Python：在lambda函数上使用lru_cache，或者使用其他方式为lambda函数创建缓存

Question

有没有办法使它工作？

func=i_want_it_to_cache_everything(lambda a,b:a+b) 

它有一行做...

UPDATE2：

我想出的解决方案（感谢大家谁回答！）。 但是...有一个有趣的现象：缓存减慢程序？

import functools,datetime,timeit 
@functools.lru_cache(maxsize=50000) 
def euclidean_distance3(p1,p2): 
    return (p1[0]-p2[0])**2+(p1[1]-p2[1])**2+(p1[2]-p2[2])**2+(p1[3]-p2[3])**2 
euclidean_distance=(functools.lru_cache(maxsize=50000)(lambda p1,p2: (p1[0]-p2[0])**2+(p1[1]-p2[1])**2+(p1[2]-p2[2])**2+(p1[3]-p2[3])**2)) 
euclidean_distance2=lambda p1,p2: (p1[0]-p2[0])**2+(p1[1]-p2[1])**2+(p1[2]-p2[2])**2+(p1[3]-p2[3])**2 
print(datetime.datetime.now()) 
def test1(): 
    for z in range(50): 
     for i in range(200): 
      for j in range(200): 
       euclidean_distance((i,i,i,i),(j,j,j,j)); 
def test2(): 
    for z in range(50): 
     for i in range(200): 
      for j in range(200): 
       euclidean_distance2((i,i,i,i),(j,j,j,j)); 
def test3(): 
    for z in range(50): 
     for i in range(200): 
      for j in range(200): 
       euclidean_distance3((i,i,i,i),(j,j,j,j)); 
t1=timeit.Timer(test1) 
print(t1.timeit(1)) 
t2=timeit.Timer(test2) 
print(t2.timeit(1)) 
t3=timeit.Timer(test3) 
print(t3.timeit(1)) 

print(euclidean_distance.cache_info()) 
print(euclidean_distance3.cache_info()) 

输出：

9.989034592910151 
4.936129879313011 
10.528836308312947 
CacheInfo(hits=1960000, misses=40000, maxsize=50000, currsize=40000) 
CacheInfo(hits=1960000, misses=40000, maxsize=50000, currsize=40000) 

Answer 1

>>> from functools import lru_cache 
>>> x = lru_cache()(lambda a,b:a+b) 
>>> x(2,3) 
5 
>>> x(4,2) 
6 
>>> x(2,3) 
5 
>>> x.cache_info() 
CacheInfo(hits=1, misses=2, maxsize=100, currsize=2)