How to split data frame with multiple delimiter using str_split_fixed?

Question

How can i split a column separated by multiple delimiter into separate columns in data frame

read.table(text = " Chr  Nm1 Nm2 Nm3
    chr10_100064111-100064134+Nfif   20  20 20
    chr10_100064115-100064138-Kitl   30 19 40
    chr10_100076865-100076888+Tert   60 440 18
    chr10_100079974-100079997-Itg    50 11 23                
    chr10_100466221-100466244+Tmtc3  55 24 53", header = TRUE)


              Chr              gene   Nm1 Nm2 Nm3
    chr10_100064111-100064134 Nfif   20  20 20
    chr10_100064115-100064138 Kitl   30 19 40
    chr10_100076865-100076888 Tert   60 440 18
    chr10_100079974-100079997 Itg    50 11 23 12                
    chr10_100466221-100466244 Tmtc3  55 24 53 12

i used

library(stringr)
df2 <- str_split_fixed(df1$name, "\\+", 2)

I would like to know how can we include both + and - delimiter

Use a regex character class in str_split with "[+-]" or use a pipe +|-. Also, I'm not sure you're gaining anything here from stringr that regular strsplit doesn't already do well. — alistaire
– alistaire, Commented Jun 16, 2016 at 14:57

alistaire · Accepted Answer · 2016-06-16 15:09:20Z

9

If you're trying to split one column into multiple, tidyr::separate is handy:

library(tidyr)

dat %>% separate(Chr, into = paste0('Chr', 1:3), sep = '[+-]')

#              Chr1      Chr2  Chr3 Nm1 Nm2 Nm3
# 1 chr10_100064111 100064134  Nfif  20  20  20
# 2 chr10_100064115 100064138  Kitl  30  19  40
# 3 chr10_100076865 100076888  Tert  60 440  18
# 4 chr10_100079974 100079997   Itg  50  11  23
# 5 chr10_100466221 100466244 Tmtc3  55  24  53

answered Jun 16, 2016 at 15:09

alistaire

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Matias Thayer · Accepted Answer · 2016-06-16 14:57:44Z

3

This should work:

str_split_fixed(a, "[-+]", 2)

answered Jun 16, 2016 at 14:57

Matias Thayer

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3 Comments

alistaire Over a year ago

You need to use str_split (or strsplit) if you're using regex, not str_split_fixed.

Matias Thayer Over a year ago

From the str_split_fixed documentation: "The default interpretation is a regular expression, as described in stringi-search-regex. Control options with regex()" Also I tested it and seems to work

alistaire Over a year ago

Hmm, you are correct! ...though that's a very confusingly named set of functions given base R regex fixed = TRUE behavior.

lmo · Accepted Answer · 2016-06-16 15:06:06Z

2

Here is a way to do this in base R with strsplit:

# split Chr into a list
tempList <- strsplit(as.character(df$Chr), split="[+-]")

# replace Chr with desired values
df$Chr <- sapply(tempList, function(i) paste(i[[1]], i[[2]], sep="-"))

# get Gene variable
df$gene <- sapply(tempList, "[[", 3)

answered Jun 16, 2016 at 15:06

lmo

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2 Comments

beginner Over a year ago

when tried df$gene <- sapply(tempList, "[[", 3) Error in FUN(X[[i]], ...) : subscript out of bounds

lmo Over a year ago

@beginner I just copied and pasted your example data.frame, naming it df, and then copied and pasted my suggested solution and did not receive this error.

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