Parse string by |

Question

I have a list of strings which look like that:

However, I only want get the first string. In this case Music(without |). I tried:

sub(categories, pattern = " |", replacement = "")

However, that does not give me the desired result. Any recommendation how to correctly parse my string?

I appreciate your answer!

UPDATE

> dput(head(df))
structure(list(data.founded_at = c("01.06.2012", "26.10.2012", 
"01.04.2011", "01.01.2012", "10.10.2011", "01.01.2007"), data.category_list = c("|Entertainment|Politics|Social Media|News|", 
"|Publishing|Education|", "|Electronics|Guides|Coffee|Restaurants|Music|iPhone|Apps|Mobile|iOS|E-Commerce|", 
"|Software|", "|Software|", "|Curated Web|")), .Names = c("data.founded_at", 
"data.category_list"), row.names = c(NA, 6L), class = "data.frame")

Several problems: 1) Your arguments for sub are in the wrong order. 2) You need to escape | with pattern = "\\|" 3) sub will only replace characters, not split strings. — Señor O
– Señor O, Commented Aug 13, 2014 at 14:37

A5C1D2H2I1M1N2O1R2T1 · Accepted Answer · 2014-08-13 15:03:21Z

3

An alternative for this could be scan:

na.omit(scan(text = categories, sep = "|", what = "", na.strings = ""))[1]
# Read 6 items
# [1] "Music"

answered Aug 13, 2014 at 15:03

A5C1D2H2I1M1N2O1R2T1

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duffymo · Accepted Answer · 2014-08-13 14:25:31Z

1

Find a function that will tokenize a string at a particular character: strsplit would be my guess.

http://stat.ethz.ch/R-manual/R-devel/library/base/html/strsplit.html

answered Aug 13, 2014 at 14:25

duffymo

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momobo · Accepted Answer · 2014-08-14 06:42:58Z

1

Note that the parameter in split is a regexp, so using split="|" will not work (unless you specify fixed=TRUE, as suggested from joran -thanks- in the comments)

strsplit(categories,split="[|]")[[1]][2]

To apply this to the data frame you could do this:

sapply(df$data.category_list, function(x) strsplit(x,split="[|]")[[1]][2])

But this is faster (see the comments):

vapply(strsplit(df$data.category_list, "|", fixed = TRUE), `[`, character(1L), 2)

(thanks to Ananda Mahto)

edited Aug 14, 2014 at 6:42

answered Aug 13, 2014 at 14:29

momobo

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7 Comments

lawyeR Over a year ago

And, if you use strsplit, which returns a list, you might want to call unlist on the expression @momobo provided. That will return a vector.

Carol.Kar Over a year ago

@momobo Thx for your answer! How to do that on a dataframe?

Carol.Kar Over a year ago

@AnandaMahto I have added my used dataframe in an update. I would appreciate an answer from you, that I will gladly mark as accepted!

momobo Over a year ago

also df$firstCat <- sapply(df$data.category_list, function(x){strsplit(x,split="[|]")[[1]][2]})

momobo Over a year ago

Ok. Yours is faster. :)

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