Python - 字符串中匹配词的索引

Question

I'm looking for an efficient way to convert an index in a string to the word the index is in.

For example, if this is my string:

This is a very stupid string

and the index I get is, let's say 10, so the output should be very . Also if the index is 11,12 or 13 - the output should be very .

One can assume that the words are separated by 1 space each time. Doing it with a for loop or something is not hard, the question is whether there is a more efficient way (as my text is HUGE and I have MANY indices to convert to words).

For the example, let the indices be 10, 13, 16 and thus the output should be:

10 very
13 very
16 stupid

Any help would be appreciated!

Answer 1

The following should perform quite well. Begin by obtaining the words in the string using split , and find the indedx where they begin using enumerate and a list comprehension:

words = s.split()
# ['This', 'is', 'a', 'very', 'stupid', 'string']
# Obtain the indices where all words begin
ix_start_word = [i+1 for i,s in enumerate(s) if s==' ']
# [5, 8, 10, 15, 22]

And now you could use NumPy's np.searchsorted to obtain a word given an index:

words[np.searchsorted(ix_start_word, ix)]

Checking with the examples above:

words[np.searchsorted(ix_start_word, 11)]
#'very'

words[np.searchsorted(ix_start_word, 13)]
# 'very'

words[np.searchsorted(ix_start_word, 16)]
# 'stupid'

Answer 2

I'm not particularly proud of how clean this is, but I think it does the trick:

from numpy import cumsum, array

sample = 'This is a very stupid string'

words = sample.split(' ')
lens = [len(_)+1 for _ in words]

ends = cumsum(lens)
starts = array([0] + list(ends[:-1]))

output = {}
for a, b, c in zip(starts, ends, words):
    for i in range(a, b):
        output[i] =  c
for a, b in output.items():
    print(a, b)

0 This
1 This
2 This
3 This
4 This
5 is
6 is
7 is
8 a
9 a
10 very
11 very
12 very
13 very
14 very
15 stupid
16 stupid
17 stupid
18 stupid
19 stupid
20 stupid
21 stupid
22 string
23 string
24 string
25 string
26 string
27 string
28 string

Answer 3

This isn't very effective, because it uses regular-expression, but is one way to solve the problem without using any loops.

import re

def stuff(pos):
    x = "This is a very stupid string"
    pattern = re.compile(r'\w+\b')
    pattern2 = re.compile(r'.*(\b\w+)')
    end = pattern.search(x, pos=pos).span()[1]
    print(pattern2.search(x, endpos=end).groups()[0])

stuff(2)
stuff(10)
stuff(11)
stuff(16)

Results:

This
very
very
stupid