检查句子中的单词

Question

I write a program in Python. The user enters a text message. It is necessary to check whether there is a sequence of words in this message. Sample. Message: "Hello world, my friend.". Check the sequence of these two words: "Hello", "world". The Result Is "True". But when checking the sequence of these words in the message: "Hello, beautiful world "the result is"false". When you need to check the presence of only two words it is possible as I did it in the code, but when combinations of 5 or more words is difficult. Is there any small solution to this problem?

s=message.text
s=s.lower()
lst = s.split()
elif "hello" in lst and "world" in lst :
    if "hello" in lst:
        c=lst.index("hello")
    if lst[c+1]=="world" or lst[c-1]=="world":
        E=True
    else:
        E=False

Answer 1

The straightforward way is to use a loop. Split your message into individual words, and then check for each of those in the sentence in general .

word_list = message.split()     # this gives you a list of words to find
word_found = True
for word in word_list:
    if word not in message2:
        word_found = False

print(word_found)

The flag word_found is True iff all words were found in the sentence. There are many ways to make this shorter and faster, especially using the all operator, and providing the word list as an in-line expression.

word_found = all(word in message2 for word in message.split())

Now, if you need to restrict your "found" property to matching exact words, you'll need more preprocessing. The above code is too forgiving of substrings, such as finding "Are you OK ?" in the sentence "your joke is only barely funny". For the more restrictive case, you should break message2 into words, strip those words of punctuation, drop them to lower-case (to make matching easier), and then look for each word (from message ) in the list of words from message2 .

Can you take it from there?

Answer 2

I don't know if that what you really need but this worked you can tested

message= 'hello world'
message2= ' hello beautiful world' 
if 'hello' in message and 'world'  in message :
  print('yes')
else :
  print('no')
if   'hello' in message2 and 'world'  in message2 :
  print('yes')  

out put : yes yes

Answer 3

I will clarify your requirement first:

• ignore case

• consecutive sequence

• match in any order, like permutation or anagram

• support duplicated words

if the number is not too large, you can try this easy-understanding but not the fastest way.

1. split all words in text message
2. join them with ' '
3. list all the permutation of words and join them with ' ' too, For example, if you want to check sequence of ['Hello', 'beautiful', 'world'] . The permutation will be 'Hello beautiful world' , 'Hello world beautiful' , 'beautiful Hello world' ... and so on.
4. and you can just find whether there is one permutation such as 'hello beautiful world' is in it.

The sample code is here:

import itertools
import re

# permutations brute-force, O(nk!)
def checkWords(text, word_list):
    # split all words without space and punctuation
    text_words= re.findall(r"[\w']+", text.lower())

    # list all the permutations of word_list, and match
    for words in itertools.permutations(word_list):
        if ' '.join(words).lower() in ' '.join(text_words):
            return True
    return False

    # or use any, just one line
    # return any(' '.join(words).lower() in ' '.join(text_words) for words in list(itertools.permutations(word_list)))
def test():
    # True
    print(checkWords('Hello world, my friend.', ['Hello', 'world', 'my']))
    # False
    print(checkWords('Hello, beautiful world', ['Hello', 'world']))
    # True
    print(checkWords('Hello, beautiful world Hello World', ['Hello', 'world', 'beautiful']))
    # True
    print(checkWords('Hello, beautiful world Hello World', ['Hello', 'world', 'world']))

But it costs a lot when words number is large, k words will generate k! permutation, the time complexity is O(nk!).

I think a more efficient solution is sliding window . The time complexity will decrease to O(n):

import itertools
import re
import collections

# sliding window, O(n)
def checkWords(text, word_list):
    # split all words without space and punctuation
    text_words = re.findall(r"[\w']+", text.lower())
    counter = collections.Counter(map(str.lower, word_list))
    start, end, count, all_indexes = 0, 0, len(word_list), []

    while end < len(text_words):
        counter[text_words[end]] -= 1
        if counter[text_words[end]] >= 0:
            count -= 1
        end += 1

        # if you want all the index of match, you can change here
        if count == 0:
            # all_indexes.append(start)
            return True

        if end - start == len(word_list):
            counter[text_words[start]] += 1
            if counter[text_words[start]] > 0:
                count += 1
            start += 1

    # return all_indexes
    return False