Question

我正在尝试遍历一个字符串列表，只保留那些与我指定的命名模板匹配的字符串。我想接受任何与模板完全匹配的列表条目，除了在变量<SCENARIO>字段中有一个整数。

支票需要一般。具体来说，字符串结构可能会发生变化，因此无法保证<SCENARIO>始终显示在字符X上（例如，使用列表推导）。

下面的代码显示了一种使用split的方法，但必须有更好的方法来进行此字符串比较。我可以在这里使用正则表达式吗？

template = 'name_is_here_<SCENARIO>_20131204.txt'

testList = ['name_is_here_100_20131204.txt',        # should accept
            'name_is_here_100_20131204.txt.NEW',    # should reject
            'other_name.txt']                       # should reject

acceptList = []

for name in testList:
    print name
    acceptFlag = True
    splitTemplate = template.split('_')
    splitName = name.split('_')
    # if lengths do not match, name cannot possibly match template
    if len(splitTemplate) == len(splitName):
        print zip(splitTemplate, splitName)
        # compare records in the split
        for t, n in zip(splitTemplate, splitName):
            if t!=n and not t=='<SCENARIO>':
                #reject if any of the "other" fields are not identical
                #(would also check that '<SCENARIO>' field is numeric - not shown here)
                print 'reject: ' + name
                acceptFlag = False
    else:
        acceptFlag = False

    # keep name if it passed checks
    if acceptFlag == True:
        acceptList.append(name)

print acceptList
# correctly prints --> ['name_is_here_100_20131204.txt']

Answer 1

尝试使用re模块在Python中使用正则表达式：

import re

template = re.compile(r'^name_is_here_(\d+)_20131204.txt$')

testList = ['name_is_here_100_20131204.txt', #accepted
            'name_is_here_100_20131204.txt.NEW', #rejected!
            'name_is_here_aabs2352_20131204.txt', #rejected!
            'other_name.txt'] #rejected!

acceptList = [item for item in testList if template.match(item)]

Answer 2

这应该这样做，我明白name_is_here只是一个字母数字字符的占位符？

import re
testList = ['name_is_here_100_20131204.txt',        # should accept
            'name_is_here_100_20131204.txt.NEW',    # should reject
            'other_name.txt', 
            'name_is_44ere_100_20131204.txt',
            'name_is_here_100_2013120499.txt', 
            'name_is_here_100_something_2013120499.txt',
            'name_is_here_100_something_20131204.txt']  


def find(scenario):
    begin  = '[a-z_]+100_' # any combinations of chars and underscores followd by 100
    end = '_[0-9]{8}.txt$' #exactly eight digits followed by .txt at the end
    pattern = re.compile("".join([begin,scenario,end]))
    result = []
    for word in testList:
        if pattern.match(word):
            result.append(word)

    return result

find('something') # returns ['name_is_here_100_something_20131204.txt']

编辑：单独变量中的场景，正则表达式现在只匹配后跟100的字符，然后是scenarion，然后是八位数后跟.txt。

Python测试字符串是否与模板值匹配

2 个答案: