# How can I compare two lists in python and return matches

## How can I compare two lists in python and return matches

### Question

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.

### Accepted Answer

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).

### Popular Answer

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

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

Read more… Read less…

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.

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

```
[x for x in a if x in b]
```

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])
```

```
>>> 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'])**
```