#Filegroup.py
from zipfile import ZipFile
from glob import glob
print "file_name","\t","file_format"
for zips in glob('*.zip'):
with ZipFile(zips) as zip:
files = zip.namelist()
filecounts = {}
for file in files:
ext = file.split('.')[-1]
if ext in filecounts:
filecounts[ext] += 1
else:
filecounts[ext] = 1
print zip.filename,'\t\t',', '.join([' '.join(map(str,elem)) for elem in filecounts.items()])
这里有一个例子。这将按字典中的扩展名对zips中的文件进行分组并打印输出。根据你的情况调整这个。在
测试:
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