参考：

参考：

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Map：映射，对于列表的每个元素进行相同的操作

filter：筛选，筛选列表中满足某一条件的所有元素

reduce：归纳，连续操作，连加、连乘等

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python 3.0以后, reduce已经不在built-in function里了, 要用它就得from functools import reduce.

reduce的用法

reduce(function, sequence[, initial]) -> value

Apply a function of two arguments cumulatively to the items of a sequence,

from left to right, so as to reduce the sequence to a single value.

For example, reduce(lambda x, y: x+y, [1, 2, 3, 4, 5]) calculates

((((1+2)+3)+4)+5).  If initial is present, it is placed before the items

of the sequence in the calculation, and serves as a default when the

sequence is empty.

意思就是对sequence连续使用function, 如果不给出initial, 则第一次调用传递sequence的两个元素, 以后把前一次调用的结果和sequence的下一个元素传递给function. 如果给出initial, 则第一次传递initial和sequence的第一个元素给function.

from functools import reducereduce(lambda x,y: x+y, [1, 2, 3])reduce(lambda x,y: x+y, [1, 2, 3], 9)reduce(lambda x,y: x*y, [1, 2, 3, 4])reduce(lambda x,y: x*y, [1, 2, 3, 4], 5)reduce(lambda x,y: x**y, [2, 3, 4])output:615241204096

Example from Ed of COMP9021

question:

For instance, dict1 = {'Lucy' : 'I am a Knight', 'Laser':'I am a Knaves'}

list1 = [(0,0), (0,1), (1,0),(1,1)]

how do I put the output like this:

{'Lucy' : 0, 'Laser':0}

{'Lucy' : 0, 'Laser':1}

{'Lucy' : 1, 'Laser':0}

{'Lucy' : 1, 'Laser':1}

answers:

dict1 = {'Lucy' : 'I am a Knight', 'Laser':'I am a Knaves'}list1 = [(0,0), (0,1), (1,0),(1,1)]answer = [i for i in map(lambda x:{[key for key in dict1][0]:x[0], [key for key in dict1][1]:x[1]}, list1)]for i in answer:	print(i)