Question

每当我运行此代码时，它只会给我一个空白列表，我想知道我做错了什么。我正在尝试打印长于n的单词列表。当我尝试运行更新的代码时，它只打印我输入的单词列表中的第一个单词。

def filterlongword(string,number):

    for i in range(len(string)):
        listwords = []
        if len(string[i]) > number:
            listwords.append(string[i])

        return listwords 


def main():
    words = input("Please input the list of words: ")
    integer = eval(input("Please input an integer: "))

    words1 = filterlongword(words,integer)

    print("The list of words greater than the integer is",words1)

main()

Answer 1

在循环

listwords

在循环

listwords

将输入字符串拆分为单词列表

def filterlongword(string,number): listwords = [] for i in range(len(string)): if len(string[i]) > number: listwords.append(string[i]) return listwords

使用list comprehension的更好的版本：

def filterlongword(string,number): return [word for word in string if len(word) > number]

要将输入字符串拆分为单词列表，请使用

words = input("Please input the list of words: ").split()

Answer 2

更好的只是

def filterlongword(string,number):
    return filter(lambda word:len(word)>number, string)
    # or: return [w for w in string if len(w) > number]

Answer 3

def listing(guess, number):

    new_list = []

    for i in range(len(guess)):
        if len(guess[i]) > number:
            new_list.append(guess[i])

    print (new_list)

list1 = input("take input: ")

list = list1.split(",")

def main():
    global list, integer1
    integer = input()
    integer1 = int(integer)
    listing(list, integer1)

main()

**尝试此代码..这样可行，使用分隔符形成输入列表**

Answer 4

您的主要问题是将单词作为单个字符串而不是可迭代的字符串传递。第二个问题是没有为缺少的.split指定单词之间的分隔符。这是我的版本。

我使longwords成为一个生成器函数，因为在实际使用中，没有必要将长字序列作为列表，我在输出格式中给出了一个例子。

def longwords(wordlist, length):
    return (word for word in wordlist if len(word) >= length)

def main():
    words = input("Enter words, separated by spaces: ").split()
    length = int(input("Minimum length of words to keep: "))
    print("Words longer than {} are {}.".format(length,
          ', '.join(longwords(words, length))))

main()

这导致例如

Enter words, separated by spaces: a bb ccc dd eeee f ggggg
Minimum length of words to keep: 3
Words longer than 3 are ccc, eeee, ggggg.

Answer 5

也许您可以将代码缩短为以下内容：

legend("topright", inset=.05, c("p(k)"), lty="solid", pch=21, col=c"black", pt.bg="red")

Answer 6

content_divs

Answer 7

尝试

*** 

def filter_long_words(n, words):
    list_of_words=[]
    a= words.split(" ")
    for x in a:
      if len(x)>n:
        list_of_words.append(x)
    return list_of_words

Answer 8

LinkEntity

Answer 9

def filter_long_words(lst,n):
    a=[]
    for i in lst:
        if n<len(i):
            a.append(i)
    return a

Answer 10

过滤单词列表

def filter_long_words(lst,n):
    
    return [word for word in lst if len(word)>n]

编写一个函数filter_long_words（），它接受一个单词列表和一个整数n，并返回长于n的单词列表

10 个答案: