Finding common elements from lists in list - python

I have a list that includes more lists (this is how it looks https://pastebin.com/BW4B9gfa). The number of lists is not constant. I need to create another list that contains only elements that are in all lists in the main list.
I made something like this as a prototype but it doesn't work:
def common_elements(list_of_lists):
lists = list_of_lists
common = lists[0].intersection(lists[1].intersection(lists[2].intersection(lists[3].intersection(lists[4].intersection(lists[5])))))
return common
I see too something like this:
A = [1,2,3,4]
B = [2,4,7,8]
commonalities = set(A) - (set(A) - set(B))
but I don't know how to use it with bigger number of lists.

You can simply do if you have a list of sets, to get a list of sets just do (lists = [set(list) for list in lists]).
lists[0].intersection(*lists)

You need to convert the first list to a set so you can use the intersection() method.
Use a loop rather than hard-coding all the indexes of the list elements.
def common_elements(lists):
if len(lists) == 0:
return []
common = set(lists[0])
for l in lists[1:]:
common = common.intersection(l)
return list(common)

using functools.reduce():
from functools import reduce
items = [[1, 2, 4], [1, 3, 4], [1, 4, 6], [1, 4, 7, 9]]
eggs = reduce(lambda x, y: set(x) & set(y), items)
print(eggs)
output:
{1, 4}
If you want to get intermediate results, you can use itertools.accumulate()
from itertools import accumulate
items = [[1, 2, 4, 5], [1, 3, 4, 5], [1, 4, 6], [1, 4, 7, 9]]
eggs = list(accumulate(items, func = lambda x, y: set(x) & set(y)))
print(eggs)
output:
[[1, 2, 4, 5], {1, 4, 5}, {1, 4}, {1, 4}]

Related

Combining two lists with different lengths to a single list [duplicate]

This question's answers are a community effort. Edit existing answers to improve this post. It is not currently accepting new answers or interactions.
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]
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, but 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.
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
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)
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.
Version Restrictions
In-Place?
Generalize to N lists?
a+b
-
No
sum([a, b, c], [])1
list(chain(a,b))2
>=2.3
No
list(chain(a, b, c))
[*a, *b]3
>=3.5
No
[*a, *b, *c]
a += b
-
Yes
No
a.extend(b)
-
Yes
No
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 from itertools import chain 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
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
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.
You can use sets to obtain merged list of unique values
mergedlist = list(set(listone + listtwo))
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, [])
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))
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]
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]
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])
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 and removes duplicates 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)
a = [1, 2, 3]
b = [4, 5, 6]
c = a + b
print(c)
Output
>>> [1, 2, 3, 4, 5, 6]
In the above code, the "+" operator is used to concatenate the two 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]
All the possible ways to join lists that I could find
import itertools
A = [1,3,5,7,9] + [2,4,6,8,10]
B = [1,3,5,7,9]
B.append([2,4,6,8,10])
C = [1,3,5,7,9]
C.extend([2,4,6,8,10])
D = list(zip([1,3,5,7,9],[2,4,6,8,10]))
E = [1,3,5,7,9]+[2,4,6,8,10]
F = list(set([1,3,5,7,9] + [2,4,6,8,10]))
G = []
for a in itertools.chain([1,3,5,7,9], [2,4,6,8,10]):
G.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))
print("G: " + str(G))
Output
A: [1, 3, 5, 7, 9, 2, 4, 6, 8, 10]
B: [1, 3, 5, 7, 9, [2, 4, 6, 8, 10]]
C: [1, 3, 5, 7, 9, 2, 4, 6, 8, 10]
D: [(1, 2), (3, 4), (5, 6), (7, 8), (9, 10)]
E: [1, 3, 5, 7, 9, 2, 4, 6, 8, 10]
F: [1, 2, 3, 4, 5, 6, 7, 8, 9, 10]
G: [1, 3, 5, 7, 9, 2, 4, 6, 8, 10]
I recommend three methods to concatenate the list, but the first method is most recommended,
# Easiest and least complexity method <= recommended
listone = [1, 2, 3]
listtwo = [4, 5, 6]
newlist = listone + listtwo
print(newlist)
# Second-easiest method
newlist = listone.copy()
newlist.extend(listtwo)
print(newlist)
In the second method, I assign newlist to a copy of the listone, because I don't want to change listone.
# Third method
newlist = listone.copy()
for j in listtwo:
newlist.append(j)
print(newlist)
This is not a good way to concatenate lists because we are using a for loop to concatenate the lists. So time complexity is much higher than with the other two methods.
The most common 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 another method called extend. It should work with structures:
list = [1,2]
list.extend([3,4])
# list = [1,2,3,4]
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]
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.
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 [], and [] + [1, 2, 3] would work as expected.
I assume you want one of the two methods:
Keep duplicate elements
It is very easy. Just concatenate like a 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 a single list. In case there are lists within a list and the merging of corresponding lists is required, the "+" operation through a 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]]

Is there a way to split a tuple containing three lists into its individual elements [duplicate]

This question's answers are a community effort. Edit existing answers to improve this post. It is not currently accepting new answers or interactions.
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]
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, but 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.
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
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)
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.
Version Restrictions
In-Place?
Generalize to N lists?
a+b
-
No
sum([a, b, c], [])1
list(chain(a,b))2
>=2.3
No
list(chain(a, b, c))
[*a, *b]3
>=3.5
No
[*a, *b, *c]
a += b
-
Yes
No
a.extend(b)
-
Yes
No
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 from itertools import chain 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
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
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.
You can use sets to obtain merged list of unique values
mergedlist = list(set(listone + listtwo))
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, [])
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))
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]
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]
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])
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 and removes duplicates 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)
a = [1, 2, 3]
b = [4, 5, 6]
c = a + b
print(c)
Output
>>> [1, 2, 3, 4, 5, 6]
In the above code, the "+" operator is used to concatenate the two 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]
All the possible ways to join lists that I could find
import itertools
A = [1,3,5,7,9] + [2,4,6,8,10]
B = [1,3,5,7,9]
B.append([2,4,6,8,10])
C = [1,3,5,7,9]
C.extend([2,4,6,8,10])
D = list(zip([1,3,5,7,9],[2,4,6,8,10]))
E = [1,3,5,7,9]+[2,4,6,8,10]
F = list(set([1,3,5,7,9] + [2,4,6,8,10]))
G = []
for a in itertools.chain([1,3,5,7,9], [2,4,6,8,10]):
G.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))
print("G: " + str(G))
Output
A: [1, 3, 5, 7, 9, 2, 4, 6, 8, 10]
B: [1, 3, 5, 7, 9, [2, 4, 6, 8, 10]]
C: [1, 3, 5, 7, 9, 2, 4, 6, 8, 10]
D: [(1, 2), (3, 4), (5, 6), (7, 8), (9, 10)]
E: [1, 3, 5, 7, 9, 2, 4, 6, 8, 10]
F: [1, 2, 3, 4, 5, 6, 7, 8, 9, 10]
G: [1, 3, 5, 7, 9, 2, 4, 6, 8, 10]
I recommend three methods to concatenate the list, but the first method is most recommended,
# Easiest and least complexity method <= recommended
listone = [1, 2, 3]
listtwo = [4, 5, 6]
newlist = listone + listtwo
print(newlist)
# Second-easiest method
newlist = listone.copy()
newlist.extend(listtwo)
print(newlist)
In the second method, I assign newlist to a copy of the listone, because I don't want to change listone.
# Third method
newlist = listone.copy()
for j in listtwo:
newlist.append(j)
print(newlist)
This is not a good way to concatenate lists because we are using a for loop to concatenate the lists. So time complexity is much higher than with the other two methods.
The most common 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 another method called extend. It should work with structures:
list = [1,2]
list.extend([3,4])
# list = [1,2,3,4]
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]
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.
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 [], and [] + [1, 2, 3] would work as expected.
I assume you want one of the two methods:
Keep duplicate elements
It is very easy. Just concatenate like a 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 a single list. In case there are lists within a list and the merging of corresponding lists is required, the "+" operation through a 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]]

List comprehension headaches

I have a nested list like this which:
list = [[1,2,3], [2,5,7,6], [1,-1], [5,7], [6,3,7,4,3], [2, 5, 1, -5]]
What I am trying to do is to remove nested lists, where the value within these lists are both positive and negative. I have tried doing it by list comprehension, but I couldn't figure it out.
def method(list):
return [obj for obj in list if (x for x in obj if -x not in obj)]
The obtained results should be like:
list = [[1,2,3], [2,5,7,6], [5,7], [6,3,7,4,3]]
Assuming you want lists where elements are either all negative or all positive you can use all predefined function to check for both possibilities
result = [L for L in x if all(y>0 for y in L) or all(y<0 for y in L)]
EDIT:
In the comments you clarified what is a valid list (e.g. [-1, 2] is valid)... with this new formulation the test should be
result = [L for L in x if all(-y not in L for y in L)]
where each single test is however now quadratic in the size of the list. Using set this problem can be removed
result = [L for L in x if all(-y not in S for S in (set(L),) for y in L)]
Using list comprehension you can do something like:
def method2(list):
return [obj for obj in list if (all(n>0 for n in obj) or all(n<0 for n in obj))]
that, with your example, give as output:
[[1, 2, 3], [2, 5, 7, 6], [5, 7], [6, 3, 7, 4, 3]]
In general is better to split the task by steps:
Given list find the positives (positives function)
Given list find the negatives and multiply them by -1 (negatives function)
If the intersection of both positives and negatives is not empty remove.
So, you could do:
def positives(ls):
return set(l for l in ls if l > 0)
def negatives(ls):
return set(-1*l for l in ls if l < 0)
list = [[1, 2, 3], [2, 5, 7, 6], [1, -1], [5, 7], [6, 3, 7, 4, 3], [2, 5, 1, -5]]
result = [l for l in list if not negatives(l) & positives(l)]
print(result)
Output
[[1, 2, 3], [2, 5, 7, 6], [5, 7], [6, 3, 7, 4, 3]]
As a side note you should not use list as a variable name as it shadows the built-int list function.
Your generator should yield whether the condition to filter an object applies.
You then feed the generator to an aggregator to determine if obj should be filtered.
the aggregator could be any or all, or something different.
# assuming obj should be filtered if both x and the inverse of x are in obj
def method_with_all(src):
return [obj for obj in src if all(-x not in obj for x in obj)]
def method_with_any(src):
return [obj for obj in src if any(-x in obj for x in obj)]
you can filter out the lists that have both negative and positive elements:
def keep_list(nested_list):
is_first_positive = nested_list[0] > 0
for element in nested_list[1:]:
if (element > 0) != is_first_positive:
return False
return True
my_list = [[1,2,3], [2,5,7,6], [1,-1], [5,7], [6,3,7,4,3], [2, 5, 1, -5]]
print(list(filter(keep_list, my_list)))
output:
[[1, 2, 3], [2, 5, 7, 6], [5, 7], [6, 3, 7, 4, 3]]
Numpy can be used as well. My solution here is similar to the "all"-operation suggested by others but coded explicitly and only needs one condition. It checks whether the sign of the all the elements equals the sign of the first element (could be any other as well).
from numpy import *
def f(b):
return [a for a in b if sum(sign(array(a)) == sign(a[0])) == len(a)]
For your case...
data = [[1,2,3], [2,5,7,6], [1,-1], [5,7], [6,3,7,4,3], [2, 5, 1, -5]]
print(f(data))
...it will return:
[[1, 2, 3], [2, 5, 7, 6], [5, 7], [6, 3, 7, 4, 3]]

function of difference between value

Is there a function in Python to get the difference between two or more values in a list? So, in those two lists:
list1 = [1, 5, 3, 7]
list2 = [4, 2, 6, 4]
I need to calculate the difference between every value in list1 and list2.
for i in list1:
for ii in list2:
print i -ii
This gives negative values, but I want the subtraction between the values of the two lists only from highest value to lowest value for not getting negative values.
For the above lists, I expect the output to be [3, 3, 3, 3].
Thanks.
Assuming you expect [3, 3, 3, 3] as the answer in your question, you can use abs and zip:
[abs(i-j) for i,j in zip(list1, list2)]
Either zip the lists, or use numpy:
>>> list1 = [1, 5, 3, 7]
>>> list2 = [4, 2, 6, 4]
>>> [a-b for a,b in zip(list1, list2)]
[-3, 3, -3, 3]
>>> import numpy as np
>>> np.array(list1) - np.array(list2)
array([-3, 3, -3, 3])
Remember to cast the array back to a list as needed.
edit:
In response to the new requirement that the absolute values are needed: you can add abs in the list comprehension:
>>> [abs(a-b) for a,b in zip(list1, list2)]
[3, 3, 3, 3]
and the numpy solution would change to:
>>> map(abs, np.array(list1) - np.array(list2))
[3, 3, 3, 3]
You may also do if else condition inside list comprehension.
>>> [i-j if i>j else j-i for i,j in zip(list1, list2)]
[3, 3, 3, 3]
You can use zip method in order combine these two lists. See the tutorials for zip method https://docs.python.org/2/library/functions.html#zip
>>> list1 = [1, 5, 3, 7]
>>> list2 = [4, 2, 6, 4]
>>> [abs(x-y) for x, y in zip(list1, list2)]
[3, 3, 3, 3]
Avinash Raj's answer is correct, or alternatively, using map():
from operator import sub
C = map(sub, A, B)

Combining two lists and removing duplicates, without removing duplicates in original list

I have two lists that i need to combine where the second list has any duplicates of the first list ignored. .. A bit hard to explain, so let me show an example of what the code looks like, and what i want as a result.
first_list = [1, 2, 2, 5]
second_list = [2, 5, 7, 9]
# The result of combining the two lists should result in this list:
resulting_list = [1, 2, 2, 5, 7, 9]
You'll notice that the result has the first list, including its two "2" values, but the fact that second_list also has an additional 2 and 5 value is not added to the first list.
Normally for something like this i would use sets, but a set on first_list would purge the duplicate values it already has. So i'm simply wondering what the best/fastest way to achieve this desired combination.
Thanks.
You need to append to the first list those elements of the second list that aren't in the first - sets are the easiest way of determining which elements they are, like this:
first_list = [1, 2, 2, 5]
second_list = [2, 5, 7, 9]
in_first = set(first_list)
in_second = set(second_list)
in_second_but_not_in_first = in_second - in_first
result = first_list + list(in_second_but_not_in_first)
print(result) # Prints [1, 2, 2, 5, 9, 7]
Or if you prefer one-liners 8-)
print(first_list + list(set(second_list) - set(first_list)))
resulting_list = list(first_list)
resulting_list.extend(x for x in second_list if x not in resulting_list)
You can use sets:
first_list = [1, 2, 2, 5]
second_list = [2, 5, 7, 9]
resultList= list(set(first_list) | set(second_list))
print(resultList)
# Results in : resultList = [1,2,5,7,9]
first_list = [1, 2, 2, 5]
second_list = [2, 5, 7, 9]
print( set( first_list + second_list ) )
You can bring this down to one single line of code if you use numpy:
a = [1,2,3,4,5,6,7]
b = [2,4,7,8,9,10,11,12]
sorted(np.unique(a+b))
>>> [1,2,3,4,5,6,7,8,9,10,11,12]
Simplest to me is:
first_list = [1, 2, 2, 5]
second_list = [2, 5, 7, 9]
merged_list = list(set(first_list+second_list))
print(merged_list)
#prints [1, 2, 5, 7, 9]
resulting_list = first_list + [i for i in second_list if i not in first_list]
You can also combine RichieHindle's and Ned Batchelder's responses for an average-case O(m+n) algorithm that preserves order:
first_list = [1, 2, 2, 5]
second_list = [2, 5, 7, 9]
fs = set(first_list)
resulting_list = first_list + [x for x in second_list if x not in fs]
assert(resulting_list == [1, 2, 2, 5, 7, 9])
Note that x in s has a worst-case complexity of O(m), so the worst-case complexity of this code is still O(m*n).
Based on the recipe :
resulting_list = list(set().union(first_list, second_list))
you can use dict.fromkeys to return a list with no duplicates:
def mergeTwoListNoDuplicates(list1, list2):
"""
Merges two lists together without duplicates
:param list1:
:param list2:
:return:
"""
merged_list = list1 + list2
merged_list = list(dict.fromkeys(merged_list))
return merged_list
This might help
def union(a,b):
for e in b:
if e not in a:
a.append(e)
The union function merges the second list into first, with out duplicating an element of a, if it's already in a. Similar to set union operator. This function does not change b. If a=[1,2,3] b=[2,3,4]. After union(a,b) makes a=[1,2,3,4] and b=[2,3,4]
first_list = [1, 2, 2, 5]
second_list = [2, 5, 7, 9]
newList=[]
for i in first_list:
newList.append(i)
for z in second_list:
if z not in newList:
newList.append(z)
newList.sort()
print newList
[1, 2, 2, 5, 7, 9]

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