Question

我有一个列有字典的列表，对于每个字典，我想要应用一个函数。什么是更有效或最佳实践，迭代列表并将dict传递给每次迭代的函数。如下所示：

dict1 = {"foo": 8, "bar": 4, "baz":2}
dict2 = {"foo": 3, "bar": 6, "baz":3}
...

dict_list = [dict1, dict2, ...]

def function_iterate(dict):
    ...

for dict in dict_list:
    function_iterate(dict)

或者最好传递列表并在函数内迭代：

dict1 = {"foo": 8, "bar": 4, "baz":2}
dict2 = {"foo": 3, "bar": 6, "baz":3}
...
dict_list = [dict1, dict2, ...]


def function_iterate(dict_list):
    for dict in dict_list:
        ...

function_iterate(dict_list)

或许你也知道另一种方法。如果两种方法都有效，你更喜欢哪一种？为什么？

Answer 1

您可以使用map：

map(function, dict_list)

map是一个内置的python函数：https://docs.python.org/2/library/functions.html#map

Answer 2

性能方面，发送整个列表会更快，因为python中的函数调用很昂贵（性能明智）。

时间测试示例 -

代码 -

l = [{1:2},{2:3},{3:4}]

def func1(d):
        for k in d:
                d[k] += 1

def func2(l):
        for d in l:
                for k in d:
                        d[k] += 1

计时结果 -

In [29]: %%timeit
   ....: for d in l:
   ....:     func1(d)
   ....:
The slowest run took 7.57 times longer than the fastest. This could mean that an intermediate result is being cached
1000000 loops, best of 3: 1.16 µs per loop

In [30]: %%timeit
   ....: func2(l)
   ....:
The slowest run took 9.25 times longer than the fastest. This could mean that an intermediate result is being cached
1000000 loops, best of 3: 988 ns per loop

In [31]: %%timeit
   ....: [func1(d) for d in l]
   ....:
100000 loops, best of 3: 1.98 µs per loop

In [33]: %%timeit
   ....: list(map(func1,l))
   ....:
The slowest run took 4.42 times longer than the fastest. This could mean that an intermediate result is being cached
100000 loops, best of 3: 2.5 µs per loop

为了便于阅读，map()会更好，因为它更具可读性。示例 -

对于Python 3.x -

list(map(function, list_of_dicts))

对于Python 2.x -

map(function, list_of_dicts)

什么更有效迭代列表并调用函数或将列表传递给函数

2 个答案: