bash: /bin/ls: Argument list too long

Question

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

Answer 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.)

Answer 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]))

Answer 3

You can use find.

Example:

find /Users/kunlun/Downloads/fu_neg/ -name "*.png" > 
/Users/kunlun/Downloads/fu_neg.txt

Answer 4

Find might help you - rather then ls use find . -name 'yourpatternhere' -print0 | xargs -0 youractionhere