Conditionally sum dynamic columns in r

Question

I am trying to conditionally sum across many columns depending on if they are greater than or less than 0. I am surprised I cannot find a dplyr or data.table work around for this. I want to calculate 4 new columns for a large data.frame (columns to calculate are at bottom of post).

dat2=matrix(nrow=10,rnorm(100));colnames(dat2)=paste0('V',rep(1:10))

dat2 %>% as.data.frame() %>%
  rowwise() %>%
  select_if(function(col){mean(col)>0}) %>%
  mutate(sum_pos=rowSums(.))  ##Obviously doesn't work

These are the simple statistics I want to calculate (yes; these apply statements work, but there are other things in my dplyr chain I want to do, so thats why I am looking for a dplyr or data.table way. The columns that are positive or negative for each given row are different, so I cannot grab a list of columns (must be done dynamically, by row).

#Calculate these, but in a dplyr chain?
n_pos=apply(dat2,1,function(x) sum((x>0)))
n_neg=apply(dat2,1,function(x) sum((x<0)))
sum_pos=apply(dat2,1,function(x) sum(x[(x>0)]))
sum_neg=apply(dat2,1,function(x) sum(x[(x<0)]))

`rowSums(replace(dat2, dat2 < 0, 0))` – Sotos Dec 17 '19 at 14:55 — Sotos, Dec 17 '19 at 14:55

akrun · Accepted Answer · 2019-12-17T15:03:15.280

2

We don't need rowwise with rowSums as rowSums can do the sum without any groupings

library(dplyr)
dat2 %>%
   as.data.frame() %>%  
   select_if(~ is.numeric(.) && mean(.) > 0) %>% 
   mutate(sum_pos = rowSums(.))

Based on the description, it seems that it is not the mean condition, but related to rowwise, sum of the positive and negative values separately

dat2 %>%
   as.data.frame %>%
   mutate(sum_pos = rowSums(. * NA^(. < 0), na.rm = TRUE),
           sum_neg = rowSums(.[1:10] * NA^(.[1:10] > 0), na.rm = TRUE) )

edited Dec 17 '19 at 15:03

answered Dec 17 '19 at 14:54

akrun

874,273
37
540
662

This isn't matching the sum_pos vector obtained from `apply` – K.J.J.K Dec 17 '19 at 14:59
@K.J.J.K In the `apply`, you are not using the `mean` condition and it is a different condition. Which one are you looking for? – akrun Dec 17 '19 at 15:00
My strategy revolved around trying to taking the mean of a column of 1 to use select_if (hence, why I used mean in the select_if statement) – K.J.J.K Dec 17 '19 at 15:03
@K.J.J.K Now, it should work with the `apply` solutiion – akrun Dec 17 '19 at 15:03
In English, is your statement multiplying by NA if `.` is greater than or less than 0, and then passing that vector (full of numbers and NAs) to rowSums? – K.J.J.K Dec 17 '19 at 15:06
@K.J.J.K It is just a way to replace the value that are negative or positive to NA when we do the sum of postive and negative. Another option is `replace` as in the comments – akrun Dec 17 '19 at 15:08

Conditionally sum dynamic columns in r

1 Answers1

Linked