字符串中短语之前的前序单词数

Question

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?

Answer 1

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}

Answer 2

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.

Answer 3

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))