Python 的子进程模块从 Unix shell 返回不同的结果 [英] Python's subprocess module returning different results from Unix shell

问题描述

我正在尝试使用 python 获取目录中的 CSV 文件列表.这在 unix 中真的很容易:

ls -l *.csv

而且，可以预见的是，我在我的目录中得到了以 .csv 结尾的文件列表.但是，当我使用 Subprocess 模块尝试 Python 等效项时:

<预><代码>>>>将子进程导入为 sp>>>sp.Popen(["ls", "-l", "*.csv"], stdout = sp.PIPE)<subprocess.Popen 对象在 0xb780e90c>>>>ls: 无法访问 *.csv: 没有那个文件或目录

有人可以解释一下这是怎么回事吗?

编辑:添加 shell = True 可消除错误，但我得到的不是仅包含 CSV 文件的列表，而是包含 all 目录中的文件.

解决方案

如果你想让它像在 shell 中那样表现，你需要传递 shell=True (你的里程可能在这里有所不同，取决于您的系统和外壳).在您的情况下，问题在于当您执行 ls -l *.csv 时，shell 正在评估 * 的含义，而不是 ls.(ls 只是格式化您的结果，但 shell 已经完成了确定哪些文件与 *.csv 匹配的繁重工作).子进程使 ls 按字面意思处理 *.csv，并查找具有该特定名称的文件，当然没有任何文件(因为这是一个很难创建的文件名)).

您真正应该做的是使用 os.listdir 并自己过滤名称.

I'm trying to get a list of the CSV files in a directory with python. This is really easy within unix:

ls -l *.csv

And, predictably, I get a list of the files that end with .csv in my directory. However, when I attempt the Python equivalent using the Subprocess module:

>>> import subprocess as sp
>>> sp.Popen(["ls", "-l", "*.csv"], stdout = sp.PIPE)
<subprocess.Popen object at 0xb780e90c>
>>> ls: cannot access *.csv: No such file or directory

Can somebody please explain what's going on?

Edit: Adding shell = True removes the error, but instead of getting a list of just CSV files, I get a list of all the files in the directory.

解决方案

If you want it to behave as it does at the shell, you need to pass shell=True (your mileage may vary here, depending on your system and shell). In your case the problem is that when you do ls -l *.csv, the shell is evaluating what * means, not ls. (ls is merely formatting your results, but the shell has done the heavy lifting to determine what files match *.csv). Subprocess makes ls treat *.csv literally, and look for a file with that specific name, which of course there aren't any (since that's a pretty hard filename to create).

What you really should be doing is using os.listdir and filtering the names yourself.

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