Replace all occurrences of a string in a data frame

I'm working on a data frame that has non-detects which are coded with '<'. Sometimes there is a space after the '<' and sometimes not e.g. '<2' or '< 2'. I'd like to remove every occurrence of the space.

Example:

data <- data.frame(name = rep(letters[1:3], each = 3), var1 = rep('< 2', 9), var2 = rep('<3', 9))

  name var1 var2 
1    a  < 2   <3
2    b  < 2   <3
3    c  < 2   <3

This is where I've got to:

I can extract all the values and make the new strings but I can't put them back in the data frame.

index <- str_detect(unlist(data), '<')
index <- matrix(index, nrow = 3)

data[index] 
#[1] "< 2" "< 2" "< 2" "<3"  "<3"  "<3" 

replacements <- str_replace_all(data[index], "<[ ]+","<") 
replacements
#[1] "<2" "<2" "<2" "<3" "<3" "<3"

data[index] <- replacements

#Error in `[<-.data.frame`(`*tmp*`, index, value = c("<2", "<2", "<2",  : 
#  unsupported matrix index in replacement

If you are only looking to replace all occurrences of "< " (with space) with "<" (no space), then you can do an lapply over the data frame, with a gsub for replacement:

> data <- data.frame(lapply(data, function(x) {
+                  gsub("< ", "<", x)
+              }))
> data
  name var1 var2
1    a   <2   <3
2    a   <2   <3
3    a   <2   <3
4    b   <2   <3
5    b   <2   <3
6    b   <2   <3
7    c   <2   <3
8    c   <2   <3
9    c   <2   <3

Equivalent to "find and replace." Don't overthink it.

Try it with one:

library(tidyverse)
df <- data.frame(name = rep(letters[1:3], each = 3), var1 = rep('< 2', 9), var2 = rep('<3', 9))

df %>% 
  mutate(var1 = str_replace(var1, " ", ""))
#>   name var1 var2
#> 1    a   <2   <3
#> 2    a   <2   <3
#> 3    a   <2   <3
#> 4    b   <2   <3
#> 5    b   <2   <3
#> 6    b   <2   <3
#> 7    c   <2   <3
#> 8    c   <2   <3
#> 9    c   <2   <3

Apply to all

df %>% 
  mutate_all(funs(str_replace(., " ", "")))
#>   name var1 var2
#> 1    a   <2   <3
#> 2    a   <2   <3
#> 3    a   <2   <3
#> 4    b   <2   <3
#> 5    b   <2   <3
#> 6    b   <2   <3
#> 7    c   <2   <3
#> 8    c   <2   <3
#> 9    c   <2   <3

If the extra space was produced by uniting columns, think about making str_trim part of your workflow.

Created on 2018-03-11 by the reprex package (v0.2.0).

To remove all spaces in every column, you can use

data[] <- lapply(data, gsub, pattern = " ", replacement = "", fixed = TRUE)

or to constrict this to just the second and third columns (i.e. every column except the first),

data[-1] <- lapply(data[-1], gsub, pattern = " ", replacement = "", fixed = TRUE)

Here is a dplyr solution

library(dplyr)
library(stringr)

Censor_consistently <-  function(x){
  str_replace(x, '^\\s*([<>])\\s*(\\d+)', '\\1\\2')
}


test_df <- tibble(x = c('0.001', '<0.002', ' < 0.003', ' >  100'),  y = 4:1)
                    
mutate_all(test_df, funs(Censor_consistently))
                    
# A tibble: 4 × 2
x     y
<chr> <chr>
1  0.001     4
2 <0.002     3
3 <0.003     2
4   >100     1

I had the problem, I had to replace "Not Available" with NA and my solution goes like this

data <- sapply(data,function(x) {x <- gsub("Not Available",NA,x)})

As an update to the answer by @Nettle, mutate_all() has been superseded by mutate( across( ... ) ):

library(tidyverse)

df <- data.frame(
    name = rep( letters[1:3], each = 3 ),
    var1 = rep( '< 2', 9 ),
    var2 = rep( '<3', 9 )
)

df %>%
    mutate( across(
        .cols = everything(),
        ~str_replace( ., " ", "" )
    ) )

#>   name var1 var2
#> 1    a   <2   <3
#> 2    a   <2   <3
#> 3    a   <2   <3
#> 4    b   <2   <3
#> 5    b   <2   <3
#> 6    b   <2   <3
#> 7    c   <2   <3
#> 8    c   <2   <3
#> 9    c   <2   <3

late to the party. but if you only want to get rid of leading/trailing white space, R base has a function trimws

For example:

data <- apply(X = data, MARGIN = 2, FUN = trimws) %>% as.data.frame()