CODEWITHSUNDEEP
rreplace
H.Hung
H.Hung

Reputation: 127

Replace values in entire data frame with column values in R

I did some search but cannot find an obvious answer of this question so hopefully it's not a duplicated question. I have a data frame looks like this:

X1 X2 V1 V2 V3 ... Vn
A  B  0  1  2      1
B  C  1  0  1      0
A  C  2  1  0      1

What I want to achieve is to replace V1 to Vn values to the "dosage" of X2. So for row 1 (each row may have different values of X1 and X2),

The expected outcome is:

X1 X2 V1 V2 V3 ... Vn
A  B  AA AB BB     AB
B  C  BC BB BC     BB
A  C  CC AC AA     AC

Here is the sample data frame:

df=data.frame(X1=c("A","B","A"),
              X2=c("B","C","C"),
              V1=c(0,1,2),
              V2=c(1,0,1),
              V3=c(2,1,0))

Thanks for the help!

Upvotes: 3

Views: 114

Answers (4)

www
www

Reputation: 39154

This is inspired from @Matt's answer. We can use mutate_at with paste0 to achieve this task.

## Load packages
library(dplyr)

dat2 <- dat %>%
  mutate_at(vars(-X1, -X2), .funs = list(
    ~case_when(
      . == 0            ~paste0(X1, X1),
      . == 1            ~paste0(X1, X2),
      . == 2            ~paste0(X2, X2),
      TRUE              ~NA_character_
    )
  ))
dat2
#   X1 X2 V1 V2 V3 Vn
# 1  A  B AA AB BB AB
# 2  B  C BC BB BC BB
# 3  A  C CC AC AA AC

DATA

dat <- read.table(text = "X1 X2 V1 V2 V3 Vn
A  B  0  1  2  1
B  C  1  0  1  0
A  C  2  1  0  1 ",
                  stringsAsFactors = FALSE, header = TRUE)

Upvotes: 4

Martin Gal
Martin Gal

Reputation: 16978

It not an elegant solution but for the sake of completeness: just nest two for-loops

for (i in 1:dim(df)[1]) {
  for (j in 3:dim(df)[2]){
    if (df[i,j] == 0){
      df[i,j] <- paste0(df[i,1], df[i,1])
    } else if (df[i,j] == 1) {
      df[i,j] <- paste0(df[i,1], df[i,2])
    } else if (df[i,j] == 2) {
      df[i,j] <- paste0(df[i,2], df[i,2])
    }
  }
}

Sorry for that.

Upvotes: 1

David T
David T

Reputation: 2143

Use spread and gather 

library(tidyverse)
df <- tibble(X1=c("A","B","A"),
              X2=c("B","C","C"),
              V1=c(0,1,2),
              V2=c(1,0,1),
              V3=c(2,1,0))

Capture your from:to translation

transl <- tibble(DOSE = c(0,1,2),
                 OUTCOME = c("AA", "AB", "AC"))

Then Gather your values into long form

longTbl <- df %>% 
  gather(key = "TheV", value = "DOSE", na.rm = TRUE,starts_with("V")) %>% 
  left_join(transl, by = "DOSE") %>% 
  select(- DOSE)


# A tibble: 9 x 4
  X1    X2    TheV  OUTCOME
  <chr> <chr> <chr> <chr>  
1 A     B     V1    AA     
2 B     C     V1    AB     
3 A     C     V1    AC     
4 A     B     V2    AB     
5 B     C     V2    AA     
6 A     C     V2    AB     
7 A     B     V3    AC     
8 B     C     V3    AB     
9 A     C     V3    AA

You might be better leaving it there. But we can pivot it back with spread.

widTbl <- longTbl %>% 
  spread(TheV, OUTCOME )
widTbl

# A tibble: 3 x 5
  X1    X2    V1    V2    V3   
  <chr> <chr> <chr> <chr> <chr>
1 A     B     AA    AB    AC   
2 A     C     AC    AB    AA   
3 B     C     AB    AA    AB

And Bob's your uncle.

Upvotes: 0

Matt
Matt

Reputation: 7385

With your actual df you can replace V1:V3 with V1:Vn.

Using your reprex, you can do:

library(dplyr)


df %>% 
  mutate_at(
    vars(V1:V3),
    funs(case_when(
      . == 0 ~ "AA",
      . == 1 ~ "AB",
      . == 2 ~ "BB"
    ))
  )

Upvotes: 3

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