问题
I need to make a list of a large number of files (40,000 files) like below:
ERR001268_1_100.fastq ERR001268_2_156.fastq ERR001753_2_78.fastq
ERR001268_1_101.fastq ERR001268_2_157.fastq ERR001753_2_79.fastq
ERR001268_1_102.fastq ERR001268_2_158.fastq ERR001753_2_7.fastq
ERR001268_1_103.fastq ERR001268_2_159.fastq ERR001753_2_80.fastq
my command is: ls ERR*_1_*.fastq |sed 's/\.fastq//g'|sort -n > masterlist
However error is: bash: /bin/ls: Argument list too long
However can I solve this problem? Any other way to make list like this by perl/python?
thx
回答1:
You should be able to replace ls ERR*_1_*.fastq
with find . -name "ERR*_1_*.fastq"
.
This way, you can avoid having the wildcard expand into a huge argument list.
(The find
output will include a leading "./", e.g. ./ERR001268_1_100.fastq
. If
that's undesirable, you can get rid of it with another sed
command later in the
pipeline.)
回答2:
If the files already all exist within your directory, python's "glob" module might have a higher limit than bash's command line.
From the command line:
python -c "import glob; print glob.glob('ERR_*_1_*.fastq')"
To do the whole thing in python, the you could try something like this:
import glob
files = glob.glob("ERR_*_1_*.fastq")
trimmedfiles = [x.replace(".fastq","") for x in files]
trimmedfiles.sort()
for f in trimmedfiles:
print f
This solution will sort the files alphabetically, and not numerically. For that you might want to add some key=lambda magic to the sort() method:
trimmedfiles.sort(key=lambda f: int(f.split("_")[2]))
回答3:
Find might help you - rather then ls use find . -name 'yourpatternhere' -print0 | xargs -0 youractionhere
回答4:
You can use find
.
Example:
find /Users/kunlun/Downloads/fu_neg/ -name "*.png" >
/Users/kunlun/Downloads/fu_neg.txt
来源:https://stackoverflow.com/questions/7030124/bash-bin-ls-argument-list-too-long