Python interview - reduce & map & filter，,Python有很多有
Python interview - reduce & map & filter，,Python有很多有
Python有很多有用有趣的内置函数，比如reduce，map，filter，lambda，zip等。已经写过了lambda和zip相关的博客。继续写关于reduce，map，filter。
Map
首先用help方法看一下map的具体用法。
help(map)# resultHelp on built-in function map in module __builtin__:map(...) map(function, sequence[, sequence, ...]) -> list Return a list of the results of applying the function to the items of the argument sequence(s). If more than one sequence is given, the function is called with an argument list consisting of the corresponding item of each sequence, substituting None for missing values when not all sequences have the same length. If the function is None, return a list of the items of the sequence (or a list of tuples if more than one sequence).
map函数，map(function, sequences)，然后返回一个list。如果超过一个sequence，那么就会根据function来访问每个list中的依次对应的元素，如果sequence不等长，那么会用None来代替不够的元素。如果function是None，就会返回一个list，或者如果是多个list的话，就会返回一个包含很多tuples的list。
l1 = [1,2,3,4,5]def func(x): return x*xprint map(func, l1)# result[1, 4, 9, 16, 25]
如果是两个sequences
l1 = [1,2,3,4,5]l2 = [6,7,8,9,10]def func(x,y): return x+yprint map(func, l1, l2)# result[7, 9, 11, 13, 15]
方法有所改变，因为是两个lists，所以方法有两个参数。
对于不等长的sequences
l1 = [1,2,3,4,5,6]l2 = [6,7,8,9,10]def func(x,y): return yprint map(func, l1, l2)# result[6, 7, 8, 9, 10, None]
我们可以看到，较短的list，输出的时候会用None代替。
map可以和lambda一起用
l1 = [1,2,3,4,5]print map((lambda x: x **2), l1)# result[1, 4, 9, 16, 25]
当我们使用lambda作为一个function的时候，我们可以传入list of function作为sequence
def square(x): return (x**2)def cube(x): return (x**3)funcs = [square, cube]for r in range(5): value = map(lambda x: x(r), funcs) print value # result[0, 0][1, 1][4, 8][9, 27][16, 64]
如果function是None的话
l1 = [1,2,3,4,5]l2 = [6,7,8,9,10]print map(None, l1, l2)# result[(1, 6), (2, 7), (3, 8), (4, 9), (5, 10)]
map就像是一个for loop，对sequence中的元素依次进行操作。
from math import sqrtdef mymap(aFunc, aSeq):result = []for x in aSeq: result.append(aFunc(x))return resultprint list(map(sqrt, [1, 2, 3]))print mymap(sqrt, [1, 2, 3])# result[1.0, 1.4142135623730951, 1.7320508075688772][1.0, 1.4142135623730951, 1.7320508075688772]
Filter
用help看一下filter的用法
help(filter)# resultHelp on built-in function filter in module __builtin__:filter(...) filter(function or None, sequence) -> list, tuple, or string Return those items of sequence for which function(item) is true. If function is None, return the items that are true. If sequence is a tuple or string, return the same type, else return a list.
filter函数，同样是filter(function，sequence)，返回list，tuple，或者string。返回经过function判断过之后是true的元素。如果function是None，那么就返回true的元素。如果sequence是tuple或者string，返回同样的类型，如果是其他类型，就返回list。
print list(range(-5,5))print list(filter((lambda x: x<0), range(-5,5)))# result[-5, -4, -3, -2, -1, 0, 1, 2, 3, 4][-5, -4, -3, -2, -1]
如果function是None
print list(filter(None, range(-5,5)))# result[-5, -4, -3, -2, -1, 1, 2, 3, 4]
只有0没有输出，因为0这里被认为是False。
其他类型
def f(x): return x != 'a'print filter(f, 'abcdef')# resultbcdef
Reduce
用help看一下reduce的用法
help(reduce)# resultHelp on built-in function reduce in module __builtin__:reduce(...) 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.
def add(x,y): return x+yprint reduce(add, range(1,11))print reduce(add, range(1,11), 20)print reduce(add, [],20)# result55 (1+2+3+4+5+6+7+8+9+10)75 (20+1+2+3+4+5+6+7+8+9+10)20 (default)<span style="font-family: Arial, Helvetica, sans-serif;"></span>
同样的也可以配合lambda一起使用
print reduce((lambda x, y: x * y), [1, 2, 3, 4])# result24 (1*2*3*4)
Python interview - reduce & map & filter
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