how to add all array's elements to one list in python

Question

with a 2 dimension array which looks like this one:

myarray = [['jacob','mary'],['jack','white'],['fantasy','clothes'],['heat','abc'],['edf','fgc']]

every elements is an array which has fixed length elements. how to become this one,

mylist = ['jacob','mary','jack','white','fantasy','clothes','heat','abc','edf','fgc']

here’s my solve

mylist = []
for x in myarray:
   mylist.extend(x)

should be more simple i guess

Asked By: user2003548

||

Source

Answer 1

>>> myarray = [['jacob','mary'],['jack','white'],['fantasy','clothes'],['heat','abc'],['edf','fgc']]
>>> sum(myarray,[])
['jacob', 'mary', 'jack', 'white', 'fantasy', 'clothes', 'heat', 'abc', 'edf', 'fgc']

Or

>>> [i for j in myarray for i in j]
['jacob', 'mary', 'jack', 'white', 'fantasy', 'clothes', 'heat', 'abc', 'edf', 'fgc']

Answered By: TerryA

Answer 2

Use itertools.chain.from_iterable:

from itertools import chain
mylist = list(chain.from_iterable(myarray))

Demo:

>>> from itertools import chain
>>> myarray = [['jacob','mary'],['jack','white'],['fantasy','clothes'],['heat','abc'],['edf','fgc']]
>>> list(chain.from_iterable(myarray))
['jacob', 'mary', 'jack', 'white', 'fantasy', 'clothes', 'heat', 'abc', 'edf', 'fgc']

However, Haidro’s sum() solution is faster for your shorter sample:

>>> timeit.timeit('f()', 'from __main__ import withchain as f')
2.858742465992691
>>> timeit.timeit('f()', 'from __main__ import withsum as f')
1.6423718839942012
>>> timeit.timeit('f()', 'from __main__ import withlistcomp as f')
2.0854451240156777

but itertools.chain wins if the input gets larger:

>>> myarray *= 100
>>> timeit.timeit('f()', 'from __main__ import withchain as f', number=25000)
1.6583486960153095
>>> timeit.timeit('f()', 'from __main__ import withsum as f', number=25000)
23.100156371016055
>>> timeit.timeit('f()', 'from __main__ import withlistcomp as f', number=25000)
2.093297885992797

Answered By: Martijn Pieters

how to add all array's elements to one list in python

Question:

Answers: