CODEWITHSUNDEEP
rdplyr
Daren Eiri
Daren Eiri

Reputation: 137

How do you evaluate a set of conditions and create a new column based on a list of columns?

I would like to take a list (or filter) a set of similarly named columns in a dataframe, evaluate the value for those columns for each row, and then create a new column with the result.

Right now I am using case_when but I am not aware of a way to provide some wildcard or list of defined columns.

I would like to take a list, or create a filter, of columns because I want to evaluate several columns in the dataframe, not just a handful. Having a long list of columns in case_when seems not efficient when the column names are quite similar to each other. 

# Dummy data
ignore1 <- c(1, 0, 0)
ignore2 <- c(1, 0, 1)
col1 <- c(0, 1, 0)
col2 <- c(0, 1, 1)
col3 <- c(0, 1, 0)

df <- data.frame(ignore1, ignore2, col1, col2, col3)

df %>% 
  mutate(evaluation = case_when(
    col1 == 0| col1 == 0 | col1 == 0  ~ "Failed",
    TRUE ~ "Passed"
    )
  )

This is the expected result: 

  ignore1 ignore2 col1 col2 col3 evaluation
1       1       1    0    0    0     Failed
2       0       0    1    1    1     Passed
3       0       1    0    1    0     Failed

where row 2 passes because col1, col2, col3 all have the value of 1.

Upvotes: 1

Views: 141

Answers (2)

akrun
akrun

Reputation: 887711

We can just use rowSums to make this efficient

i1 <- startsWith(names(df), 'col')
c( "Failed",  "Passed")[(rowSums(df[i1] == 1) == 3) + 1]
#[1] "Failed" "Passed" "Failed"

Or another base R efficient option is Reduce

c("Failed", "Passed")[Reduce(`&`, df[i1]) +1]
#[1] "Failed" "Passed" "Failed"

NOTE: Both base R solutions are compact and are very efficient

Or with &

library(dplyr)
df %>% 
     mutate(evaluation =  c('Failed', 'Passed')[1 + (col1 & col2 & col3)])
#  ignore1 ignore2 col1 col2 col3 evaluation
#1       1       1    0    0    0     Failed
#2       0       0    1    1    1     Passed
#3       0       1    0    1    0     Failed

Or we can have rowSums within dplyr

df %>%
   mutate(evaluation = c("Failed", "Passed")[(rowSums(.[i1] == 1) == 3) + 1])

NOTE: Both the solutions are very efficient and doesn't use any packages that are not really needed

Or if we need some packages, then use magrittr with purrr

library(magrittr)
library(purrr)
df %>% 
   mutate(evaluation = select(., starts_with('col')) %>% 
                          reduce(`&`) %>%
                          add(1) %>%
                          c("Failed", "Passed")[.])
#  ignore1 ignore2 col1 col2 col3 evaluation
#1       1       1    0    0    0     Failed
#2       0       0    1    1    1     Passed
#3       0       1    0    1    0     Failed

NOTE: Here also, we are not looping over rows, so it should be efficient

Upvotes: 1

Ronak Shah
Ronak Shah

Reputation: 389215

As you want to perform a row-wise operation we can use pmap variants

library(dplyr)
library(purrr)

df %>% mutate(result =c("Passed","Failed")[pmap_lgl(select(., starts_with('col')),
                       ~any(c(...) == 0)) + 1])

#  ignore1 ignore2 col1 col2 col3 result
#1       1       1    0    0    0 Failed
#2       0       0    1    1    1 Passed
#3       0       1    0    1    0 Failed

In base R, we can use apply row-wise : 

cols <- startsWith(names(df), 'col')
df$Result <- c("Passed", "Failed")[apply(df[cols] == 0, 1, any) + 1]

Upvotes: 0

Related Questions