Correspondence analysis basics - R software and data mining - Easy Guides - Wiki (2024)

The data is a contingency table containing 13 housetasks and their repartition in the couple :

• rows are the different tasks
• values are the frequencies of the tasks done:
• by the wife only
• alternatively
• by the husband only
• or jointly

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)

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 sumscol.sum <- apply(housetasks, 2, sum)# average row profile = Column sums / grand totalaverage.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:

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

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 profileslaundry.p <- row.profile["Laundry",]main_meal.p <- row.profile["Main_meal",]# Distance between Laundry and Main_meald2 <- sum(((laundry.p - main_meal.p)^2) / average.rp)d2

[1] 0.03684787

The distance between Laundry and Driving is:

# Driving profiledriving.p <- row.profile["Driving",]# Distance between Laundry and Drivingd2 <- 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

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

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 1.329Main_meal 1.034Dinner 0.618Breakfeast 0.512Tidying 0.353Dishes 0.302Shopping 0.218Official 0.968Driving 1.274Finances 0.456Insurance 0.727Repairs 3.307Holidays 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 profiledist.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 matrixdist.mat <- dist.matrix(row.profile, average.rp)dist.mat <-round(dist.mat, 2)# Visualize the matrixlibrary("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.totalhead(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:

\[row.inertia = row.mass * d^2(row)\]

# Row inertiarow.inertia <- row.mass * d2.rowhead(row.inertia)

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

# Total inertiasum(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 inertiaLaundry 1.329 0.101 0.134Main_meal 1.034 0.088 0.091Dinner 0.618 0.062 0.038Breakfeast 0.512 0.080 0.041Tidying 0.353 0.070 0.025Dishes 0.302 0.065 0.020Shopping 0.218 0.069 0.015Official 0.968 0.055 0.053Driving 1.274 0.080 0.102Finances 0.456 0.065 0.030Insurance 0.727 0.080 0.058Repairs 3.307 0.095 0.313Holidays 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:

\[col.profile = \frac{col}{col.sum}\]

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

col.profile <- t(housetasks)/col.sumcol.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:

\[average.cp = row.sum/grand.total\]

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

# Row sumsrow.sum <- apply(housetasks, 1, sum)# average column profile= row sums/grand totalaverage.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:

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

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 profileswife.p <- col.profile[, "Wife"]husband.p <- col.profile[, "Husband"]# Distance between Wife and Husbandd2 <- 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

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

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 matrixdist.mat <- dist.matrix(t(col.profile), average.cp)dist.mat <-round(dist.mat, 2)dist.mat

 Wife Alternating Husband JointlyWife 0.00 1.71 4.05 2.93Alternating 1.71 0.00 2.67 2.58Husband 4.05 2.67 0.00 3.70Jointly 2.93 2.58 3.70 0.00

# Visualize the matrixlibrary("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.totalhead(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:

\[col.inertia = col.mass * d^2(col)\]

col.inertia <- col.mass * d2.colhead(col.inertia)

 Wife Alternating Husband Jointly 0.3010185 0.1178242 0.3813729 0.3147248 

# total inertiasum(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 inertiaWife 0.875 0.344 0.301Alternating 0.809 0.146 0.118Husband 1.746 0.218 0.381Jointly 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:

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

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 testdata: housetasksX-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 countschisq$observed

 Wife Alternating Husband JointlyLaundry 156 14 2 4Main_meal 124 20 5 4Dinner 77 11 7 13Breakfeast 82 36 15 7Tidying 53 11 1 57Dishes 32 24 4 53Shopping 33 23 9 55Official 12 46 23 15Driving 10 51 75 3Finances 13 13 21 66Insurance 8 1 53 77Repairs 0 3 160 2Holidays 0 1 6 153

# Expected countsround(chisq$expected,2)

 Wife Alternating Husband JointlyLaundry 60.55 25.63 38.45 51.37Main_meal 52.64 22.28 33.42 44.65Dinner 37.16 15.73 23.59 31.52Breakfeast 48.17 20.39 30.58 40.86Tidying 41.97 17.77 26.65 35.61Dishes 38.88 16.46 24.69 32.98Shopping 41.28 17.48 26.22 35.02Official 33.03 13.98 20.97 28.02Driving 47.82 20.24 30.37 40.57Finances 38.88 16.46 24.69 32.98Insurance 47.82 20.24 30.37 40.57Repairs 56.77 24.03 36.05 48.16Holidays 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 JointlyLaundry 12.266 -2.298 -5.878 -6.609Main_meal 9.836 -0.484 -4.917 -6.084Dinner 6.537 -1.192 -3.416 -3.299Breakfeast 4.875 3.457 -2.818 -5.297Tidying 1.702 -1.606 -4.969 3.585Dishes -1.103 1.859 -4.163 3.486Shopping -1.289 1.321 -3.362 3.376Official -3.659 8.563 0.443 -2.459Driving -5.469 6.836 8.100 -5.898Finances -4.150 -0.852 -0.742 5.750Insurance -5.758 -4.277 4.107 5.720Repairs -7.534 -4.290 20.646 -6.651Holidays -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:

\[contrib = \frac{r^2}{\chi^2}\]

• r is the residual of the cell

# Contibution in percentage (%)contrib <- 100*chisq$residuals^2/chisq$statisticround(contrib, 3)

 Wife Alternating Husband JointlyLaundry 7.738 0.272 1.777 2.246Main_meal 4.976 0.012 1.243 1.903Dinner 2.197 0.073 0.600 0.560Breakfeast 1.222 0.615 0.408 1.443Tidying 0.149 0.133 1.270 0.661Dishes 0.063 0.178 0.891 0.625Shopping 0.085 0.090 0.581 0.586Official 0.688 3.771 0.010 0.311Driving 1.538 2.403 3.374 1.789Finances 0.886 0.037 0.028 1.700Insurance 1.705 0.941 0.868 1.683Repairs 2.919 0.947 21.921 2.275Holidays 2.831 1.098 1.233 12.445

# Visualize the contributioncorrplot(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:

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

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)

• 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

• 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 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:

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

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-testdata: 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):

\[ratio = \frac{o}{e}\]

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$observed/chisq$expectedround(ratio,3)

 Wife Alternating Husband JointlyLaundry 2.576 0.546 0.052 0.078Main_meal 2.356 0.898 0.150 0.090Dinner 2.072 0.699 0.297 0.412Breakfeast 1.702 1.766 0.490 0.171Tidying 1.263 0.619 0.038 1.601Dishes 0.823 1.458 0.162 1.607Shopping 0.799 1.316 0.343 1.570Official 0.363 3.290 1.097 0.535Driving 0.209 2.519 2.470 0.074Finances 0.334 0.790 0.851 2.001Insurance 0.167 0.049 1.745 1.898Repairs 0.000 0.125 4.439 0.042Holidays 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)

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

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

Eigenvalues and screeplot

# Eigenvalueseig <- sv^2# Variances in percentagevariance <- eig*100/sum(eig)# Cumulative variancescumvar <- c*msum(variance)eig<- data.frame(eig = eig, variance = variance, cumvariance = cumvar)head(eig)

 eig variance cumvariance1 0.5428893 48.69222 48.692222 0.4450028 39.91269 88.604913 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 plotlines(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.

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 sumrow.sum <- apply(housetasks, 1, sum)# row massrow.mass <- row.sum/n# row coord = sv * u /sqrt(row.mass)cc <- t(apply(u, 1, '*', sv)) # each row X svrow.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.3Laundry -0.992 -0.495 -0.317Main_meal -0.876 -0.490 -0.164Dinner -0.693 -0.308 -0.207Breakfeast -0.509 -0.453 0.220Tidying -0.394 0.434 -0.094Dishes -0.189 0.442 0.267Shopping -0.118 0.403 0.203Official 0.227 -0.254 0.923Driving 0.742 -0.653 0.544Finances 0.271 0.618 0.035Insurance 0.647 0.474 -0.289Repairs 1.529 -0.864 -0.472Holidays 0.252 1.435 -0.130

# plotplot(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 columnscol.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 Dim3Wife -0.83762154 -0.3652207 -0.19991139Alternating -0.06218462 -0.2915938 0.84858939Husband 1.16091847 -0.6019199 -0.18885924Jointly 0.14942609 1.0265791 -0.04644302

# plotplot(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]))*1.1ylim <- range(c(row.coord[,2], col.coord[,2]))*1.1# Plot of rowsplot(row.coord, pch=19, col = "blue", xlim = xlim, ylim = ylim)text(row.coord, labels =rownames(row.coord), pos = 3, col ="blue")# plot off columnspoints(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/eigenvaluecc <- apply(row.coord^2, 2, "*", row.mass)row.contrib <- t(apply(cc, 1, "/", eig[1:nb.axes,1])) *100round(row.contrib, 2)

 Dim.1 Dim.2 Dim.3Laundry 18.29 5.56 7.97Main_meal 12.39 4.74 1.86Dinner 5.47 1.32 2.10Breakfeast 3.82 3.70 3.07Tidying 2.00 2.97 0.49Dishes 0.43 2.84 3.63Shopping 0.18 2.52 2.22Official 0.52 0.80 36.94Driving 8.08 7.65 18.60Finances 0.88 5.56 0.06Insurance 6.15 4.02 5.25Repairs 40.73 15.88 16.60Holidays 1.08 42.45 1.21

corrplot(row.contrib, is.cor = FALSE)

Contribution of columns in %

# contrib <- col.mass * col.coord^2/eigenvaluecc <- apply(col.coord^2, 2, "*", col.mass)col.contrib <- t(apply(cc, 1, "/", eig[1:nb.axes,1])) *100round(col.contrib, 2)

 Dim1 Dim2 Dim3Wife 44.46 10.31 10.82Alternating 0.10 2.78 82.55Husband 54.23 17.79 6.13Jointly 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:

\[row.cos2 = \frac{row.coord^2}{d^2}\]

• 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.sumhead(round(row.profile, 3))

 Wife Alternating Husband JointlyLaundry 0.886 0.080 0.011 0.023Main_meal 0.810 0.131 0.033 0.026Dinner 0.713 0.102 0.065 0.120Breakfeast 0.586 0.257 0.107 0.050Tidying 0.434 0.090 0.008 0.467Dishes 0.283 0.212 0.035 0.469

average.profile <- col.sum/nhead(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.3Laundry 0.740 0.185 0.075Main_meal 0.742 0.232 0.026Dinner 0.777 0.154 0.070Breakfeast 0.505 0.400 0.095Tidying 0.440 0.535 0.025Dishes 0.118 0.646 0.236Shopping 0.064 0.748 0.189Official 0.053 0.066 0.881Driving 0.432 0.335 0.233Finances 0.161 0.837 0.003Insurance 0.576 0.309 0.115Repairs 0.707 0.226 0.067Holidays 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.sumcol.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 Dim3Wife 0.802 0.152 0.046Alternating 0.005 0.105 0.890Husband 0.772 0.208 0.020Jointly 0.021 0.977 0.002

visualize the cos2:

corrplot(col.cos2, is.cor = FALSE)

Supplementary rows/columns

# Supplementary rowsup.row <- as.data.frame(housetasks["Dishes",, drop = FALSE])# Supplementary row profilesup.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 coordsup.row.coord <- as.matrix(sup.row.profile) %*% vvsup.row.coord

 [,1] [,2] [,3]Dishes -0.1889641 0.4419662 0.2669493

## COS2 = coor^2/Distance from average profiled2.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 = "/")