Question

变量游戏中有一个对象列表。

for g in games:
  if g.clam ==5: g.new_var=1
  if g.clam ==4: g.new_var=0

如何使用map（）函数获得上述功能？我尝试了类似下面的内容，但我认为它并不接近正确的方式。

def assign_var(clam):
  if clam==5: return 1
  if clam==4: return 0

games.new_var = map(assign_var, games.clam)

Answer 1

在assign_var功能

中创建新属性

>>> def assign_var(instance):
...     if instance.clam == 5:
...         instance.new_var = 1
...     elif instance.clam == 4:
...         instance.new_var = 0
... 
>>> map(assign_var, games)
[None, None, None] # Intentional. It will modify the list "games" in place.
>>> for ins in games:
...     print ins.new_var
... 
0
1
0

但实际上，这不是应该使用map()的内容。 map()应该用于可以使用已更改的数据返回的列表，而您无法使用类的属性执行此操作。

一个简单的for循环应该绝对正确：

for ins in games:
    if ins.clam == 5:
        instance.new_var = 1
    elif instance.clam == 4:
        instance.new_var = 0

请注意，请记住稀疏优于密集;）。

Answer 2

目前还不清楚为什么要在这里使用map()。如果你正在改变games一个简单的for - 循环就足够了。

map()更适合创建新的list（即使这样，列表理解也会更受欢迎）。

我还认为dict是定义映射的一种更简洁的方法，因为它不需要维护。这可以包含在您的Game课程中。

以下是一个例子：

#!/usr/bin/env python


class Game(object):
    clam_to_new_var = {
        4: 0,
        5: 1,
    }

    def __init__(self, clam):
        self.clam = clam

    @property
    def new_var(self):
        return Game.clam_to_new_var.get(self.clam, None)

    def __str__(self):
        return 'Game with clam "{}" has new_var "{}"'.format(
            self.clam,
            self.new_var,
        )

if __name__ == '__main__':
    games = map(Game, xrange(10))
    for g in games:
        print g

示例输出：

Game with clam "0" has new_var "None"
Game with clam "1" has new_var "None"
Game with clam "2" has new_var "None"
Game with clam "3" has new_var "None"
Game with clam "4" has new_var "0"
Game with clam "5" has new_var "1"
Game with clam "6" has new_var "None"
Game with clam "7" has new_var "None"
Game with clam "8" has new_var "None"
Game with clam "9" has new_var "None"

我在该示例中留下map()，但在生产代码中，我更愿意：

games = [Game(clam) for clam in range(10)]

为什么呢？请参阅The fate of reduce() in Python 3000。

Python - map（）和对象列表

2 个答案: