Count how many ID's have more than one value in columns -R-CodePudding

Hi i have this dataset:

>   id  s1 s2 s3 s4 
    1   "A"  "A"  "NA" "A"
    2   "NA"  "A"  "NA" "A"
    3   "Na"  "NA"  "NA" "A"
    4   "A"  "NA"  "NA" "Na"
    5   "A"  "A"  "NA" "A"

I want to count how many ID's have only one value of "A" in either s1, s2, s3, s4. In this case it is only 2 persons (ID 3 and 4). But if i have a large dataset how can I count this ?

CodePudding user response：

You can use

which(rowSums(!is.na(df[-1])) == 1)

# [1] 3 4

Replace which() with sum() to get the number of ID that have only 1 non-missing value.

Update

If unfortunately, you store all NA as "NA", "Na", or "na", then use the following code to convert them back to regular NA in advance.

df[] <- lapply(df, \(x) { x[x %in% c('NA', 'Na', 'na')] <- NA; x })

CodePudding user response：

Checks for "A" in the string

library(tidyverse)

df %>%
  filter(rowSums(across(s1:s4, ~ str_detect(.x, "A")), na.rm = TRUE) == 1)

# A tibble: 2 × 5
     id s1    s2    s3    s4   
  <dbl> <chr> <chr> <lgl> <chr>
1     3 Na    NA    NA    A    
2     4 A     NA    NA    Na