Python: how to replace substrings in a string given list of indices

Question

I have a string:

"A XYZ B XYZ C"

and a list of index-tuples:

((2, 5), (8, 11))

I would like to apply a replacement of each substring defined by indices by the sum of them:

A 7 B 19 C

I can't do it using string replace as it will match both instances of XYZ. Replacing using index information will break on the second and forth iterations as indices are shifting throughout the process.

Is there a nice solution for the problem?

UPDATE. String is given for example. I don't know its contents a priori nor can I use them in the solution.

My dirty solution is:

text = "A XYZ B XYZ C"
replace_list = ((2, 5), (8, 11))

offset = 0
for rpl in replace_list:
    l = rpl[0] + offset
    r = rpl[1] + offset

    replacement = str(r + l)
    text = text[0:l] + replacement + text[r:]

    offset += len(replacement) - (r - l)

Which counts on the order of index-tuples to be ascending. Could it be done nicer?

Answer 1

Here's a reversed-order list-slice assignment solution:

text = "A XYZ B XYZ C"
indices = ((2, 5), (8, 11))
chars = list(text)

for start, end in reversed(indices):
    chars[start:end] = str(start + end)

text = ''.join(chars) # A 7 B 19 C

Answer 2

Anticipating that if these pairs-of-integer selections are useful here, they will also be useful in other places, then I would proably do something like this:

def make_selections(data, selections):
    start = 0
    # sorted(selections) if you don't want to require the caller to provide them in order
    for selection in selections:
        yield None, data[start:selection[0]]
        yield selection, data[selection[0]:selection[1]]
        start = selection[1]
    yield None, data[start:]

def replace_selections_with_total(data, selections):
    return ''.join(
        str(selection[0] + selection[1]) if selection else value
        for selection, value in make_selections(data, selections)
    )

This still relies on the selections not overlapping, but I'm not sure what it would even mean for them to overlap.

You could then make the replacement itself more flexible too:

def replace_selections(data, selections, replacement):
    return ''.join(
        replacement(selection, value) if selection else value
        for selection, value in make_selections(data, selections)
    )

def replace_selections_with_total(data, selections):
    return replace_selections(data, selections, lambda s,_: str(s[0]+s[1]))

Answer 3

You can use re.sub():

In [17]: s = "A XYZ B XYZ C"

In [18]: ind = ((2, 5), (8, 11))

In [19]: inds = map(sum, ind)

In [20]: re.sub(r'XYZ', lambda _: str(next(inds)), s)
Out[20]: 'A 7 B 19 C'

But note that if the number of matches is larger than your index pairs it will raise a StopIteration error. In that case you can pass a default argument to the next() to replace the sub-string with.

If you want to use the tuples of indices for finding the sub strings, here is another solution:

In [81]: flat_ind = tuple(i for sub in ind for i in sub)
# Create all the pairs with respect to your intended indices. 
In [82]: inds = [(0, ind[0][0]), *zip(flat_ind, flat_ind[1:]), (ind[-1][-1], len(s))]
# replace the respective slice of the string with sum of indices of they exist in intended pairs, otherwise just the sub-string itself.
In [85]: ''.join([str(i+j) if (i, j) in ind else s[i:j] for i, j in inds])
Out[85]: 'A 7 B 19 C'

Answer 4

Assuming there are no overlaps then you could do it in reverse order

text = "A XYZ B XYZ C"
replace_list = ((2, 5), (8, 11))

for start, end in reversed(replace_list):
    text = f'{text[:start]}{start + end}{text[end:]}'

# A 7 B 19 C

Answer 5

Here's a quick and slightly dirty solution using string formatting and tuple unpacking:

s = 'A XYZ B XYZ C'
reps = ((2, 5), (8, 11))
totals = (sum(r) for r in reps)
print s.replace('XYZ','{}').format(*totals)

This prints:

A 7 B 19 C

First, we use a generator expression to find the totals for each of our replacements. Then, by replacing 'XYZ' with '{}' we can use string formatting - *totals will ensure we get the totals in the correct order.

Edit

I didn't realise the indices were actually string indices - my bad. To do this, we could use re.sub as follows:

import re
s = 'A XYZ B XYZ C'

reps = ((2, 5), (8, 11))
for a, b in reps:
    s = s[:a] + '~'*(b-a) + s[b:]
totals = (sum(r) for r in reps)
print re.sub(r'(~+)', r'{}', s).format(*totals)

Assuming there are no tildes (~) used in your string - if there are, replace with a different character. This also assumes none of the "replacement" groups are consecutive.

Answer 6

Yet another itertools solution

from itertools import *

s = "A XYZ B XYZ C"
inds = ((2, 5), (8, 11))
res = 'A 7 B 19 C'


inds = list(chain([0], *inds, [len(s)]))
res_ = ''.join(s[i:j] if k % 2 == 0 else str(i + j)
        for k, (i,j) in enumerate(zip(inds, inds[1:])))

assert res == res_

Answer 7

Imperative and stateful:

s = 'A XYZ B XYZ C'
indices = ((2, 5), (8, 11))
res = []
i = 0
for start, end in indices:
    res.append(s[i:start] + str(start + end))
    i = end
res.append(s[end:])
print(''.join(res))

Result:

A 7 B 19 C

Answer 8

One way to do this using itertools.groupby.

from itertools import groupby


indices = ((2, 5), (8, 11))
data = list("A XYZ B XYZ C")

We start with replacing the range of matched items with equal number of None.

for a, b in indices:
    data[a:b] = [None] * (b - a)

print(data)
# ['A', ' ', None, None, None, ' ', 'B', ' ', None, None, None, ' ', 'C']

The we loop over the grouped data and replace the None groups with the sum from indices list.

it = iter(indices)
output = []
for k, g in groupby(data, lambda x: x is not None):
    if k:
        output.extend(g)
    else:
        output.append(str(sum(next(it))))

print(''.join(output))
# A 7 B 19 C

Answer 9

There is also a solution which does exactly what you want. I have not worked it out completely, but you may want to use: re.sub() from the re library.

Look here, and look for the functions re.sub() or re.subn(): https://docs.python.org/2/library/re.html

If I have time, I will work out your example later today.