R - Reverse order with cut

Question

I want to split a vector according to given cutoffs (or breaks) into bins in R.

This can be done as followed:

numbers <- 1:10
cutoffs <- c(0,2,6,7,10, Inf)

data.frame(data = numbers, bins = as.integer(cut(numbers, breaks = cutoffs, right = FALSE)))

This will result in the following:

   data bins
1     1    1
2     2    2
3     3    2
4     4    2
5     5    2
6     6    3
7     7    4
8     8    4
9     9    4
10   10    5

However, what I want is for the lowest value to be in the highest bin and vice versa. So somehow I want to achieve the following instead:

   data bins
1     1    5
2     2    4
3     3    4
4     4    4
5     5    4
6     6    3
7     7    2
8     8    2
9     9    2
10   10    1

I have tried many combinations with 'rev()' to reverse the sequence. But keep in mind I cannot just reverse the results of the earlier cut-command as they are asymmetrical.

I realise there might be a simple solution to this, but for some reason it keeps escaping me. Any suggestions in which direction I should look?

Answer 1

Since you wanted the left interval boundaries closed and wanted an integer result then findInterval (rather than cut) is the natural choice:

data.frame(data = numbers, bins = 6L - findInterval(numbers, vec = cutoffs)) 
   data bins
1     1    5
2     2    4
3     3    4
4     4    4
5     5    4
6     6    3
7     7    2
8     8    2
9     9    2
10   10    1
>