Replace all occurrences of a string in a data frame

Question

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\

Answer 1

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

Answer 2

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

Answer 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).

Answer 4

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)})

Answer 5

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)

Answer 6

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()