I need to rename a bunch of files, by searching for string matches in a
list. Each list element containings a character with the old filename that
I want to match to, and the new file name that I want to rename by.
For instance, here filename '1001.xls' should match to list[[1]]$oldname
and I want to rename it to 'newa.xls' from list[[1]]$newname.
The actual new file names I have will feature a random alphanumeric number
and the list will have a length of ~600.
# files I want to rename
files with old file name =c('1001.xls', '1002.xls')
# list with old file names and new file names
oldnames=c('1001', '1002', '1003')
newnames=c('newa', 'newb', 'newc')
df=data.frame(oldnames,newnames)
list <- split(df, rownames(df))
# turn list elements in character
for(i in 1:length(list)) list[[i]]$oldnames=as.character(list[[i]]$oldnames)
for(i in 1:length(list)) list[[i]]$newnames=as.character(list[[i]]$newnames)
I heard that it would be better to vectorize this than trying to do a
double loop so if someone could give me a hint about how to do this I would
be very grateful!
Suggestions for vectorizing/double loop
6 messages · Thierry Onkelinx, Michael Dewey, Luanna Dixson +1 more
Dear Luanna, Assuming that oldnames and newnames are character (and not factor), the just use stringsAsFactors = FALSE. That will save you from having to convert the factors back to character. data.frame(oldnames, newnames, stringsAsFactor = FALSE) Best regards, ir. Thierry Onkelinx Instituut voor natuur- en bosonderzoek / Research Institute for Nature and Forest team Biometrie & Kwaliteitszorg / team Biometrics & Quality Assurance Kliniekstraat 25 1070 Anderlecht Belgium To call in the statistician after the experiment is done may be no more than asking him to perform a post-mortem examination: he may be able to say what the experiment died of. ~ Sir Ronald Aylmer Fisher The plural of anecdote is not data. ~ Roger Brinner The combination of some data and an aching desire for an answer does not ensure that a reasonable answer can be extracted from a given body of data. ~ John Tukey 2017-01-23 13:28 GMT+01:00 Luanna Dixson <justanotherdigression at gmail.com>:
I need to rename a bunch of files, by searching for string matches in a
list. Each list element containings a character with the old filename that
I want to match to, and the new file name that I want to rename by.
For instance, here filename '1001.xls' should match to list[[1]]$oldname
and I want to rename it to 'newa.xls' from list[[1]]$newname.
The actual new file names I have will feature a random alphanumeric number
and the list will have a length of ~600.
# files I want to rename
files with old file name =c('1001.xls', '1002.xls')
# list with old file names and new file names
oldnames=c('1001', '1002', '1003')
newnames=c('newa', 'newb', 'newc')
df=data.frame(oldnames,newnames)
list <- split(df, rownames(df))
# turn list elements in character
for(i in 1:length(list)) list[[i]]$oldnames=as.
character(list[[i]]$oldnames)
for(i in 1:length(list)) list[[i]]$newnames=as.
character(list[[i]]$newnames)
I heard that it would be better to vectorize this than trying to do a
double loop so if someone could give me a hint about how to do this I would
be very grateful!
[[alternative HTML version deleted]]
______________________________________________ R-help at 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.
Dear Luanna It is not compulsory to avoid for loops but see below
On 23/01/2017 12:28, Luanna Dixson wrote:
I need to rename a bunch of files, by searching for string matches in a
list. Each list element containings a character with the old filename that
I want to match to, and the new file name that I want to rename by.
For instance, here filename '1001.xls' should match to list[[1]]$oldname
and I want to rename it to 'newa.xls' from list[[1]]$newname.
The actual new file names I have will feature a random alphanumeric number
and the list will have a length of ~600.
# files I want to rename
files with old file name =c('1001.xls', '1002.xls')
# list with old file names and new file names
oldnames=c('1001', '1002', '1003')
newnames=c('newa', 'newb', 'newc')
df=data.frame(oldnames,newnames)
list <- split(df, rownames(df))
Both df and list are already in existience and you are overwriting them. It is best practice not to do that.
# turn list elements in character for(i in 1:length(list)) list[[i]]$oldnames=as.character(list[[i]]$oldnames)
Something like lapply(list, function(x) x$oldnames <- as.character(x$oldnames)) would probably work. You might want sapply instead of lapply
for(i in 1:length(list)) list[[i]]$newnames=as.character(list[[i]]$newnames) I heard that it would be better to vectorize this than trying to do a double loop so if someone could give me a hint about how to do this I would be very grateful! [[alternative HTML version deleted]]
______________________________________________ R-help at 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.
Hi In your case if you want to rename abot 600 items and your doble loop works it is not worth to vectorize it. However if you want to repeat such task often and you expect much bigger number of files vectorizing can speed things up and make them cleaner. Although I must admit that I do not understand what your commands should actually do. AFAICU the list is the same before and after your "double" loop. Cheers Petr
-----Original Message-----
From: R-help [mailto:r-help-bounces at r-project.org] On Behalf Of Luanna
Dixson
Sent: Monday, January 23, 2017 1:28 PM
To: r-help at r-project.org
Subject: [R] Suggestions for vectorizing/double loop
I need to rename a bunch of files, by searching for string matches in a list.
Each list element containings a character with the old filename that I want to
match to, and the new file name that I want to rename by.
For instance, here filename '1001.xls' should match to list[[1]]$oldname and I
want to rename it to 'newa.xls' from list[[1]]$newname.
The actual new file names I have will feature a random alphanumeric number
and the list will have a length of ~600.
# files I want to rename
files with old file name =c('1001.xls', '1002.xls')
# list with old file names and new file names
oldnames=c('1001', '1002', '1003')
newnames=c('newa', 'newb', 'newc')
df=data.frame(oldnames,newnames)
list <- split(df, rownames(df))
# turn list elements in character
for(i in 1:length(list)) list[[i]]$oldnames=as.character(list[[i]]$oldnames)
for(i in 1:length(list)) list[[i]]$newnames=as.character(list[[i]]$newnames)
I heard that it would be better to vectorize this than trying to do a double
loop so if someone could give me a hint about how to do this I would be very
grateful!
[[alternative HTML version deleted]]
______________________________________________ R-help at 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.
________________________________ Tento e-mail a jak?koliv k n?mu p?ipojen? dokumenty jsou d?v?rn? a jsou ur?eny pouze jeho adres?t?m. Jestli?e jste obdr?el(a) tento e-mail omylem, informujte laskav? neprodlen? jeho odes?latele. Obsah tohoto emailu i s p??lohami a jeho kopie vyma?te ze sv?ho syst?mu. Nejste-li zam??len?m adres?tem tohoto emailu, nejste opr?vn?ni tento email jakkoliv u??vat, roz?i?ovat, kop?rovat ?i zve?ej?ovat. Odes?latel e-mailu neodpov?d? za eventu?ln? ?kodu zp?sobenou modifikacemi ?i zpo?d?n?m p?enosu e-mailu. V p??pad?, ?e je tento e-mail sou??st? obchodn?ho jedn?n?: - vyhrazuje si odes?latel pr?vo ukon?it kdykoliv jedn?n? o uzav?en? smlouvy, a to z jak?hokoliv d?vodu i bez uveden? d?vodu. - a obsahuje-li nab?dku, je adres?t opr?vn?n nab?dku bezodkladn? p?ijmout; Odes?latel tohoto e-mailu (nab?dky) vylu?uje p?ijet? nab?dky ze strany p??jemce s dodatkem ?i odchylkou. - trv? odes?latel na tom, ?e p??slu?n? smlouva je uzav?ena teprve v?slovn?m dosa?en?m shody na v?ech jej?ch n?le?itostech. - odes?latel tohoto emailu informuje, ?e nen? opr?vn?n uzav?rat za spole?nost ??dn? smlouvy s v?jimkou p??pad?, kdy k tomu byl p?semn? zmocn?n nebo p?semn? pov??en a takov? pov??en? nebo pln? moc byly adres?tovi tohoto emailu p??padn? osob?, kterou adres?t zastupuje, p?edlo?eny nebo jejich existence je adres?tovi ?i osob? j?m zastoupen? zn?m?. This e-mail and any documents attached to it may be confidential and are intended only for its intended recipients. If you received this e-mail by mistake, please immediately inform its sender. Delete the contents of this e-mail with all attachments and its copies from your system. If you are not the intended recipient of this e-mail, you are not authorized to use, disseminate, copy or disclose this e-mail in any manner. The sender of this e-mail shall not be liable for any possible damage caused by modifications of the e-mail or by delay with transfer of the email. In case that this e-mail forms part of business dealings: - the sender reserves the right to end negotiations about entering into a contract in any time, for any reason, and without stating any reasoning. - if the e-mail contains an offer, the recipient is entitled to immediately accept such offer; The sender of this e-mail (offer) excludes any acceptance of the offer on the part of the recipient containing any amendment or variation. - the sender insists on that the respective contract is concluded only upon an express mutual agreement on all its aspects. - the sender of this e-mail informs that he/she is not authorized to enter into any contracts on behalf of the company except for cases in which he/she is expressly authorized to do so in writing, and such authorization or power of attorney is submitted to the recipient or the person represented by the recipient, or the existence of such authorization is known to the recipient of the person represented by the recipient.
Hi all,
Thanks very much for your help! You are correct in thinking the list is the
same as before, actually, my question was more about how to do the next
steps, where I needed to match the filenames of the files in my directory
with old (i.e current) and new file name prefixes in my list. For each
match I then wanted to rename the original file using the corresponding new
filename prefix from the list.
Sorry for being a bit confusing, I didn't post my first attempt to do this
at first as my code just didn't work at all, but I have had another go
using a matrix instead and I think this does the job.
old_path="./old/"
new_path="./new/"
old_names <- list.files(old_path)
head(old_names)
oldnames=c('1002', '1003')
newnames=c('1002_new', '1003_new')
mapping=cbind(oldnames,newnames)
head(mapping)
for (i in old_names){
temp <- unlist(strsplit(i, "[.]"))[1]
n <- which(is.element(mapping,temp))
if(length(n)>0) {
#copy the file to the new folder
#re name it and the name is paste(mapping[n,2], '.xls', sep="")
print(paste(mapping[n,2], '.xls', sep=""))
newnames <- paste(mapping[n,2], '.xls', sep="")
file.copy(from = paste(old_path, i, sep=""),
to = paste(new_path, newnames))
}
}
On 23 January 2017 at 13:28, Luanna Dixson <justanotherdigression at gmail.com>
wrote:
I need to rename a bunch of files, by searching for string matches in a
list. Each list element containings a character with the old filename that
I want to match to, and the new file name that I want to rename by.
For instance, here filename '1001.xls' should match to list[[1]]$oldname
and I want to rename it to 'newa.xls' from list[[1]]$newname.
The actual new file names I have will feature a random alphanumeric number
and the list will have a length of ~600.
# files I want to rename
files with old file name =c('1001.xls', '1002.xls')
# list with old file names and new file names
oldnames=c('1001', '1002', '1003')
newnames=c('newa', 'newb', 'newc')
df=data.frame(oldnames,newnames)
list <- split(df, rownames(df))
# turn list elements in character
for(i in 1:length(list)) list[[i]]$oldnames=as.
character(list[[i]]$oldnames)
for(i in 1:length(list)) list[[i]]$newnames=as.
character(list[[i]]$newnames)
I heard that it would be better to vectorize this than trying to do a
double loop so if someone could give me a hint about how to do this I would
be very grateful!
Well, your second post is rather confusing too, as is your not reproducible code. If I understand correctly, you just want to change names of files in a directory according to some rules, which you did not clerly specify. First I am not sure if R is the best tool for it. Nevertheless, you do not change a contents of any file, just change the name. I found this with simple answer questioning internet. http://stackoverflow.com/questions/10758965/how-do-i-rename-files-using-r Is this what you want? Here is another possibility https://www.r-bloggers.com/operating-on-files-with-r-copy-and-rename/ Anyway the concept I would use is. #read old names old_names <- list.files(old_path) #setting new names (you did not explain how do you want to do this) new_names <- some vector of new names If order of names is correct file.rename(old_names, new_names) othervise it would be needed correctly order both vectors before file.rename. The crucial information is how do you want set the new names vector. You shold specify it by reproducible way if you want more precise answer. Copying files is in this case best done by OS itself. Cheers Petr
-----Original Message-----
From: R-help [mailto:r-help-bounces at r-project.org] On Behalf Of Luanna
Dixson
Sent: Monday, January 23, 2017 6:39 PM
To: r-help at r-project.org
Subject: Re: [R] Suggestions for vectorizing/double loop
Hi all,
Thanks very much for your help! You are correct in thinking the list is the
same as before, actually, my question was more about how to do the next
steps, where I needed to match the filenames of the files in my directory
with old (i.e current) and new file name prefixes in my list. For each match I
then wanted to rename the original file using the corresponding new
filename prefix from the list.
Sorry for being a bit confusing, I didn't post my first attempt to do this at first
as my code just didn't work at all, but I have had another go using a matrix
instead and I think this does the job.
old_path="./old/"
new_path="./new/"
old_names <- list.files(old_path)
head(old_names)
oldnames=c('1002', '1003')
newnames=c('1002_new', '1003_new')
mapping=cbind(oldnames,newnames)
head(mapping)
for (i in old_names){
temp <- unlist(strsplit(i, "[.]"))[1]
n <- which(is.element(mapping,temp))
if(length(n)>0) {
#copy the file to the new folder
#re name it and the name is paste(mapping[n,2], '.xls', sep="")
print(paste(mapping[n,2], '.xls', sep=""))
newnames <- paste(mapping[n,2], '.xls', sep="")
file.copy(from = paste(old_path, i, sep=""),
to = paste(new_path, newnames))
}
}
On 23 January 2017 at 13:28, Luanna Dixson
<justanotherdigression at gmail.com>
wrote:
I need to rename a bunch of files, by searching for string matches in a list. Each list element containings a character with the old filename that I want to match to, and the new file name that I want to
rename by.
For instance, here filename '1001.xls' should match to list[[1]]$oldname and I want to rename it to 'newa.xls' from
list[[1]]$newname.
The actual new file names I have will feature a random alphanumeric
number and the list will have a length of ~600.
# files I want to rename
files with old file name =c('1001.xls', '1002.xls')
# list with old file names and new file names
oldnames=c('1001', '1002', '1003')
newnames=c('newa', 'newb', 'newc')
df=data.frame(oldnames,newnames)
list <- split(df, rownames(df))
# turn list elements in character
for(i in 1:length(list)) list[[i]]$oldnames=as.
character(list[[i]]$oldnames)
for(i in 1:length(list)) list[[i]]$newnames=as.
character(list[[i]]$newnames)
I heard that it would be better to vectorize this than trying to do a
double loop so if someone could give me a hint about how to do this I
would be very grateful!
[[alternative HTML version deleted]]
______________________________________________ R-help at 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.
________________________________ Tento e-mail a jak?koliv k n?mu p?ipojen? dokumenty jsou d?v?rn? a jsou ur?eny pouze jeho adres?t?m. Jestli?e jste obdr?el(a) tento e-mail omylem, informujte laskav? neprodlen? jeho odes?latele. Obsah tohoto emailu i s p??lohami a jeho kopie vyma?te ze sv?ho syst?mu. Nejste-li zam??len?m adres?tem tohoto emailu, nejste opr?vn?ni tento email jakkoliv u??vat, roz?i?ovat, kop?rovat ?i zve?ej?ovat. Odes?latel e-mailu neodpov?d? za eventu?ln? ?kodu zp?sobenou modifikacemi ?i zpo?d?n?m p?enosu e-mailu. V p??pad?, ?e je tento e-mail sou??st? obchodn?ho jedn?n?: - vyhrazuje si odes?latel pr?vo ukon?it kdykoliv jedn?n? o uzav?en? smlouvy, a to z jak?hokoliv d?vodu i bez uveden? d?vodu. - a obsahuje-li nab?dku, je adres?t opr?vn?n nab?dku bezodkladn? p?ijmout; Odes?latel tohoto e-mailu (nab?dky) vylu?uje p?ijet? nab?dky ze strany p??jemce s dodatkem ?i odchylkou. - trv? odes?latel na tom, ?e p??slu?n? smlouva je uzav?ena teprve v?slovn?m dosa?en?m shody na v?ech jej?ch n?le?itostech. - odes?latel tohoto emailu informuje, ?e nen? opr?vn?n uzav?rat za spole?nost ??dn? smlouvy s v?jimkou p??pad?, kdy k tomu byl p?semn? zmocn?n nebo p?semn? pov??en a takov? pov??en? nebo pln? moc byly adres?tovi tohoto emailu p??padn? osob?, kterou adres?t zastupuje, p?edlo?eny nebo jejich existence je adres?tovi ?i osob? j?m zastoupen? zn?m?. This e-mail and any documents attached to it may be confidential and are intended only for its intended recipients. If you received this e-mail by mistake, please immediately inform its sender. Delete the contents of this e-mail with all attachments and its copies from your system. If you are not the intended recipient of this e-mail, you are not authorized to use, disseminate, copy or disclose this e-mail in any manner. The sender of this e-mail shall not be liable for any possible damage caused by modifications of the e-mail or by delay with transfer of the email. In case that this e-mail forms part of business dealings: - the sender reserves the right to end negotiations about entering into a contract in any time, for any reason, and without stating any reasoning. - if the e-mail contains an offer, the recipient is entitled to immediately accept such offer; The sender of this e-mail (offer) excludes any acceptance of the offer on the part of the recipient containing any amendment or variation. - the sender insists on that the respective contract is concluded only upon an express mutual agreement on all its aspects. - the sender of this e-mail informs that he/she is not authorized to enter into any contracts on behalf of the company except for cases in which he/she is expressly authorized to do so in writing, and such authorization or power of attorney is submitted to the recipient or the person represented by the recipient, or the existence of such authorization is known to the recipient of the person represented by the recipient.