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本帖最后由 substr函数 于 2015-05-21 17:22 编辑
回复 4# qidunhu
统计里面那个列表索引为0的元素重复次数 ,不用第三方模块
你可能会感兴趣:有不止 10 种方法来做到这一点。
究竟有多少种写法?
- #!/usr/bin/python2
- l = [['222', 'das'], ['123', 'asd'], ['222', 'llkl']]
- # method 1
- c1 = [i[0] for i in l].count('222')
- print c1
- # method 2
- c2 = sum([1 for i in l if i[0] == '222'])
- print c2
- # method 3
- c3 = reduce(lambda x, y: x + (y[0] == '222'), l, 0)
- print c3
- # method 4
- c4 = filter(lambda x: x[0] == '222', l).__len__()
- print c4
- # method 5
- c5 = sum([i[0] == '222' for i in l])
- print c4
- # method 6
- c6 = sum(map(lambda x: x[0] == '222', l))
- print c6
- # method 7
- c7 = map(lambda x: x[0], l).count('222')
- print c7
- # method 8
- c8 = zip(*l)[0].count('222')
- print c8
- # method 9
- c9 = 0
- for i in l: c9 += (i[0] == '222')
- print c9
- # method 10
- c10 = reduce(lambda x, y: x + [y[0]], l, []).count('222')
- print c10
- # method 11
- # method 12, 13, .....
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