Calculate correlation with cor(), only for numerical columns

Question

I have a dataframe and would like to calculate the correlation (with Spearman, data is categorical and ranked) but only for a subset of columns. I tried with all, but R\'s c

Answer 1

I found an easier way by looking at the R script generated by Rattle. It looks like below:

correlations <- cor(mydata[,c(1,3,5:87,89:90,94:98)], use="pairwise", method="spearman")

Answer 2

if you have a dataframe where some columns are numeric and some are other (character or factor) and you only want to do the correlations for the numeric columns, you could do the following:

set.seed(10)

x = as.data.frame(matrix(rnorm(100), ncol = 10))
x$L1 = letters[1:10]
x$L2 = letters[11:20]

cor(x)

Error in cor(x) : 'x' must be numeric

but

cor(x[sapply(x, is.numeric)])

             V1         V2          V3          V4          V5          V6          V7
V1   1.00000000  0.3025766 -0.22473884 -0.72468776  0.18890578  0.14466161  0.05325308
V2   0.30257657  1.0000000 -0.27871430 -0.29075170  0.16095258  0.10538468 -0.15008158
V3  -0.22473884 -0.2787143  1.00000000 -0.22644156  0.07276013 -0.35725182 -0.05859479
V4  -0.72468776 -0.2907517 -0.22644156  1.00000000 -0.19305921  0.16948333 -0.01025698
V5   0.18890578  0.1609526  0.07276013 -0.19305921  1.00000000  0.07339531 -0.31837954
V6   0.14466161  0.1053847 -0.35725182  0.16948333  0.07339531  1.00000000  0.02514081
V7   0.05325308 -0.1500816 -0.05859479 -0.01025698 -0.31837954  0.02514081  1.00000000
V8   0.44705527  0.1698571  0.39970105 -0.42461411  0.63951574  0.23065830 -0.28967977
V9   0.21006372 -0.4418132 -0.18623823 -0.25272860  0.15921890  0.36182579 -0.18437981
V10  0.02326108  0.4618036 -0.25205899 -0.05117037  0.02408278  0.47630138 -0.38592733
              V8           V9         V10
V1   0.447055266  0.210063724  0.02326108
V2   0.169857120 -0.441813231  0.46180357
V3   0.399701054 -0.186238233 -0.25205899
V4  -0.424614107 -0.252728595 -0.05117037
V5   0.639515737  0.159218895  0.02408278
V6   0.230658298  0.361825786  0.47630138
V7  -0.289679766 -0.184379813 -0.38592733
V8   1.000000000  0.001023392  0.11436143
V9   0.001023392  1.000000000  0.15301699
V10  0.114361431  0.153016985  1.00000000

Answer 3

Another option would be to just use the excellent corrr package https://github.com/drsimonj/corrr and do

require(corrr)
require(dplyr)

myData %>% 
   select(x,y,z) %>%  # or do negative or range selections here
   correlate() %>%
   rearrange() %>%  # rearrange by correlations
   shave() # Shave off the upper triangle for a cleaner result

Steps 3 and 4 are entirely optional and are just included to demonstrate the usefulness of the package.

Answer 4

For numerical data you have the solution. But it is categorical data, you said. Then life gets a bit more complicated...

Well, first : The amount of association between two categorical variables is not measured with a Spearman rank correlation, but with a Chi-square test for example. Which is logic actually. Ranking means there is some order in your data. Now tell me which is larger, yellow or red? I know, sometimes R does perform a spearman rank correlation on categorical data. If I code yellow 1 and red 2, R would consider red larger than yellow.

So, forget about Spearman for categorical data. I'll demonstrate the chisq-test and how to choose columns using combn(). But you would benefit from a bit more time with Agresti's book : http://www.amazon.com/Categorical-Analysis-Wiley-Probability-Statistics/dp/0471360937

set.seed(1234)
X <- rep(c("A","B"),20)
Y <- sample(c("C","D"),40,replace=T)

table(X,Y)
chisq.test(table(X,Y),correct=F)
# I don't use Yates continuity correction

#Let's make a matrix with tons of columns

Data <- as.data.frame(
          matrix(
            sample(letters[1:3],2000,replace=T),
            ncol=25
          )
        )

# You want to select which columns to use
columns <- c(3,7,11,24)
vars <- names(Data)[columns]

# say you need to know which ones are associated with each other.
out <-  apply( combn(columns,2),2,function(x){
          chisq.test(table(Data[,x[1]],Data[,x[2]]),correct=F)$p.value
        })

out <- cbind(as.data.frame(t(combn(vars,2))),out)

Then you should get :

> out
   V1  V2       out
1  V3  V7 0.8116733
2  V3 V11 0.1096903
3  V3 V24 0.1653670
4  V7 V11 0.3629871
5  V7 V24 0.4947797
6 V11 V24 0.7259321

Where V1 and V2 indicate between which variables it goes, and "out" gives the p-value for association. Here all variables are independent. Which you would expect, as I created the data at random.