Python - 如何使用函数删除与用户输入匹配的任何列表元素

Question

基本上我希望用户输入一个数字，然后扫描从 dictionary.txt 导入的列表中的单词。然后它删除任何长度等于用户输入的单词，然后更新函数之外的列表。

def read_and_delete(x):
        count = 0
        i = 0
        for delete in x:
            if x[i] != word_length:
                del x[i]
                i += 1
                return
            elif x[i] == word_length:
                count += 1
                i += 1
                return
            else:
                break
        print(len(list_of_words))

word_length = int(input("Please enter a word length: "))
dictionary = open("dictionary.txt", "r")
list_of_words = [word.rstrip('\n').rstrip('\r') for word in dictionary]
read_and_delete(list_of_words)

Answer 1

首先，我想回顾一下你已经拥有的功能：

def read_and_delete(x):
    count = 0
    i = 0

    # for each word in list, store word in the "delete" variable
    for delete in x:

        # you are iterating over elements, but also using index
        if x[i] != word_length:  # if word not equals length
            del x[i]
            i += 1
            return  # exit from function

        elif x[i] == word_length:  # again, comparing a word and a number
            count += 1
            i += 1
            return  # exit from function
        else:  # never happens, because word is either
               # equal or not, there is no other case
            break  # exit from loop
    # if we actually visited the for loop (the list was not empty),
    # this line would never be executed
    print(len(list_of_words))

接下来，如何解决：

1. 如果您想遍历 all 列表，则不应使用 return 或 break（认为这些在其他情况下很有用）。
2. 如果你确实想从列表中删除元素，你应该做的有点不同：How to remove list elements in a for loop in Python?
3. 要获取单词（和任何其他字符串）的长度，您可以使用len()。
4. 另外，检查条件的逻辑。

总而言之，这里是固定函数（创建一个新列表）：

def read_and_delete(x):
    result = []

    for word in x:
        if len(word) != word_length:
            result.append(word)  # add word to the end of "result"

    print(len(result))

如您所见，循环现在非常简单。因此，您可以将其重写为列表推导：

# instead of function call
filtered_list = [word for word in list_of_words if len(word) != word_length]
print(len(filtered_list))

你可以这样理解：

[word                          # put each "word" into a list
 for word in list_of_words     # get them from "list_of_words"
 if len(word) != word_length]  # leave only those which length != "word_length"

编辑：函数中的固定条件。

Answer 2

假设您的dictionary.txt 每行列出一个单词：

word_length = int(input("Please enter a word length: "))

with open("dictionary.txt", "r", encoding="utf-8") as word_list:
    words = [line.strip() for line in word_list if len(line.strip()) != word_length]

print("%i words remain" % len(words))
print(words)

我们的想法不是创建一个列表，然后在单独的步骤中从中删除项目，而是在我们构建列表时过滤项目。

Answer 3

您可以使用内置的“过滤器”功能：

ls = ['word1','word','w0']
len_delete = 4
new_ls = filter(lambda x:len(x)!=len_delete,ls)

new_ls = ['word1', 'w0']

Answer 4

def read_and_delete(x):
    resulting_list = []          
    for delete in x:
        if len(delete) == word_length: # len(delete) returns length of element
            continue  # goes to the next element in the list                             
        else:
            resulting_list.append(delete)             
    print(resulting_list)

word_length = int(input("Please enter a word length: "))
dictionary = open("dictionary.txt", "r")
list_of_words = [word.rstrip('\n').rstrip('\r') for word in dictionary]
read_and_delete(list_of_words)

附：学习更多关于循环是如何工作的:)

Answer 5

我的六便士值：

def read_and_delete(x,word_length):
    x = [elem for elem in x if len(elem) !=word_length]
    return x

a=['123','1234','123','123','abcdef','abc','1234']
a=read_and_delete(a,3)
a
['1234', 'abcdef', '1234']

对不起@Meloman！刚刚看到你的帖子，我几乎复制了它，而且解释得少得多。

Answer 6

我认为您来自 C 或类似语言：

def read_and_delete(x):
    for ix, delete in enumerate(x):
        if len(delete) == word_length:
            del x[ix]

    return x

word_length = int(input("Please enter a word length: "))
dictionary = open("dictionary.txt", "r")
list_of_words = [word.rstrip('\n').rstrip('\r') for word in dictionary]
read_and_delete(list_of_words)
print(list_of_words)

这个较新版本的代码将就地更新列表，因此在read_and_delete 中对x（传递给函数的列表）所做的任何更改都会影响该列表的所有引用，包括list_of_words。此函数还返回受影响的列表，但是您可以轻松地将其删除，因为 list_of_words 会通过函数内部对其进行的任何更改进行更新。当然，有一种更简单的方法可以在 Python 中实现这一点，为了清楚起见，我使用了您的代码，以免让您感到困惑。

编辑： 感谢 @RolfofSaxony 指出上述代码中的歧义，因此我编辑了这个问题来解决这个问题：

word_length = int(input("Please enter a word length: "))
dictionary = open("dictionary.txt", "r")
list_of_words = [word.rstrip('\n').rstrip('\r') for word in dictionary]

def read_and_delete(x):
    for delete in x[:]:
        if len(delete) == word_length:
            ix = x.index(delete)
            del x[ix]
    return x


read_and_delete(list_of_words)
print(list_of_words)