Question

我想从一对文件名生成通配符字符串。反种类的反面。例如：

file1 = 'some foo file.txt'
file2 = 'some bar file.txt'
assert 'some * file.txt' == inverse_glob(file1, file2)

也许使用difflib？这已经解决了吗？

应用程序是一组具有相似名称的大量数据文件。我想比较每对文件名，然后比较具有“相似”名称的文件对。我想如果我可以在每对上做反向球，那么那些带有“好”通配符（例如不是lots*of*stars*.txt或*）的对是比较好的候选者。因此，我可能会获取此假定inverse_glob()的输出并拒绝具有多个*的通配符，或者glob()不会生成两个文件的通配符。

Answer 1

例如：

<强>文件名：

names = [('some foo file.txt','some bar file.txt', 'some * file.txt'),
         ("filename.txt", "filename2.txt", "filenam*.txt"),
         ("1filename.txt", "filename2.txt", "*.txt"),
         ("inverse_glob", "inverse_glob2", "inverse_glo*"),
         ("the 24MHz run new.sr", "the 16MHz run old.sr", "the *MHz run *.sr")]

def inverse_glob（...）：

    import re
    def inverse_glob(f1, f2, force_single_asterisk=None):
        def adjust_name(pp, diff):
            if len(pp) == 2:
                return pp[0][:-diff] + '?'*(diff+1) + '.' + pp[1]
            else:
                return pp[0][:-diff] + '?' * (diff + 1)

        l1 = len(f1); l2 = len(f2)
        if l1 > l2:
            f2 = adjust_name(f2.split('.'), l1-l2)
        elif l2 > l1:
            f1 = adjust_name(f1.split('.'), l2-l1)

        result = ['?' for n in range(len(f1))]
        for i, c in enumerate(f1):
            if c == f2[i]:
                result[i] = c

        result = ''.join(result)
        result = re.sub(r'\?{2,}', '*', result)
        if force_single_asterisk:
            result = re.sub(r'\*.+\*', '*', result)
        return result

<强>用法：

for name in names:
    result = inverse_glob(name[0], name[1])
    print('{:20} <=> {:20} = {}'.format(name[0], name[1], result))
    assert name[2] == result

输出：

some foo file.txt    <=> some bar file.txt    = some * file.txt  
filename.txt         <=> filename2.txt        = filenam*.txt  
1filename.txt        <=> filename2.txt        = *.txt  
inverse_glob         <=> inverse_glob2        = inverse_glo*
the 24MHz run new.sr <=> the 16MHz run old.sr = the *MHz run *.sr

使用Python测试：3.4.2

反向全局 - 从文件名反向工程通配符字符串

1 个答案: