参考:4. Map, Filter and Reduce — Python Tips 0.1 documentation
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
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 reduce reduce(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: 6 15 24 120 4096
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)