How to replace specific values in a column with other values?-CodePudding

I have a data frame and want to add a new column to it based on another column and then replace its values.

For example column ID_old is what I have:

df1 <- structure(list(ID.old=c(1,1,1,  2,2,  3,3,3,3,  4,4,  5,5,5,5,5,  6,6,6,  7,7,7,7,  8,8,  9, 10,10,10, 11,11,  12,12,12, 13,13,  14,14,14,14, 15,15,  16, 17,17, 18, 19,19,19, 20,20,20)),
                 class = "data.frame", row.names = c(NA,-52L))

and now column ID_new is what I need:

df2 <- structure(list(ID.old=c(1,1,1,  2,2,  3,3,3,3,  4,4,  5,5,5,5,5,  6,6,6,  7,7,7,7,  8,8,  9, 10,10,10, 11,11,  12,12,12, 13,13,  14,14,14,14, 15,15,  16, 17,17, 18, 19,19,19, 20,20,20),
                      ID.new=c('a1','a1','a1', 'a2','a2', 'a3','a3','a3','a3', 'a4','a4', 'a5','a5','a5','a5','a5', 'a1','a1','a1', 'a2','a2','a2','a2', 'a3','a3', 'a4', 'a5','a5','a5', 'a1','a1', 'a2','a2','a2', 'a3','a3', 'a4','a4','a4','a4', 'a5','a5', 'a1', 'a2','a2', 'a3', 'a4','a4','a4', 'a5','a5','a5')),
                 class = "data.frame", row.names = c(NA,-52L))

I thought that I can use str_replace_all from stringer, but it produces something different,

library(stringr)
df1<- df1 %>% 
  mutate(ID.new = ID.old)
replace = c("1"="a1", "2"="a2", "3"="a3", "4"="a4", "5"="a5",
            "6"="a1", "7"="a2", "8"="a3", "9"="a4", "10"="a5",
            "11"="a1", "12"="a2", "13"="a3", "14"="a4", "15"="a5",
            "16"="a1", "17"="a2", "18"="a3", "19"="a4", "20"="a5")

df1$ID.new<- str_replace_all(df1$ID.new, replace)

in my original data frame, I have many rows, and specifically, I need wherever it is 1,6,11,16 to be "a1".

2,7,12,17 to be "a2" etc.

How can I get a column like what we have in df2 ID.new Thanks

CodePudding user response：

You could use modulo %% and replace zeros with 5.

res <- transform(df1, ID.new=paste0('a', ID.old %% 5 |> {\(.) replace(., .  == 0, 5)}()))

head(res, 17)
#    ID.old ID.new
# 1       1     a1
# 2       1     a1
# 3       1     a1
# 4       2     a2
# 5       2     a2
# 6       3     a3
# 7       3     a3
# 8       3     a3
# 9       3     a3
# 10      4     a4
# 11      4     a4
# 12      5     a5
# 13      5     a5
# 14      5     a5
# 15      5     a5
# 16      5     a5
# 17      6     a1

Data:

df1 <- structure(list(ID.old = c(1, 1, 1, 2, 2, 3, 3, 3, 3, 4, 4, 5, 
5, 5, 5, 5, 6, 6, 6, 7, 7, 7, 7, 8, 8, 9, 10, 10, 10, 11, 11, 
12, 12, 12, 13, 13, 14, 14, 14, 14, 15, 15, 16, 17, 17, 18, 19, 
19, 19, 20, 20, 20)), class = "data.frame", row.names = c(NA, 
-52L))

CodePudding user response：

stringr::str_replace_all is based on regex. For example, with your 'replace' dictionnary, it replaces every 1 it encounters with "a1", so the number '11' is replaced by "a1a1", as it contains two successive 1. Since you have already designed a dictionary, you should simply add 'start' (^) and end ($) regex tags, as I suggest below:

Simply add this line of code after the creation of your actual 'replace' dictionnary:

names(replace) = paste0("^", names(replace), "$")

And know the replacement is correct if you proceed again df1$ID.new<- str_replace_all(df1$ID.new, replace)