给定一个文件目录,例如:
mydir/
test1.abc
set123.abc
jaja98.abc
test1.xyz
set123.xyz
jaja98.xyz我需要检查每个.abc文件都有一个等价的.xyz文件。我可以这样做:
>>> filenames = ['test1.abc', 'set123.abc', 'jaja98.abc', 'test1.xyz', 'set123.xyz', 'jaja98.xyz']
>>> suffixes = ('.abc', '.xyz')
>>> assert all( os.path.splitext(_filename)[0]+suffixes[1] in filenames for _filename in filenames if _filename.endswith(suffixes[0]) )上面的代码应该传递断言,而类似的内容可能会失败:
>>> filenames = ['test1.abc', 'set123.abc', 'jaja98.abc', 'test1.xyz', 'set123.xyz']
>>> suffixes = ('.abc', '.xyz') >>> assert all(os.path.splitext(_filename)[0]+suffixes[1] in filenames for _filename in filenames if _filename.endswith(suffixes[0]))
Traceback (most recent call last):
File "<stdin>", line 1, in <module>
AssertionError但这有点太冗长了。
还有更好的办法做同样的检查吗?
发布于 2016-10-07 02:13:34
您可以定义帮助函数,该函数将返回文件名的set,而不需要与给定后缀匹配的扩展名。然后,您可以轻松地检查带有后缀的is文件,.abc是带有.xyz后缀的文件的子集。
filenames = ['test1.abc', 'set123.abc', 'jaja98.abc', 'test1.xyz', 'set123.xyz', 'jaja98.xyz']
filenames2 = ['test1.abc', 'set123.abc', 'jaja98.abc', 'test1.xyz', 'set123.xyz']
suffixes = ('.abc', '.xyz')
def filter_ext(names, ext):
return {n[:-len(ext)] for n in names if n.endswith(ext)}
assert filter_ext(filenames, suffixes[0]) <= filter_ext(filenames, suffixes[1])
assert filter_ext(filenames2, suffixes[0]) <= filter_ext(filenames2, suffixes[1]) # fail上述方法也会更有效,因为它具有O(n)时间复杂度,其中原始的是O(n^2)。当然,如果列表很小,这并不重要。
https://stackoverflow.com/questions/39908156
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