replacing string with numeric character in a column through function in R

Question

I have been trying to write a function in R that can replace strings in particular column by numerals. Following is my example:

d <- data.frame(A = c("D",1,2,3,"D",1,2,"B","D",3,5),
                  B = c(7,8,9,4,5,8,9,1,6,7,8))
func <- function(dat,rep_val_col,rep_val_col_change,new_val)
{
  dat[dat[,rep_val_col] == rep_val_col_change[1],],rep_val_col] = new_val[1]
  dat[dat[,rep_val_col] == rep_val_col_change[2],],rep_val_col] = new_val[2]
}

func(d,"A",c("D","B"),new_val = c(9,10))

I want to replace "D" and "B" in column A by 9 and 10 respectively.

Adam Quek · Accepted Answer · 2017-06-01 08:41:45Z

4

levels(d$A)[levels(d$A) %in% c("B", "D")] <- c(9, 10)
d$A <- as.numeric(as.character(d$A))

answered Jun 1, 2017 at 8:41

Adam Quek

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1 Comment

Sana Ali Over a year ago

Thanks. That worked I have modified it for function and it worked perfectly. code func <- function(dat,rep_val_col,rep_val_col_change,new_val) { levels(dat[,rep_val_col])[levels(dat[,rep_val_col]) %in% rep_val_col_change] <- new_val dat[,rep_val_col] <- as.numeric(as.character(dat[,rep_val_col])) return(dat) } func(d,"A",c("D","B"),new_val = c(9,10)) code

akrun · Accepted Answer · 2017-06-01 09:10:23Z

2

We can do this with tidyverse

library(dplyr)
d %>%
     mutate(A = case_when(A =="D" ~"9",
                          A=="B" ~ "10",
                         TRUE ~as.character(A)), 
           A = as.integer(A))    #    A B
#1   9 7
#2   1 8
#3   2 9
#4   3 4
#5   9 5
#6   1 8
#7   2 9
#8  10 1
#9   9 6
#10  3 7
#11  5 8

answered Jun 1, 2017 at 9:10

akrun

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Comments

LAP · Accepted Answer · 2017-06-01 08:41:48Z

0

You can use nested ifelse():

d$A <- with(d, ifelse(A == "D", 9, ifelse(A == "B", 10, A)))

> d
    A B
1   9 7
2   1 8
3   2 9
4   3 4
5   9 5
6   1 8
7   2 9
8  10 1
9   9 6
10  3 7
11  4 8

answered Jun 1, 2017 at 8:41

LAP

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Comments

theSZ · Accepted Answer · 2017-06-01 08:42:05Z

0

Since you have a factor variable you can change the levels of this factor via

 func <- function(dat,rep_val_col,rep_val_col_change,new_val)
 {
    levels(dat[,rep_val_col])[levels(dat[,rep_val_col]) == rep_val_col_change[1]] <- new_val[1]
    levels(dat[,rep_val_col])[levels(dat[,rep_val_col]) == rep_val_col_change[2]] <- new_val[2]
    return(dat)
 }

 func(d,"A",c("D","B"),new_val = c(9,10))

answered Jun 1, 2017 at 8:42

theSZ

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2 Comments

Sana Ali Over a year ago

This one only add NAs at places of D and B.

theSZ Over a year ago

Really? It seems to work fine for me though. I get [1] 9 1 2 3 9 1 2 10 9 3 5 with Levels: 1 2 3 5 10 9 for column A.

hbabbar · Accepted Answer · 2017-06-01 08:44:41Z

0

Not sure if you are looking for something in general but you can do a simple replacement using

d$A <- gsub("D", 9, d$A)
d$A <- gsub("B", 10, d$A)

answered Jun 1, 2017 at 8:44

hbabbar

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Comments

ssokolen · Accepted Answer · 2017-06-01 09:00:00Z

0

First, note that having a mixture of strings and numerical values in a column will autmatically convert the whole column to a factor (or character in some cases).

In this case, however, having a factor is actually useful as a factor keeps track of unique values as levels that we can change using match and replace.

First, check the levels of d$A:

levels(d$A)
[1] "1" "2" "3" "5" "B" "D"

We can then find the indexes of "B" and "D" using match:

match(c('D','B'), levels(d$A))
[1] 6 5

And replace them using replace:

replace(levels(d$A), match(c('D','B'), levels(d$A)), c(9, 10))
[1] "1"  "2"  "3"  "5"  "10" "9"

Note that the levels are still of type character. Save the new levels and convert the d$A to numeric:

levels(d$A) <- replace(levels(d$A), match(c('D','B'), levels(d$A)), c(9, 10))
d$A <- as.numeric(as.character(d$A))

answered Jun 1, 2017 at 9:00

ssokolen

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1 Comment

Sana Ali Over a year ago

Thanks for the explanation. It also explains Adam's answer

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