Replace column based on column names - r

Question

I have 65 columns, but a sample of data could be as follows:
df<-read.table (text=" Name D A D E
Rose D D C B
Smith B A D D
Lora A A D D
Javid A D D B
Ahmed C A E A
Helen B A D D
Nadia A A D A
", header=TRUE)
I want to get the following table:
Name D A D E
Rose 2 1 1 1
Smith 1 2 2 1
Lora 1 2 2 1
Javid 1 1 2 1
Ahmed 1 2 1 1
Helen 1 2 2 1
Nadia 1 2 2 1
The numbers follow the first raw. For example, the second column is D, so all Ds should read 2 and else should read 1. Or in the third column, which is A, all As should read 2 and else should read 1 and so on. Please consider I have 65 columns. I understand I should have different names for the columns, but In this case, I cannot change them as you understand it.

Answer 1

With ifelse and sapply:
df[2:ncol(df)] <- sapply(2:ncol(df), function(i) ifelse(df[i] == colnames(df[i]), 2, 1))
output
#> df
Name D A D E
1 Rose 2 1 1 1
2 Smith 1 2 2 1
3 Lora 1 2 2 1
4 Javid 1 1 2 1
5 Ahmed 1 2 1 1
6 Helen 1 2 2 1
7 Nadia 1 2 2 1
data
df <- structure(list(Name = c("Rose", "Smith", "Lora", "Javid", "Ahmed",
"Helen", "Nadia"), D = c("D", "B", "A", "A", "C", "B", "A"),
A = c("D", "A", "A", "D", "A", "A", "A"), D = c("C", "D",
"D", "D", "E", "D", "D"), E = c("B", "D", "D", "B", "A",
"D", "A")), class = "data.frame", row.names = c(NA, -7L))

Answer 2

cols = names(df)[-1]
df[cols] = lapply(cols, \(x) (df[[x]] == x) + 1L)
# Name D A
# 1 Rose 2 1
# 2 Smith 1 2
# 3 Lora 1 2
# 4 Javid 1 1
# 5 Ahmed 1 2
# 6 Helen 1 2
# 7 Nadia 1 2
Simplified data (without repeated column names)
df <- data.frame(
Name = c("Rose", "Smith", "Lora", "Javid", "Ahmed", "Helen", "Nadia"),
D = c("D", "B", "A", "A", "C", "B", "A"),
A = c("D", "A", "A", "D", "A", "A", "A")
)

Answer 3

Another approach, you can stack, replace and unstack, i.e
stack_df <- stack(df[-1])
stack_df$values <- ifelse(stack_df$values == stack_df$ind, 2, 1)
cbind.data.frame(Name = df$Name, unstack(stack_df))
# Name D A E
#1 Rose 2 1 1
#2 Smith 1 2 1
#3 Lora 1 2 1
#4 Javid 1 1 1
#5 Ahmed 1 2 1
#6 Helen 1 2 1
#7 Nadia 1 2 1
DATA
structure(list(Name = c("Rose", "Smith", "Lora", "Javid", "Ahmed",
"Helen", "Nadia"), D = c("D", "B", "A", "A", "C", "B", "A"),
A = c("D", "A", "A", "D", "A", "A", "A"), E = c("B", "D",
"D", "B", "A", "D", "A")), row.names = c(NA, -7L), class = "data.frame")

Answer 4

dplyr option with ifelse like this:
library(dplyr)
df %>%
mutate(across(D:E, ~ifelse(. == cur_column(), 2, 1)))
#> Name D A D.1 E
#> 1 Rose 2 1 1 1
#> 2 Smith 1 2 1 1
#> 3 Lora 1 2 1 1
#> 4 Javid 1 1 1 1
#> 5 Ahmed 1 2 1 1
#> 6 Helen 1 2 1 1
#> 7 Nadia 1 2 1 1
Created on 2022-09-19 with reprex v2.0.2

Answer 5

Using dplyr:
The data:
df <- read.table (
text = " Name A B C D
Rose D D C B
Smith B A D D
Lora A A D D
Javid A D D B
Ahmed C A E A
Helen B A D D
Nadia A A D A
",
header = TRUE
)
> df
Name A B C D
1 Rose D D C B
2 Smith B A D D
3 Lora A A D D
4 Javid A D D B
5 Ahmed C A E A
6 Helen B A D D
7 Nadia A A D A
Note that i changed the column names
df %>%
mutate(across(!c(Name),
.fns = ~ ifelse(.x == cur_column(), 2, 1)))
Name A B C D
1 Rose 1 1 2 1
2 Smith 1 1 1 2
3 Lora 2 1 1 2
4 Javid 2 1 1 1
5 Ahmed 1 1 1 1
6 Helen 1 1 1 2
7 Nadia 2 1 1 1
The mutate-command modifies columns in dataframes. Using the across()-function we specify that the mutation should be applied to more than one column. inside the across-call, we select every column but the name column using !c(Name). We then specify a function that compares the name of the column cur_column() with the values in the column .x. If they are the same, set the value to 2, else to 1.
EDIT: used ifelse instead of case_when as there is only one condition to check

Answer 6

You can compare each row with the column names. Adding 1 to the logical values converts FALSE and TRUE into 1 and 2 respectively.
df[-1] <- t((t(df[-1]) == names(df)[-1]) + 1)
df
# Name D A D E
# 1 Rose 2 1 1 1
# 2 Smith 1 2 2 1
# 3 Lora 1 2 2 1
# 4 Javid 1 1 2 1
# 5 Ahmed 1 2 1 1
# 6 Helen 1 2 2 1
# 7 Nadia 1 2 2 1