从R数据框中的列替换部分值

Question

I have a dataframe, example as follows.

        chr   start2     end2 value
88  chrom16 56063633 56063634 0.238
78  chrom12 83039622 83039623 0.429
50  chrom12 73209081 73209082 0.313
68  chrom12 75138610 75138611 0.679
45  chrom12 67566601 67566602 0.859
120 chrom16 57694245 57694246 0.438

I would like to change partial values from a column. In this dataframe, I would like to change "chrom" to "chr" in column 2.

OUTPUT should look like

     chr   start2     end2 value
88  chr16 56063633 56063634 0.238
78  chr12 83039622 83039623 0.429
50  chr12 73209081 73209082 0.313
68  chr12 75138610 75138611 0.679
45  chr12 67566601 67566602 0.859
120 chr16 57694245 57694246 0.438

Answer 1

If it the pattern is as simple as in your example you can just replace the "om" in "chrom" with an empty string.

df <- read.table(text = " chr   start2     end2 value
                              88  chrom16 56063633 56063634 0.238
                              78  chrom12 83039622 83039623 0.429
                              50  chrom12 73209081 73209082 0.313
                              68  chrom12 75138610 75138611 0.679
                              45  chrom12 67566601 67566602 0.859
                              120 chrom16 57694245 57694246 0.438", header = TRUE)

df$chr <- sub("om", "", df$chr)

df
#          chr   start2     end2 value
#    88  chr16 56063633 56063634 0.238
#    78  chr12 83039622 83039623 0.429
#    50  chr12 73209081 73209082 0.313
#    68  chr12 75138610 75138611 0.679
#    45  chr12 67566601 67566602 0.859
#    120 chr16 57694245 57694246 0.438

Answer 2

We can use sub to capture the first 3 characters as a group followed by matching 2 characters and then capture the numbers, replace it with the backreferences ( \\\\1 and \\\\2 for the captured groups).

df1$chr <- sub("(.{3}).{2}(\\d+)", "\\1\\2", df1$chr)
df1$chr
#[1] "chr16" "chr12" "chr12" "chr12" "chr12" "chr16"

Or we can also use lookarounds

sub(".{2}(?=\\d)", "", df1$chr, perl = TRUE)
#[1] "chr16" "chr12" "chr12" "chr12" "chr12" "chr16"

This should also work if the string elements are changed.

Or another faster option is substr based on the position of character.

df1$chr <- with(df1, paste0(substr(chr, 1, 3), substr(chr, 6,7)))
df1$chr
#[1] "chr16" "chr12" "chr12" "chr12" "chr12" "chr16"

Answer 3

Two possible ways:

data <- read.table(text = 'chr     start2   end2     value
                           chrom16 56063633 56063634 0.238
                           chrom12 83039622 83039623 0.429
                           chrom12 73209081 73209082 0.313
                           chrom12 75138610 75138611 0.679
                           chrom12 67566601 67566602 0.859
                           chrom16 57694245 57694246 0.438', 
                   stringsAsFactors = FALSE, 
                   header = TRUE)

# stringr package + base R for assignment
library(stringr)
data['chr'] <- str_replace(data[['chr']], "chrom", "chr")

data
#   chr   start2     end2 value
# 1 chr16 56063633 56063634 0.238
# 2 chr12 83039622 83039623 0.429
# 3 chr12 73209081 73209082 0.313
# 4 chr12 75138610 75138611 0.679
# 5 chr12 67566601 67566602 0.859
# 6 chr16 57694245 57694246 0.438

# with stringr and dplyr packages
library(dplyr)
data <- 
  data %>% 
  mutate(chr = str_replace(chr, "chrom", "chr"))

data
#   chr   start2     end2 value
# 1 chr16 56063633 56063634 0.238
# 2 chr12 83039622 83039623 0.429
# 3 chr12 73209081 73209082 0.313
# 4 chr12 75138610 75138611 0.679
# 5 chr12 67566601 67566602 0.859
# 6 chr16 57694245 57694246 0.438