Hi
attachement did not went through, only limited attachement types are allowed -
see Posting guide.
I am not sure if R is the best possibility to remove some characters. If " is
at the end of all your strings
> dput(test)
structure(list(V1 = c("adfvadfg\"", "sdfasd\"", "vafdv\"", "hjk/tiuk\""
)), class = "data.frame", row.names = c(NA, -4L))
> test2 <- sapply(test, function(x) substr(x, 1, as.numeric(apply(test, 2,
> nchar))-1))
combination of sapply, apply and substr could remove trailing ".
> test2
V1
[1,] "adfvadfg"
[2,] "sdfasd"
[3,] "vafdv"
[4,] "hjk/tiuk"
>
And splitting acccording to / is simpler but it ends in list
> sapply(test2, function(x) strsplit(x, "/"))
$adfvadfg
[1] "adfvadfg"
$sdfasd
[1] "sdfasd"
$vafdv
[1] "vafdv"
$`hjk/tiuk`
[1] "hjk" "tiuk"
Changing to data frame you could find yourself, I believe it is mentioned
several times on Stackexchange, simple as.data.frame is not a best option.
Cheers
Petr
> -----Original Message-----
> From: R-help <r-help-boun...@r-project.org> 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 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 transactional format - then I
> can
> run the algorithm and keep learning. This little problem has caused a barrier
> in my path - I can alienate the string in excel - but wanted to do in R - so
> researching I tried doing this:
> x<- substr(x, 1, nchar(x) - 1) // but I wasn't successful and I tried many
> other
> things - its not coming in the transactional format. *** Hence now reached
> out to the experts.**** Many Thanks.
>
> Hello Dear R Community,
>
> I would ask a little bit of help from you please:I have a dataset, which is
> in a
> CSV file – I have read it into R as follows:
>
> V1
> tropical fruit"
> whole milk"
> pip fruit"
> other vegetables"
> whole milk"
> rolls/buns"
>
> The issue is: the data set in csv file also appears with the quotation marks
> “. I
> can’t get rid of the quotation marks. I want to do it in R.
> The Quotes only appear at the end of the string. The dataset has many rows –
> this is just a copy. My intention is to be able to get rid of the quotes and
> then
> want to separate the strings with a ‘/’. i.e.
> rolls/buns should be rolls in one column and buns in another.
>
> I know this is something very simple I am lacking – but if you could please
> show me how to do this? If someone could throw some light please. I read
> the data in with a simple read.csv statement:
>
> > x <- read.csv("Fight.csv", stringsAsFactors = F, header = F)
> > str(x)
> Output:
> > str(calc)
> 'data.frame': 38765 obs. of 1 variable:
> $ V1: chr "tropical fruit\"" "whole milk\"" "pip fruit\"" "other
> vegetables\"" ...
>
> Many Thanks in advance for your help.
> Kind Regards,
> Sam.
> ______________________________________________
> R-help@r-project.org mailing list -- To UNSUBSCRIBE and more, see
> https://stat.ethz.ch/mailman/listinfo/r-help
> PLEASE do read the posting guide http://www.R-project.org/posting-
> guide.html
> and provide commented, minimal, self-contained, reproducible code.
______________________________________________
R-help@r-project.org mailing list -- To UNSUBSCRIBE and more, see
https://stat.ethz.ch/mailman/listinfo/r-help
PLEASE do read the posting guide http://www.R-project.org/posting-guide.html
and provide commented, minimal, self-contained, reproducible code.
