Assuming I have a list of phrases:
list = ['new york', 'school', 'new']
and a string
text = 'i am going to a school in new york and therefore i have to buy a new uniform to go to new york'
I would like to find the number of words preceeding each phrase (just for first appearance) ie output should be:
new york = 7
school = 5
new = 7
Any idea how I can effectively achieve this?
Naive approach, without any performance or NLP considerations:
lst = ['new york', 'school', 'new'] # do not use 'list' as a name
text = 'i am going to a school in new york and therefore i have to buy a new uniform to go to new york'
{p: len(text[:text.find(p)].strip().split()) for p in lst}
# {'new york': 7, 'school': 5, 'new': 7}
Using count
and index
:
lst = ['new york', 'school', 'new']
text = 'i am going to a school in new york and therefore i have to buy a new uniform to go to new york'
for x in lst:
print(f"{x} = {text.count(' ', 0, text.index(x))}")
# new york = 7
# school = 5
# new = 7
count
counts whitespaces in text
from start till you meet the first appearance of phrase which is same as the number of words preceding that phrase.
lst = ['new york', 'school', 'new']
text = 'i am going to a school in new york and therefore i have to buy a new uniform to go to new york'
This will give you the string whose count you are searching and count of string
for x in lst:
print(x +": "+str(len(text[0:text.index(x)].split(' ')) -1))
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