Replace given value in vector

Question

I m looking for a function which will replace all occurrences of one value with another value  For example I d like to replace all zeros with ones  I don t want to have to store the result in a variable  but want to be able to use the vector anonymously as part of a larger expression   I know how to write a suitable function myself    gt  vrepl  lt - function haystack  needle  replacement        haystack haystack    needle   lt - replacement     return haystack       gt    gt  vrepl c 3  2  1  0  4  0   0  1   1  3 2 1 1 4 1   But I m wondering whether there is some standard function to do this job  preferrably from the base package  as an alternative from some other commonly used package  I believe that using such a standard will likely make my code more readable  and I won t have to redefine that function wherever I need it

User · Answer

The ifelse function would be a quick and easy way to do this

User · Answer

A simple way to do this is using ifelse  which is vectorized  If the condition is satisfied  we use a replacement value  otherwise we use the original value   v  lt - c 3  2  1  0  4  0  ifelse v    0  1  v    We can avoid a named variable by using a pipe   c 3  2  1  0  4  0    gt   ifelse      0  1       A common task is to do multiple replacements  Instead of nested ifelse statements  we can use case when from dplyr   case when v    0   1            v    1   2            TRUE   v      Old answer   For factor or character vectors  we can use revalue from plyr    gt  revalue c  a    b    c    c  b     B     1   a   B   c    This has the advantage of only specifying the input vector once  so we can use a pipe like   x   gt   revalue c  b     B

User · Answer

Another simpler option is to do     gt  x   c 1  1  2  4  5  2  1  3  2    gt  x x  1   lt - 0   gt  x   1  0 0 2 4 5 2 0 3 2

User · Answer

Why the fuss   replace haystack  haystack  in  needles  replacements    Demo   haystack  lt - c  q    w    e    r    t    y   needles  lt - c  q    w   replacements  lt - c  a    z    replace haystack  haystack  in  needles  replacements    gt   1   a   z   e   r   t   y

User · Answer

If you want to replace lot of values in single go  you can use  library car     Example  library car   x  lt - rep 1 5 3   xr  lt - recode x   3 1  4 2    x     1  1 2 3 4 5 1 2 3 4 5 1 2 3 4 5 xr     1  1 2 1 2 5 1 2 1 2 5 1 2 1 2 5

User · Answer

To replace more than one number   vec  lt - 1 10 replace vec  vec   c 2 6   c 0 9    2 and 6 will be replaced by 0 and 9    Edit   for a continous series  you can do this vec  lt - c 1 10   replace vec  vec  in  c 2 6   c 0 9    but for vec  lt - c 1 10 2 2 2   replace vec  vec  in  c 2 6   0  we can replace multiple values with one value

User · Answer

Perhaps replace is what you are looking for    gt  x   c 3  2  1  0  4  0   gt  replace x  x  0  1   1  3 2 1 1 4 1   Or  if you don t have x  any specific reason why not     replace c 3  2  1  0  4  0   c 3  2  1  0  4  0   0  1    Many people are familiar with gsub  so you can also try either of the following   as numeric gsub 0  1  x   as numeric gsub 0  1  c 3  2  1  0  4  0      Update  After reading the comments  perhaps with is an option   with data frame x   c 3  2  1  0  4  0    replace x  x    0  1

[r] Replace given value in vector

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