I'm very confused by the phrase "string alienation".
You mention two problems:
(1) remove " from a string
sub('"', '', vector.of.strings)
will do that. See
?grep
for details.
(2) split a string at occurrences of /
strsplit(vector.of.strings, "/")
will do that.
ssage-
> From: R-help On Behalf Of Soumyadip
> Bhattacharyya
> Sent: Tuesday, April 14, 2020 12:55 PM
> To: r-help-requ...@r-project.org; r-help-ow...@r-project.org; r-help@r-
> project.org; ericjber...@gmail.com
> Subject: [R] A simple string alienation problem
>
> ***Dear
***Dear Eric,*
sending from gmail following the way you suggested. Hope now everyone
can see this email. I have also attached the first 50 rows of the
FIght.csv.***
***Output - I will try to do Market basket analysis on this to find
out rules that I am learning. so once I have the data in
Hi Sam,
My code below adds new columns to your data frame so you have the original
columns in order to compare.
(Also this could help in case there are a few rows that don't work in the
full set.)
> x <- read.csv("Fight.csv", stringsAsFactors = F, header = F)
> x$V3 <- sub("\\\"","",x$V1) # remo
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