Python - 测试所有行是否包含与其他矩阵的任何公共元素

给定两个矩阵，检查所有行是否包含至少一个与其他矩阵的同一索引行相同的元素。

Input : test_list1 = [[5, 6, 1], [2, 4], [9, 3, 5]], test_list2 = [[9, 1, 2], [9, 8, 2], [3, 7, 10]] 
Output : True 
Explanation : 1, 2 and 3 are common elements in rows.

Input : test_list1 = [[5, 6, 1], [2, 4], [9, 2, 6]], test_list2 = [[9, 1, 2], [9, 8, 2], [3, 7, 6]] 
Output : True 
Explanation : 1, 2 and 6 are common elements in rows. 
 

方法 #1：使用循环 + in运算符

在这里，我们迭代矩阵的每一行，并使用 in运算符检查其他列表中的任何元素匹配，如果找到任何元素，我们将其标记为真，如果所有行都为真，则返回真。

Python3

# Python3 code to demonstrate working of
# Test if all rows contain any common element with other Matrix
# Using loop "+" in operator
 
# initializing lists
test_list1 = [[5, 6, 1], [2, 4], [9, 3, 5]]
test_list2 = [[9, 1, 2], [9, 8, 2], [3, 7, 10]]
 
# printing original lists
print("The original list 1 is : " + str(test_list1))
print("The original list 2 is : " + str(test_list2))
 
res = True
for idx in range(0, len(test_list1)):
     
    temp = False
     
    # checking for any common element in 2nd list
    for ele in test_list1[idx]:
        if ele in test_list2[idx]:
            temp = True
            break
         
    # if any element not found, Result is false   
    if not temp :
        res = False
        break
 
# printing result
print("All row contain common elements with other Matrix : " + str(res))

输出：

The original list 1 is : [[5, 6, 1], [2, 4], [9, 3, 5]]
The original list 2 is : [[9, 1, 2], [9, 8, 2], [3, 7, 10]]
All row contain common elements with other Matrix : True

方法#2：使用 any() + 循环

在这种情况下，避免了一个嵌套循环，并且使用 any() 计算结果以获取第二个矩阵中的任何公共元素。

蟒蛇3

# Python3 code to demonstrate working of
# Test if all rows contain any common element with other Matrix
# Using loop any() "+" loop
 
# initializing lists
test_list1 = [[5, 6, 1], [2, 4], [9, 3, 5]]
test_list2 = [[9, 1, 2], [9, 8, 2], [3, 7, 10]]
 
# printing original lists
print("The original list 1 is : " + str(test_list1))
print("The original list 2 is : " + str(test_list2))
 
res = True
for idx in range(0, len(test_list1)):
     
    # checking for common element in list 2 in same index
    temp = any(ele in test_list2[idx] for ele in test_list1[idx])
         
    # if any element not found, Result is false   
    if not temp :
        res = False
        break
 
# printing result
print("All row contain common elements with other Matrix : " + str(res))

输出：

The original list 1 is : [[5, 6, 1], [2, 4], [9, 3, 5]]
The original list 2 is : [[9, 1, 2], [9, 8, 2], [3, 7, 10]]
All row contain common elements with other Matrix : True