我的清单中包含重复的元素，我不知道为什么

Question

I have a list of strings in a list called "texts". I am trying to scan each string for each of the words in a list called "key_words". If any of the key words is in the string, it goes into "list1". If none of the key words is in the string, it goes into "list2". My goal is for each string to be in its appropriate list once. The problem is that because I have three words in "key_words", a string with any of the words will go into list1 three times. I don't know why this is happening and I've been stuck working on this for an hour even though this seems pretty simple. Any help greatly appreciated.

I have a list of strings in a list called "texts".

list1 = []
list2 = []
key_words = ['must', 'should', 'wish']

for text in texts:

    for word in key_words:

        if text not in list1 and text not in list2:

            if word in text:
                 list1.append(text)

            else:
                list2.append(text)

Answer 1

Firstly, your code has a bug:

If any of the keywords are in text , it should go to list1

However in your code, you immediately copy the text to list2 , even if the first keyword is not present. The trick to solve this simply is in the quote above. Here is a simple and efficient solution:

import re

keyword_regex = '|'.join(key_words)  # Compile the regex if you have to use many times

for text in texts:
    if re.search(keyword_regex, text):  # Success if any keyword is in text
        list1.append(text)
    else:
        list2.append(text)

Answer 2

When you are looping over the keywords you are adding the text to lists multiple times.

list1 = []
list2 = []
key_words = ['must', 'should', 'wish']
texts = ["must the a hooray", "hooray should the a", "a the an"]

for text in texts:

    found = False
    if text not in list1 and text not in list2:
        for word in key_words:

                if word in text:
                     found = True
                     break
        if found:
            list1.append(text)
        else:
            list2.append(text)

print(list1)
print(list2)

Generates:

['must the a hooray', 'hooray should the a']

['a the an']

Answer 3

It will scan through the entire text document and add the "text" word to a respective list if the "text" word has not been previously inserted.

list1 = []
list2 = []
key_words = ['must', 'should', 'wish']

for text in texts:
    for word in key_words:
       if (word in text and not in list1):
          list1.append(text)
       elif (word not in list2):
          list2.append(text)

Answer 4

You need to scan the words in the text , not in the key_words list. The latter is just used to check the condition and decide for list1 or list2 .

This is an option using re.findall library for splitting the text in words, without punctuation . Once you have the list of words, you can iterate over it and check if each word in in key_words .

In the following example I'm using just one text string, you can extend the code for a list of texts.

This is what happen to text when applying the re.findal l method:

text = 'Must the show go on? I wish, it should! It must.'
print(re.findall(r'\w+',text))
#=> ['Must', 'the', 'show', 'go', 'on', 'I', 'wish', 'it', 'should', 'It', 'must']

The lookup on the text is performed once as the loop starts, here the complete code:

for txt_word in re.findall(r'\w+',text):
  if txt_word.lower() in key_words: # <- note .lower()
    list1.append(txt_word) # just add if not in list1 if you don't want duplicates
  else:
    list2.append(txt_word)

This is the output:

print(list1) #=>['Must', 'wish', 'should', 'must']
print(list2) #=> ['the', 'show', 'go', 'on', 'I', 'it', 'It']