How can I compare two lists in python and return matches

The easiest way to do that is to use sets:

>>> a = [1, 2, 3, 4, 5]
>>> b = [9, 8, 7, 6, 5]
>>> set(a) & set(b)
set([5])

Use set.intersection(), it’s fast and readable.

>>> set(a).intersection(b)
set([5])

Quick way:

list(set(a).intersection(set(b)))

Do you want duplicates? If not maybe you should use sets instead:

>>> set([1, 2, 3, 4, 5]).intersection(set([9, 8, 7, 6, 5]))
set([5])

You can use

def returnMatches(a,b):
       return list(set(a) & set(b))

>>> s = ['a','b','c']   
>>> f = ['a','b','d','c']  
>>> ss= set(s)  
>>> fs =set(f)  
>>> print ss.intersection(fs)   
   **set(['a', 'c', 'b'])**  
>>> print ss.union(fs)        
   **set(['a', 'c', 'b', 'd'])**  
>>> print ss.union(fs)  - ss.intersection(fs)   
   **set(['d'])**

I prefer the set based answers, but here’s one that works anyway

[x for x in a if x in b]

Also you can try this,by keeping common elements in a new list.

new_list = []
for element in a:
    if element in b:
        new_list.append(element)

A quick performance test showing Lutz’s solution is the best:

import time

def speed_test(func):
    def wrapper(*args, **kwargs):
        t1 = time.time()
        for x in xrange(5000):
            results = func(*args, **kwargs)
        t2 = time.time()
        print '%s took %0.3f ms' % (func.func_name, (t2-t1)*1000.0)
        return results
    return wrapper

@speed_test
def compare_bitwise(x, y):
    set_x = frozenset(x)
    set_y = frozenset(y)
    return set_x & set_y

@speed_test
def compare_listcomp(x, y):
    return [i for i, j in zip(x, y) if i == j]

@speed_test
def compare_intersect(x, y):
    return frozenset(x).intersection(y)

# Comparing short lists
a = [1, 2, 3, 4, 5]
b = [9, 8, 7, 6, 5]
compare_bitwise(a, b)
compare_listcomp(a, b)
compare_intersect(a, b)

# Comparing longer lists
import random
a = random.sample(xrange(100000), 10000)
b = random.sample(xrange(100000), 10000)
compare_bitwise(a, b)
compare_listcomp(a, b)
compare_intersect(a, b)

These are the results on my machine:

# Short list:
compare_bitwise took 10.145 ms
compare_listcomp took 11.157 ms
compare_intersect took 7.461 ms

# Long list:
compare_bitwise took 11203.709 ms
compare_listcomp took 17361.736 ms
compare_intersect took 6833.768 ms

Obviously, any artificial performance test should be taken with a grain of salt, but since the set().intersection() answer is at least as fast as the other solutions, and also the most readable, it should be the standard solution for this common problem.

You can use:

a = [1, 3, 4, 5, 9, 6, 7, 8]
b = [1, 7, 0, 9]
same_values = set(a) & set(b)
print same_values

Output:

set([1, 7, 9])

another a bit more functional way to check list equality for list 1 (lst1) and list 2 (lst2) where objects have depth one and which keeps the order is:

all(i == j for i, j in zip(lst1, lst2))   

Can use itertools.product too.

>>> common_elements=[]
>>> for i in list(itertools.product(a,b)):
...     if i[0] == i[1]:
...         common_elements.append(i[0])

If you want a boolean value:

>>> a = [1, 2, 3, 4, 5]
>>> b = [9, 8, 7, 6, 5]
>>> set(b) == set(a)  & set(b) and set(a) == set(a) & set(b)
False
>>> a = [3,1,2]
>>> b = [1,2,3]
>>> set(b) == set(a)  & set(b) and set(a) == set(a) & set(b)
True

Using __and__ attribute method also works.

>>> a = [1, 2, 3, 4, 5]
>>> b = [9, 8, 7, 6, 5]
>>> set(a).__and__(set(b))
set([5])

or simply

>>> set([1, 2, 3, 4, 5]).__and__(set([9, 8, 7, 6, 5]))
set([5])
>>>    

a = [1, 2, 3, 4, 5]
b = [9, 8, 7, 6, 5]

lista =set(a)
listb =set(b)   
print listb.intersection(lista)   
returnMatches = set(['5']) #output 

print " ".join(str(return) for return in returnMatches ) # remove the set()   

 5        #final output 

The following solution works for any order of list items and also supports both lists to be different length.

import numpy as np
def getMatches(a, b):
    matches = []
    unique_a = np.unique(a)
    unique_b = np.unique(b)
    for a in unique_a:
        for b in unique_b:
            if a == b:
                matches.append(a)
    return matches
print(getMatches([1, 2, 3, 4, 5], [9, 8, 7, 6, 5, 9])) # displays [5]
print(getMatches([1, 2, 3], [3, 4, 5, 1])) # displays [1, 3]

you can | for set union and & for set intersection.
for example:

    set1={1,2,3}
    set2={3,4,5}
    print(set1&set2)
    output=3

    set1={1,2,3}
    set2={3,4,5}
    print(set1|set2)
    output=1,2,3,4,5

curly braces in the answer.

I just used the following and it worked for me:

group1 = [1, 2, 3, 4, 5]
group2 = [9, 8, 7, 6, 5]

for k in group1:
    for v in group2:
        if k == v:
            print(k)

this would then print 5 in your case. Probably not great performance wise though.

This is for someone who might what to return a certain string or output,
here is the code, hope it helps:

lis =[]
#convert to list
a = list(data)
b = list(data)
def make_list():
    c = "greater than"
    d = "less_than"
    e = "equal"
    for first, first_te in zip(a, b):
        if first < first_te:
            lis.append(d)
        elif first > first_te:
            lis.append(c)
        else:
            lis.append(e)
    return lis

make_list()

One more way to find common values:

a = [1, 2, 3, 4, 5]
b = [9, 8, 7, 6, 5]
matches = [i for i in a if i in b]