2024-03-29 10:29:52 发布
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例如,我想检查int 2是在另一个2之后还是之前:
list = [2, 2, 3] if 2 and 2 in list: print "True"
如果清单是这样的:
泰铢
a = [0, 2, 15, 13, 13, 5, 12, 3, 1, 3, 14, 14, 17, 20, 0, 14, 2, 0, 1, 10, 5, 13, 15, 13, 19, 11, 20, 5, 4, 17, 20, 7, 4, 20, 20, 19, 3, 20, 20, 0, 14, 8, 18, 0, 18, 2, 5, 15, 12, 20, 5, 7, 12, 6, 17, 18, 15, 12, 2, 16, 6, 11, 14, 20, 8, 4, 11, 10, 15, 6, 10, 19, 1, 20, 6, 19, 2, 5, 13, 7, 5, 18, 11, 11, 3, 0, 11, 3, 14, 18, 15, 11, 2, 9, 12, 0, 18, 1, 4, 2]
浏览列表一次,找出所有的对并将它们放入一个集合中
然后像这样使用它:
>>> n = 2 >>> n in z False >>> n = 13 >>> n in z True >>>
zip(a, a[1:])
将生成(a[0], a[1]), (a[1], a[2]), (a[2], a[3]), ...的元组
(a[0], a[1]), (a[1], a[2]), (a[2], a[3]), ...
pairwiseItertools Recipe将执行相同的操作。在
pairwise
试试这个:
def check(list_): last = None for element in list_: if element == last: return True else: last = element return False
浏览列表一次,找出所有的对并将它们放入一个集合中
^{pr2}$然后像这样使用它:
将生成
(a[0], a[1]), (a[1], a[2]), (a[2], a[3]), ...
的元组pairwise
Itertools Recipe将执行相同的操作。在试试这个:
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