问题
I want to be able to match a pattern in glob format to a list of strings, rather than to actual files in the filesystem. Is there any way to do this, or convert a glob pattern easily to a regex?
回答1:
Good artists copy; great artists steal.
I stole ;)
fnmatch.translate translates globs ? and * to regex . and .* respectively. I tweaked it not to.
import re
def glob2re(pat):
"""Translate a shell PATTERN to a regular expression.
There is no way to quote meta-characters.
"""
i, n = 0, len(pat)
res = ''
while i < n:
c = pat[i]
i = i+1
if c == '*':
#res = res + '.*'
res = res + '[^/]*'
elif c == '?':
#res = res + '.'
res = res + '[^/]'
elif c == '[':
j = i
if j < n and pat[j] == '!':
j = j+1
if j < n and pat[j] == ']':
j = j+1
while j < n and pat[j] != ']':
j = j+1
if j >= n:
res = res + '\\['
else:
stuff = pat[i:j].replace('\\','\\\\')
i = j+1
if stuff[0] == '!':
stuff = '^' + stuff[1:]
elif stuff[0] == '^':
stuff = '\\' + stuff
res = '%s[%s]' % (res, stuff)
else:
res = res + re.escape(c)
return res + '\Z(?ms)'
This one à la fnmatch.filter, both re.match and re.search work.
def glob_filter(names,pat):
return (name for name in names if re.match(glob2re(pat),name))
Glob patterns and strings found on this page pass test.
pat_dict = {
'a/b/*/f.txt': ['a/b/c/f.txt', 'a/b/q/f.txt', 'a/b/c/d/f.txt','a/b/c/d/e/f.txt'],
'/foo/bar/*': ['/foo/bar/baz', '/spam/eggs/baz', '/foo/bar/bar'],
'/*/bar/b*': ['/foo/bar/baz', '/foo/bar/bar'],
'/*/[be]*/b*': ['/foo/bar/baz', '/foo/bar/bar'],
'/foo*/bar': ['/foolicious/spamfantastic/bar', '/foolicious/bar']
}
for pat in pat_dict:
print('pattern :\t{}\nstrings :\t{}'.format(pat,pat_dict[pat]))
print('matched :\t{}\n'.format(list(glob_filter(pat_dict[pat],pat))))
回答2:
The glob module uses the fnmatch module for individual path elements.
That means the path is split into the directory name and the filename, and if the directory name contains meta characters (contains any of the characters [, * or ?) then these are expanded recursively.
If you have a list of strings that are simple filenames, then just using the fnmatch.filter() function is enough:
import fnmatch
matching = fnmatch.filter(filenames, pattern)
but if they contain full paths, you need to do more work as the regular expression generated doesn't take path segments into account (wildcards don't exclude the separators nor are they adjusted for cross-platform path matching).
You can construct a simple trie from the paths, then match your pattern against that:
import fnmatch
import glob
import os.path
from itertools import product
# Cross-Python dictionary views on the keys
if hasattr(dict, 'viewkeys'):
# Python 2
def _viewkeys(d):
return d.viewkeys()
else:
# Python 3
def _viewkeys(d):
return d.keys()
def _in_trie(trie, path):
"""Determine if path is completely in trie"""
current = trie
for elem in path:
try:
current = current[elem]
except KeyError:
return False
return None in current
def find_matching_paths(paths, pattern):
"""Produce a list of paths that match the pattern.
* paths is a list of strings representing filesystem paths
* pattern is a glob pattern as supported by the fnmatch module
"""
if os.altsep: # normalise
pattern = pattern.replace(os.altsep, os.sep)
pattern = pattern.split(os.sep)
# build a trie out of path elements; efficiently search on prefixes
path_trie = {}
for path in paths:
if os.altsep: # normalise
path = path.replace(os.altsep, os.sep)
_, path = os.path.splitdrive(path)
elems = path.split(os.sep)
current = path_trie
for elem in elems:
current = current.setdefault(elem, {})
current.setdefault(None, None) # sentinel
matching = []
current_level = [path_trie]
for subpattern in pattern:
if not glob.has_magic(subpattern):
# plain element, element must be in the trie or there are
# 0 matches
if not any(subpattern in d for d in current_level):
return []
matching.append([subpattern])
current_level = [d[subpattern] for d in current_level if subpattern in d]
else:
# match all next levels in the trie that match the pattern
matched_names = fnmatch.filter({k for d in current_level for k in d}, subpattern)
if not matched_names:
# nothing found
return []
matching.append(matched_names)
current_level = [d[n] for d in current_level for n in _viewkeys(d) & set(matched_names)]
return [os.sep.join(p) for p in product(*matching)
if _in_trie(path_trie, p)]
This mouthful can quickly find matches using globs anywhere along the path:
>>> paths = ['/foo/bar/baz', '/spam/eggs/baz', '/foo/bar/bar']
>>> find_matching_paths(paths, '/foo/bar/*')
['/foo/bar/baz', '/foo/bar/bar']
>>> find_matching_paths(paths, '/*/bar/b*')
['/foo/bar/baz', '/foo/bar/bar']
>>> find_matching_paths(paths, '/*/[be]*/b*')
['/foo/bar/baz', '/foo/bar/bar', '/spam/eggs/baz']
回答3:
On Python 3.4+ you can just use PurePath.match.
pathlib.PurePath(path_string).match(pattern)
On Python 3.3 or earlier (including 2.x), get pathlib from PyPI.
Note that to get platform-independent results (which will depend on why you're running this) you'd want to explicitly state PurePosixPath or PureWindowsPath.
回答4:
While fnmatch.fnmatch can be used directly to check whether a pattern matches a filename or not, you can also use the fnmatch.translate method to generate the regex out of the given fnmatch pattern:
>>> import fnmatch
>>> fnmatch.translate('*.txt')
'.*\\.txt\\Z(?ms)'
From the documenation:
fnmatch.translate(pattern)Return the shell-style pattern converted to a regular expression.
回答5:
never mind, I found it. I want the fnmatch module.
回答6:
An extension to @Veedrac PurePath.match answer that can be applied to a lists of strings:
# Python 3.4+
from pathlib import Path
path_list = ["foo/bar.txt", "spam/bar.txt", "foo/eggs.txt"]
# convert string to pathlib.PosixPath / .WindowsPath, then apply PurePath.match to list
print([p for p in path_list if Path(p).match("ba*")]) # "*ba*" also works
# output: ['foo/bar.txt', 'spam/bar.txt']
print([p for p in path_list if Path(p).match("*o/ba*")])
# output: ['foo/bar.txt']
It is preferable to use pathlib.Path() over pathlib.PurePath(), because then you don't have to worry about the underlying filesystem.
来源：https://stackoverflow.com/questions/27726545/python-glob-but-against-a-list-of-strings-rather-than-the-filesystem