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How do I concatenate two lists in Python?
Example:
listone = [1, 2, 3]
listtwo = [4, 5, 6]
Expected outcome:
>>> joinedlist
[1, 2, 3, 4, 5, 6]
[1,2,5] and [2,4,5,6]
? Do you want the duplicates included, excluded, or don't-care?
Use the +
operator to combine the lists:
listone = [1, 2, 3]
listtwo = [4, 5, 6]
joinedlist = listone + listtwo
Output:
>>> joinedlist
[1, 2, 3, 4, 5, 6]
Python >= 3.5
alternative: [*l1, *l2]
Another alternative has been introduced via the acceptance of PEP 448
which deserves mentioning.
The PEP, titled Additional Unpacking Generalizations, generally reduced some syntactic restrictions when using the starred *
expression in Python; with it, joining two lists (applies to any iterable) can now also be done with:
>>> l1 = [1, 2, 3]
>>> l2 = [4, 5, 6]
>>> joined_list = [*l1, *l2] # unpack both iterables in a list literal
>>> print(joined_list)
[1, 2, 3, 4, 5, 6]
This functionality was defined for Python 3.5
it hasn't been backported to previous versions in the 3.x
family. In unsupported versions a SyntaxError
is going to be raised.
As with the other approaches, this too creates as shallow copy of the elements in the corresponding lists.
The upside to this approach is that you really don't need lists in order to perform it, anything that is iterable will do. As stated in the PEP:
This is also useful as a more readable way of summing iterables into a list, such as my_list + list(my_tuple) + list(my_range) which is now equivalent to just [*my_list, *my_tuple, *my_range].
So while addition with +
would raise a TypeError
due to type mismatch:
l = [1, 2, 3]
r = range(4, 7)
res = l + r
The following won't:
res = [*l, *r]
because it will first unpack the contents of the iterables and then simply create a list
from the contents.
res = [*l1, *reversed(l2)]
. Since reversed
returns an iterator, res = l1 + reversed(l2)
would throw an error.
**
syntax only supports string keys.
It's also possible to create a generator that simply iterates over the items in both lists using itertools.chain()
. This allows you to chain lists (or any iterable) together for processing without copying the items to a new list:
import itertools
for item in itertools.chain(listone, listtwo):
# Do something with each list item
chain
is on the slower side (but not by much) for two lists, but is the fastest solution for chaining multiple lists (n >> 2).
chain
to make an iterator over all the elements but the converts the result to a list. Sometimes that's exactly what you want, but if you simply want to iterate over all the elements you can simply use the iterator from chain
. That's probably a lot faster.
You could also use the list.extend()
method in order to add a list
to the end of another one:
listone = [1,2,3]
listtwo = [4,5,6]
listone.extend(listtwo)
If you want to keep the original list intact, you can create a new list
object, and extend
both lists to it:
mergedlist = []
mergedlist.extend(listone)
mergedlist.extend(listtwo)
None
in my case?
listone = [1,2,3]; listtwo = [4,5,6]; listone.extend(listtwo)
this returns me None
listone
. So check that is in the list listone
return list1.extend(list2)
and the this expression returns None
to me.
You can use sets to obtain merged list of unique values
mergedlist = list(set(listone + listtwo))
listone + [x for x in listtwo if x not in listone]
import collections; mergedlist = list(collections.OrderedDict.fromkeys(listone + listtwo))
will do the trick.
How do I concatenate two lists in Python?
As of 3.9, these are the most popular stdlib methods for concatenating two (or more) lists in python.
https://i.stack.imgur.com/cRqfa.png
Footnotes This is a slick solution because of its succinctness. But sum performs concatenation in a pairwise fashion, which means this is a quadratic operation as memory has to be allocated for each step. DO NOT USE if your lists are large. See chain and chain.from_iterable from the docs. You will need to import itertools first. Concatenation is linear in memory, so this is the best in terms of performance and version compatibility. chain.from_iterable was introduced in 2.6. This method uses Additional Unpacking Generalizations (PEP 448), but cannot generalize to N lists unless you manually unpack each one yourself. a += b and a.extend(b) are more or less equivalent for all practical purposes. += when called on a list will internally call list.__iadd__, which extends the first list by the second.
Performance
2-List Concatenation1
https://i.stack.imgur.com/mfQTe.png
There's not much difference between these methods but that makes sense given they all have the same order of complexity (linear). There's no particular reason to prefer one over the other except as a matter of style.
N-List Concatenation
https://i.stack.imgur.com/XcX7A.png
Plots have been generated using the perfplot module. Code, for your reference.
1. The iadd
(+=
) and extend
methods operate in-place, so a copy has to be generated each time before testing. To keep things fair, all methods have a pre-copy step for the left-hand list which can be ignored.
Comments on Other Solutions
DO NOT USE THE DUNDER METHOD list.__add__ directly in any way, shape or form. In fact, stay clear of dunder methods, and use the operators and operator functions like they were designed for. Python has careful semantics baked into these which are more complicated than just calling the dunder directly. Here is an example. So, to summarise, a.__add__(b) => BAD; a + b => GOOD.
Some answers here offer reduce(operator.add, [a, b]) for pairwise concatenation -- this is the same as sum([a, b], []) only more wordy.
Any method that uses set will drop duplicates and lose ordering. Use with caution.
for i in b: a.append(i) is more wordy, and slower than a.extend(b), which is single function call and more idiomatic. append is slower because of the semantics with which memory is allocated and grown for lists. See here for a similar discussion.
heapq.merge will work, but its use case is for merging sorted lists in linear time. Using it in any other situation is an anti-pattern.
yielding list elements from a function is an acceptable method, but chain does this faster and better (it has a code path in C, so it is fast).
operator.add(a, b) is an acceptable functional equivalent to a + b. It's use cases are mainly for dynamic method dispatch. Otherwise, prefer a + b which is shorter and more readable, in my opinion. YMMV.
"There's not much difference between these methods but that makes sense given they all have the same order of complexity (linear). There's no particular reason to prefer one over the other except as a matter of style.
" Solutions not listed in my answer, or critized in "Comments" I recommend to not use.
This is quite simple, and I think it was even shown in the tutorial:
>>> listone = [1,2,3]
>>> listtwo = [4,5,6]
>>>
>>> listone + listtwo
[1, 2, 3, 4, 5, 6]
This question directly asks about joining two lists. However it's pretty high in search even when you are looking for a way of joining many lists (including the case when you joining zero lists).
I think the best option is to use list comprehensions:
>>> a = [[1,2,3], [4,5,6], [7,8,9]]
>>> [x for xs in a for x in xs]
[1, 2, 3, 4, 5, 6, 7, 8, 9]
You can create generators as well:
>>> map(str, (x for xs in a for x in xs))
['1', '2', '3', '4', '5', '6', '7', '8', '9']
Old Answer
Consider this more generic approach:
a = [[1,2,3], [4,5,6], [7,8,9]]
reduce(lambda c, x: c + x, a, [])
Will output:
[1, 2, 3, 4, 5, 6, 7, 8, 9]
Note, this also works correctly when a
is []
or [[1,2,3]]
.
However, this can be done more efficiently with itertools
:
a = [[1,2,3], [4,5,6], [7,8,9]]
list(itertools.chain(*a))
If you don't need a list
, but just an iterable, omit list()
.
Update
Alternative suggested by Patrick Collins in the comments could also work for you:
sum(a, [])
reduce
is now in functools
so you'll need to import it first.
You could simply use the +
or +=
operator as follows:
a = [1, 2, 3]
b = [4, 5, 6]
c = a + b
Or:
c = []
a = [1, 2, 3]
b = [4, 5, 6]
c += (a + b)
Also, if you want the values in the merged list to be unique you can do:
c = list(set(a + b))
list(dict.fromkeys(a + b))
It's worth noting that the itertools.chain
function accepts variable number of arguments:
>>> l1 = ['a']; l2 = ['b', 'c']; l3 = ['d', 'e', 'f']
>>> [i for i in itertools.chain(l1, l2)]
['a', 'b', 'c']
>>> [i for i in itertools.chain(l1, l2, l3)]
['a', 'b', 'c', 'd', 'e', 'f']
If an iterable (tuple, list, generator, etc.) is the input, the from_iterable
class method may be used:
>>> il = [['a'], ['b', 'c'], ['d', 'e', 'f']]
>>> [i for i in itertools.chain.from_iterable(il)]
['a', 'b', 'c', 'd', 'e', 'f']
For cases with a low number of lists you can simply add the lists together or use in-place unpacking (available in Python-3.5+):
In [1]: listone = [1, 2, 3]
...: listtwo = [4, 5, 6]
In [2]: listone + listtwo
Out[2]: [1, 2, 3, 4, 5, 6]
In [3]: [*listone, *listtwo]
Out[3]: [1, 2, 3, 4, 5, 6]
As a more general way for cases with more number of lists you can use chain.from_iterable()
1 function from itertools
module. Also, based on this answer this function is the best; or at least a very good way for flatting a nested list as well.
>>> l=[[1, 2, 3], [4, 5, 6], [7, 8, 9]]
>>> import itertools
>>> list(itertools.chain.from_iterable(l))
[1, 2, 3, 4, 5, 6, 7, 8, 9]
1. Note that `chain.from_iterable()` is available in Python 2.6 and later. In other versions, use `chain(*l)`.
With Python 3.3+ you can use yield from:
listone = [1,2,3]
listtwo = [4,5,6]
def merge(l1, l2):
yield from l1
yield from l2
>>> list(merge(listone, listtwo))
[1, 2, 3, 4, 5, 6]
Or, if you want to support an arbitrary number of iterators:
def merge(*iters):
for it in iters:
yield from it
>>> list(merge(listone, listtwo, 'abcd', [20, 21, 22]))
[1, 2, 3, 4, 5, 6, 'a', 'b', 'c', 'd', 20, 21, 22]
itertools.chain
(which is equivalent) instead of defining your own function.
If you want to merge the two lists in sorted form, you can use the merge
function from the heapq
library.
from heapq import merge
a = [1, 2, 4]
b = [2, 4, 6, 7]
print list(merge(a, b))
If you can't use the plus operator (+
), you can use the operator
import:
import operator
listone = [1,2,3]
listtwo = [4,5,6]
result = operator.add(listone, listtwo)
print(result)
>>> [1, 2, 3, 4, 5, 6]
Alternatively, you could also use the __add__
dunder function:
listone = [1,2,3]
listtwo = [4,5,6]
result = list.__add__(listone, listtwo)
print(result)
>>> [1, 2, 3, 4, 5, 6]
+
is off the table, use operator.add
.
If you need to merge two ordered lists with complicated sorting rules, you might have to roll it yourself like in the following code (using a simple sorting rule for readability :-) ).
list1 = [1,2,5]
list2 = [2,3,4]
newlist = []
while list1 and list2:
if list1[0] == list2[0]:
newlist.append(list1.pop(0))
list2.pop(0)
elif list1[0] < list2[0]:
newlist.append(list1.pop(0))
else:
newlist.append(list2.pop(0))
if list1:
newlist.extend(list1)
if list2:
newlist.extend(list2)
assert(newlist == [1, 2, 3, 4, 5])
heapq.merge
.
If you are using NumPy, you can concatenate two arrays of compatible dimensions with this command:
numpy.concatenate([a,b])
Use a simple list comprehension:
joined_list = [item for list_ in [list_one, list_two] for item in list_]
It has all the advantages of the newest approach of using Additional Unpacking Generalizations - i.e. you can concatenate an arbitrary number of different iterables (for example, lists, tuples, ranges, and generators) that way - and it's not limited to Python 3.5 or later.
Another way:
>>> listone = [1, 2, 3]
>>> listtwo = [4, 5, 6]
>>> joinedlist = [*listone, *listtwo]
>>> joinedlist
[1, 2, 3, 4, 5, 6]
>>>
list(set(listone) | set(listtwo))
The above code, does not preserve order, removes duplicate from each list (but not from the concatenated list)
As already pointed out by many, itertools.chain()
is the way to go if one needs to apply exactly the same treatment to both lists. In my case, I had a label and a flag which were different from one list to the other, so I needed something slightly more complex. As it turns out, behind the scenes itertools.chain()
simply does the following:
for it in iterables:
for element in it:
yield element
(see https://docs.python.org/2/library/itertools.html), so I took inspiration from here and wrote something along these lines:
for iterable, header, flag in ( (newList, 'New', ''), (modList, 'Modified', '-f')):
print header + ':'
for path in iterable:
[...]
command = 'cp -r' if os.path.isdir(srcPath) else 'cp'
print >> SCRIPT , command, flag, srcPath, mergedDirPath
[...]
The main points to understand here are that lists are just a special case of iterable, which are objects like any other; and that for ... in
loops in python can work with tuple variables, so it is simple to loop on multiple variables at the same time.
You could use the append()
method defined on list
objects:
mergedlist =[]
for elem in listone:
mergedlist.append(elem)
for elem in listtwo:
mergedlist.append(elem)
I recommend three methods to concatenate the list but 1st method is most recommended,
# easiest and least complexity method <= recommended
listone = [1, 2, 3]
listtwo = [4, 5, 6]
newlist = listone + listtwo
print(newlist)
# 2nd easiest method
newlist = listone.copy()
newlist.extend(listtwo)
print(newlist)
In the 2nd method, I assign newlist
to a copy of the listone
because I don't want to change listone
.
# 3rd method
newlist = listone.copy()
for j in listtwo:
newlist.append(j)
print(newlist)
This is not a good way to concatenate lists because we using for loop to concatenate the lists. So time complexity is much higher than with the other two methods.
A really concise way to combine a list of lists is
list_of_lists = [[1,2,3], [4,5,6], [7,8,9]]
reduce(list.__add__, list_of_lists)
which gives us
[1, 2, 3, 4, 5, 6, 7, 8, 9]
list.__add__
, use operator.add
instead. This is the more wordy equivalent of sum(list_of_lists, [])
which is just as bad. DO NOT USE!
obj.__class__
and obj.__dict__
.
__add__
seems too low-level and unstable (prone to change), you can use np.union1d
instead.
The most commom method used to concatenate lists are the plus operator and the built-in method append, for example:
list = [1,2]
list = list + [3]
# list = [1,2,3]
list.append(3)
# list = [1,2,3]
list.append([3,4])
# list = [1,2,[3,4]]
For most of the cases this will work but the append function will not extend a list if one was added. Because that is not expected you can use an another method called extend should work with structures:
list = [1,2]
list.extend([3,4])
# list = [1,2,3,4]
So there are two easy ways.
Using +: It creates a new list from provided lists
Example:
In [1]: a = [1, 2, 3]
In [2]: b = [4, 5, 6]
In [3]: a + b
Out[3]: [1, 2, 3, 4, 5, 6]
In [4]: %timeit a + b
10000000 loops, best of 3: 126 ns per loop
Using extend: It appends new list to existing list. That means it does not create a separate list.
Example:
In [1]: a = [1, 2, 3]
In [2]: b = [4, 5, 6]
In [3]: %timeit a.extend(b)
10000000 loops, best of 3: 91.1 ns per loop
Thus we see that out of two of most popular methods, extend
is efficient.
chain.from_iterable
).
You could also just use sum
.
>>> a = [1, 2, 3]
>>> b = [4, 5, 6]
>>> sum([a, b], [])
[1, 2, 3, 4, 5, 6]
>>>
This works for any length and any element type of list:
>>> a = ['a', 'b', 'c', 'd']
>>> b = [1, 2, 3, 4]
>>> c = [1, 2]
>>> sum([a, b, c], [])
['a', 'b', 'c', 'd', 1, 2, 3, 4, 1, 2]
>>>
The reason I add []
, is because the start
argument is set to 0
by default, so it loops through the list and adds to start
, but 0 + [1, 2, 3]
would give an error, so if we set the start
to []
, it would add to []
, [] + [1, 2, 3]
would work as expected.
a=[1,2,3]
b=[4,5,6]
c=a+b
print(c)
OUTPUT:
>>> [1, 2, 3, 4, 5, 6]
In the above code "+" operator is used to concatenate the 2 lists into a single list.
ANOTHER SOLUTION:
a=[1,2,3]
b=[4,5,6]
c=[] #Empty list in which we are going to append the values of list (a) and (b)
for i in a:
c.append(i)
for j in b:
c.append(j)
print(c)
OUTPUT:
>>> [1, 2, 3, 4, 5, 6]
I assume you are wanting one of the two methods:
Keep Duplicate elements
It is very easy, just concatenate like string:
def concat_list(l1,l2):
l3 = l1+l2
return l3
Next if you want to eliminate duplicate elements
def concat_list(l1,l2):
l3 = []
for i in [l1,l2]:
for j in i:
if j not in l3:
#Check if element exists in final list, if no then add element to list
l3.append(j)
return l3
The solutions provided are for single list. In case there are lists within a list and the merging of corresponding lists is required. "+" operation through for loop does the work.
a=[[1,2,3],[4,5,6]]
b=[[0,1,2],[7,8,9]]
for i in range(len(a)):
cc.append(a[i]+b[i])
output: [[1, 2, 3, 0, 1, 2], [4, 5, 6, 7, 8, 9]]
import itertools
A = list(zip([1,3,5,7,9],[2,4,6,8,10]))
B = [1,3,5,7,9]+[2,4,6,8,10]
C = list(set([1,3,5,7,9] + [2,4,6,8,10]))
D = [1,3,5,7,9]
D.append([2,4,6,8,10])
E = [1,3,5,7,9]
E.extend([2,4,6,8,10])
F = []
for a in itertools.chain([1,3,5,7,9], [2,4,6,8,10]):
F.append(a)
print ("A: " + str(A))
print ("B: " + str(B))
print ("C: " + str(C))
print ("D: " + str(D))
print ("E: " + str(E))
print ("F: " + str(F))
Output:
A: [(1, 2), (3, 4), (5, 6), (7, 8), (9, 10)]
B: [1, 3, 5, 7, 9, 2, 4, 6, 8, 10]
C: [1, 2, 3, 4, 5, 6, 7, 8, 9, 10]
D: [1, 3, 5, 7, 9, [2, 4, 6, 8, 10]]
E: [1, 3, 5, 7, 9, 2, 4, 6, 8, 10]
F: [1, 3, 5, 7, 9, 2, 4, 6, 8, 10]
Success story sharing
listone += listtwo
results inlistone == [1, 2, 3, 4, 5, 6]
list3 = listone
listone+=listtwo
Is list3 changed as well?