Group list of 4-strings into list of pairs

I have following list of strings:

['word1 word2 word3 word4', 'word5 word6 word7 word8']

(I have shown only two strings, but there can be many.)
I want to create new list which should look like this:

['word1 word2', 'word3 word4', 'word5 word6', 'word7 word8']

I tried following:

lines = ['word1 word2 word3 word4', 'word5 word6 word7 word8']
[[word1 + ' ' + word2, word3 + ' ' + word4] for line in lines for word1, word2, word3, word4 in line.split()]

But it gives following error:

ValueError: too many values to unpack (expected 4)

How do I do this in most pythonic way?

With short regex matching:

import re

lst = ['word1 word2 word3 word4', 'word5 word6 word7 word8']
res = [pair for words in lst for pair in re.findall(r'\S+ \S+', words)]

• \S+ \S+ - matches 2 consecutive "words"

['word1 word2', 'word3 word4', 'word5 word6', 'word7 word8']

Pythonic doesn't mean "fewer lines". This is easily done with a simple for loop:

result = []
for line in lines:
    words = line.split()
    result.append(' '.join(words[:2]))
    result.append(' '.join(words[2:]))

Which gives your desired result:

['word1 word2', 'word3 word4', 'word5 word6', 'word7 word8']

Try it online!

If you want to make this more general for strings with more words, you can write a function that will yield chunks of the desired size, and use that with str.join:

def chunks(iterable, chunk_size):
    c = []
    for item in iterable:
        c.append(item)
        if len(c) == chunk_size:
            yield c
            c = []

    if c: yield c

result = []
for line in lines:
    words = line.split()
    for chunk in chunks(words, 2):
        result.append(' '.join(chunk))

Try it online!

Modified @jsbueno's ealier answer which was slightly incorrect:

>>> words = [item for line in lines for item in line.split()]
>>> words
['word1', 'word2', 'word3', 'word4', 'word5', 'word6', 'word7', 'word8']
>>> [l[i] + ' ' + l[i+1] for i in range(0, len(words), 2)]
['word1 word2', 'word3 word4', 'word5 word6', 'word7 word8']

An optimized solution that pushes all the per-item work to the C layer:

from itertools import chain

lines = ['word1 word2 word3 word4', 'word5 word6 word7 word8']
words = chain.from_iterable(map(str.split, lines))
paired = list(map('{} {}'.format, words, words))
print(paired)

[Try it online!][TIO-ldt8uubn]

chain.from_iterable(map(str.split, lines)) creates an iterator of the individual words. map('{} {}'.format, words, words) maps the same iterator twice to put them back together in pairs (map(' '.join, zip(words, words)) would get the same effect, but with an additional intermediate product; feel free to test which is faster in practice). The list wrapper consumes it to produce the final result.

This beats the existing answers by avoiding all per-item work at the Python layer (no additional bytecode executed as the input grows), and avoids one of the weirdly high overhead aspects of Python (indexing and simple integer math).

[TIO-ldt8uubn]: https://tio.run/##LY3LCsIwEEX3@YrZpYVS0Pra@CUiEm1KB/JiMiBS@u2xGd2cyzD3kT48xzCUMlH0gGyJY3QZ0KdIDK/ZYFDKYbAZrnDT70jjDir3wkF40B3I6yjnSXgWXvRdVa1xaevr0qMumaezjTepyUx9Tg65A1lqW5UMkh23jMPMYtLLCsuq@ymSN5tTSv9SA4SBm1@sLeUL "Python 3 – Try It Online"