Question

我有以下代码可以正常工作。然而，虽然我正在学习我试图找到不断改进我的代码的方法。是否可以以任何方式使下面的代码更整洁？

def isValidWord(word, hand, wordList):
   """
   Returns True if word is in the wordList and is entirely
    composed of letters in the hand. Otherwise, returns False.

    Does not mutate hand or wordList.

    word: string
    hand: dictionary (string -> int)
    wordList: list of lowercase strings
    """
    x=''
    if word in wordList:
        x=True
    else:
        x=False
    #dont mutate the hand so copy
    hand2=hand.copy()
    for letter in word:
        if letter in hand2.keys():
            hand2[letter]-=1
            if hand2[letter]<0:
                x=False
                break

        else:
            x = False
    return x

运行代码：

hand = {'r': 1, 'a': 3, 'p': 2, 'e': 1, 't': 1, 'u':1}

word = "rapture"

wordList = ['rapture','alex']
isValidWord(word, hand, wordList)

Answer 1

使用all和Counter：

from collections import Counter

def isValidWord(word, hand, wordList):
    wc = Counter(word)
    return word in wordList and all(wc[c] <= hand.get(c, -1) for c in wc)

如果您有非常大的单词列表，请考虑使用集合代替O（1）查找时间。

Answer 2

正确处理计数：

from collections import Counter

def isValidWord(word, hand, wordList):
    # Check for membership first and early out if fails
    # Otherwise, see if the count of any letter in word exceeds that of the hand
    return word in wordList and all(hand.get(let, -1) >= cnt for let, cnt in Counter(word).items())

Python - Neaten up Code - 循环和递归功能

2 个答案: