简单的肖特基二极管电路（两个元祖T1='a', 'b',T2='c', 'd'使用匿名函数将其）

简单的肖特基二极管电路

两个元祖T1='a', 'b',T2='c', 'd'使用匿名函数将其

一道Python面试题的几种解答: 两个元祖T1=('a', 'b'), T2=('c', 'd')，请使用匿名函数将其转变成[{'a': 'c'}, {'b': 'd'}]

方法一:

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>>> T1 = ('a', 'b') >>> T2 = ('c', 'd') >>> list(map(lambda x:{x[0]:x[1]}, zip(T1, T2))) [{'a': 'c'}, {'b': 'd'}]

方法二:

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>>> T1 = ('a', 'b') >>> T2 = ('c', 'd') >>> [{v1:v2} for (i1,v1) in enumerate(T1) for (i2,v2) in enumerate(T2) if i1==i2] [{'a': 'c'}, {'b': 'd'}]

方法三:

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>>> T1 = ('a', 'b') >>> T2 = ('c', 'd') >>> ret = lambda t1,t2:[{x:y} for x in t1 for y in t2 if t1.index(x) == t2.index(y)] >>> ret(T1, T2) [{'a': 'c'}, {'b': 'd'}]

方法四:

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>>> T1 = ('a', 'b') >>> T2 = ('c', 'd') >>> ret = lambda t1,t2:[{x,y} for (x,y) in zip(t1, t2)] >>> ret(T1, T2) [{'a', 'c'}, {'d', 'b'}]

方法五:

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>>> T1 = ('a', 'b') >>> T2 = ('c', 'd') >>> ret = lambda t1,t2:[{t1[i]:t2[i]} for i in range(len(t1))] >>> ret(T1, T2) [{'a': 'c'}, {'b': 'd'}]

方法六:

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>>> T1 = ('a', 'b') >>> T2 = ('c', 'd') >>> list(map(lambda x,y:{x:y}, T1, T2)) [{'a': 'c'}, {'b': 'd'}]

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原文链接：https://blog.csdn.net/Jerry_1126/article/details/86375725