Question

我目前正在尝试使用列表做一个信访柜台。该函数的目标是计算每个字符串中有多少个字符，然后打印出多少个字符串的长度为4。我遇到了一些麻烦，希望能有所帮助。我对python很陌生，所以如果答案很简单，或者错过了冒号或其他内容，我深表歉意。

def count_4let(letter_list):

    if letter_list == len(4):
                counter+=1

    print("The number of words with the length of 4 is", counter)

letter_list = ["Whos","the","leader","of","the","club","thats","made","for","you","and","me","M I C K E Y","M O U S E"]
#main
print(count_4let(letter_list))

Answer 1

您需要遍历列表的元素。您在每个元素上调用len()，而不是len(4)。

在添加变量之前，还需要初始化counter变量。

该函数应仅返回计数，而不输出任何内容。打印由调用方完成。

def count_4let(word_list):
    counter = 0
    for word in word_list:
        if len(word) == 4:
            counter += 1
    return counter

letter_list = ["Whos","the","leader","of","the","club","thats","made","for","you","and","me","M I C K E Y","M O U S E"]
#main
print("The number of words with length 4 is", count_4let(letter_list))

Answer 2

您可以将sum函数与生成器表达式一起使用，该表达式可输出列表中给定单词的长度是否为4：

def count_4let(letter_list):
    return sum(len(word) == 4 for word in letter_list)

Answer 3

使用map()和count()的另一种解决方案

>>> l = ["whos", "the", "leader", "of", "the", "club", "thats", "made", "for", "you", "and", "me", "M I C K E Y", "M O U S E"]
>>> list(map(len, l)).count(4)
3

Answer 4

首先，您不应该写数字的长度，请考虑一下这是没有意义的：

def count_4let(letter_list):
  if len(letter_list) == 4:
    counter += 1
print("The number of words with the length of 4 is", counter)

letter_list ["Whos", "the", "leader", "of", "the", "club", "thats", "made", "for", "you"," and", "me", "M I C K E Y", "M O U S E"]
print(count_4let(letter_list))

带列表的字母柜台

4 个答案: