I have to do the following.
given a text I have to find the preceding words to all numbers up to a stop word belonging to a check_words list (kind of stopwrods)
My code:
check_words = ['the', 'a', 'with','to']
mystring = 'the code to find the beautiful words 78 that i have to nicely check 45 with the snippet'
list_of_words = mystring.split()
In that particular text I would check before '78' and '45' and will go backwards up to the point where I find any of the words check_words (but not more than 8 words).
The code for doing that might be:
preceding_chunks = []
for i,word in enumerate(list_of_words):
if any(char.isdigit() for char in word):
# 8 precedent words (taking into account that I can not slice with 8 back at the beginning)
preceding_words = list_of_words[max(0,i-8):i]
preceding_words[::-1]
# I check from the end of the list towards the start
for j,sub_word in enumerate(preceding_words[::-1]):
if sub_word in check_words:
# printing out j for checking
myposition = j
print(j)
real_preceding_chunk = preceding_words[len(preceding_words)-j:]
print(real_preceding_chunk)
preceding_chunks.append(real_preceding_chunk)
break
This code works. But I have the impression (perhaps I am wrong) that it can be achieved with a couple of one liners and hence without loops. Any idea?
NOTE: This question is about improvement the readability of the code, trying to get rid of the loops to make the code faster, and trying to make the code nicer, which is part of the zen of python. Somehow "making things nicer" is sometimes not welcome in stackoverflow, please just dont downgrade because that.
NOTE 2: Some previous checks that I did:
Finding the position of an item in another list from a number in a different list
