集合是另一种解法

a = [3,2,4]
b = [2,3,5]
set(a)&set(b)
{2, 3}

1) Method1 保存list1是字典，然后迭代list2中的每个elem

def findarrayhash(a,b):
    h1={k:1 for k in a}
    for val in b:
        if val in h1:
            print("common found",val)
            del h1[val]
        else:
            print("different found",val)
    for key in h1.iterkeys():
        print ("different found",key)

找到共同和不同的元素:

2) Method2 使用设置

def findarrayset(a,b):
    common = set(a)&set(b)
    diff=set(a)^set(b)
    print list(common)
    print list(diff) 

使用集合交叉，set(list1) & set(list2)

>>> def common_elements(list1, list2):
...     return list(set(list1) & set(list2))
...
>>>
>>> common_elements([1,2,3,4,5,6], [3,5,7,9])
[3, 5]
>>>
>>> common_elements(['this','this','n','that'],['this','not','that','that'])
['this', 'that']
>>>
>>>

注意，结果列表可能与原始列表顺序不同。

list1=[123,324523,5432,311,23]
list2=[2343254,34234,234,322123,123,234,23]
common=[]
def common_elements(list1,list2):
    for x in range(0,len(list1)):
        if list1[x] in list2:
            common.append(list1[x])
            
common_elements(list1,list2)
print(common)

你可以使用numpy来解决这个问题:

import numpy as np

list1 = [1, 2, 3, 4, 5, 6]
list2 = [3, 5, 7, 9]

common_elements = np.intersect1d(list1, list2)
print(common_elements)

Common_elements将是numpy数组:[3 5]。

我比较了每个答案提到的每一种方法。目前，我使用python 3.6.3来实现这个功能。这是我使用的代码:

import time
import random
from decimal import Decimal


def method1():
    common_elements = [x for x in li1_temp if x in li2_temp]
     print(len(common_elements))


def method2():
    common_elements = (x for x in li1_temp if x in li2_temp)
    print(len(list(common_elements)))


def method3():
    common_elements = set(li1_temp) & set(li2_temp)
    print(len(common_elements))


def method4():
    common_elements = set(li1_temp).intersection(li2_temp)
    print(len(common_elements))


if __name__ == "__main__":
    li1 = []
    li2 = []
    for i in range(100000):
        li1.append(random.randint(0, 10000))
        li2.append(random.randint(0, 10000))

    li1_temp = list(set(li1))
    li2_temp = list(set(li2))

    methods = [method1, method2, method3, method4]
    for m in methods:
        start = time.perf_counter()
        m()
        end = time.perf_counter()
        print(Decimal((end - start)))

如果你运行这段代码，你可以看到如果你使用列表或生成器(如果你迭代生成器，而不是仅仅使用它)。当我强迫生成器打印它的长度时，我这样做了)，你会得到几乎相同的性能。但是如果你使用set，你会得到更好的性能。如果你使用交叉方法你会得到更好的性能。在我的电脑中，每种方法的结果如下:

method1: 0.8150673999999999974619413478649221360683441 method2: 0.8329545000000001531148541289439890533685684 method3: 0.0016547000000000089414697868051007390022277 method4: 0.0010262999999999244948867271887138485908508