list bash: /bin/ls: 参数列表太长
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bash: /bin/ls: Argument list too long
提问by LookIntoEast
I need to make a list of a large number of files (40,000 files) like below:
我需要列出大量文件（40,000 个文件），如下所示：
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 > masterlistHowever error is: bash: /bin/ls: Argument list too long
我的命令是：ls ERR*_1_*.fastq |sed 's/\.fastq//g'|sort -n > masterlist但是错误是：bash: /bin/ls: Argument list too long
However can I solve this problem? Any other way to make list like this by perl/python?
但是我能解决这个问题吗？任何其他方式通过 perl/python 制作这样的列表？
thx
谢谢
回答by Jim Lewis
You should be able to replace ls ERR*_1_*.fastqwith find . -name "ERR*_1_*.fastq".
This way, you can avoid having the wildcard expand into a huge argument list.
您应该可以替换ls ERR*_1_*.fastq为find . -name "ERR*_1_*.fastq".
这样，您可以避免将通配符扩展为一个巨大的参数列表。
(The findoutput will include a leading "./", e.g. ./ERR001268_1_100.fastq. If
that's undesirable, you can get rid of it with another sedcommand later in the
pipeline.)
（find输出将包括一个前导“./”，例如./ERR001268_1_100.fastq。如果这是不受欢迎的，您可以sed在管道中稍后使用另一个命令来摆脱它。）
回答by Ben G.
If the files already all exist within your directory, python's "glob" module might have a higher limit than bash's command line.
如果文件已经全部存在于您的目录中，python 的“glob”模块可能比 bash 的命令行具有更高的限制。
From the command line:
从命令行：
python -c "import glob; print glob.glob('ERR_*_1_*.fastq')"
To do the whole thing in python, the you could try something like this:
要在 python 中完成整个事情，你可以尝试这样的事情：
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:
此解决方案将按字母顺序而不是数字顺序对文件进行排序。为此，您可能希望向 sort() 方法添加一些 key=lambda 魔法：
trimmedfiles.sort(key=lambda f: int(f.split("_")[2]))
回答by j03m
Find might help you - rather then ls use find . -name 'yourpatternhere' -print0 | xargs -0 youractionhere
Find 可能会帮助您 - 而不是 ls 使用 find . -name 'yourpatternhere' -print0 | xargs -0 youractionhere
回答by logan lu
You can use find.
您可以使用find.
Example:
例子：
find /Users/kunlun/Downloads/fu_neg/ -name "*.png" >
/Users/kunlun/Downloads/fu_neg.txt
