How to combine two elements of a list based on a given condition

Question

I want to combine two elements in a list based on a given condition.

For example if I encounter the character 'a' in a list, I would like to combine it with the next element. The list:

['a', 'b', 'c', 'a', 'd']

becomes

['ab', 'c', 'ad']

Is there any quick way to do this?

One solution I have thought of is to create a new empty list and iterate through the first list. As we encounter the element 'a' in list 1, we join list1[index of a] and list1[index of a + 1] and append the result to list 2. However I wanted to know if there is any way to do it without creating a new list and copying values into it.

You can modify the list on-the-fly with your algorithm. – ForceBru Apr 26 '17 at 15:00 — ForceBru, Apr 26 '17 at 15:00

EyuelDK · Accepted Answer · 2017-04-26T15:11:35.033

4

This does not create a new list, just modifies the existing one.

l = ['a', 'b', 'c', 'a', 'd']
for i in range(len(l)-2, -1, -1):
    if l[i] == 'a':
        l[i] = l[i] + l.pop(i+1)
print(l)

edited Apr 26 '17 at 15:11

answered Apr 26 '17 at 15:05

EyuelDK

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The crucial detail here seems to be that the list is processed in reverse. You might want to explain why this is a good idea. – mkrieger1 Jan 02 '22 at 19:49
@mkrieger1 you get out index in l.pop(i+1) if you start with `for i in range(len(l))` – Javad Bahoosh Feb 02 '22 at 15:14

MassPikeMike · Answer 2 · 2017-04-26T15:15:50.797

3

If you don't want to use list comprehension to create a new list (maybe because your input list is huge) you could modify the list in-place:

i=0
while i < len(l):
 if l[i]=='a':
  l[i] += l.pop(i+1)
  i += 1

edited Apr 26 '17 at 15:15

answered Apr 26 '17 at 15:07

MassPikeMike

672
3
12

Elegant solution but you are not showing a key part of your code: what is `i` initialized to at the beginning? – EyuelDK Apr 26 '17 at 15:12
you are missing `i++` simply put, you code doesn't work. – EyuelDK Apr 26 '17 at 15:14
Thanks again! bad copy-paste on my part. Note though that python has no `++` operator, so you have to use `i += 1` – MassPikeMike Apr 26 '17 at 15:16

Moses Koledoye · Answer 3 · 2017-04-26T15:07:56.950

2

Use a list comprehension with an iterator on your list. When the current iteratee is a simply join it with the next item from the iterator using next:

l = ['a', 'b', 'c', 'a', 'd']

it = iter(l)
l[:] =  [i+next(it) if i == 'a' else i for i in it]
print l
# ['ab', 'c', 'ad']

edited Apr 26 '17 at 15:07

answered Apr 26 '17 at 15:02

Moses Koledoye

77,341
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List comprehension is creating a new list, thus contradicting with the user's requirements `without creating a new list ` – EyuelDK Apr 26 '17 at 15:06
@EyuelDK Well about a full slice assignment to the original list? – Moses Koledoye Apr 26 '17 at 15:08
It Still creates a new list, but you just reassigned the elements to the slice. It still does not modify in place. To summarize, if you are using a list comprehension, it is creating a new list. – EyuelDK Apr 26 '17 at 15:10
@EyuelDK Depends on what create a new list really means in this context. The `range` you've used in your solution creates a new list in Python 2. – Moses Koledoye Apr 26 '17 at 15:15
You are right, in python 2.7 but I think that isn't relevant for this problem. It is implied that the user didn't want to create a new list of the original list, my use of range is irrelevant - can be fixed by using Python 3 or just a while statement, it isn't an issue. But what you are doing is creating a new list and then reassigning to the original list - just in a fancy syntax. That's why I think it doesn't meet requirement. – EyuelDK Apr 26 '17 at 15:18

score 0 · Answer 4 · answered Apr 26 '17 at 15:17

Well, if you don't want to create a new list so much, here we go:

from itertools import islice

a = list("abcdabdbac")

i = 0
for x, y in zip(a, islice(a, 1, None)):
    if x == 'a':
        a[i] = x + y
        i += 1
    elif y != 'a':
        a[i] = y
        i += 1

try:
    del a[i:]
except:
    pass

score 0 · Answer 5 · answered Apr 26 '17 at 15:44

you could use itertools.groupby and group by:

letter follows a or
letter is not a

using enumerate to generate the current index, which allows to fetch the previous element from the list (creating a new list but one-liner)

import itertools

l = ['a', 'b', 'c', 'a', 'd']

print(["".join(x[1] for x in v) for _,v in itertools.groupby(enumerate(l),key=lambda t: (t[0] > 0 and l[t[0]-1]=='a') or t[1]=='a')])

result:

['ab', 'c', 'ad']

score -1 · Answer 6 · answered Apr 26 '17 at 15:05

This is easy way. Mb not pythonic way.

l1 = ['a', 'b', 'c', 'a', 'd']

do_combine = False
combine_element = None
for el in l1:
    if do_combine:
        indx = l1.index(el)
        l1[indx] = combine_element + el 
        do_combine = False
        l1.remove(combine_element)
    if el == 'a':
        combine_element = el
        do_combine = True

print(l1)
    # ['ab', 'c', 'ad']

How to combine two elements of a list based on a given condition

6 Answers6

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