R loop to generate multiple variables based on a condition - r

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!

Answer 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).

Answer 2

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

Answer 3

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

Answer 4

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)