如何比较python中的两个列表并返回匹配项

问题:如何比较python中的两个列表并返回匹配项

我想获取两个列表并查找两个列表中都出现的值。

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

returnMatches(a, b)

[5]例如,将返回。

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 0

不是最有效的方法,但到目前为止,最明显的方法是:

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

如果订单很重要,您可以使用以下列表理解方法:

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

(仅适用于大小相等的列表,这意味着顺序意义)。

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 1

使用set.intersection(),它快速且可读。

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

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

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

回答 2

快速的性能测试表明Lutz的解决方案是最好的:

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)

这些是我机器上的结果:

# 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

显然,任何人工性能测试都应以一粒盐进行,但由于 set().intersection()答案至少与其他解决方案一样快,并且也是最易读的,因此它应该是解决此常见问题的标准解决方案。

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 3

我更喜欢基于集合的答案,但这还是可行的

[x for x in a if x in b]

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

[x for x in a if x in b]

回答 4

最简单的方法是使用set

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

快速方式:

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

Quick way:

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

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'])**
>>> 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

您也可以通过将公共元素保留在新列表中来尝试此操作。

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

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 8

您要重复吗?如果不是,您应该使用集合:


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

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 9

检查对象的深度为1且保持顺序的列表1(lst1)和列表2(lst2)的列表相等性的另一种实用方式是:

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

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 10

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 
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 11

也可以使用itertools.product。

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

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 12

您可以使用

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

You can use

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

回答 13

您可以使用:

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

输出:

set([1, 7, 9])

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 14

如果要布尔值:

>>> 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

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 15

以下解决方案适用于任何顺序的列表项,并且还支持两个列表的长度不同。

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]

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 16

使用__and__属性方法也可以。

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

或简单地

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

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 17

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.
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 18

我只使用了以下内容,它对我有用:

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)

这将在您的情况下打印5。可能不是明智的表现。

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.