get n words before a character number in a string WITHOUT cut words

user_7168098 0
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I am given a string and a character position in that string. I want to get the n words before that position in a way that it does not include the last word if the character positon is in the middle of a word

text = 'the house is big the house is big the house is big'
char_nr = 19
list_of_words_before = text[:char_nr-1].split()
print(list_of_words_before) # we see that the string is splited in "the" I dont want hence the t in the list
nr_words = 3
if nr_words >len(list_of_words_before):
    nr_words = len(list_of_words_before)
    
list_of_words_before[-nr_words:]

this gives:

['the', 'house', 'is', 'big', 't']
['is', 'big', 't']

but actually what I really want is ['house', 'is','big'] since t is just part of a word.

How would you make sure in the first place to divide by a space between words? Is any other solution?

list python string
5 Answers

Using regex:

>>> import re
>>> text = 'the house is big the house is big the house is big'
>>> result = re.match(r".{0,18}\b", text).group(0).split()
>>> result
['the', 'house', 'is', 'big']
>>> result[-3:]
['house', 'is', 'big']

Explanation:

  • . any character
  • {0,18} match the preceding (.) 0 to 18 times, as many as possible
  • \b the match ends in a beginning or ending of a word, so we don't get partial words
user_98358720

Maybe something like this:

text = 'the house is big the house is big the house is big'
char_nr = 19
list_of_words_before = text[:char_nr - 1]
splitted = list_of_words_before.split()

if list_of_words_before[-1] != ' ':
    splitted = splitted[:-1]

nr_words = 3
print(splitted[-nr_words:])

Output:

['house', 'is', 'big']
user_99267210

You can check the character at char_nr and if it's a non-word character then the splitting was correct, otherwise you need to remove the last item from the list. Assuming that " " is the only character that can occur between words:

if text[char_nr] != " ":
    list_of_words_before = list_of_words_before[:-1]
user_37672390

I think this is what you're looking for:

def get_n_words(text, char_nr, nr_words):
    if text[char_nr-1] == " ":
        list_of_words_before = text[:char_nr-1].split()
    else:
        list_of_words_before = text[:char_nr-1].split()[:-1]
    print(list_of_words_before)
    if nr_words >len(list_of_words_before):
        nr_words = len(list_of_words_before)
        
    print(list_of_words_before[-nr_words:])

text_1 = 'the house is big the house is big the house is big'
text_2 = 'the house is big a house is big the house is big'

print("Last word truncated:")
get_n_words(text_1, 19, 3)
print("\nLast word not truncated:")
get_n_words(text_2, 19, 3)

That has the following output:

Last word truncated:
['the', 'house', 'is', 'big']
['house', 'is', 'big']

Last word not truncated:
['the', 'house', 'is', 'big', 'a']
['is', 'big', 'a']
user_196676750

You might use a pattern starting the match with a non whitespace character using \S and then match 0-18 times any character using .{0,18} while asserting not a non whitespace character to the right using a negative lookahead (?!\S)

\S.{0,18}(?!\S)

Regex demo | Python demo

import re

text = 'the house is big the house is big the house is big'
char_nr = 19
pattern = rf"\S.{{0,{char_nr - 1}}}(?!\S)"

strings = re.findall(pattern, text)

print(strings)

list_of_words_before = strings[1].split()
print(list_of_words_before)

nr_words = 3
lenOfWordsBefore = len(list_of_words_before)
if nr_words > lenOfWordsBefore:
    nr_words = lenOfWordsBefore

print(list_of_words_before[-nr_words:])

Output

['the house is big', 'the house is big', 'the house is big']
['the', 'house', 'is', 'big']
['house', 'is', 'big']
user_54249880
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