Basically,i want to copy files from one directory to another.The condition is that the file must not have as first letter a capital and must end in (.txt).
(This is my first question in stackoverflow so be gentle )
This do the trick cp [a-z]*.txt directory
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I have am writing an R script that performs a function for all files in a series of subdirectories. I have ran into a problem where several files in these subdirectories are being recognized by my glob2rx function, and I need help refining my pattern so I can select the file I want.
Here is an example of my directory structure:
subdir1
file1_aaa_111_subdir1.txt
file1_bbb_111_subdir1.txt
file1_aaa_subdir1.txt
subdir2
file1_aaa_111_subdir2.txt
file1_bbb_111_subdir2.txt
file1_aaa_subdir2.txt
I want to select for the last file in each directory, although in my actual directory its position is varied. I want to use something like:
inFilePaths = list.files(path=".", pattern=glob2rx("*aaa*.txt"), full.names=TRUE)
but I dont get any files. In looking at this pattern, I would in theory get both the first and last file in each directory. Meaning I need to write an exception to exclude the aaa_111 files, and keep the aaa_subdir files.
There is a second option I have been thinking about, but lack the ability to realize. Notice the name of the subdirectory is at the end of each file name. Is it possible to extract the directory name, and then combine it with a glob2rx pattern, and then directly specify which file I want? Like this:
#list all the subdirectories
subDirsPaths = list.dirs(path=".", full.names=TRUE)
#perform a function on these directories one by one
for (subDirsPath in subDirsPaths){
#make the subdirectory the working directory
setwd("/home/phil/Desktop/working")
setwd(paste(subDirsPath, sep=""))
# get the working directory name, and trim the "./" from it
directory <- gsub("./", "", paste(subDirsPath, sep=""))
# attempt to the get the desired file by pasting the directory name into the glob2rx funtion
inFilePaths = list.files(path=".", pattern=glob2rx("*aaa_", print(directory,".txt")), full.names=TRUE)
for (inFilePath in inFilePaths)
{
inFileData <- read_tsv(inFilePath, col_names=TRUE)
}
}
With some modification the second option worked well. I ended up using paste in combination with print as follows:
inFilePaths = list.files(path=".", pattern=glob2rx(print(paste("*", "aaa_", directory, ".txt", sep=""))), full.names=TRUE)
The paste function combined the text into a single string, which also preserved the wildcard. The print function added this to the list.files function as the glob2rx pattern.
While this doesn't allow me to place a wild card in the middle of an expression, which I believe is done use an escape character, and it doesn't address the need to place exceptions on the wild card, it works for my purposes.
I hope this helps others in my position.
I have 1,500 7zip archives, each archive contains 2 to 10 files, with no subdirectories.
Each file has the same extension, however the filename varies.
I only want one file out of each archive, but I'd like to perform this in bulk. I do not care which file is taken out, as long as only one file is taken out. It can be the first file, the newest, the biggest, the smallest, it doesn't matter.
Here's an example:
aa.7z {blah 56.smc, blah 57.smc, 1 blah 58.smc}
ab.7z {xx.smc, xx 1.smc, xx_2.smc}
ac.7z {1.smc}
I want to run something equivalent to:
7z e *.7z # But somehow only extract one file
Thank you!
Ultimately my solution was to extract all files and run the following in the directory:
for n in *; do echo "$n"; done > files.txt
I then imported that list into excel, and split the files by a special character that divided the title of the file with the qualifying data inside the filename (for example: Some Title (V1) [X2].smc), specifically I used a brackets delimiter.
Then I removed all duplicates, leaving me with only one edition of each from the zip. I finally remerged the columns (unfortunately the bracket was deleted during the splitting so wrote a function to add it back on the condition of whether there was content in the next column) and then resaved files.txt, after a bit of reviewing StackOverflow for answers, deleted files based on an input file (files.txt). A word of warning on this, spaces in filenames cause problems with rm and xargs so I had to encapsulate the variable with quotes.
Ultimately this still didn't serve me well enough so I just used a different resource entirely.
Posting this answer so others who find themselves in a similar predicament find an alternative resolution.
I would like to get the list of files with a specific extention in a folder. However, these files has space in the name. So for example, imagining I have files named file test1.txt, file test2.txt, file test3.txt, file test4.txt, if I do
list.files(pattern="file test*.txt")
I got
character(0)
NOTA: Apparentely, using simply pattern="file test*" it works fine but I need the extention file as well.
Try:
list.files(pattern="file test.*.txt")
Actually, what this says is:
list.files(pattern="file test(.*).txt")
(which also works). . refers to any character and * refers to the idea that this character should be present 0 or more times (see ?regex).
In your kast example you said that using pattern="file test*" works but you need a way to search for the extension as well.
All you have to do is Change your code to pattern="file test.*.txt". This would make your code search for any filename that matched "file testX.txt" with any one character in place of X.
TLDNR: How do I use Sys.glob () within unzip ()?
I have multiple .zip files and I want to extract only one file from each archive.
For example, one of the archives contains the following files:
[1] "cmc-20150531.xml" "cmc-20150531.xsd" "cmc-20150531_cal.xml" "cmc-20150531_def.xml" "cmc-20150531_lab.xml"
[6] "cmc-20150531_pre.xml"
I want to extract the first file because it matches a pattern. In order to do that I use the following command:
unzip("zip-archive.zip", files=Sys.glob("[a-z][a-z][a-z][-][0-9][0-9][0-9][0-9][0-9][0-9][0-9][0-9][.][x][m][l]"))
However, the command doesn't work, and I don't know why. R just extracts all files in the archive.
On the other hand, the following command works:
unzip("zip-archive.zip", files="cmc-20150531.xml")
How do I use Sys.glob() within unzip()?
Sys.glob expands files that already exist. So the parameter to your unzip call will depend on what files are in your working directory.
Perhaps you want to do unzip with list=TRUE to return the list of files in the zip first, and then use some pattern matching to select the files you want.
See ?grep for info on matching strings with patterns. These patterns are "regular expressions" rather than "glob" expansions, but you should be able to work with that.
Here's a concrete example:
# whats in the zip?
files = unzip("c.zip", list=TRUE)$Name
files
[1] "l_spatial.dbf" "l_spatial.shp" "l_spatial.shx" "ls_polys_bin.dbf"
[5] "ls_polys_bin.shp" "ls_polys_bin.shx" "rast_jan90.tif"
# what files have "dbf" in them:
files[grepl("dbf",files)]
[1] "l_spatial.dbf" "ls_polys_bin.dbf"
# extract just those:
unzip("c.zip", files=files[grepl("dbf",files)])
The regular expression for your glob
"[a-z][a-z][a-z][-][0-9][0-9][0-9][0-9][0-9][0-9][0-9][0-9][.][x][m][l]"
would be
"^[a-z]{3}-[0-9]{8}\\.xml$"
that's a match of start of string ("^"), 3 a-z (lower case only), a dash, eight digits, a dot (backslashes are needed, one because dot means "any one char" in regexps and another because R needs a backslash to escape a backslash), "xml", and the end of the string ("$").
Just with any other collections do an itertive loop through the results from Sys.glob and supply the itertive holding variable to unzip. This is achieved by using a for-loop
While unzip() takes an argument for the path, and files is an arugment for what files within that zip file.
Mined you I'm more a full stack programmer not so much so on the R lang, but the concepts are the same; so the code should something like:
files <- Sys.glob(path_expand(".","*.zip"))
for (idx in 1:length(files)) {
results = unzip(files[idx], "*.xml")
}
As for using regex in unzip() that is something one should read the documentation. I could only advise doing another for-loop to compare the contest of the zip file to your regex then preforming the extraction. Psudocode follows:
files ::= glob(*.zip)
regex ::=
for idx1 in length(files); do
regex="[a-z]{3}\-[0-9]{8}\.xml"
content = unzip(files[idx1])
for idx2 in length(content); do
if content[idx2].name ~= regex.expand(); then
# do something with found file
end if
end for
end for
Basically your just looping through your list of zip files, then through the list of files within the zip file and comparing the filename from inside your zipfile agenst the regex and extracting/preforming operations on only that file.
When I organize my directories I often want certain directories to stand out in ls. For example, I will sometimes have a directory called #backup# and this will end up in the top of the list of directories, rather than in between all directories starting in "b". Sometimes, though, I want a directory to be at the bottom of the list, but I haven't found any symbol that achieves this. (The closest I've come is z#name#z, but this doesn't quite cut it.) So: Are there any symbols that come after all letters in Unix's alphabetical sorting?
You can use any (e.g. ASCII or Unicode [it depends upon your encoding and localization]) character except NULL (used as the ending of filepath) and / (used to separate directories in file path). See path_resolution(7). You might consider using ~ because several utilities (see indent(1), mv(1)....) adopt the convention to backup file /home/nag/foo as /home/nag/foo~. AFAIK #foo# could be used by emacs to backup temporarily an edited (but unsaved) file foo.