有什么方法可以更有效地检查单词中的所有字母是否都在列表中？

可以实现的一种加速方法是重新组织逻辑，以便在遇到不匹配的字母时逐字母循环结束。平均而言，这将与不允许的字母数成正比，从而加快代码的速度。

def wordsearch(board, sowpods):
    length = []
    for word in sowpods:
        testboard = list(board)
        count = 0
        for letter in word:
            if letter in testboard:
                testboard.remove(letter)
                count += 1
        if len(word) == count:
            length.append(len(word))
    if len(length) == 0:
        return 0
    return max(length)

def find_matches(allowed, dictionary):
    allowed_list = list(allowed)
    match_words = []
    for word in dictionary:    
        good = True
        for letter in word:
            if letter not in allowed_list:
                good = False
                break
        if good == True:
            match_words.append(len(word))    
    return max(match_words)

import timeit
start_time = timeit.default_timer()
allowed = 'iptneazol'
result = wordsearch(allowed, sb_list)
# code you want to evaluate
elapsed = timeit.default_timer() - start_time
print(elapsed)
>>>0.6867701730000135

start_time = timeit.default_timer()
allowed = 'iptneazol'
result = find_matches(allowed, sb_list)
elapsed = timeit.default_timer() - start_time
print(elapsed)
>>>0.10806877499999246

由于代码不会转换允许列表，因此这也解决了允许字母出现两次的问题。为了获得更高的效率，您可以将for循环重写为调用服务函数的列表理解，或者使用迭代器/生成器。

回答来源：stackoverflow