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) 

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) 

您还可以使用集合并在一行中获得共性:从其中一个集合中减去包含差异的集合。

A = [1,2,3,4]
B = [2,4,7,8]
commonalities = set(A) - (set(A) - set(B))

使用Python的集合交集:

>>> list1 = [1,2,3,4,5,6]
>>> list2 = [3, 5, 7, 9]
>>> list(set(list1).intersection(list2))
[3, 5]

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]。