Python 使用 re.match 过滤字符串列表时失败
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Failure when filtering string list with re.match
提问by miluz
I'd like to filter a list of strings in python by using regex. In the following case, keeping only the files with a '.npy' extension.
我想使用正则表达式过滤 python 中的字符串列表。在以下情况下，仅保留具有“.npy”扩展名的文件。
The code that doesn't work:
不起作用的代码：
import re
files = [ '/a/b/c/la_seg_x005_y003.png',
'/a/b/c/la_seg_x005_y003.npy',
'/a/b/c/la_seg_x004_y003.png',
'/a/b/c/la_seg_x004_y003.npy',
'/a/b/c/la_seg_x003_y003.png',
'/a/b/c/la_seg_x003_y003.npy', ]
regex = re.compile(r'_x\d+_y\d+\.npy')
selected_files = filter(regex.match, files)
print(selected_files)
The same regex works for me in Ruby:
相同的正则表达式在 Ruby 中对我有用：
selected = files.select { |f| f =~ /_x\d+_y\d+\.npy/ }
What's wrong with the Python code?
Python 代码有什么问题？
采纳答案by Kevin Guan
selected_files = filter(regex.match, files)
re.match('regex')is equal to re.search('^regex')or text.startswith('regex')but regex version. It only check if the string starts with the regex.
re.match('regex')等于re.search('^regex')或text.startswith('regex')但是正则表达式版本。它只检查字符串是否以 regex 开头。
So, use re.search()instead:
因此，请re.search()改用：
import re
files = [ '/a/b/c/la_seg_x005_y003.png',
'/a/b/c/la_seg_x005_y003.npy',
'/a/b/c/la_seg_x004_y003.png',
'/a/b/c/la_seg_x004_y003.npy',
'/a/b/c/la_seg_x003_y003.png',
'/a/b/c/la_seg_x003_y003.npy', ]
regex = re.compile(r'_x\d+_y\d+\.npy')
selected_files = list(filter(regex.search, files))
# The list call is only required in Python 3, since filter was changed to return a generator
print(selected_files)
Output:
输出：
['/a/b/c/la_seg_x005_y003.npy',
'/a/b/c/la_seg_x004_y003.npy',
'/a/b/c/la_seg_x003_y003.npy']
And if you just want to get all of the .npyfiles, just use str.endswith():
如果您只想获取所有.npy文件，只需使用str.endswith()：
files = [ '/a/b/c/la_seg_x005_y003.png',
'/a/b/c/la_seg_x005_y003.npy',
'/a/b/c/la_seg_x004_y003.png',
'/a/b/c/la_seg_x004_y003.npy',
'/a/b/c/la_seg_x003_y003.png',
'/a/b/c/la_seg_x003_y003.npy', ]
selected_files = list(filter(lambda x: x.endswith('.npy'), files))
print(selected_files)
回答by SIslam
Just use search- since match starts matching from the beginning to end (i.e. entire) of string and search matches anywhere in the string.
只需使用search- 因为匹配从字符串的开头到结尾（即整个）开始匹配，并在字符串中的任何位置搜索匹配项。
import re
files = [ '/a/b/c/la_seg_x005_y003.png',
'/a/b/c/la_seg_x005_y003.npy',
'/a/b/c/la_seg_x004_y003.png',
'/a/b/c/la_seg_x004_y003.npy',
'/a/b/c/la_seg_x003_y003.png',
'/a/b/c/la_seg_x003_y003.npy', ]
regex = re.compile(r'_x\d+_y\d+\.npy')
selected_files = filter(regex.search, files)
print(selected_files)
Output-
输出-
['/a/b/c/la_seg_x005_y003.npy', '/a/b/c/la_seg_x004_y003.npy', '/a/b/c/la_seg_x003_y003.npy']
回答by miku
If you match, the pattern must cover the entireinput. Either extend you regular expression:
如果匹配，则模式必须覆盖整个输入。要么扩展你的正则表达式：
regex = re.compile(r'.*_x\d+_y\d+\.npy')
Which would match:
哪个匹配：
['/a/b/c/la_seg_x005_y003.npy',
'/a/b/c/la_seg_x004_y003.npy',
'/a/b/c/la_seg_x003_y003.npy']
Or use re.search, which
或者使用re.search，其中
scans through string looking for the first locationwhere the regular expression pattern produces a match [...]
扫描字符串以查找正则表达式模式产生匹配的第一个位置[...]
回答by Vlad
re.match()looks for a match at the beginning of the string. You can use re.search()instead.
re.match()在字符串的开头查找匹配项。你可以re.search()改用。
