在列表中查找具有值的最小字典鍵的 Pythonic 方法？

Question

我正在 Python 中編寫 Astar 算法的實現，在某些時候我想要 select 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是一個元組(x, y)的列表，表示一個點在 2D 中的坐標， f_score是一個字典{(x, y): score} ，它為每個元組(x, y)分配其f_score值。

我對min_f_score的第一個實現如下：

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

它工作正常，但我想知道我是否可以想出一些更精簡、更 pythonic 的代碼。 經過一些研究，我想出了另外兩個實現：

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")))

兩者似乎都有效並給我相同的結果，但比第一個實施慢得多。

出於好奇，我想知道是否有更好的方法來實現min_f_score ？

編輯：按照@azro（謝謝）的建議，我添加了一個要執行的代碼示例：

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)

Output: (0, 6)

Answer 1

V1

帶有dict.get()的版本比你的 2 min()版本快得多（慢了 10 倍以上），但仍然慢了大約 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建議，那個比經典迭代慢一點

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

---

筆記

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:

---

基准代碼

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