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:

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)

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:

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 
  0.3440367   0.1456422   0.2184633   0.2918578 

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:

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] 0.03684787

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] 3.772028

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

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.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 
0.10091743 0.08772936 0.06192661 0.08027523 0.06995413 0.06479358 

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.inertia <- row.mass * d2.row
head(row.inertia)

   Laundry  Main_meal     Dinner Breakfeast    Tidying     Dishes 
0.13415976 0.09069235 0.03824633 0.04112368 0.02466697 0.01958732 

# Total inertia
sum(row.inertia)

[1] 1.11494

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 profiles

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

The profile of a given column is computed as follow:

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)

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:

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 
0.10091743 0.08772936 0.06192661 0.08027523 0.06995413 0.06479358 

Distance (similarity) between column profiles

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

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] 4.050311

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

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.sum <- apply(housetasks, 2, sum)
grand.total <- sum(housetasks)
col.mass <- col.sum/grand.total
head(col.mass)

       Wife Alternating     Husband     Jointly 
  0.3440367   0.1456422   0.2184633   0.2918578 

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 * d2.col
head(col.inertia)

       Wife Alternating     Husband     Jointly 
  0.3010185   0.1178242   0.3813729   0.3147248 

# total inertia
sum(col.inertia)

[1] 1.11494

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

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:

The Chi-square statistic is calculated as follow:

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 = 1944.456, df = 36, p-value < 2.2e-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    60.55       25.63   38.45   51.37
Main_meal  52.64       22.28   33.42   44.65
Dinner     37.16       15.73   23.59   31.52
Breakfeast 48.17       20.39   30.58   40.86
Tidying    41.97       17.77   26.65   35.61
Dishes     38.88       16.46   24.69   32.98
Shopping   41.28       17.48   26.22   35.02
Official   33.03       13.98   20.97   28.02
Driving    47.82       20.24   30.37   40.57
Finances   38.88       16.46   24.69   32.98
Insurance  47.82       20.24   30.37   40.57
Repairs    56.77       24.03   36.05   48.16
Holidays   55.05       23.30   34.95   46.70

As mentioned above the Chi-square statistic is 1944.456196.

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    12.266      -2.298  -5.878  -6.609
Main_meal   9.836      -0.484  -4.917  -6.084
Dinner      6.537      -1.192  -3.416  -3.299
Breakfeast  4.875       3.457  -2.818  -5.297
Tidying     1.702      -1.606  -4.969   3.585
Dishes     -1.103       1.859  -4.163   3.486
Shopping   -1.289       1.321  -3.362   3.376
Official   -3.659       8.563   0.443  -2.459
Driving    -5.469       6.836   8.100  -5.898
Finances   -4.150      -0.852  -0.742   5.750
Insurance  -5.758      -4.277   4.107   5.720
Repairs    -7.534      -4.290  20.646  -6.651
Holidays   -7.419      -4.620  -4.897  15.556

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.

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:

• 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

Chi-square statistic and the 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:

phi2 <- as.numeric(chisq$statistic/sum(housetasks))
phi2

[1] 1.11494

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)

Likelihood ratio test in R

The functions likelihood.test()[in Deducer package] or G.test()[in 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() work as chisq.test():

library("RVAideMemoire")
gtest <- G.test(as.matrix(housetasks))
gtest

    G-test

data:  as.matrix(housetasks)
G = 1907.658, df = 36, p-value < 2.2e-16

Interpret the association between rows and columns using likelihood ratio

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

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)

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 0.5428893 48.69222    48.69222
2 0.4450028 39.91269    88.60491
3 0.1270484 11.39509   100.00000

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 :

r and c are respectively the number of rows and columns in the table.

2. 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        -0.83762154 -0.3652207 -0.19991139
Alternating -0.06218462 -0.2915938  0.84858939
Husband      1.16091847 -0.6019199 -0.18885924
Jointly      0.14942609  1.0265791 -0.04644302

# 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:

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 :

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        44.46 10.31 10.82
Alternating  0.10  2.78 82.55
Husband     54.23 17.79  6.13
Jointly      1.20 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.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:

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.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

# 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 -0.1889641 0.4419662 0.2669493

## 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 = "/")