How to create a for loop to obtain correlations from a data frame in R?

Question

I have the following data frame:

Gene <- c("1","2","3","4","5","6")
A1.1 <- c(1,1,2,4,3,5)
B1.1 <- c(1,2,3,4,5,6)
C1.1 <- c(2,2,3,5,5,5)
A1.2 <- c(1,2,3,5,5,5)
B1.2 <- c(3,2,5,6,6,6)
C1.2 <- c(1,1,2,2,4,6)
df <- data.frame(Gene, A1.1, B1.1, C1.1, A1.2, B1.2, C1.2)

   Gene A1.1 B1.1 C1.1 A1.2 B1.2 C1.2
1    1    1    1    2    1    3    1
2    2    1    2    2    2    2    1
3    3    2    3    3    3    5    2
4    4    4    4    5    5    6    2
5    5    3    5    5    5    6    4
6    6    5    6    5    5    6    6

So I need to obtain correlation values between columns of the same letter. So obtain the correlation values for A1.1 and A1.2, B1.1 and B1.2, and C1.1 and C1.2 for a total of 3 correlation values.

I can do this by using the cor() function for each (eg. cor(df$A1.1, df$A1.2) ), but is there a for loop I could create that could obtain the correlations for all these at once?

Answer 1

You could use split.default :

sapply(split.default(df[-1], sub('.\\d+$', '', names(df)[-1])), 
              function(x)cor(x[[1]], x[[2]]))

       A1        B1        C1 
0.9042908 0.8546548 0.7656415 

If there are many columns with the same names:

 lapply(split.default(df[-1], sub('.\\d+$', '', names(df)[-1])), cor)
$A1
          A1.1      A1.2
A1.1 1.0000000 0.9042908
A1.2 0.9042908 1.0000000

$B1
          B1.1      B1.2
B1.1 1.0000000 0.8546548
B1.2 0.8546548 1.0000000

$C1
          C1.1      C1.2
C1.1 1.0000000 0.7656415
C1.2 0.7656415 1.0000000

Answer 2

If you have more columns with the same letter, correlation matrices might be more convenient to use, as you need to compare all columns with all the other columns

cor_list <- list()

col_names <- colnames(df[-1])
column_letters <- unique(substr(col_names, 1, 1))

for (let in column_letters){
  indices <- substr(colnames(df),1, 1) == let
  cor_list[[let]] <- cor(df[indices])
}

Answer 3

When your columns are as nicely arranged as shown, you can correlate the pairs.

sapply(0:2, function(i) cor(df[, 2 + i], df[, 5 + i]))
# [1] 0.9042908 0.8546548 0.7656415