我想从一对文件名中生成一个通配符字符串。有点像反球。示例:
file1 = 'some foo file.txt'
file2 = 'some bar file.txt'
assert 'some * file.txt' == inverse_glob(file1, file2)
使用difflib也许?这个问题已经解决了吗?
应用程序是一组具有相似名称的数据文件。我想比较每对文件名,然后比较具有“相似”名称的文件对。我想如果我可以对每一对做一个反向 glob,那么那些带有“好”通配符的对(例如不是 lots*of*stars*.txt
也不是 *
)是比较好的候选者。所以我可能会采用这个假定的 inverse_glob()
的输出并拒绝具有多个 *
或 glob()
没有的通配符'恰好生成两个文件。
最佳答案
例如:
Filenames:
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
Usage:
for name in names:
result = inverse_glob(name[0], name[1])
print('{:20} <=> {:20} = {}'.format(name[0], name[1], result))
assert name[2] == result
Output:
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:3.4.2 测试
关于python - Inverse glob - 从文件名反向工程通配符字符串,我们在Stack Overflow上找到一个类似的问题: https://stackoverflow.com/questions/43808808/