How can I compare two lists in python and return m

问题:

I want to take two lists and find the values that appear in both.

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

returnMatches(a, b)

would return [5], for instance.

回答1:

Not the most efficient one, but by far the most obvious way to do it is:

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

if order is significant you can do it with list comprehensions like this:

>>> [i for i, j in zip(a, b) if i == j]
[5]

(only works for equal-sized lists, which order-significance implies).

回答2:

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

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

回答3:

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.

回答4:

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

[x for x in a if x in b]

回答5:

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])

回答6:

>>> 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\'])**

回答7:

Quick way:

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

回答8:

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)

回答9:

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])

回答10:

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])

回答11:

You can use

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

回答12:

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])

回答13:

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))   

回答14:

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 

回答15:

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

回答16:

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]

回答17:

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])
>>>    

回答18:

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.

回答19:

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.