Create a new variable from conditions on multiple variables in R using ifelse condition-CodePudding

I am really new to R. I have a table named RWA2010LONG containing 65 variables. I want to create a new variable named NEWVAR from the 30:49th variable of the table RWA2010LONG and from another variable of the same table named (BIRTH) based on the following condition: for each row, if one of the values of the variables 30:49 of the table RWA2010LONG is equal to the value of the variable BIRTH, NEWVAR takes the value 1. Otherwise, NEWVAR takes the value 0 This is what I tried to do:

RWA2010LONG$ NEWVAR <- for (i in colnames(RWA2010LONG[30:49])){ifelse(i==RWA2010LONG$BIRTH, 1,0)}

CodePudding user response：

Here is an approach. Since you didn't provide data, I am making up some sample data and checking if any values in columns 5 through 10 are the same as BIRTH:

#Example data
df <- data.frame(matrix(rnbinom(100, mu = 5, size = 0.5), ncol = 10,
             dimnames = list(c(sprintf("obs_%s", 1:10)),
                             c("BIRTH",sprintf("col_%s", 2:10)))))

df$newvar <- apply(df[,5:10] == df$BIRTH, 1, any)*1

The apply statement checks for the condition df[,5:10] == df$BIRTH by row (the 1 indicates apply the condition by row, if for future reference, if you put a 2 it will check the condition by column) and returns TRUE or FALSE if the condition is met. The *1 converts those logicals to a numerical value.

Output:

#            BIRTH col_2 col_3 col_4 col_5 col_6 col_7 col_8 col_9 col_10  newvar
# obs_1      0     3     4     0     6    18     0    10     5      7      1
# obs_2      5     1     0     7     5     0     2     2     2      3      1
# obs_3      1     2     4     2     1    13    14     1     2      8      1
# obs_4      1     0     0     0    11     0     0     0    15      0      0
# obs_5      1     9     1     0     4    27     2     1     0      0      1
# obs_6      5     1     0     2     0     7     2     4     0      0      0
# obs_7      1     0     0     6     0     0     1     5     0      0      1
# obs_8     44     0    15     1     1     0     1     1     0      6      0
# obs_9      6     6     0     0     0     4    17     0     6      1      1
# obs_10     0     2     0     0     2    11     2     1     9      2      0

CodePudding user response：

An option with if_any

library(dplyr)
df <- df %>%
       mutate(newvar =  (if_any(5:10, ~ .x == BIRTH)))