有比mapKeysWith更好的python等价物吗

Question

Haskell有一个很棒的函数mapKeysWith。在地图的关键点上，它应用一些变换，如果现在存在碰撞，则将数据对象与给定函数组合。我创建了下面的示例代码，它非常详细。有没有更像蟒蛇的方式？

def mapKeysWith(combineF,op,mp):
ret = {}
for it in mp:
if op(it) in ret:
ret[op(it)] = combineF(ret[op(it)],mp[it])
else:
ret[op(it)] = mp[it]
return ret
z = {1:2,3:4,10:11}
mapKeysWith(lambda x,y: x+y,lambda x: math.floor(x/10),z)

Answer 1

两种可能的替代方案如下：

import math
from functools import reduce
from collections import defaultdict

def map_keys_with(combine_f, op, mp):
ret = {}
for it in mp:
if op(it) in ret:
ret[op(it)] = combine_f(ret[op(it)], mp[it])
else:
ret[op(it)] = mp[it]
return ret

def map_keys_with_setdefault(combine_f, op, mp):
ret = {}
for key, value in mp.items():
ret.setdefault(op(key), []).append(value)
return {key: reduce(combine_f, values) for key, values in ret.items()}

def map_keys_with_defaultdict(combine_f, op, mp):
ret = defaultdict(list)
for key, value in mp.items():
ret[op(key)].append(value)
return {key: reduce(combine_f, values) for key, values in ret.items()}

z = {1: 2, 3: 4, 10: 11}
print(map_keys_with(lambda x, y: x + y, lambda x: math.floor(x / 10), z))
print(map_keys_with_setdefault(lambda x, y: x + y, lambda x: math.floor(x / 10), z))
print(map_keys_with_defaultdict(lambda x, y: x + y, lambda x: math.floor(x / 10), z))

输出

{0: 6, 1: 11}
{0: 6, 1: 11}
{0: 6, 1: 11}