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根据list元素包含某个字符的多少对list排序

2020-10-27 58 10/27
a=['/','/aass/dda','/sdfsdf/dsfs/aa/s','/s','/sdf/fd/sdf/f/dsf/sdf/sdf/sd/fs/d']
def takeSecond(elem):
    elem.count('/')
a.sort(key=takeSecond)
print (a)

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