How to get the number of elements separated by an underscore

user_11652655 0
Viewed 64
strings<-c("A_A_A","B", "C_C_C_C", "D_D_D_D_D")

I have this vector of strings. I want to have the number of elements in each string so that to have:

[1] 3  1  4  5
r
7 Answers
stringr::str_count(c("A_A_A","B", "C_C_C_C", "D_D_D_D_D"), '_') + 1
user_111577530

For a base R option, we could use string replacement:

strings <- c("A_A_A","B", "C_C_C_C", "D_D_D_D_D")
nchar(strings) - nchar(gsub("_", "", strings, fixed=TRUE)) + 1

[1] 3 1 4 5
user_18632290

You can use lengths with strsplit.

strings <- c("A_A_A", "B", "C_C_C_C", "D_D_D_D_D")

lengths(strsplit(strings, split = "_"))
# [1] 3 1 4 5
user_134606020

Yet another possible solution:

library(tidyverse)

strings<-c("A_A_A","B", "C_C_C_C", "D_D_D_D_D")

strings %>% 
  str_remove_all("_") %>% 
  str_count

#> [1] 3 1 4 5
user_115644870

Another option with nchar + gsub

> nchar(gsub("[^_]", "", strings)) + 1
[1] 3 1 4 5

or

> nchar(gsub("_", "", strings))
[1] 3 1 4 5
user_121587570

If your structure is always like your sample data, that means 1 character seperated by one length seperator you can simply do

ceiling(nchar(strings)/2)

# [1] 3 1 4 5

or

nchar(strings)%/%2+1

# [1] 3 1 4 5
user_104157490

Another option is to specify the types of characters that you want to count. So, in this case, we can just count the alphabetic characters (i.e., "[[:alpha:]]"). Or if you also have numbers, then could use "[[:alnum:]]".

library(stringr)

str_count(strings, "[[:alpha:]]")
# [1] 3 1 4 5

Or in base R:

nchar(gsub("([^[:alpha:]])+", "", strings))
# [1] 3 1 4 5
user_152931910
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