Python - 测试连续元素矩阵

给定一个矩阵，测试它是否由连续元素组成。

Input : test_list = [[4, 5, 6], [7], [8, 9, 10], [11, 12]] 
Output : True 
Explanation : Elements in Matrix range from 4 to 12.

Input : test_list = [[4, 5, 6], [7], [8, 18, 10], [11, 12]] 
Output : False 
Explanation : Elements not consecutive and not in a range. 
 

方法#1：使用循环

在这里，我们检查行内的连续性，并在每行之前，检查第一个元素是否与上一行的最后一个元素相邻。

Python3

# Python3 code to demonstrate working of
# Test Consecutive Element Matrix
# Using loop
 
# initializing list
test_list = [[4, 5, 6], [7], [8, 9, 10], [11, 12]]
 
# printing original list
print("The original list is : " + str(test_list))
 
res = True
mem_list = []
for sub in test_list:
    flag = True
    for ele in sub:
 
        # checking for last element to be Consecutive
        if len(mem_list) != 0:
            if mem_list[-1] != ele - 1:
                flag = False
                break
        mem_list.append(ele)
 
    if not flag:
        res = False
        break
 
# printing result
print("Is Matrix Consecutive ? : " + str(res))

输出

The original list is : [[4, 5, 6], [7], [8, 9, 10], [11, 12]]
Is Matrix Consecutive ? : True

方法#2：使用chain.from_iterable() + all()

在这里，我们将矩阵展平到列表，然后使用 all() 检查所有连续的元素，是否比展平后的 Matrix 中的前一个元素大 1。

蟒蛇3

# Python3 code to demonstrate working of
# Test Consecutive Element Matrix
# Using chain.from_iterable() + all()
from itertools import chain
 
# initializing list
test_list = [[4, 5, 6], [7], [8, 9, 10], [11, 12]]
 
# printing original list
print("The original list is : " + str(test_list))
 
# flattening matrix
test_list = list(chain.from_iterable(test_list))
 
# checking for boolean true
res = all(test_list[idx - 1] == test_list[idx] -
          1 for idx in range(1, len(test_list)))
 
# printing result
print("Is Matrix Consecutive ? : " + str(res))

输出

The original list is : [[4, 5, 6], [7], [8, 9, 10], [11, 12]]
Is Matrix Consecutive ? : True