R loop to generate multiple variables based on a condition

Question

The data is given as below.

x1<- c(1,0,0,0,1,1,1,0)
x2<- c(1,0,0,0,1,1,1,0)
x3<- c(1,0,0,0,1,1,1,0)
x4<- c(1,0,0,0,1,1,1,0)
x5<- c(1,0,0,0,1,1,1,0)
x6<- c(1,0,0,0,1,1,1,0)

my_data <- as.data.frame(cbind(x1, x2, x3, x4, x5, x6))

I want to use a loop to automate the following process:

my_data$a1 = ifelse(my_data$x1> 0 & is.na(my_data$x1) != T, 1, 0)
my_data$a2 = ifelse(my_data$x2> 0 & is.na(my_data$x2) != T, 1, 0)
my_data$a3 = ifelse(my_data$x3> 0 & is.na(my_data$x3) != T, 1, 0)
my_data$a4 = ifelse(my_data$x4> 0 & is.na(my_data$x4) != T, 1, 0)
my_data$a5 = ifelse(my_data$x5> 0 & is.na(my_data$x5) != T, 1, 0)
my_data$a6 = ifelse(my_data$x6> 0 & is.na(my_data$x6) != T, 1, 0)

Any help would be appreciated, thanks!

Ronak Shah · Accepted Answer · 2021-06-06 12:24:04Z

1

You can use the following code -

my_data[paste0('a', seq_along(my_data))] <-  +(my_data > 0 & !is.na(my_data))
my_data

#  x1 x2 x3 x4 x5 x6 a1 a2 a3 a4 a5 a6
#1  1  1  1  1  1  1  1  1  1  1  1  1
#2  0  0  0  0  0  0  0  0  0  0  0  0
#3  0  0  0  0  0  0  0  0  0  0  0  0
#4  0  0  0  0  0  0  0  0  0  0  0  0
#5  1  1  1  1  1  1  1  1  1  1  1  1
#6  1  1  1  1  1  1  1  1  1  1  1  1
#7  1  1  1  1  1  1  1  1  1  1  1  1
#8  0  0  0  0  0  0  0  0  0  0  0  0

This will assign 1 where the value is greater than 0 and is not NA. my_data > 0 & !is.na(my_data) returns a logical value (TRUE/FALSE) adding + ahead of it turns them to integers (1/0).

answered Jun 6, 2021 at 12:24

Ronak Shah

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MaxMiak · Accepted Answer · 2021-06-06 11:42:57Z

0

You can use following for loop

for (i in 1:ncol(my_data)) {
 
   my_data[,paste0("a",i)] <- ifelse(my_data[,i] > 0 & !is.na(my_data[,i]),1,0)
  
}

Output

  x1 x2 x3 x4 x5 x6 a1 a2 a3 a4 a5 a6
1  1  1  1  1  1  1  1  1  1  1  1  1
2  0  0  0  0  0  0  0  0  0  0  0  0
3  0  0  0  0  0  0  0  0  0  0  0  0
4  0  0  0  0  0  0  0  0  0  0  0  0
5  1  1  1  1  1  1  1  1  1  1  1  1
6  1  1  1  1  1  1  1  1  1  1  1  1
7  1  1  1  1  1  1  1  1  1  1  1  1
8  0  0  0  0  0  0  0  0  0  0  0  0

answered Jun 6, 2021 at 11:42

MaxMiak

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Comments

akrun · Accepted Answer · 2021-06-06 19:22:00Z

0

We can use tidyverse

library(dplyr)
library(stringr)
df <-  my_data %>% 
    mutate(across(everything(), ~ +(. > 0 & !is.na(.)), 
     .names = "a{.col}")) %>% 
    rename_with(~ str_remove(., 'x'), starts_with('a'))
df
  x1 x2 x3 x4 x5 x6 a1 a2 a3 a4 a5 a6
1  1  1  1  1  1  1  1  1  1  1  1  1
2  0  0  0  0  0  0  0  0  0  0  0  0
3  0  0  0  0  0  0  0  0  0  0  0  0
4  0  0  0  0  0  0  0  0  0  0  0  0
5  1  1  1  1  1  1  1  1  1  1  1  1
6  1  1  1  1  1  1  1  1  1  1  1  1
7  1  1  1  1  1  1  1  1  1  1  1  1
8  0  0  0  0  0  0  0  0  0  0  0  0

answered Jun 6, 2021 at 19:22

akrun

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Comments

Yuriy Saraykin · Accepted Answer · 2021-06-06 19:24:57Z

data.table

x1<- c(1,0,0,0,1,1,1,0)
x2<- c(1,0,0,0,1,1,1,0)
x3<- c(1,0,0,0,1,1,1,0)
x4<- c(1,0,0,0,1,1,1,0)
x5<- c(1,0,0,0,1,1,1,0)
x6<- c(1,0,0,0,1,1,1,0)

my_data <- as.data.frame(cbind(x1, x2, x3, x4, x5, x6))
library(data.table)
setDT(my_data)[, (paste0("a", seq_len(length(names(my_data))))) := lapply(.SD, function(x) ifelse(x > 0 & !is.na(x), 1, 0))][]
#>    x1 x2 x3 x4 x5 x6 a1 a2 a3 a4 a5 a6
#> 1:  1  1  1  1  1  1  1  1  1  1  1  1
#> 2:  0  0  0  0  0  0  0  0  0  0  0  0
#> 3:  0  0  0  0  0  0  0  0  0  0  0  0
#> 4:  0  0  0  0  0  0  0  0  0  0  0  0
#> 5:  1  1  1  1  1  1  1  1  1  1  1  1
#> 6:  1  1  1  1  1  1  1  1  1  1  1  1
#> 7:  1  1  1  1  1  1  1  1  1  1  1  1
#> 8:  0  0  0  0  0  0  0  0  0  0  0  0

^{Created on 2021-06-06 by the reprex package (v2.0.0)}

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