Python，While-True循环，获取整数并返回相同长度的单词

Question

我不确定如何开始这个函数。从概念上讲，我认为它应该查找单词列表，直到找到与给定整数长度相同的单词，才返回该单词。

lenword(num)示例

def lenword(4):
   wrdlst = [ 'i' , 'to', 'two', 'four']
while True:
#stuck here

#returns 'four'

请帮帮我！

Answer 1

def lenword(n):
    wrdlst = [ 'i' , 'to', 'two', 'four']
    # for every item in the list, if the length of that item is
    # equal to n (in this case 4) then print the item, and return it.
    for word in wrdlst:
        if len(word) == n:
            print word
            return word

# call the function
four_letter_word = lenword(4)

或者如果你的列表中有一个以上的四个字母的单词。

def lenword(n):
    wrdlst = [ 'i' , 'to', 'two', 'four']
    found_words = []
    # for every item in the list, if the length of that item is
    # equal to n (in this case 4) then print the item, and return it.
    for word in wrdlst:
        if len(word) == n:
            found_words.append(word)
    return found_words

# call the function
four_letter_words = lenword(4)

# print your words from the list
for item in four_letter_words:
    print item 

Answer 2

你可能想要这样的东西：

def lenword(word_length):
    wordlist = [ 'i' , 'to', 'two', 'four']
    for word in wordlist:
        if len(word) == word_length:
            # found word with matching length
            return word
    # not found, returns None

使用为特定的单词列表创建专门的搜索器可以使其更好：

def create_searcher(words):
    def searcher(word_length, words=words):
        for word in words:
            if len(word) == word_length:
                # found word with matching length
                return word
        # not found, returns None
    return searcher

并像这样使用它：

# create search function specialized for a list of words
words_searcher = create_searcher(['list', 'of', 'words'])
# use it
words_searcher(4) # returns 'list'
words_searcher(3) # returns None
words_searcher(2) # returns 'of'

Answer 3

如果你的列表是有序的，这样第n个元素的长度是n+1，那么你只需要：

wrdlist = [ 'i' , 'to', 'two', 'four']

def lenword(n):
   return wrdlist[n-1]

lenword(4)