两个元祖T1=('a', 'b'),T2=(&#

一道python/' target='_blank'>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:

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

方法二:

>>> 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:

>>> 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:

>>> 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:

>>> 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:

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

总结

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