CODEWITHSUNDEEP
rtidyverse
small_lebowski
small_lebowski

Reputation: 773

Replace a specific character in all of the variables in data frame

I have this data, where every cell consists of characters, 

x1 <- c(100, 0, 120)
x2 <- c(0, 0, 0)
x3 <- c(110, 0, 0)
data<- data.frame(x1, x2, x3)
testdata <- lapply(data, as.character)
testdata
$`x1`
[1] "100" "0"   "120"
$x2
[1] "0" "0" "0"
$x3
[1] "110" "0"   "0"

I want to replace the 0-only string entries to 000. That means, my data should look like, 

> str(testdata)
    List of 3
     $ x1: chr [1:3] "100" "000" "120"
     $ x2: chr [1:3] "000" "000" "000"
     $ x3: chr [1:3] "110" "000" "000"

Following this, I can write this,

testdata2 <- data.frame(lapply(testdata, function(x) {gsub("0", "000", x)}))

Or this,

testdata %>% mutate_all(funs(str_replace_all(., "0", "000")))

In both cases, it replaces ALL 0s with 000. And the resulting data looks like this,

> testdata
       x1  x2    x3
1 1000000 000 11000
2     000 000   000
3   12000 000   000

which is not what I am looking for. Any idea how to solve this problem?

Upvotes: 4

Views: 238

Answers (5)

Sotos
Sotos

Reputation: 51622

You can also use sprintf, i.e.

lapply(testdata, function(i)sprintf('%03d', as.numeric(i)))
#$`x1`
#[1] "100" "000" "120"

#$x2
#[1] "000" "000" "000"

#$x3
#[1] "110" "000" "000"

Upvotes: 3

jdobres
jdobres

Reputation: 11957

x1 <- c(100, 0, 120)
x2 <- c(0, 0, 0)
x3 <- c(110, 0, 0)
data<- data.frame(x1, x2, x3)
testdata <- lapply(data, as.character)

If it's possible to keep your data in data.frame format, the following would work:

testdata <- as.data.frame(testdata, stringsAsFactors = F)

testdata[testdata == '0'] <- '000'

   x1  x2  x3
1 100 000 110
2 000 000 000
3 120 000 000

Upvotes: 2

Rui Barradas
Rui Barradas

Reputation: 76683

In base R, there is sub with the appropriate regex.

lapply(testdata, function(x) sub("^0$", "000", x))
#$x1
#[1] "100" "000" "120"
#
#$x2
#[1] "000" "000" "000"
#
#$x3
#[1] "110" "000" "000"

Explanation: "^" marks the beginning and "$" marks the end of the string. So the pattern "^0$" is comprised of the character "0" and only of that one character.

Upvotes: 1

akrun
akrun

Reputation: 887981

We can use ifelse with strrep in base R

lapply(testdata, function(x) ifelse(x == 0, strrep(x, 3), x))
#$x1
#[1] "100" "000" "120"

#$x2
#[1] "000" "000" "000"

#$x3
#[1] "110" "000" "000"

In the OP's post, it is replacing "0"s with gsub or str_replace_all which matches all the "0" digit instead of just checking whether the value is 0 or not

Upvotes: 1

r.user.05apr
r.user.05apr

Reputation: 5456

Or:

library(tidyverse)

testdata %>%
  map_df(~if_else(.x == "0", "000", .x))

# A tibble: 3 x 3
#x1    x2    x3   
#<chr> <chr> <chr>
#  1 100   000   110  
#2 000   000   000  
#3 120   000   000

Upvotes: 2

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