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   lt     Sometimes there is a space after the   lt   and sometimes not e g    lt 2  or   lt  2    I d like to remove every occurrence of the space   Example   data  lt - data frame name   rep letters 1 3   each   3   var1   rep   lt  2   9   var2   rep   lt 3   9      name var1 var2  1    a   lt  2    lt 3 2    b   lt  2    lt 3 3    c   lt  2    lt 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  lt - str detect unlist data     lt    index  lt - matrix index  nrow   3   data index     1    lt  2    lt  2    lt  2    lt 3     lt 3     lt 3    replacements  lt - str replace all data index     lt         lt     replacements   1    lt 2    lt 2    lt 2    lt 3    lt 3    lt 3   data index   lt - replacements   Error in    lt - data frame    tmp    index  value   c   lt 2     lt 2     lt 2          unsupported matrix index in replacement

User · Answer

I had  the problem  I had to replace  Not Available  with NA and my solution goes like this  data  lt - sapply data function x   x  lt - gsub  Not Available  NA x

User · Answer

If you are only looking to replace all occurrences of   lt     with space  with   lt    no space   then you can do an lapply over the data frame  with a gsub for replacement    gt  data  lt - data frame lapply data  function x                       gsub   lt       lt    x                      gt  data   name var1 var2 1    a    lt 2    lt 3 2    a    lt 2    lt 3 3    a    lt 2    lt 3 4    b    lt 2    lt 3 5    b    lt 2    lt 3 6    b    lt 2    lt 3 7    c    lt 2    lt 3 8    c    lt 2    lt 3 9    c    lt 2    lt 3

User · Answer

To remove all spaces in every column  you can use  data    lt - 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   lt - lapply data -1   gsub  pattern        replacement       fixed   TRUE

User · Answer

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  lt - apply X   data  MARGIN   2  FUN   trimws    gt   as data frame

User · Answer

Equivalent to  find and replace   Don t overthink it    Try it with one   library tidyverse  df  lt - data frame name   rep letters 1 3   each   3   var1   rep   lt  2   9   var2   rep   lt 3   9    df   gt      mutate var1   str replace var1              gt    name var1 var2   gt  1    a    lt 2    lt 3   gt  2    a    lt 2    lt 3   gt  3    a    lt 2    lt 3   gt  4    b    lt 2    lt 3   gt  5    b    lt 2    lt 3   gt  6    b    lt 2    lt 3   gt  7    c    lt 2    lt 3   gt  8    c    lt 2    lt 3   gt  9    c    lt 2    lt 3   Apply to all  df   gt      mutate all funs str replace                 gt    name var1 var2   gt  1    a    lt 2    lt 3   gt  2    a    lt 2    lt 3   gt  3    a    lt 2    lt 3   gt  4    b    lt 2    lt 3   gt  5    b    lt 2    lt 3   gt  6    b    lt 2    lt 3   gt  7    c    lt 2    lt 3   gt  8    c    lt 2    lt 3   gt  9    c    lt 2    lt 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

User · Answer

Here is a dplyr solution  library dplyr  library stringr   Censor consistently  lt -  function x     str replace x      s    lt  gt     s    d        1  2       test df  lt - tibble x   c  0 001     lt 0 002      lt  0 003      gt   100     y   4 1   mutate all test df  funs Censor consistently      A tibble  4    2 x     y  lt chr gt   lt chr gt  1  0 001     4 2  lt 0 002     3 3  lt 0 003     2 4    gt 100     1

[r] Replace all occurrences of a string in a data frame

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