大于当前键值时在python中设置字典值的最快方法

Question

我经常遇到一种情况，即当且仅当新值大于旧值时，才必须将字典键设置为一个值（始终为整数）。 分析揭示了我当前正在执行的方式正在占用某种功能的全部时间的1/3。 这是我目前的操作方式：

some_dict[k] = max(new_val, some_dict.get(k, 0))

有没有更快的方法？

Answer 1

比较我能想到的三种不同方法-

In [12]: def foo():
   ....:     d = {1:2 , 3:4}
   ....:     d[1] = max(3, d.get(1,0))
   ....:

In [13]: def foo1():
   ....:     d = {1:2, 3:4}
   ....:     if d.get(1,0) < 3:
   ....:         d[1] = 3
   ....:

In [14]: def foo2():
   ....:     d = {1:2, 3:4}
   ....:     d[1] = 3 if d.get(1,0) < 3 else d.get(1)
   ....:

In [15]: %timeit foo()
The slowest run took 4.18 times longer than the fastest. This could mean that an intermediate result is being cached
1000000 loops, best of 3: 1 µs per loop

In [16]: %timeit foo1()
The slowest run took 11.46 times longer than the fastest. This could mean that an intermediate result is being cached
1000000 loops, best of 3: 564 ns per loop

In [17]: %timeit foo2()
The slowest run took 4.79 times longer than the fastest. This could mean that an intermediate result is being cached
1000000 loops, best of 3: 556 ns per loop

In [18]: %timeit foo()
The slowest run took 10.17 times longer than the fastest. This could mean that an intermediate result is being cached
1000000 loops, best of 3: 861 ns per loop

In [19]: %timeit foo1()
The slowest run took 5.90 times longer than the fastest. This could mean that an intermediate result is being cached
1000000 loops, best of 3: 645 ns per loop

In [20]: %timeit foo2()
The slowest run took 8.01 times longer than the fastest. This could mean that an intermediate result is being cached
1000000 loops, best of 3: 523 ns per loop

如果我们认为%timeit的结果似乎是最快的-

some_dict[k] = new_val if some_dict.get(k, 0) < new_val else some_dict.get(k)

假定new_val始终大于0，因此在else部分中不需要some_dict.get(k,0) 。 尽管即使没有太大的区别。

Answer 2

好吧，显而易见的另一种方法是不总是执行分配，并且不调用额外的函数调用：

if some_dict.get(k, 0) < new_val:
    some_dict[k] = new_val

这实际上是否更快是另一个问题。

Answer 3

还有另一种尚未提及的方法：

get = d.get
val = get(k, 0)
d[k] = (val, new_val)[val<new_val]

此方法仅在内存中播放，根本不使用分支。

有多快？ 嗯，应该很快。

get = d.get

将指针保存在字典之外，以便更快地访问该方法。

您也可以使用dict.setdefault（）。 这是一个有趣的小方法。

参见help（dict.setdefault）。