在列表中查找具有值的最小字典键的 Pythonic 方法？

Question

我正在用 Python 编写 Astar 算法的实现，有时我想选择 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 是元组 (x, y) 的列表，表示 2D 中点的坐标，f_score 是字典 {(x, y): score}，它为每个元组 (x, y) 分配其 f_score 值。

我对@9​​87654330@ 的第一个实现如下：

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，但我想知道我是否可以想出一些更精简、更 Python 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]))

和：

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，但比第一个实现慢得多。

出于好奇，我想知道是否有更好的方法来实现min_f_score？

编辑：正如@a​​zro 所建议的（谢谢），我正在添加一个要执行的代码示例：

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

带有dict.get() 的版本比您的 2 个min() 版本快得多（慢了 10 倍以上），但仍然慢了大约 2 倍

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

@stefan-pochmann 建议，比经典迭代慢一点

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:

---

基准代码

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