sum columns with similar names in R

Question

I\'ve got a dataframe that\'s got lots of columns that are something like this:

data <- data.frame (a.1 = 1:5, a.2b = 3:7, a.5 = 5:9, bt.16 = 4:8, bt.1234         


        

Answer 1

You can try

library(tidyverse)
data.frame (a.1 = 1:5, a.2b = 3:7, a.5 = 5:9, bt.16 = 4:8, bt.12342 = 7:11) %>% 
  rownames_to_column() %>% 
  gather(k, v, -rowname) %>% 
  separate(k, letters[1:2]) %>% 
  group_by(rowname, a) %>% 
  summarise(Sum=sum(v)) %>% 
  spread(a, Sum)
#> # A tibble: 5 x 3
#> # Groups:   rowname [5]
#>   rowname     a    bt
#>   <chr>   <int> <int>
#> 1 1           9    11
#> 2 2          12    13
#> 3 3          15    15
#> 4 4          18    17
#> 5 5          21    19

Created on 2018-04-16 by the reprex package (v0.2.0).

Answer 2

Here's another tidyverse solution:

library(tidyverse)

t(data) %>%
  data.frame() %>%
  group_by(., id = gsub('\\..*', '', rownames(.))) %>%
  summarise_all(sum) %>%
  data.frame() %>%
  column_to_rownames(var = 'id') %>%
  t()

Result:

    a bt
X1  9 11
X2 12 13
X3 15 15
X4 18 17
X5 21 19

Answer 3

Here a solution with base R:

> prefixes = unique(sub("\\..*", "", colnames(data)))
> sapply(prefixes, function(x)rowSums(data[,startsWith(colnames(data), x)]))
      a bt
[1,]  9 11
[2,] 12 13
[3,] 15 15
[4,] 18 17
[5,] 21 19

Answer 4

data <- data.frame (a.1 = 1:5, a.2b = 3:7, a.5 = 5:9, bt.16 = 4:8, bt.12342 = 7:11)
i <- grepl("a.", names(data), fixed = TRUE)
result <- data.frame(a=rowSums(data[, i]), bt=rowSums(data[, !i]))
result
# > result
#    a bt
# 1  9 11
# 2 12 13
# 3 15 15
# 4 18 17
# 5 21 19

If you have more than two prefixes you can do something like:

prefs <- c("a.", "bt.")
as.data.frame(lapply(prefs, function(p) rowSums(data[, grepl(p, names(data), fixed = TRUE)]) ))