How do I flatten a list of lists/nested lists? [duplicate]

Question

I have python code like this:

newlist =[[52, None, None], [129, None, None], [56, None, None], [111, None, None],  
          [22, None, None], [33, None, None], [28, None, None], [52, None, None],  
          [52, None, None], [52, None, None], [129, None, None], [56, None, None],  
          [111, None, None], [22, None, None], [33, None, None], [28, None, None]]

I want the newlist like:

newlist =[52, None, None,129, None, None,56, None, None,111, None, None,22, 
          None, None,33, None, None,28, None, None,52, None, None,52, None,  
          None,52, None, None,129, None, None,56, None, None, 111, None,  
          None,22, None, None,33, None, None,28, None, None]

Is there any way to work around ?

Botond · Accepted Answer · 2023-01-20 18:51:36Z

131

What you are trying to do is called flattening the list. And according to the Zen of Python, you are trying to do the right thing. Quoting from that

Flat is better than nested.

So you can use list comprehension like this

 newlist = [item for items in newlist for item in items]

Or you can use chain from itertools like this

 from itertools import chain
 newlist = list(chain(*newlist))

Or you can use chain.from_iterable, where unpacking of the list is not necessary
```
 from itertools import chain
 newlist = list(chain.from_iterable(newlist))
```
Using sum function
```
 newlist = sum(newlist, [])
```

Using reduce function

 newlist = reduce(lambda x,y: x+y, newlist)

Using operator.add. This will be faster than the reduce with lambda version.

 import operator
 from functools import reduce
 newlist = reduce(operator.add, newlist)

Edit: For the sake of completeness, included the answers found in How do I make a flat list out of a list of lists? as well.

I tried to time all of them in Python 2.7, like this

from timeit import timeit
print(timeit("[item for items in newlist for item in items]", "from __main__ import newlist"))
print(timeit("sum(newlist, [])", "from __main__ import newlist"))
print(timeit("reduce(lambda x,y: x+y, newlist)", "from __main__ import newlist"))
print(timeit("reduce(add, newlist)", "from __main__ import newlist; from operator import add"))
print(timeit("list(chain(*newlist))", "from __main__ import newlist; from itertools import chain"))
print(timeit("list(chain.from_iterable(newlist))", "from __main__ import newlist; from itertools import chain"))

Output on my machine

2.26074504852
2.45047688484
3.50180387497
2.56596302986
1.78825688362
1.61612296104

So, the most efficient way to do this is to use list(chain.from_iterable(newlist)), in Python 2.7. Ran the same test in Python 3.3

from timeit import timeit
print(timeit("[item for items in newlist for item in items]", "from __main__ import newlist"))
print(timeit("sum(newlist, [])", "from __main__ import newlist"))
print(timeit("reduce(lambda x,y: x+y, newlist)", "from __main__ import newlist; from functools import reduce"))
print(timeit("reduce(add, newlist)", "from __main__ import newlist; from operator import add; from functools import reduce"))
print(timeit("list(chain(*newlist))", "from __main__ import newlist; from itertools import chain"))
print(timeit("list(chain.from_iterable(newlist))", "from __main__ import newlist; from itertools import chain"))

Output on my machine

2.26074504852
2.45047688484
3.50180387497
2.56596302986
1.78825688362
1.61612296104

So, be it Python 2.7 or 3.3, use list(chain.from_iterable(newlist)) to flatten the nested lists.

edited Jan 20, 2023 at 18:51

Botond

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answered Nov 21, 2013 at 4:55

thefourtheye

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10 Comments

tmj Over a year ago

Got a question. Why does this work [i for j in l for i in j] and this [i for i in j for j in l] doesn't?

thefourtheye Over a year ago

@tMJ In the second form, the sublists are being comprehended again, not the individual elements in the sublists. Just break the comprehension and print the values of i in both the cases.

DSM Over a year ago

@tMJ: think about the equivalent for loops. The first is for j in l: for i in j: newlist.append(i), and the second is for i in j: for j in l: newlist.append(i).

DSM Over a year ago

@tMJ: In Python 2, list comprehensions leak the loop variables. Try the same thing again, but del j between the two, and you'll see NameError: name 'j' is not defined.

aIKid Over a year ago

An extremely detailed answer, the best i've seen in a while. +1

|

lennon310 · Accepted Answer · 2013-11-21 05:18:48Z

21

Just the easiest one:

newlist = sum(newlist, [])
print newlist

answered Nov 21, 2013 at 5:18

lennon310

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Comments

Christian Tapia · Accepted Answer · 2013-11-21 04:55:56Z

6

Try:

newlist = [j for i in newlist for j in i]

answered Nov 21, 2013 at 4:55

Christian Tapia

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Comments

tmj · Accepted Answer · 2013-11-21 04:54:05Z

2

temp = []
for small_list in newlist:
    temp += small_list
newlist = temp

This should do it.

answered Nov 21, 2013 at 4:54

tmj

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4 Comments

dstromberg Over a year ago

Is this quadratic? It may be better to use temp.extend(small_list).

tmj Over a year ago

what do you mean by quadratic?

tmj Over a year ago

stackoverflow.com/questions/3653298/… It's slightly better to use += than .extend()

dstromberg Over a year ago

Quadratic means O(n^2) time.

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