python列表推导式嵌套

列表推导式提供了一个更简单的创建列表的方法。常见的用法是把某种操作应用于序列或可迭代对象的每个元素上，然后使用其结果来创建列表，或者通过满足某些特定条件元素来创建子序列。

例如，假设我们想创建一个平方列表，像这样

>>> squares = []
>>> for x in range(10):
...     squares.append(x**2)
...
>>> squares
[0, 1, 4, 9, 16, 25, 36, 49, 64, 81]

 

注意这里创建（或被重写）的名为 x 的变量在for循环后仍然存在。我们可以计算平方列表的值而不会产生任何副作用

squares = list(map(lambda x: x**2, range(10)))

 

或者，等价于

squares = [x**2 for x in range(10)]

 

上面这种写法更加简洁易读。

Python列表推导式的结构是由一对方括号所包含的以下内容：一个表达式，后面跟一个 for 子句，然后是零个或多个 for或 if 子句。 其结果将是一个新列表，由对表达式依据后面的 for 和 if 子句的内容进行求值计算而得出。 举例来说，以下列表推导式会将两个列表中不相等的元素组合起来:

>>> [(x, y) for x in [1,2,3] for y in [3,1,4] if x != y]
[(1, 3), (1, 4), (2, 3), (2, 1), (2, 4), (3, 1), (3, 4)]

而它等价于

>>> combs = []
>>> for x in [1,2,3]:
...     for y in [3,1,4]:
...         if x != y:
...             combs.append((x, y))
...
>>> combs
[(1, 3), (1, 4), (2, 3), (2, 1), (2, 4), (3, 1), (3, 4)]

注意在上面两个代码片段中， for 和 if 的顺序是相同的。

如果表达式是一个Python元组（例如上面的 (x, y)），那么就必须加上括号

>>>

>>> vec = [-4, -2, 0, 2, 4]
>>> # create a new list with the values doubled
>>> [x*2 for x in vec]
[-8, -4, 0, 4, 8]
>>> # filter the list to exclude negative numbers
>>> [x for x in vec if x >= 0]
[0, 2, 4]
>>> # apply a function to all the elements
>>> [abs(x) for x in vec]
[4, 2, 0, 2, 4]
>>> # call a method on each element
>>> freshfruit = ['  banana', '  loganberry ', 'passion fruit  ']
>>> [weapon.strip() for weapon in freshfruit]
['banana', 'loganberry', 'passion fruit']
>>> # create a list of 2-tuples like (number, square)
>>> [(x, x**2) for x in range(6)]
[(0, 0), (1, 1), (2, 4), (3, 9), (4, 16), (5, 25)]
>>> # the tuple must be parenthesized, otherwise an error is raised
>>> [x, x**2 for x in range(6)]
  File "<stdin>", line 1, in <module>
    [x, x**2 for x in range(6)]
               ^
SyntaxError: invalid syntax
>>> # flatten a list using a listcomp with two 'for'
>>> vec = [[1,2,3], [4,5,6], [7,8,9]]
>>> [num for elem in vec for num in elem]
[1, 2, 3, 4, 5, 6, 7, 8, 9]

 

列表推导式可以使用复杂的表达式和嵌套函数

>>> from math import pi
>>> [str(round(pi, i)) for i in range(1, 6)]
['3.1', '3.14', '3.142', '3.1416', '3.14159']

嵌套的列表推导式

列表推导式中的初始表达式可以是任何表达式，包括另一个列表推导式。

考虑下面这个 3x4的矩阵，它由3个长度为4的列表组成

>>> matrix = [
...     [1, 2, 3, 4],
...     [5, 6, 7, 8],
...     [9, 10, 11, 12],
... ]

 

下面的列表推导式将交换其行和列

>>> [[row[i] for row in matrix] for i in range(4)]
[[1, 5, 9], [2, 6, 10], [3, 7, 11], [4, 8, 12]]

如上节所示，嵌套的列表推导式是基于跟随其后的 for 进行求值的，所以这个例子等价于:

>>> transposed = []
>>> for i in range(4):
...     transposed.append([row[i] for row in matrix])
...
>>> transposed
[[1, 5, 9], [2, 6, 10], [3, 7, 11], [4, 8, 12]]

反过来说，也等价于

>>> transposed = []
>>> for i in range(4):
...     # the following 3 lines implement the nested listcomp
...     transposed_row = []
...     for row in matrix:
...         transposed_row.append(row[i])
...     transposed.append(transposed_row)
...
>>> transposed
[[1, 5, 9], [2, 6, 10], [3, 7, 11], [4, 8, 12]]

实际应用中，你应该会更喜欢使用内置函数去组成复杂的流程语句。 zip() 函数将会很好地处理这种情况

>>> list(zip(*matrix))
[(1, 5, 9), (2, 6, 10), (3, 7, 11), (4, 8, 12)]