I could not find any valid example on the internet where I can see the difference between them and why to choose one over the other.
6 Answers
The first takes 0 or more arguments, each an iterable, the second one takes one argument which is expected to produce the iterables:
from itertools import chain
chain(list1, list2, list3)
iterables = [list1, list2, list3]
chain.from_iterable(iterables)
but iterables
can be any iterator that yields the iterables:
def gen_iterables():
for i in range(10):
yield range(i)
itertools.chain.from_iterable(gen_iterables())
Using the second form is usually a case of convenience, but because it loops over the input iterables lazily, it is also the only way you can chain an infinite number of finite iterators:
def gen_iterables():
while True:
for i in range(5, 10):
yield range(i)
chain.from_iterable(gen_iterables())
The above example will give you a iterable that yields a cyclic pattern of numbers that will never stop, but will never consume more memory than what a single range()
call requires.

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4i still can't get it. can you give me the output differnce and use case in practical situation where to use what – user1994660 Feb 21 '13 at 23:18
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16@user1994660: there is no output difference. It's an *input* difference. It makes it easier to use certain inputs. – Martijn Pieters Feb 21 '13 at 23:51
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@user1994660: I use the second form in [this answer](http://stackoverflow.com/questions/12900444/trying-to-add-to-dictionary-values-by-counting-occurrences-in-a-list-of-lists-p/12900577#12900577). – Martijn Pieters Feb 21 '13 at 23:52
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@user1994660: good usecase for the second form: [Python idiom to chain (flatten) an infinite iterable of finite iterables?](http://stackoverflow.com/a/120886) – Martijn Pieters Feb 22 '13 at 00:58
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1@user1994660: Run this code: `# Return an iterator of iterators` `def it_it(): return iter( [iter( [11, 22] ), iter( [33, 44] )] )` `print( list(itertools.chain.from_iterable(it_it())) )` `print( list(itertools.chain(it_it())) )` `print( list(itertools.chain(*it_it())) )` The first one is best; the second one doesn't get at the nested iterators, it returns iterators, instead of the desired numbers; the third one produces the correct output BUT it isn't fully lazy: the "*" forced all the iterators to be created. For this dumb input that doesn't matter. – ToolmakerSteve Dec 20 '13 at 19:21
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2Note that if the iterables is not too big, you can also do `itertools.chain(*iterables)` – balki Feb 14 '14 at 21:14
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@MartijnPieters would you mind if I quoted you on *"the only way you can chain a infinite number of finite iterators"*? – Ryan Haining Jun 09 '14 at 00:55
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I recently added chain.from_iterable to [a c++ project](https://github.com/ryanhaining/cppitertools) – Ryan Haining Jun 09 '14 at 01:08
I could not find any valid example ... where I can see the difference between them [
chain
andchain.from_iterable
] and why to choose one over the other
The accepted answer is thorough. For those seeking a quick application, consider flattening several lists:
list(itertools.chain(["a", "b", "c"], ["d", "e"], ["f"]))
# ['a', 'b', 'c', 'd', 'e', 'f']
You may wish to reuse these lists later, so you make an iterable of lists:
iterable = (["a", "b", "c"], ["d", "e"], ["f"])
Attempt
However, passing in an iterable to chain
gives an unflattened result:
list(itertools.chain(iterable))
# [['a', 'b', 'c'], ['d', 'e'], ['f']]
Why? You passed in one item (a tuple). chain
needs each list separately.
Solutions
When possible, you can unpack an iterable:
list(itertools.chain(*iterable))
# ['a', 'b', 'c', 'd', 'e', 'f']
list(itertools.chain(*iter(iterable)))
# ['a', 'b', 'c', 'd', 'e', 'f']
More generally, use .from_iterable
(as it also works with infinite iterators):
list(itertools.chain.from_iterable(iterable))
# ['a', 'b', 'c', 'd', 'e', 'f']
g = itertools.chain.from_iterable(itertools.cycle(iterable))
next(g)
# "a"

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They do very similar things. For small number of iterables itertools.chain(*iterables)
and itertools.chain.from_iterable(iterables)
perform similarly.
The key advantage of from_iterables
lies in the ability to handle large (potentially infinite) number of iterables since all of them need not be available at the time of the call.

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1@Rotareti, yes it does unpack lazily (one at a time) but in this case `itertools.chain(*iterables)` is a function call. All arguments must be present at the time of call. – BiGYaN Mar 27 '18 at 07:30
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Is this true? From the CPython code, it seems to be the same https://stackoverflow.com/a/62513808/610569 – alvas Jun 22 '20 at 12:49
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@alvas Try changing the number of elements to very large; in the range of 10_000 to 1_000_000 and you will see the `from_iterables` becoming faster. – BiGYaN Jun 29 '20 at 22:38
Extending @martijn-pieters answer
Although the access to the inner items in the iterable remains the same, and implementation wise,
itertools_chain_from_iterable
(i.e.chain.from_iterable
in Python) andchain_new
(i.e.chain
in Python)
in the CPython implementation, are both duck-types of chain_new_internal
Are there any optimization benefits from using chain.from_iterable(x)
, where x
is an iterable of iterable; and the main purpose is to ultimately consume the flatten list of items?
We can try benchmarking it with:
import random
from itertools import chain
from functools import wraps
from time import time
from tqdm import tqdm
def timing(f):
@wraps(f)
def wrap(*args, **kw):
ts = time()
result = f(*args, **kw)
te = time()
print('func:%r args:[%r, %r] took: %2.4f sec' % (f.__name__, args, kw, te-ts))
return result
return wrap
def generate_nm(m, n):
# Creates m generators of m integers between range 0 to n.
yield iter(random.sample(range(n), n) for _ in range(m))
def chain_star(x):
# Stores an iterable that will unpack and flatten the list of list.
chain_x = chain(*x)
# Consumes the items in the flatten iterable.
for i in chain_x:
pass
def chain_from_iterable(x):
# Stores an iterable that will unpack and flatten the list of list.
chain_x = chain.from_iterable(x)
# Consumes the items in the flatten iterable.
for i in chain_x:
pass
@timing
def versus(f, n, m):
f(generate_nm(n, m))
P/S: Benchmark running... Waiting for the results.
Results
chain_star, m=1000, n=1000
for _ in range(10):
versus(chain_star, 1000, 1000)
[out]:
func:'versus' args:[(<function chain_star at 0x7f5c7188ef28>, 1000, 1000), {}] took: 0.6494 sec
func:'versus' args:[(<function chain_star at 0x7f5c7188ef28>, 1000, 1000), {}] took: 0.6603 sec
func:'versus' args:[(<function chain_star at 0x7f5c7188ef28>, 1000, 1000), {}] took: 0.6367 sec
func:'versus' args:[(<function chain_star at 0x7f5c7188ef28>, 1000, 1000), {}] took: 0.6350 sec
func:'versus' args:[(<function chain_star at 0x7f5c7188ef28>, 1000, 1000), {}] took: 0.6296 sec
func:'versus' args:[(<function chain_star at 0x7f5c7188ef28>, 1000, 1000), {}] took: 0.6399 sec
func:'versus' args:[(<function chain_star at 0x7f5c7188ef28>, 1000, 1000), {}] took: 0.6341 sec
func:'versus' args:[(<function chain_star at 0x7f5c7188ef28>, 1000, 1000), {}] took: 0.6381 sec
func:'versus' args:[(<function chain_star at 0x7f5c7188ef28>, 1000, 1000), {}] took: 0.6343 sec
func:'versus' args:[(<function chain_star at 0x7f5c7188ef28>, 1000, 1000), {}] took: 0.6309 sec
chain_from_iterable, m=1000, n=1000
for _ in range(10):
versus(chain_from_iterable, 1000, 1000)
[out]:
func:'versus' args:[(<function chain_from_iterable at 0x7f5c7188eb70>, 1000, 1000), {}] took: 0.6416 sec
func:'versus' args:[(<function chain_from_iterable at 0x7f5c7188eb70>, 1000, 1000), {}] took: 0.6315 sec
func:'versus' args:[(<function chain_from_iterable at 0x7f5c7188eb70>, 1000, 1000), {}] took: 0.6535 sec
func:'versus' args:[(<function chain_from_iterable at 0x7f5c7188eb70>, 1000, 1000), {}] took: 0.6334 sec
func:'versus' args:[(<function chain_from_iterable at 0x7f5c7188eb70>, 1000, 1000), {}] took: 0.6327 sec
func:'versus' args:[(<function chain_from_iterable at 0x7f5c7188eb70>, 1000, 1000), {}] took: 0.6471 sec
func:'versus' args:[(<function chain_from_iterable at 0x7f5c7188eb70>, 1000, 1000), {}] took: 0.6426 sec
func:'versus' args:[(<function chain_from_iterable at 0x7f5c7188eb70>, 1000, 1000), {}] took: 0.6287 sec
func:'versus' args:[(<function chain_from_iterable at 0x7f5c7188eb70>, 1000, 1000), {}] took: 0.6353 sec
func:'versus' args:[(<function chain_from_iterable at 0x7f5c7188eb70>, 1000, 1000), {}] took: 0.6297 sec
chain_star, m=10000, n=1000
func:'versus' args:[(<function chain_star at 0x7f5c7188ef28>, 10000, 1000), {}] took: 6.2659 sec
func:'versus' args:[(<function chain_star at 0x7f5c7188ef28>, 10000, 1000), {}] took: 6.2966 sec
func:'versus' args:[(<function chain_star at 0x7f5c7188ef28>, 10000, 1000), {}] took: 6.2953 sec
func:'versus' args:[(<function chain_star at 0x7f5c7188ef28>, 10000, 1000), {}] took: 6.3141 sec
func:'versus' args:[(<function chain_star at 0x7f5c7188ef28>, 10000, 1000), {}] took: 6.2802 sec
func:'versus' args:[(<function chain_star at 0x7f5c7188ef28>, 10000, 1000), {}] took: 6.2799 sec
func:'versus' args:[(<function chain_star at 0x7f5c7188ef28>, 10000, 1000), {}] took: 6.2848 sec
func:'versus' args:[(<function chain_star at 0x7f5c7188ef28>, 10000, 1000), {}] took: 6.3299 sec
func:'versus' args:[(<function chain_star at 0x7f5c7188ef28>, 10000, 1000), {}] took: 6.2730 sec
func:'versus' args:[(<function chain_star at 0x7f5c7188ef28>, 10000, 1000), {}] took: 6.3052 sec
chain_from_iterable, m=10000, n=1000
func:'versus' args:[(<function chain_from_iterable at 0x7f5c7188eb70>, 10000, 1000), {}] took: 6.3129 sec
func:'versus' args:[(<function chain_from_iterable at 0x7f5c7188eb70>, 10000, 1000), {}] took: 6.3064 sec
func:'versus' args:[(<function chain_from_iterable at 0x7f5c7188eb70>, 10000, 1000), {}] took: 6.3071 sec
func:'versus' args:[(<function chain_from_iterable at 0x7f5c7188eb70>, 10000, 1000), {}] took: 6.2660 sec
func:'versus' args:[(<function chain_from_iterable at 0x7f5c7188eb70>, 10000, 1000), {}] took: 6.2837 sec
func:'versus' args:[(<function chain_from_iterable at 0x7f5c7188eb70>, 10000, 1000), {}] took: 6.2877 sec
func:'versus' args:[(<function chain_from_iterable at 0x7f5c7188eb70>, 10000, 1000), {}] took: 6.2756 sec
func:'versus' args:[(<function chain_from_iterable at 0x7f5c7188eb70>, 10000, 1000), {}] took: 6.2939 sec
func:'versus' args:[(<function chain_from_iterable at 0x7f5c7188eb70>, 10000, 1000), {}] took: 6.2715 sec
func:'versus' args:[(<function chain_from_iterable at 0x7f5c7188eb70>, 10000, 1000), {}] took: 6.2877 sec
chain_star, m=100000, n=1000
func:'versus' args:[(<function chain_star at 0x7f5c7188ef28>, 100000, 1000), {}] took: 62.7874 sec
func:'versus' args:[(<function chain_star at 0x7f5c7188ef28>, 100000, 1000), {}] took: 63.3744 sec
func:'versus' args:[(<function chain_star at 0x7f5c7188ef28>, 100000, 1000), {}] took: 62.5584 sec
func:'versus' args:[(<function chain_star at 0x7f5c7188ef28>, 100000, 1000), {}] took: 63.3745 sec
func:'versus' args:[(<function chain_star at 0x7f5c7188ef28>, 100000, 1000), {}] took: 62.7982 sec
func:'versus' args:[(<function chain_star at 0x7f5c7188ef28>, 100000, 1000), {}] took: 63.4054 sec
func:'versus' args:[(<function chain_star at 0x7f5c7188ef28>, 100000, 1000), {}] took: 62.6769 sec
func:'versus' args:[(<function chain_star at 0x7f5c7188ef28>, 100000, 1000), {}] took: 62.6476 sec
func:'versus' args:[(<function chain_star at 0x7f5c7188ef28>, 100000, 1000), {}] took: 63.7397 sec
func:'versus' args:[(<function chain_star at 0x7f5c7188ef28>, 100000, 1000), {}] took: 62.8980 sec
chain_from_iterable, m=100000, n=1000
for _ in range(10):
versus(chain_from_iterable, 100000, 1000)
[out]:
func:'versus' args:[(<function chain_from_iterable at 0x7f5c7188eb70>, 100000, 1000), {}] took: 62.7227 sec
func:'versus' args:[(<function chain_from_iterable at 0x7f5c7188eb70>, 100000, 1000), {}] took: 62.7717 sec
func:'versus' args:[(<function chain_from_iterable at 0x7f5c7188eb70>, 100000, 1000), {}] took: 62.7159 sec
func:'versus' args:[(<function chain_from_iterable at 0x7f5c7188eb70>, 100000, 1000), {}] took: 62.7569 sec
func:'versus' args:[(<function chain_from_iterable at 0x7f5c7188eb70>, 100000, 1000), {}] took: 62.7906 sec
func:'versus' args:[(<function chain_from_iterable at 0x7f5c7188eb70>, 100000, 1000), {}] took: 62.6211 sec
func:'versus' args:[(<function chain_from_iterable at 0x7f5c7188eb70>, 100000, 1000), {}] took: 62.7294 sec
func:'versus' args:[(<function chain_from_iterable at 0x7f5c7188eb70>, 100000, 1000), {}] took: 62.8260 sec
func:'versus' args:[(<function chain_from_iterable at 0x7f5c7188eb70>, 100000, 1000), {}] took: 62.8356 sec
func:'versus' args:[(<function chain_from_iterable at 0x7f5c7188eb70>, 100000, 1000), {}] took: 62.9738 sec
chain_star, m=500000, n=1000
for _ in range(3):
versus(chain_from_iterable, 500000, 1000)
[out]:
func:'versus' args:[(<function chain_star at 0x7f5c7188ef28>, 500000, 1000), {}] took: 314.5671 sec
func:'versus' args:[(<function chain_star at 0x7f5c7188ef28>, 500000, 1000), {}] took: 313.9270 sec
func:'versus' args:[(<function chain_star at 0x7f5c7188ef28>, 500000, 1000), {}] took: 313.8992 sec
chain_from_iterable, m=500000, n=1000
for _ in range(3):
versus(chain_from_iterable, 500000, 1000)
[out]:
func:'versus' args:[(<function chain_from_iterable at 0x7f5c7188eb70>, 500000, 1000), {}] took: 313.8301 sec
func:'versus' args:[(<function chain_from_iterable at 0x7f5c7188eb70>, 500000, 1000), {}] took: 313.8104 sec
func:'versus' args:[(<function chain_from_iterable at 0x7f5c7188eb70>, 500000, 1000), {}] took: 313.9440 sec

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Another way to see it:
chain(iterable1, iterable2, iterable3, ...)
is for when you already know what iterables you have, so you can write them as these comma-separated arguments.
chain.from_iterable(iterable)
is for when your iterables (like iterable1, iterable2, iterable3) are obtained from another iterable.

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Another way to look at it is to use chain.from_iterable
when you have an iterable of iterables like a nested iterable(or a compound iterbale) and use chain
for simple iterables.

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