Selecting multiple odd or even columns/rows for dataframe

Question

Is there a way in R to select many non-consecutive i.e. odd or even rows/columns?

I'm plotting the loadings for my Principal Components Analysis. I have 84 rows of data ordered like this: x_1 y_1 x_2..... x_42 y_42

And at the moment I am creating the dataframes for the x and y loadings figures like this:

data.pc = princomp(as.matrix(data))

x.loadings <- data.frame(x=data.pc$loadings[c(1, 3, 5, 7, 9, 11, 13 ,15, 17, 19, 
21, 23, 25, 27, 29, 31, 33, 35, 37, 39, 41), 1])

yloadings <- data.frame(y=data.pc$loadings[c(2, 4, 6, 8, 10, 12, 14, 16, 18, 20, 
22, 24, 26, 28, 30, 32, 34, 36, 38, 40, 42), 1])

Surely there's an easier way to do this?

Answer 1

You can always generate sequences with seq:

even_indexes<-seq(2,42,2)
odd_indexes<-seq(1,41,2)

Then,

  x.loadings <- data.frame(x=data.pc$loadings[odd_indexes,1])

Answer 2

When logical vectors are used for indexing, they are recycled so this gets you odd columns or odd rows

 calld[ c(TRUE,FALSE), ]  # rows
 calld[ , c(TRUE,FALSE) ] #columns

Even rows or columns:

 calld[ !c(TRUE,FALSE), ]  # rows
 calld[ , !c(TRUE,FALSE) ] #columns

Every third column:

  calld[ , c(TRUE,FALSE, FALSE) ]   #columns 1,4,7 , ....

Answer 3

I wish to add the tidyverse style approach to this problem, using the %% operator.

library(dplyr)
df <- data.frame(V1 = seq(26), V2 = letters)
df %>% dplyr::filter(row_number() %% 2 == 0) ## Select even rows
df %>% dplyr::filter(row_number() %% 2 == 1) ## Select odd rows
df %>% dplyr::filter(row_number() %% 3 == 1) 
## Select every 3rd row starting from first row

You can use the same idea to delete every n-th row, of course. See here.

Answer 4

Use %% in combination with seq_len to create vector for indexing your data frame to find even and odds columns/rows

Try something like this:

even <- seq_len(ncol(data.pc)) %% 2   # index
x.loadings <- data.frame(x=data.pc$loadings[even, ])
y.loadings <- data.frame(x=data.pc$loadings[!even, ] )