python方式找到一个最小的字典键值在一个列表?

Question

我正在用Python编写Astar算法的实现，在某些时候我想选择open_set中具有最低f_score value的元素:

# current is the node in open_set having the lowest f_score value
current = min_f_score(open_set, f_score)

open_set是一个元组的列表(x, y)，表示二维中一个点的坐标，f_score是一个字典{(x, y): score}，它给每个元组(x, y)分配f_score的值。

我的第一个min_f_score实现如下:

def min_f_score(open_set, f_score):
# First we initialize min_score_element and min_score
min_score_element = open_set[0]
min_score = f_score[open_set[0]]
# Then we look for the element from f_score keys with the lowest value
for element in open_set:
if element in f_score.keys() and f_score[element] < min_score:
min_score = f_score[element]
min_score_element = element
return min_score_element

它工作得很好，但我想知道我是否能想出一些更精简，更python化的代码。经过一番研究，我想出了另外两个实现:

def min_f_score(open_set, f_score):   
# We filter the elements from open_set and then find the min f_score value
return min(filter(lambda k: k in open_set, f_score), key=(lambda k: f_score[k]))

:

def min_f_score(open_set, f_score):   
# We look for the min while assigning inf value to elements not in open_set
return min(f_score, key=(lambda k: f_score[k] if k in open_set else float("inf")))

两者似乎都可以工作，并给我相同的结果，但明显比第一个实现慢。

出于好奇，我想知道是否有更好的方法来实现min_f_score?

编辑:根据@azro(谢谢)的建议，我正在添加一个执行代码示例:

open_set = [(1, 2), (1, 3), (2, 1), (2, 2), (3, 1), (1, 4), (4, 1), (1, 5), (5, 0), (5, 1), (0, 6), (1, 6)]
f_score = {(0, 0): 486.0, (0, 1): 308.0, (1, 0): 308.0, (1, 1): 265.0, (0, 2): 265.0, (1, 2): 338.0, (0, 3): 284.0, (1, 3): 450.0, (2, 0): 265.0, (2, 1): 338.0, (2, 2): 629.0, (3, 0): 284.0, (3, 1): 450.0, (0, 4): 310.0, (1, 4): 550.0, (4, 0): 310.0, (4, 1): 564.0, (0, 5): 316.0, (1, 5): 588.0, (5, 0): 316.0, (5, 1): 606.0, (0, 6): 298.0, (1, 6): 534.0}
min_f_score(open_set, f_score)

输出:(0, 6)

Answer 1

V1

有dict.get()的版本比你的2个min()版本快得多(慢10倍以上)，但一直是大约2倍慢

def min_f_score_d(open_set, f_score):
inf = float("inf")
return min(open_set, key=lambda k: f_score.get(k, inf))

V2

由@stefan-pochmann建议，这个比经典迭代稍微慢一点

def min_f_score(open_set, f_score):
return min(f_score.keys() & open_set, key=f_score.get)

---

注意

if element in f_score        and f_score[element] < min_score:
# if faster than 
if element in f_score.keys() and f_score[element] < min_score:

<<BK_HR>

工作台代码/h3>

from timeit import timeit
A = """
def min_f_score(open_set, f_score):
min_score_element, min_score = open_set[0], f_score[open_set[0]]
for element in open_set:
if element in f_score and f_score[element] < min_score:
min_score = f_score[element]
min_score_element = element
return min_score_element
"""
B = """
def min_f_score(open_set, f_score):
inf = float("inf")
return min(open_set, key=lambda k: f_score.get(k, inf))
"""
C = """
def min_f_score(open_set, f_score):
return min(f_score.keys() & open_set, key=f_score.get)
"""
SETUP = """
from random import randrange
open_set = [(randrange(1000), randrange(1000)) for _ in range(1000)]
f_score = {pair: randrange(1000) for pair in open_set[:len(open_set) // 2]}
"""
NB = 20_000
print(timeit(setup=SETUP + A, stmt="min_f_score(open_set, f_score)", number=NB))  # ~2.7
print(timeit(setup=SETUP + B, stmt="min_f_score(open_set, f_score)", number=NB))  # ~4.8
print(timeit(setup=SETUP + C, stmt="min_f_score(open_set, f_score)", number=NB))  # ~3.1