Python的递推式构造列表（List comprehension）

介绍

在数学上，自然数的平方数是：{ x² | x ∈ ? } 或者复数：{ (x,y) | x ∈ ? ∧ y ∈ ? }.

在Python里，递推式构造列表（List comprehension）是一种定义和创建列表的优雅方式，这些列表通常是有一些约束的集合，并不是所有案例的集合。

对于函数map(), filter(), 和reduce()，递推式构造列表（List comprehension）是一个完整的lambda替代者。对于大部分人们，递推式构造列表（List comprehension）的语法更容易被人们掌握。

举列

>>> Celsius = [39.2, 36.5, 37.3, 37.8]
>>> Fahrenheit = [ ((float(9)/5)*x + 32) for x in Celsius ]
>>> print Fahrenheit
[102.56, 97.700000000000003, 99.140000000000001, 100.03999999999999]
>>> 

下面的递推式构造列表（list comprehension）创建了毕达哥拉斯三元组：

>>> [(x,y,z) for x in range(1,30) for y in range(x,30) for z in range(y,30) if x**2 + y**2 == z**2]
[(3, 4, 5), (5, 12, 13), (6, 8, 10), (7, 24, 25), (8, 15, 17), (9, 12, 15), (10, 24, 26), (12, 16, 20), (15, 20, 25), (20, 21, 29)]
>>> 

两个集合的交叉乘积：

>>> colours = [ "red", "green", "yellow", "blue" ]
>>> things = [ "house", "car", "tree" ]
>>> coloured_things = [ (x,y) for x in colours for y in things ]
>>> print coloured_things
[(‘red‘, ‘house‘), (‘red‘, ‘car‘), (‘red‘, ‘tree‘), (‘green‘, ‘house‘), (‘green‘, ‘car‘), (‘green‘, ‘tree‘), (‘yellow‘, ‘house‘), (‘yellow‘, ‘car‘), (‘yellow‘, ‘tree‘), (‘blue‘, ‘house‘), (‘blue‘, ‘car‘), (‘blue‘, ‘tree‘)]
>>> 

递推式构造生成器(Generator Comprehension)

>>> x = (x **2 for x in range(20))
>>> print(x)
 at 0xb7307aa4>
>>> x = list(x)
>>> print(x)
[0, 1, 4, 9, 16, 25, 36, 49, 64, 81, 100, 121, 144, 169, 196, 225, 256, 289, 324, 361]

一些更高级的例子

>>> noprimes = [j for i in range(2, 8) for j in range(i*2, 100, i)]
>>> primes = [x for x in range(2, 100) if x not in noprimes]
>>> print primes
[2, 3, 5, 7, 11, 13, 17, 19, 23, 29, 31, 37, 41, 43, 47, 53, 59, 61, 67, 71, 73, 79, 83, 89, 97]
>>> 

我们把前面的例子写成更通俗的格式，所以我们可以计算到任意数n的质数的列表：

>>> from math import sqrt
>>> n = 100
>>> sqrt_n = int(sqrt(n))
>>> no_primes = [j for i in range(2,sqrt_n) for j in range(i*2, n, i)]

如果我们去看no_primes的内容，就会发现一个问题。这里有很多重复的元素在这个列表里：

>>> no_primes
[4, 6, 8, 10, 12, 14, 16, 18, 20, 22, 24, 26, 28, 30, 32, 34, 36, 38, 40, 42, 44, 46, 48, 50, 52, 54, 56, 58, 60, 62, 64, 66, 68, 70, 72, 74, 76, 78, 80, 82, 84, 86, 88, 90, 92, 94, 96, 98, 6, 9, 12, 15, 18, 21, 24, 27, 30, 33, 36, 39, 42, 45, 48, 51, 54, 57, 60, 63, 66, 69, 72, 75, 78, 81, 84, 87, 90, 93, 96, 99, 8, 12, 16, 20, 24, 28, 32, 36, 40, 44, 48, 52, 56, 60, 64, 68, 72, 76, 80, 84, 88, 92, 96, 10, 15, 20, 25, 30, 35, 40, 45, 50, 55, 60, 65, 70, 75, 80, 85, 90, 95, 12, 18, 24, 30, 36, 42, 48, 54, 60, 66, 72, 78, 84, 90, 96, 14, 21, 28, 35, 42, 49, 56, 63, 70, 77, 84, 91, 98, 16, 24, 32, 40, 48, 56, 64, 72, 80, 88, 96, 18, 27, 36, 45, 54, 63, 72, 81, 90, 99]
>>>

这个无法容忍的问题将会在递推式构造集合（set comprehension)中被解决，我们将会在下一节中讲解

递推式构造集合（set comprehension)

>>> from math import sqrt
>>> n = 100
>>> sqrt_n = int(sqrt(n))
>>> no_primes = {j for i in range(2,sqrt_n) for j in range(i*2, n, i)}
>>> no_primes
{4, 6, 8, 9, 10, 12, 14, 15, 16, 18, 20, 21, 22, 24, 25, 26, 27, 28, 30, 32, 33, 34, 35, 36, 38, 39, 40, 42, 44, 45, 46, 48, 49, 50, 51, 52, 54, 55, 56, 57, 58, 60, 62, 63, 64, 65, 66, 68, 69, 70, 72, 74, 75, 76, 77, 78, 80, 81, 82, 84, 85, 86, 87, 88, 90, 91, 92, 93, 94, 95, 96, 98, 99}
>>> primes = {i for i in range(n) if i not in no_primes}
>>> print(primes)
{0, 1, 2, 3, 5, 7, 11, 13, 17, 19, 23, 29, 31, 37, 41, 43, 47, 53, 59, 61, 67, 71, 73, 79, 83, 89, 97}
>>> 

利用递归函数来计算质数

下面的Python脚本使用递归函数来创建质数，他的功能完全可以去检查一个到n的平方根的多个质数：

from math import sqrt
def primes(n):
    if n == 0:
        return []
    elif n == 1:
        return [1]
    else:
        p = primes(int(sqrt(n)))
        no_p = {j for i in p for j in range(i*2, n, i)}
        p = {x for x in range(2, n) if x not in no_p}
    return p

print(primes(40))

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译文（中文）：http://www.cnblogs.com/reanote/p/python_list_comprehension.html