字符串中的 Python 通配符搜索

Do you mean any specific syntax for wildcards? Usually * stands for "one or many" characters and ? stands for one.

The simplest way probably is to translate a wildcard expression into a regular expression, then use that for filtering the results.

Use fnmatch:

import fnmatch
lst = ['this','is','just','a','test']
filtered = fnmatch.filter(lst, 'th?s')

If you want to allow _ as a wildcard, just replace all underscores with '?' (for one character) or * (for multiple characters).

If you want your users to use even more powerful filtering options, consider allowing them to use regular expressions.

Regular expressions are probably the easiest solution to this problem:

import re
regex = re.compile('th.s')
l = ['this', 'is', 'just', 'a', 'test']
matches = [string for string in l if re.match(regex, string)]

You could try the fnmatch module, it's got a shell-like wildcard syntax

or can use regular expressions

import re

Same idea as Yuushi in using regular expressions, but this uses the findall method within the re library instead of a list comprehension:

import re
regex = re.compile('th.s')
l = ['this', 'is', 'just', 'a', 'test']
matches = re.findall(regex, string)

Easy method is try os.system:

import os
text = 'this is text'
os.system("echo %s | grep 't*'" % text)

Why don't you just use the join function? In a regex findall() or group() you will need a string so:

import re
regex = re.compile('th.s')
l = ['this', 'is', 'just', 'a', 'test']
matches = re.findall(regex, ' '.join(l)) #Syntax option 1
matches = regex.findall(' '.join(l)) #Syntax option 2

The join() function allows you to transform a list in a string. The single quote before join is what you will put in the middle of each string on list. When you execute this code part (' '.join(l)) you'll receive this:

'this is just a test'

So you can use the findal() function.

I know i am 7 years late, but i recently create an account because I'm studying and other people could have the same question. I hope this help you and others.

Update After @FélixBrunet comments:

import re
regex = re.compile(r'th.s')
l = ['this', 'is', 'just', 'a', 'test','th','s', 'this is']


matches2=[] #declare a list
for i in range(len(l)): #loop with the iterations = list l lenght. This avoid the first item commented by @Felix
if regex.findall(l[i]) != []: #if the position i is not an empty list do the next line. PS: remember regex.findall() command return a list.
if l[i]== ''.join(regex.findall(l[i])): # If the string of i position of l list = command findall() i position so it'll allow the program do the next line - this avoid the second item commented by @Félix
matches2.append(''.join(regex.findall(l[i]))) #adds in the list just the string in the matches2 list


print(matches2)