Laundry and Main_meal

Note that, correspondence analysis is just the singular value decomposition of the standardized residuals. This will be explained in the next section.

The contribution (in %) of a given cell to the total Chi-square score is calculated as follow:

$contrib = \frac{r^2}{\chi^2}$

• r is the residual of the cell
# Contibution in percentage (%)
contrib <- 100*chisq$residuals^2/chisq$statistic
round(contrib, 3)
##             Wife Alternating Husband Jointly
## Laundry    7.738       0.272   1.777   2.246
## Main_meal  4.976       0.012   1.243   1.903
## Dinner     2.197       0.073   0.600   0.560
## Breakfeast 1.222       0.615   0.408   1.443
## Tidying    0.149       0.133   1.270   0.661
## Dishes     0.063       0.178   0.891   0.625
## Shopping   0.085       0.090   0.581   0.586
## Official   0.688       3.771   0.010   0.311
## Driving    1.538       2.403   3.374   1.789
## Finances   0.886       0.037   0.028   1.700
## Insurance  1.705       0.941   0.868   1.683
## Repairs    2.919       0.947  21.921   2.275
## Holidays   2.831       1.098   1.233  12.445
# Visualize the contribution
corrplot(contrib, is.cor = FALSE)

The relative contribution of each cell to the total Chi-square score give some indication of the nature of the dependency between rows and columns of the contingency table.

It can be seen that:

1. The column “Wife” is strongly associated with Laundry, Main_meal, Dinner
2. The column “Husband” is strongly associated with the row Repairs
3. The column jointly is frequently associated with the row Holidays

From the image above, it can be seen that the most contributing cells to the Chi-square are Wife/Laundry (7.74%), Wife/Main_meal (4.98%), Husband/Repairs (21.9%), Jointly/Holidays (12.44%).

These cells contribute about 47.06% to the total Chi-square score and thus account for most of the difference between expected and observed values.

This confirms the earlier visual interpretation of the data. As stated earlier, visual interpretation may be complex when the contingency table is very large. In this case, the contribution of one cell to the total Chi-square score becomes a useful way of establishing the nature of dependency.

### Total inertia

As mentioned above, the total inertia is the amount of the information contained in the data table.

It’s called $$\phi^2$$ (squared phi) and is calculated as follow:

$\phi^2 = \frac{\chi^2}{grand.total}$

phi2 <- as.numeric(chisq$statistic/sum(housetasks)) phi2 ## [1] 1.11 The square root of $$\phi^2$$ are called trace and may be interpreted as a correlation coefficient (Bendixen, 2003). Any value of the trace > 0.2 indicates a significant dependency between rows and columns (Bendixen M., 2003) ### Mosaic plot Mosaic plot is used to visualize a contingency table in order to examine the association between categorical variables. The function mosaicplot() [graphics package] can be used. library("graphics") # Mosaic plot of observed values mosaicplot(housetasks, las=2, col="steelblue", main = "housetasks - observed counts") # Mosaic plot of expected values mosaicplot(chisq$expected,  las=2, col = "gray",
main = "housetasks - expected counts")

In these plots, column variables are firstly spitted (vertical split) and then row variables are spited(horizontal split). For each cell, the height of bars is proportional to the observed relative frequency it contains:

$\frac{cell.value}{column.sum}$

The blue plot, is the mosaic plot of the observed values. The gray one is the mosaic plot of the expected values under null hypothesis.

If row and column variables were completely independent the mosaic bars for the observed values (blue graph) would be aligned as the mosaic bars for the expected values (gray graph).

It’s also possible to color the mosaic plot according to the value of the standardized residuals:

mosaicplot(housetasks, shade = TRUE,   # Color the graph
las = 2,                    # produces vertical labels
main = "housetasks")

• This plot clearly show you that Laundry, Main_meal, Dinner and Breakfeast are more often done by the “Wife”.
• Repairs are done by the Husband

### G-test: Likelihood ratio test

The G–test of independence is an alternative to the chi-square test of independence, and they will give approximately the same conclusion.

The test is based on the likelihood ratio defined as follow:

$ratio = \frac{o}{e}$

• o is the observed value
• e is the expected value under null hypothesis

This likelihood ratio, or its logarithm, can be used to compute a p-value. When the logarithm of the likelihood ratio is used, the statistic is known as a log-likelihood ratio statistic.

This test is called G-test (or likelihood ratio test or maximum likelihood statistical significance test) and can be used in situations where Chi-square tests were previously recommended.

The G-test is generally defined as follow:

$G = 2 * \sum{o * log(\frac{o}{e})}$

• o is the observed frequency in a cell
• e is the expected frequency under the null hypothesis
• log is the natural logarithm
• The sum is taken over all non-empty cells.

The distribution of G is approximately a chi-squared distribution, with the same number of degrees of freedom as in the corresponding chi-squared test:

$df = (r - 1)(c - 1)$

• r is the number of rows in the contingency table
• c is the number of column in the contingency table

The commonly used Pearson Chi-square test is, in fact, just an approximation of the log-likelihood ratio on which the G-tests are based.

Remember that, the Chi-square formula is:

$\chi^2 = \sum{\frac{(o - e)^2}{e}}$

The functions likelihood.test() [Deducer package] or G.test() [RVAideMemoire] can be used to perform a G-test on a contingency table.

We’ll use the package RVAideMemoire which can be installed as follow : install.packages("RVAideMemoire").

The function G.test() works as chisq.test():

library("RVAideMemoire")
gtest <- G.test(as.matrix(housetasks))
gtest
##
##  G-test
##
## data:  as.matrix(housetasks)
## G = 2000, df = 40, p-value <2e-16

### Interpret the association between rows and columns

To interpret the association between the rows and the columns of the contingency table, the likelihood ratio can be used as an index (i):

$ratio = \frac{o}{e}$

For a given cell,

• If ratio > 1, there is an “attraction” (association) between the corresponding column and row
• If ratio < 1, there is a “repulsion” between the corresponding column and row

The ratio can be calculated as follow:

ratio <- chisq$observed/chisq$expected
round(ratio,3)
##             Wife Alternating Husband Jointly
## Laundry    2.576       0.546   0.052   0.078
## Main_meal  2.356       0.898   0.150   0.090
## Dinner     2.072       0.699   0.297   0.412
## Breakfeast 1.702       1.766   0.490   0.171
## Tidying    1.263       0.619   0.038   1.601
## Dishes     0.823       1.458   0.162   1.607
## Shopping   0.799       1.316   0.343   1.570
## Official   0.363       3.290   1.097   0.535
## Driving    0.209       2.519   2.470   0.074
## Finances   0.334       0.790   0.851   2.001
## Insurance  0.167       0.049   1.745   1.898
## Repairs    0.000       0.125   4.439   0.042
## Holidays   0.000       0.043   0.172   3.276

Note that, you can also use the R code : gtest\$observed/gtest\$expected

The package corrplot can be used to make a graph of the likelihood ratio:

corrplot(ratio, is.cor = FALSE)

The image above confirms our previous observations:

• The rows Laundry, Main_meal and Dinner are associated with the column Wife
• Repairs are done more often by the Husband
• Holidays are taken Jointly

Let’s take the log(ratio) to see the attraction and the repulsion in different colors:

• If ratio < 1 => log(ratio) < 0 (negative values) => red color
• If ratio > 1 = > log(ratio) > 0 (positive values) => blue color

We’ll also add a small value (0.5) to all cells to avoid log(0):

corrplot(log2(ratio + 0.5), is.cor = FALSE)

## Correspondence analysis

Correspondence analysis (CA) is required for large contingency table.

It used to graphically visualize row points and column points in a low dimensional space.

CA is a dimensional reduction method applied to a contingency table. The information retained by each dimension is called eigenvalue.

The total information (or inertia) contained in the data is called phi ($$\phi^2$$) and can be calculated as follow:

$\phi^2 = \frac{\chi^2}{grand.total}$

For a given axis, the eigenvalue ($$\lambda$$) is computed as follow:

$\lambda_{axis} = \sum{\frac{row.sum}{grand.total} * row.coord^2}$

Or equivalently

$\lambda_{axis} = \sum{\frac{col.sum}{grand.total} * col.coord^2}$

• row.coord and col.coord are the coordinates of row and column variables on the axis.

The association index between a row and column for the principal axes can be computed as follow:

$i = 1 + \sum{\frac{row.coord * col.coord}{\sqrt{\lambda}}}$

• $$\lambda$$ is the eigenvalue of the axes
• The sum denotes the sum for all axis

If there is an attraction the corresponding row and column coordinates have the same sign on the axes. If there is a repulsion the corresponding row and column coordinates have different signs on the axes. A high value indicates a strong attraction or repulsion

## CA - Singular value decomposition of the standardized residuals

Correspondence analysis (CA) is used to represent graphically the table of distances between row variables or between column variables.

CA approach includes the following steps:

STEP 1. Compute the standardized residuals

The standardized residuals (S) is:

$S = \frac{o - e}{\sqrt{e}}$

In fact, S is just the square roots of the terms comprising $$\chi^2$$ statistic.

STEP II. Compute the singular value decomposition (SVD) of the standardized residuals.

Let M be: $$M = \frac{1}{sqrt(grand.total)} \times S$$

SVD means that we want to find orthogonal matrices U and V, together with a diagonal matrix $$\Delta$$, such that:

$M = U \Delta V^T$

(Phillip M. Yelland, 2010)

• $$U$$ is a matrix containing row eigenvectors
• $$\Delta$$ is the diagonal matrix. The numbers on the diagonal of the matrix are called singular values (SV). The eigenvalues are the squared SV.
• $$V$$ is a matrix containing column eigenvectors

The eigenvalue of a given axis is:

$\lambda = \delta^2$

• $$\delta$$ is the singular value

The coordinates of row variables on a given axis are:

$row.coord = \frac{U * \delta }{\sqrt{row.mass}}$

The coordinates of columns are:

$col.coord = \frac{V * \delta }{\sqrt{col.mass}}$

Compute SVD in R:

# Grand total
n <- sum(housetasks)
# Standardized residuals
residuals <- chisq$residuals/sqrt(n) # Number of dimensions nb.axes <- min(nrow(residuals)-1, ncol(residuals)-1) # Singular value decomposition res.svd <- svd(residuals, nu = nb.axes, nv = nb.axes) res.svd ##$d
## [1] 7.37e-01 6.67e-01 3.56e-01 1.01e-16
##
## $u ## [,1] [,2] [,3] ## [1,] -0.4276 -0.2359 -0.2823 ## [2,] -0.3520 -0.2176 -0.1363 ## [3,] -0.2339 -0.1149 -0.1448 ## [4,] -0.1956 -0.1923 0.1752 ## [5,] -0.1414 0.1722 -0.0699 ## [6,] -0.0653 0.1686 0.1906 ## [7,] -0.0419 0.1586 0.1491 ## [8,] 0.0722 -0.0892 0.6078 ## [9,] 0.2842 -0.2765 0.4312 ## [10,] 0.0935 0.2358 0.0248 ## [11,] 0.2479 0.2005 -0.2292 ## [12,] 0.6382 -0.3985 -0.4074 ## [13,] 0.1038 0.6516 -0.1101 ## ##$v
##         [,1]   [,2]    [,3]
## [1,] -0.6668 -0.321 -0.3290
## [2,] -0.0322 -0.167  0.9086
## [3,]  0.7364 -0.422 -0.2477
## [4,]  0.1096  0.831 -0.0704
sv <- res.svd$d[1:nb.axes] # singular value u <-res.svd$u
v <- res.svd$v ### Eigenvalues and screeplot # Eigenvalues eig <- sv^2 # Variances in percentage variance <- eig*100/sum(eig) # Cumulative variances cumvar <- cumsum(variance) eig<- data.frame(eig = eig, variance = variance, cumvariance = cumvar) head(eig) ## eig variance cumvariance ## 1 0.543 48.7 48.7 ## 2 0.445 39.9 88.6 ## 3 0.127 11.4 100.0 barplot(eig[, 2], names.arg=1:nrow(eig), main = "Variances", xlab = "Dimensions", ylab = "Percentage of variances", col ="steelblue") # Add connected line segments to the plot lines(x = 1:nrow(eig), eig[, 2], type="b", pch=19, col = "red") How many dimensions to retain?: 1. The maximum number of axes in the CA is : $nb.axes = min( r-1, c-1)$ r and c are respectively the number of rows and columns in the table. 1. Use elbow method ### Row coordinates We can use the function apply to perform arbitrary operations on the rows and columns of a matrix. A simplified format is: apply(X, MARGIN, FUN, ...) • x: a matrix • MARGIN: allowed values can be 1 or 2. 1 specifies that we want to operate on the rows of the matrix. 2 specifies that we want to operate on the column. • FUN: the function to be applied • ...: optional arguments to FUN # row sum row.sum <- apply(housetasks, 1, sum) # row mass row.mass <- row.sum/n # row coord = sv * u /sqrt(row.mass) cc <- t(apply(u, 1, '*', sv)) # each row X sv row.coord <- apply(cc, 2, '/', sqrt(row.mass)) rownames(row.coord) <- rownames(housetasks) colnames(row.coord) <- paste0("Dim.", 1:nb.axes) round(row.coord,3) ## Dim.1 Dim.2 Dim.3 ## Laundry -0.992 -0.495 -0.317 ## Main_meal -0.876 -0.490 -0.164 ## Dinner -0.693 -0.308 -0.207 ## Breakfeast -0.509 -0.453 0.220 ## Tidying -0.394 0.434 -0.094 ## Dishes -0.189 0.442 0.267 ## Shopping -0.118 0.403 0.203 ## Official 0.227 -0.254 0.923 ## Driving 0.742 -0.653 0.544 ## Finances 0.271 0.618 0.035 ## Insurance 0.647 0.474 -0.289 ## Repairs 1.529 -0.864 -0.472 ## Holidays 0.252 1.435 -0.130 # plot plot(row.coord, pch=19, col = "blue") text(row.coord, labels =rownames(row.coord), pos = 3, col ="blue") abline(v=0, h=0, lty = 2) ### Column coordinates # Coordinates of columns col.sum <- apply(housetasks, 2, sum) col.mass <- col.sum/n # coordinates sv * v /sqrt(col.mass) cc <- t(apply(v, 1, '*', sv)) col.coord <- apply(cc, 2, '/', sqrt(col.mass)) rownames(col.coord) <- colnames(housetasks) colnames(col.coord) <- paste0("Dim", 1:nb.axes) head(col.coord) ## Dim1 Dim2 Dim3 ## Wife -0.8376 -0.365 -0.1999 ## Alternating -0.0622 -0.292 0.8486 ## Husband 1.1609 -0.602 -0.1889 ## Jointly 0.1494 1.027 -0.0464 # plot plot(col.coord, pch=17, col = "red") text(col.coord, labels =rownames(col.coord), pos = 3, col ="red") abline(v=0, h=0, lty = 2) ### Biplot of rows and columns to view the association xlim <- range(c(row.coord[,1], col.coord[,1]))*1.1 ylim <- range(c(row.coord[,2], col.coord[,2]))*1.1 # Plot of rows plot(row.coord, pch=19, col = "blue", xlim = xlim, ylim = ylim) text(row.coord, labels =rownames(row.coord), pos = 3, col ="blue") # plot off columns points(col.coord, pch=17, col = "red") text(col.coord, labels =rownames(col.coord), pos = 3, col ="red") abline(v=0, h=0, lty = 2) You can interpret the distance between rows points or between column points but the distance between column points and row points are not meaningful. ### Diagnostic Recall that, the total inertia contained in the data is: $\phi^2 = \frac{\chi^2}{n} = 1.11494$ Our two-dimensional plot captures about 88% of the total inertia of the table. ### Contribution of rows and columns The contributions of a rows/columns to the definition of a principal axis are : $row.contrib = \frac{row.mass * row.coord^2}{eigenvalue}$ $col.contrib = \frac{col.mass * col.coord^2}{eigenvalue}$ Contribution of rows in % # contrib <- row.mass * row.coord^2/eigenvalue cc <- apply(row.coord^2, 2, "*", row.mass) row.contrib <- t(apply(cc, 1, "/", eig[1:nb.axes,1])) *100 round(row.contrib, 2) ## Dim.1 Dim.2 Dim.3 ## Laundry 18.29 5.56 7.97 ## Main_meal 12.39 4.74 1.86 ## Dinner 5.47 1.32 2.10 ## Breakfeast 3.82 3.70 3.07 ## Tidying 2.00 2.97 0.49 ## Dishes 0.43 2.84 3.63 ## Shopping 0.18 2.52 2.22 ## Official 0.52 0.80 36.94 ## Driving 8.08 7.65 18.60 ## Finances 0.88 5.56 0.06 ## Insurance 6.15 4.02 5.25 ## Repairs 40.73 15.88 16.60 ## Holidays 1.08 42.45 1.21 corrplot(row.contrib, is.cor = FALSE) Contribution of columns in % # contrib <- col.mass * col.coord^2/eigenvalue cc <- apply(col.coord^2, 2, "*", col.mass) col.contrib <- t(apply(cc, 1, "/", eig[1:nb.axes,1])) *100 round(col.contrib, 2) ## Dim1 Dim2 Dim3 ## Wife 44.5 10.31 10.82 ## Alternating 0.1 2.78 82.55 ## Husband 54.2 17.79 6.13 ## Jointly 1.2 69.12 0.50 corrplot(col.contrib, is.cor = FALSE) ### Quality of the representation The quality of the representation is called COS2. The quality of the representation of a row on an axis is: $row.cos2 = \frac{row.coord^2}{d^2}$ • row.coord is the coordinate of the row on the axis • $$d^2$$ is the squared distance from the average profile Recall that the distance between each row profile and the average row profile is: $d^2(row_i, average.profile) = \sum{\frac{(row.profile_i - average.profile)^2}{average.profile}}$ row.profile <- housetasks/row.sum head(round(row.profile, 3)) ## Wife Alternating Husband Jointly ## Laundry 0.886 0.080 0.011 0.023 ## Main_meal 0.810 0.131 0.033 0.026 ## Dinner 0.713 0.102 0.065 0.120 ## Breakfeast 0.586 0.257 0.107 0.050 ## Tidying 0.434 0.090 0.008 0.467 ## Dishes 0.283 0.212 0.035 0.469 average.profile <- col.sum/n head(round(average.profile, 3)) ## Wife Alternating Husband Jointly ## 0.344 0.146 0.218 0.292 The R code below computes the distance from the average profile for all the row variables d2.row <- apply(row.profile, 1, function(row.p, av.p){sum(((row.p - av.p)^2)/av.p)}, average.rp) head(round(d2.row,3)) ## Laundry Main_meal Dinner Breakfeast Tidying Dishes ## 1.329 1.034 0.618 0.512 0.353 0.302 The cos2 of rows on the factor map are: row.cos2 <- apply(row.coord^2, 2, "/", d2.row) round(row.cos2, 3) ## Dim.1 Dim.2 Dim.3 ## Laundry 0.740 0.185 0.075 ## Main_meal 0.742 0.232 0.026 ## Dinner 0.777 0.154 0.070 ## Breakfeast 0.505 0.400 0.095 ## Tidying 0.440 0.535 0.025 ## Dishes 0.118 0.646 0.236 ## Shopping 0.064 0.748 0.189 ## Official 0.053 0.066 0.881 ## Driving 0.432 0.335 0.233 ## Finances 0.161 0.837 0.003 ## Insurance 0.576 0.309 0.115 ## Repairs 0.707 0.226 0.067 ## Holidays 0.030 0.962 0.008 visualize the cos2: corrplot(row.cos2, is.cor = FALSE) ### Cos2 of columns $col.cos2 = \frac{col.coord^2}{d^2}$ col.profile <- t(housetasks)/col.sum col.profile <- t(col.profile) #head(round(col.profile, 3)) average.profile <- row.sum/n #head(round(average.profile, 3)) The R code below computes the distance from the average profile for all the column variables d2.col <- apply(col.profile, 2, function(col.p, av.p){sum(((col.p - av.p)^2)/av.p)}, average.profile) #round(d2.col,3) The cos2 of columns on the factor map are: col.cos2 <- apply(col.coord^2, 2, "/", d2.col) round(col.cos2, 3) ## Dim1 Dim2 Dim3 ## Wife 0.802 0.152 0.046 ## Alternating 0.005 0.105 0.890 ## Husband 0.772 0.208 0.020 ## Jointly 0.021 0.977 0.002 visualize the cos2: corrplot(col.cos2, is.cor = FALSE) ### Supplementary rows/columns #### The supplementary row coordinates $sup.row.coord = sup.row.profile * \frac{v}{\sqrt{col.mass}}$ # Supplementary row sup.row <- as.data.frame(housetasks["Dishes",, drop = FALSE]) # Supplementary row profile sup.row.sum <- apply(sup.row, 1, sum) sup.row.profile <- sweep(sup.row, 1, sup.row.sum, "/") # V/sqrt(col.mass) vv <- sweep(v, 1, sqrt(col.mass), FUN = "/") # Supplementary row coord sup.row.coord <- as.matrix(sup.row.profile) %*% vv sup.row.coord ## [,1] [,2] [,3] ## Dishes -0.189 0.442 0.267 ## COS2 = coor^2/Distance from average profile d2.row <- apply(sup.row.profile, 1, function(row.p, av.p){sum(((row.p - av.p)^2)/av.p)}, average.rp) sup.row.cos2 <- sweep(sup.row.coord^2, 1, d2.row, FUN = "/") ## Practice in R library(FactoMineR) res.ca <- CA(housetasks, graph = F) # print res.ca ## **Results of the Correspondence Analysis (CA)** ## The row variable has 13 categories; the column variable has 4 categories ## The chi square of independence between the two variables is equal to 1944 (p-value = 0 ). ## *The results are available in the following objects: ## ## name description ## 1 "$eig"            "eigenvalues"
## 2  "$col" "results for the columns" ## 3 "$col$coord" "coord. for the columns" ## 4 "$col$cos2" "cos2 for the columns" ## 5 "$col$contrib" "contributions of the columns" ## 6 "$row"            "results for the rows"
## 7  "$row$coord"      "coord. for the rows"
## 8  "$row$cos2"       "cos2 for the rows"
## 9  "$row$contrib"    "contributions of the rows"
## 10 "$call" "summary called parameters" ## 11 "$call$marge.col" "weights of the columns" ## 12 "$call$marge.row" "weights of the rows" # eigenvalue head(res.ca$eig)[, 1:2]
##       eigenvalue percentage of variance
## dim 1      0.543                   48.7
## dim 2      0.445                   39.9
## dim 3      0.127                   11.4
# barplot of percentage of variance
barplot(res.ca$eig[,2], names.arg = rownames(res.ca$eig))

# Plot row points
plot(res.ca, invisible ="col")

# Plot column points
plot(res.ca, invisible ="col")

# Biplot of rows and columns
plot(res.ca)