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Use itertools.chain() with the * operator:

import itertools
 
list2d = [[1,2,3], [4,5,6], [7], [8,9]]
merged = list(itertools.chain(*list2d))

Or use itertools.chain.from_iterable():

merged = list(itertools.chain.from_iterable(list2d))

You can use itertools.chain():

>>> import itertools
>>> list2d = [[1,2,3], [4,5,6], [7], [8,9]]
>>> merged = list(itertools.chain(*list2d))

Or you can use itertools.chain.from_iterable() which doesn't require unpacking the list with the * operator:

>>> import itertools
>>> list2d = [[1,2,3], [4,5,6], [7], [8,9]]
>>> merged = list(itertools.chain.from_iterable(list2d))

This approach is arguably more readable than [item for sublist in l for item in sublist] and appears to be faster too:

$ python3 -mtimeit -s'l=[[1,2,3],[4,5,6], [7], [8,9]]*99;import itertools' 'list(itertools.chain.from_iterable(l))'
20000 loops, best of 5: 10.8 usec per loop
$ python3 -mtimeit -s'l=[[1,2,3],[4,5,6], [7], [8,9]]*99' '[item for sublist in l for item in sublist]'
10000 loops, best of 5: 21.7 usec per loop
$ python3 -mtimeit -s'l=[[1,2,3],[4,5,6], [7], [8,9]]*99' 'sum(l, [])'
1000 loops, best of 5: 258 usec per loop
$ python3 -mtimeit -s'l=[[1,2,3],[4,5,6], [7], [8,9]]*99;from functools import reduce' 'reduce(lambda x,y: x+y,l)'
1000 loops, best of 5: 292 usec per loop
$ python3 --version
Python 3.7.5rc1

Use itertools.chain():

import itertools

list2d = [[1,2,3], [4,5,6], [7], [8,9]]
merged = list(itertools.chain(*list2d))

Or you can use itertools.chain.from_iterable() which doesn't require unpacking the list with the * operator:

merged = list(itertools.chain.from_iterable(list2d))

Use itertools.chain() with the * operator:

import itertools

list2d = [[1,2,3], [4,5,6], [7], [8,9]]
merged = list(itertools.chain(*list2d))

Or use itertools.chain.from_iterable():

merged = list(itertools.chain.from_iterable(list2d))

You can use itertools.chain():

import itertools

list2d = [[1,2,3], [4,5,6], [7], [8,9]]
merged = list(itertools.chain(*list2d))

Or you can use itertools.chain.from_iterable() which doesn't require unpacking the list with the * operator:

merged = list(itertools.chain.from_iterable(list2d))

Use itertools.chain():

import itertools

list2d = [[1,2,3], [4,5,6], [7], [8,9]]
merged = list(itertools.chain(*list2d))

Or you can use itertools.chain.from_iterable() which doesn't require unpacking the list with the * operator:

merged = list(itertools.chain.from_iterable(list2d))