Adding lists to a sequence [duplicate] - python

This question already has answers here:
How do I iterate through two lists in parallel?
(8 answers)
Apply function to each element of a list
(4 answers)
Closed 8 months ago.
I want to perform an element wise multiplication, to multiply two lists together by value in Python, like we can do it in Matlab.
This is how I would do it in Matlab.
a = [1,2,3,4]
b = [2,3,4,5]
a .* b = [2, 6, 12, 20]
A list comprehension would give 16 list entries, for every combination x * y of x from a and y from b. Unsure of how to map this.
If anyone is interested why, I have a dataset, and want to multiply it by Numpy.linspace(1.0, 0.5, num=len(dataset)) =).


Use a list comprehension mixed with zip():.
[a*b for a,b in zip(lista,listb)]


Since you're already using numpy, it makes sense to store your data in a numpy array rather than a list. Once you do this, you get things like element-wise products for free:
In [1]: import numpy as np
In [2]: a = np.array([1,2,3,4])
In [3]: b = np.array([2,3,4,5])
In [4]: a * b
Out[4]: array([ 2, 6, 12, 20])


Use np.multiply(a,b):
import numpy as np
a = [1,2,3,4]
b = [2,3,4,5]
np.multiply(a,b)


You can try multiplying each element in a loop. The short hand for doing that is
ab = [a[i]*b[i] for i in range(len(a))]


Yet another answer:
-1 ... requires import
+1 ... is very readable
import operator
a = [1,2,3,4]
b = [10,11,12,13]
list(map(operator.mul, a, b))
outputs [10, 22, 36, 52]
Edit
For Python3.6+ map does not automatically unpack values.
Use itertools.starmap
import itertools
import operator
itertools.starmap(operator.mul, zip(a, b)))


you can multiplication using lambda
foo=[1,2,3,4]
bar=[1,2,5,55]
l=map(lambda x,y:x*y,foo,bar)


Fairly intuitive way of doing this:
a = [1,2,3,4]
b = [2,3,4,5]
ab = [] #Create empty list
for i in range(0, len(a)):
ab.append(a[i]*b[i]) #Adds each element to the list


gahooa's answer is correct for the question as phrased in the heading, but if the lists are already numpy format or larger than ten it will be MUCH faster (3 orders of magnitude) as well as more readable, to do simple numpy multiplication as suggested by NPE. I get these timings:
0.0049ms -> N = 4, a = [i for i in range(N)], c = [a*b for a,b in zip(a, b)]
0.0075ms -> N = 4, a = [i for i in range(N)], c = a * b
0.0167ms -> N = 4, a = np.arange(N), c = [a*b for a,b in zip(a, b)]
0.0013ms -> N = 4, a = np.arange(N), c = a * b
0.0171ms -> N = 40, a = [i for i in range(N)], c = [a*b for a,b in zip(a, b)]
0.0095ms -> N = 40, a = [i for i in range(N)], c = a * b
0.1077ms -> N = 40, a = np.arange(N), c = [a*b for a,b in zip(a, b)]
0.0013ms -> N = 40, a = np.arange(N), c = a * b
0.1485ms -> N = 400, a = [i for i in range(N)], c = [a*b for a,b in zip(a, b)]
0.0397ms -> N = 400, a = [i for i in range(N)], c = a * b
1.0348ms -> N = 400, a = np.arange(N), c = [a*b for a,b in zip(a, b)]
0.0020ms -> N = 400, a = np.arange(N), c = a * b
i.e. from the following test program.
import timeit
init = ['''
import numpy as np
N = {}
a = {}
b = np.linspace(0.0, 0.5, len(a))
'''.format(i, j) for i in [4, 40, 400]
for j in ['[i for i in range(N)]', 'np.arange(N)']]
func = ['''c = [a*b for a,b in zip(a, b)]''',
'''c = a * b''']
for i in init:
for f in func:
lines = i.split('\n')
print('{:6.4f}ms -> {}, {}, {}'.format(
timeit.timeit(f, setup=i, number=1000), lines[2], lines[3], f))


For large lists, we can do it the iter-way:
product_iter_object = itertools.imap(operator.mul, [1,2,3,4], [2,3,4,5])
product_iter_object.next() gives each of the element in the output list.
The output would be the length of the shorter of the two input lists.


create an array of ones;
multiply each list times the array;
convert array to a list
import numpy as np
a = [1,2,3,4]
b = [2,3,4,5]
c = (np.ones(len(a))*a*b).tolist()
[2.0, 6.0, 12.0, 20.0]


Can use enumerate.
a = [1, 2, 3, 4]
b = [2, 3, 4, 5]
ab = [val * b[i] for i, val in enumerate(a)]


The map function can be very useful here.
Using map we can apply any function to each element of an iterable.
Python 3.x
>>> def my_mul(x,y):
... return x*y
...
>>> a = [1,2,3,4]
>>> b = [2,3,4,5]
>>>
>>> list(map(my_mul,a,b))
[2, 6, 12, 20]
>>>
Of course:
map(f, iterable)
is equivalent to
[f(x) for x in iterable]
So we can get our solution via:
>>> [my_mul(x,y) for x, y in zip(a,b)]
[2, 6, 12, 20]
>>>
In Python 2.x map() means: apply a function to each element of an iterable and construct a new list.
In Python 3.x, map construct iterators instead of lists.
Instead of my_mul we could use mul operator
Python 2.7
>>>from operator import mul # import mul operator
>>>a = [1,2,3,4]
>>>b = [2,3,4,5]
>>>map(mul,a,b)
[2, 6, 12, 20]
>>>
Python 3.5+
>>> from operator import mul
>>> a = [1,2,3,4]
>>> b = [2,3,4,5]
>>> [*map(mul,a,b)]
[2, 6, 12, 20]
>>>
Please note that since map() constructs an iterator we use * iterable unpacking operator to get a list.
The unpacking approach is a bit faster then the list constructor:
>>> list(map(mul,a,b))
[2, 6, 12, 20]
>>>


To maintain the list type, and do it in one line (after importing numpy as np, of course):
list(np.array([1,2,3,4]) * np.array([2,3,4,5]))
or
list(np.array(a) * np.array(b))


you can use this for lists of the same length
def lstsum(a, b):
c=0
pos = 0
for element in a:
c+= element*b[pos]
pos+=1
return c


import ast,sys
input_str = sys.stdin.read()
input_list = ast.literal_eval(input_str)
list_1 = input_list[0]
list_2 = input_list[1]
import numpy as np
array_1 = np.array(list_1)
array_2 = np.array(list_2)
array_3 = array_1*array_2
print(list(array_3))

Related

How to efficiently create nested loops and append the output (as a multi-dimensional array) from all the loops?

I'm looking for an 'efficient' way in Python 3.x to create various nested loops, and then append the result of each inner loop (a multi-dimensional array).
For instance, the function model_A() has 3 parameters (a, b, c), and I want to enumerate all of the possibilities to test the model. The casual way is:
result_a = []
for a_value in a:
result_a_b = []
for b_value in b:
result_a_b_c = []
for c_value in c:
result = model_A(a, b, c)
result_a_b_c.append(result)
result_a_b.append(result_a_b_c)
result_a.append(result_a_b)
I think there should be a way to 'efficiently' create the nested loop, and append the result, without having to create an empty list before each loop, and append the result at the end of each inner loop.

Probably itertools.product() has what you need
for instance:
a = [1, 2]
b = [3, 4]
c = [5, 6]
for element in itertools.product(a,b,c):
print(element)
Results in:
(1, 3, 5)
(1, 3, 6)
(1, 4, 5)
(1, 4, 6)
(2, 3, 5)
(2, 3, 6)
(2, 4, 5)
(2, 4, 6)
This might enable you to avoid the deeply nested for loops.

One small improvement can be had by using list comprehension (even though it is not so readable in my opinion in this case):
A_LIST = list(range(10))
B_LIST = list(range(10))
C_LIST = list(range(10))
def model_A(a, b, c):
# mock implementation that does some calculation
return a + b * c
def f1():
# your version
result_a = []
for a_value in A_LIST:
result_a_b = []
for b_value in B_LIST:
result_a_b_c = []
for c_value in C_LIST:
result = model_A(a_value, b_value, c_value)
result_a_b_c.append(result)
result_a_b.append(result_a_b_c)
result_a.append(result_a_b)
return result_a
def f2():
# some improvement
return [
[
[model_A(a_value, b_value, c_value)
for c_value in C_LIST]
for b_value in B_LIST]
for a_value in A_LIST]
Comparing these 2 versions, list comprehension is slightly faster:
>>> f1() == f2()
True
>>> import timeit
>>> timeit.timeit('f()', 'from __main__ import f1 as f', number=10000)
1.905
>>> timeit.timeit('f()', 'from __main__ import f2 as f', number=10000)
1.503

Use multiple lists as input arguments of a function (Python)

I know that the property map(function,list) applies a function to each element of a single list. But how would it be if my function requires more than one list as input arguments?.
For example I tried:
def testing(a,b,c):
result1=a+b
result2=a+c
return (result1,result2)
a=[1,2,3,4]
b=[1,1,1,1]
c=[2,2,2,2]
result1,result2=testing(a,b,c)
But this only concatenates the arrays:
result1=[1,2,3,4,1,1,1,1]
result2=[1, 2, 3, 4, 2, 2, 2, 2]
and what I need is the following result:
result1=[2,3,4,5]
result2=[3,4,5,6]
I would be grateful if somebody could let me know how would this be possible or point me to a link where my question could be answered in a similar case.

You can use operator.add
from operator import add
def testing(a,b,c):
result1 = map(add, a, b)
result2 = map(add, b, c)
return (result1, result2)

You can use zip:
def testing(a,b,c):
result1=[x + y for x, y in zip(a, b)]
result2=[x + y for x, y in zip(a, c)]
return (result1,result2)
a=[1,2,3,4]
b=[1,1,1,1]
c=[2,2,2,2]
result1,result2=testing(a,b,c)
print result1 #[2, 3, 4, 5]
print result2 #[3, 4, 5, 6]

Quick and simple:
result1 = [a[i] + b[i] for i in range(0,len(a))]
result2 = [a[i] + c[i] for i in range(0,len(a))]
(Or for safety you can use range(0, min(len(a), len(b)))

Instead of list, use array from numpy instead.
list concatenate while arrays add corresponding elements. Here I converted the input to numpy arrays. You can feed the function numpy arrays and avoid the conversion steps.
def testing(a,b,c):
a = np.array(a)
b = np.array(b)
c = np.array(c)
result1=a+b
result2=a+c
return (result1,result2)
a=[1,2,3,4]
b=[1,1,1,1]
c=[2,2,2,2]
result1,result2=testing(a,b,c)
print(result1, result2)

Combining two Python lists in an interleaved manner

I want to interleave two lists. For example:
arr1 = [1,2,3,4,5,6]
arr2 = [9,8,7,6]
I't like to get an output like
[1,9,2,8,3,7,4,6,5,6]
I have created the following script, but it's not working for some reason:
arr1 = [1,2,3,4,5,6]
arr2 = [9,8,7,6]
x = 0
for a in arr2:
x = x + 2
arr1.insert(x, a)
Where am I going wrong? I am not looking for random shuffling. I am using python2.x

You can use zip_longest and chain.from_iterable from the itertools module:
>>> arr1 = [1,2,3,4,5,6]
>>> arr2 = [9,8,7,6]
>>> from itertools import chain, zip_longest
>>> [i for i in chain.from_iterable(zip_longest(arr1, arr2)) if i is not None]
[1, 9, 2, 8, 3, 7, 4, 6, 5, 6]
You need to use izip_longest instead of zip_longest in python-2.x

I have figured out a different way to do this, without imports.
mylist = []
a = [1,2,3,4,5,6]
b = [9,8,7,6]
for x in range(max(len(a), len(b))):
if x < len(a):
mylist.append(a[x])
if x < len(b):
mylist.append(b[x])

if you actually want to shuffle them in a randomized fashion the key would be to combine them into one list and use shuffle().
from random import shuffle
import itertools as it
arr1 = [1,2,3,4,5,6]
arr2 = [9,8,7,6]
x = list(it.chain(arr1,arr2))
shuffle(x)
print(x)
EDIT question was updated to reflect the order desired wasn't random.

How to exempt a list item from iteration?

I have two lists of integers of equal length. I would like to add each element of the first list to its corresponding element of the second list using the line:
complete_list = [first_list[i] + second_list[i] for i in range(len(first_list))]
However, some elements of first_list are special numbers and I would like to exempt them from the above operation whilst having the other elements added to make complete_list. Thanks!

You can stick an if at the end of your listcomp:
>>> first_list = [1,2,3,10]
>>> second_list = [10,20,30,50]
>>> special = {2, 3}
>>> [first_list[i]+second_list[i] for i in range(len(first_list)) if first_list[i] not in special]
[11, 60]
Although I'd probably use zip here:
>>> [a+b for a,b in zip(first_list, second_list) if a not in special]
[11, 60]
If you want a to pass through unadjusted, you could move the if and use the X if Y else Z ternary syntax:
>>> [a+b if a not in special else a for a,b in zip(first_list, second_list)]
[11, 2, 3, 60]

I would tend to write your original example as:
a = [1, 2, 3]
b = [4, 5, 6]
from operator import add
c = map(add, a, b)
# [5, 7, 9]
Then, if you only want elements to "not add" depending on criteria on a, then build a generator over b referencing the corresponding element on a and make it 0 for a no-op.
special = {2}
b2 = (j if a[i] not in special else 0 for i, j in enumerate(b))
map(add, a, b2)
# [5, 2, 9]

How to find list intersection?

a = [1,2,3,4,5]
b = [1,3,5,6]
c = a and b
print c
actual output: [1,3,5,6]
expected output: [1,3,5]
How can we achieve a boolean AND operation (list intersection) on two lists?

If order is not important and you don't need to worry about duplicates then you can use set intersection:
>>> a = [1,2,3,4,5]
>>> b = [1,3,5,6]
>>> list(set(a) & set(b))
[1, 3, 5]

Using list comprehensions is a pretty obvious one for me. Not sure about performance, but at least things stay lists.
[x for x in a if x in b]
Or "all the x values that are in A, if the X value is in B".

If you convert the larger of the two lists into a set, you can get the intersection of that set with any iterable using intersection():
a = [1,2,3,4,5]
b = [1,3,5,6]
set(a).intersection(b)

Make a set out of the larger one:
_auxset = set(a)
Then,
c = [x for x in b if x in _auxset]
will do what you want (preserving b's ordering, not a's -- can't necessarily preserve both) and do it fast. (Using if x in a as the condition in the list comprehension would also work, and avoid the need to build _auxset, but unfortunately for lists of substantial length it would be a lot slower).
If you want the result to be sorted, rather than preserve either list's ordering, an even neater way might be:
c = sorted(set(a).intersection(b))

Here's some Python 2 / Python 3 code that generates timing information for both list-based and set-based methods of finding the intersection of two lists.
The pure list comprehension algorithms are O(n^2), since in on a list is a linear search. The set-based algorithms are O(n), since set search is O(1), and set creation is O(n) (and converting a set to a list is also O(n)). So for sufficiently large n the set-based algorithms are faster, but for small n the overheads of creating the set(s) make them slower than the pure list comp algorithms.
#!/usr/bin/env python
''' Time list- vs set-based list intersection
See http://stackoverflow.com/q/3697432/4014959
Written by PM 2Ring 2015.10.16
'''
from __future__ import print_function, division
from timeit import Timer
setup = 'from __main__ import a, b'
cmd_lista = '[u for u in a if u in b]'
cmd_listb = '[u for u in b if u in a]'
cmd_lcsa = 'sa=set(a);[u for u in b if u in sa]'
cmd_seta = 'list(set(a).intersection(b))'
cmd_setb = 'list(set(b).intersection(a))'
reps = 3
loops = 50000
def do_timing(heading, cmd, setup):
t = Timer(cmd, setup)
r = t.repeat(reps, loops)
r.sort()
print(heading, r)
return r[0]
m = 10
nums = list(range(6 * m))
for n in range(1, m + 1):
a = nums[:6*n:2]
b = nums[:6*n:3]
print('\nn =', n, len(a), len(b))
#print('\nn = %d\n%s %d\n%s %d' % (n, a, len(a), b, len(b)))
la = do_timing('lista', cmd_lista, setup)
lb = do_timing('listb', cmd_listb, setup)
lc = do_timing('lcsa ', cmd_lcsa, setup)
sa = do_timing('seta ', cmd_seta, setup)
sb = do_timing('setb ', cmd_setb, setup)
print(la/sa, lb/sa, lc/sa, la/sb, lb/sb, lc/sb)
output
n = 1 3 2
lista [0.082171916961669922, 0.082588911056518555, 0.0898590087890625]
listb [0.069530963897705078, 0.070394992828369141, 0.075379848480224609]
lcsa [0.11858987808227539, 0.1188349723815918, 0.12825107574462891]
seta [0.26900982856750488, 0.26902294158935547, 0.27298116683959961]
setb [0.27218389511108398, 0.27459001541137695, 0.34307217597961426]
0.305460649521 0.258469975867 0.440838458259 0.301898526833 0.255455833892 0.435697630214
n = 2 6 4
lista [0.15915989875793457, 0.16000485420227051, 0.16551494598388672]
listb [0.13000702857971191, 0.13060092926025391, 0.13543915748596191]
lcsa [0.18650484085083008, 0.18742108345031738, 0.19513416290283203]
seta [0.33592700958251953, 0.34001994132995605, 0.34146714210510254]
setb [0.29436492919921875, 0.2953648567199707, 0.30039691925048828]
0.473793098554 0.387009751735 0.555194537893 0.540689066428 0.441652573672 0.633583767462
n = 3 9 6
lista [0.27657914161682129, 0.28098297119140625, 0.28311991691589355]
listb [0.21585917472839355, 0.21679902076721191, 0.22272896766662598]
lcsa [0.22559309005737305, 0.2271728515625, 0.2323150634765625]
seta [0.36382699012756348, 0.36453008651733398, 0.36750602722167969]
setb [0.34979605674743652, 0.35533690452575684, 0.36164689064025879]
0.760194128313 0.59330170819 0.62005595016 0.790686848184 0.61710008036 0.644927481902
n = 4 12 8
lista [0.39616990089416504, 0.39746403694152832, 0.41129183769226074]
listb [0.33485794067382812, 0.33914685249328613, 0.37850618362426758]
lcsa [0.27405810356140137, 0.2745978832244873, 0.28249192237854004]
seta [0.39211201667785645, 0.39234519004821777, 0.39317893981933594]
setb [0.36988520622253418, 0.37011313438415527, 0.37571001052856445]
1.01034878821 0.85398540833 0.698928091731 1.07106176249 0.905302334456 0.740927452493
n = 5 15 10
lista [0.56792402267456055, 0.57422614097595215, 0.57740211486816406]
listb [0.47309303283691406, 0.47619009017944336, 0.47628307342529297]
lcsa [0.32805585861206055, 0.32813096046447754, 0.3349759578704834]
seta [0.40036201477050781, 0.40322518348693848, 0.40548801422119141]
setb [0.39103078842163086, 0.39722800254821777, 0.43811702728271484]
1.41852623806 1.18166313332 0.819398061028 1.45237674242 1.20986133789 0.838951479847
n = 6 18 12
lista [0.77897095680236816, 0.78187918663024902, 0.78467702865600586]
listb [0.629547119140625, 0.63210701942443848, 0.63321495056152344]
lcsa [0.36563992500305176, 0.36638498306274414, 0.38175487518310547]
seta [0.46695613861083984, 0.46992206573486328, 0.47583580017089844]
setb [0.47616910934448242, 0.47661614418029785, 0.4850609302520752]
1.66818870637 1.34819326075 0.783028414812 1.63591241329 1.32210827369 0.767878297495
n = 7 21 14
lista [0.9703209400177002, 0.9734041690826416, 1.0182771682739258]
listb [0.82394003868103027, 0.82625699043273926, 0.82796716690063477]
lcsa [0.40975093841552734, 0.41210508346557617, 0.42286920547485352]
seta [0.5086359977722168, 0.50968098640441895, 0.51014018058776855]
setb [0.48688101768493652, 0.4879908561706543, 0.49204087257385254]
1.90769222837 1.61990115188 0.805587768483 1.99293236904 1.69228211566 0.841583309951
n = 8 24 16
lista [1.204819917678833, 1.2206029891967773, 1.258256196975708]
listb [1.014998197555542, 1.0206191539764404, 1.0343101024627686]
lcsa [0.50966787338256836, 0.51018595695495605, 0.51319599151611328]
seta [0.50310111045837402, 0.50556015968322754, 0.51335406303405762]
setb [0.51472997665405273, 0.51948785781860352, 0.52113485336303711]
2.39478683834 2.01748351664 1.01305257092 2.34068341135 1.97190418975 0.990165516871
n = 9 27 18
lista [1.511646032333374, 1.5133969783782959, 1.5639569759368896]
listb [1.2461750507354736, 1.254518985748291, 1.2613379955291748]
lcsa [0.5565330982208252, 0.56119203567504883, 0.56451296806335449]
seta [0.5966339111328125, 0.60275578498840332, 0.64791703224182129]
setb [0.54694414138793945, 0.5508568286895752, 0.55375313758850098]
2.53362406013 2.08867620074 0.932788243907 2.76380331728 2.27843203069 1.01753187594
n = 10 30 20
lista [1.7777848243713379, 2.1453688144683838, 2.4085969924926758]
listb [1.5070111751556396, 1.5202279090881348, 1.5779800415039062]
lcsa [0.5954139232635498, 0.59703707695007324, 0.60746097564697266]
seta [0.61563014984130859, 0.62125110626220703, 0.62354087829589844]
setb [0.56723213195800781, 0.57257509231567383, 0.57460403442382812]
2.88774814689 2.44791645689 0.967161734066 3.13413984189 2.6567803378 1.04968299523
Generated using a 2GHz single core machine with 2GB of RAM running Python 2.6.6 on a Debian flavour of Linux (with Firefox running in the background).
These figures are only a rough guide, since the actual speeds of the various algorithms are affected differently by the proportion of elements that are in both source lists.

A functional way can be achieved using filter and lambda operator.
list1 = [1,2,3,4,5,6]
list2 = [2,4,6,9,10]
>>> list(filter(lambda x:x in list1, list2))
[2, 4, 6]
Edit: It filters out x that exists in both list1 and list, set difference can also be achieved using:
>>> list(filter(lambda x:x not in list1, list2))
[9,10]
Edit2: python3 filter returns a filter object, encapsulating it with list returns the output list.

a = [1,2,3,4,5]
b = [1,3,5,6]
c = list(set(a).intersection(set(b)))
Should work like a dream. And, if you can, use sets instead of lists to avoid all this type changing!

You can also use numpy.intersect1d(ar1, ar2).
It return the unique and sorted values that are in both of two arrays.

This way you get the intersection of two lists and also get the common duplicates.
>>> from collections import Counter
>>> a = Counter([1,2,3,4,5])
>>> b = Counter([1,3,5,6])
>>> a &= b
>>> list(a.elements())
[1, 3, 5]

This is an example when you need Each element in the result should appear as many times as it shows in both arrays.
def intersection(nums1, nums2):
#example:
#nums1 = [1,2,2,1]
#nums2 = [2,2]
#output = [2,2]
#find first 2 and remove from target, continue iterating
target, iterate = [nums1, nums2] if len(nums2) >= len(nums1) else [nums2, nums1] #iterate will look into target
if len(target) == 0:
return []
i = 0
store = []
while i < len(iterate):
element = iterate[i]
if element in target:
store.append(element)
target.remove(element)
i += 1
return store

In the case you have a list of lists map comes handy:
>>> lists = [[1, 2, 3], [2, 3, 4], [2, 3, 5]]
>>> set(lists.pop()).intersection(*map(set, lists))
{2, 3}
would work for similar iterables:
>>> lists = ['ash', 'nazg']
>>> set(lists.pop()).intersection(*map(set, lists))
{'a'}
pop will blow if the list is empty so you may want to wrap in a function:
def intersect_lists(lists):
try:
return set(lists.pop()).intersection(*map(set, lists))
except IndexError: # pop from empty list
return set()

It might be late but I just thought I should share for the case where you are required to do it manually (show working - haha) OR when you need all elements to appear as many times as possible or when you also need it to be unique.
Kindly note that tests have also been written for it.
from nose.tools import assert_equal
'''
Given two lists, print out the list of overlapping elements
'''
def overlap(l_a, l_b):
'''
compare the two lists l_a and l_b and return the overlapping
elements (intersecting) between the two
'''
#edge case is when they are the same lists
if l_a == l_b:
return [] #no overlapping elements
output = []
if len(l_a) == len(l_b):
for i in range(l_a): #same length so either one applies
if l_a[i] in l_b:
output.append(l_a[i])
#found all by now
#return output #if repetition does not matter
return list(set(output))
else:
#find the smallest and largest lists and go with that
sm = l_a if len(l_a) len(l_b) else l_b
for i in range(len(sm)):
if sm[i] in lg:
output.append(sm[i])
#return output #if repetition does not matter
return list(set(output))
## Test the Above Implementation
a = [1, 1, 2, 3, 5, 8, 13, 21, 34, 55, 89]
b = [1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13]
exp = [1, 2, 3, 5, 8, 13]
c = [4, 4, 5, 6]
d = [5, 7, 4, 8 ,6 ] #assuming it is not ordered
exp2 = [4, 5, 6]
class TestOverlap(object):
def test(self, sol):
t = sol(a, b)
assert_equal(t, exp)
print('Comparing the two lists produces')
print(t)
t = sol(c, d)
assert_equal(t, exp2)
print('Comparing the two lists produces')
print(t)
print('All Tests Passed!!')
t = TestOverlap()
t.test(overlap)

Most of the solutions here don't take the order of elements in the list into account and treat lists like sets. If on the other hand you want to find one of the longest subsequences contained in both lists, you can try the following code.
def intersect(a, b):
if a == [] or b == []:
return []
inter_1 = intersect(a[1:], b)
if a[0] in b:
idx = b.index(a[0])
inter_2 = [a[0]] + intersect(a[1:], b[idx+1:])
if len(inter_1) <= len(inter_2):
return inter_2
return inter_1
For a=[1,2,3] and b=[3,1,4,2] this returns [1,2] rather than [1,2,3]. Note that such a subsequence is not unique as [1], [2], [3] are all solutions for a=[1,2,3] and b=[3,2,1].

If, by Boolean AND, you mean items that appear in both lists, e.g. intersection, then you should look at Python's set and frozenset types.

You can also use a counter! It doesn't preserve the order, but it'll consider the duplicates:
>>> from collections import Counter
>>> a = [1,2,3,4,5]
>>> b = [1,3,5,6]
>>> d1, d2 = Counter(a), Counter(b)
>>> c = [n for n in d1.keys() & d2.keys() for _ in range(min(d1[n], d2[n]))]
>>> print(c)
[1,3,5]

when we used tuple and we want to intersection
a=([1,2,3,4,5,20], [8,3,9,5,1,4,20])
for i in range(len(a)):
b=set(a[i-1]).intersection(a[i])
print(b)
{1, 3, 4, 5, 20}

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