python 链表表达式 mapfilter易读版

Posted 2020-10-18 苏轶然

链表推导式

[x for x in x] 

链表推导式提供了一个创建链表的简单途径，无需使用 map()， filter() 以及 lambda。返回链表的定义通常要比创建这些链表更清晰。每一个链表推导式包括在一个for语句之后的表达式，零或多个for或if语句。返回值是由for或if子句之后的表达式得到的元素组成的链表。如果想要得到一个元组，必须要加上括号。

>>> freshfruit = [‘  banana‘, ‘  loganberry ‘, ‘passion fruit  ‘]
>>> [weapon.strip() for weapon in freshfruit]
[‘banana‘, ‘loganberry‘, ‘passion fruit‘]
>>> vec = [2, 4, 6]
>>> [3*x for x in vec]
[6, 12, 18]
>>> [3*x for x in vec if x > 3]
[12, 18]
>>> [3*x for x in vec if x < 2]
[]
>>> [[x,x**2] for x in vec]
[[2, 4], [4, 16], [6, 36]]
>>> [x, x**2 for x in vec]      # error - parens required for tuples
  File "<stdin>", line 1, in ?
    [x, x**2 for x in vec]
               ^
SyntaxError: invalid syntax
>>> [(x, x**2) for x in vec]
[(2, 4), (4, 16), (6, 36)]
>>> vec1 = [2, 4, 6]
>>> vec2 = [4, 3, -9]
>>> [x*y for x in vec1 for y in vec2]
[8, 6, -18, 16, 12, -36, 24, 18, -54]
>>> [x+y for x in vec1 for y in vec2]
[6, 5, -7, 8, 7, -5, 10, 9, -3]
>>> [vec1[i]*vec2[i] for i in range(len(vec1))]
[8, 12, -54]

 

为使链表推导式匹配for循环的行为，可以在推导之外保留循环变量：

>>> x = 100                     # this gets overwritten
>>> [x**3 for x in range(5)]
[0, 1, 8, 27, 64]
>>> x                           # the final value for range(5)

 

摘自
Python handbook
Guido van Rossum 
Fred L. Drake, Jr., editor