Python - K 差连续元素

给定一个整数元素列表，检查每个元素与连续元素的差是否为 K。

Input : test_list = [5, 6, 3, 2, 5, 3, 4], K = 1
Output : [True, False, True, False, False, True]
Explanation : 5, 6; 3, 2; and 3, 4 have 1 diff. between them.

Input : test_list = [5, 6, 3, 2, 5, 3, 4], K = 2
Output : [False, False, False, False, True, False]
Explanation : Only 5, 3 has 2 diff between it.

方法#1：使用列表推导

这是可以执行此任务的方式之一。在此，我们检查每个元素与下一个元素，如果差异为 K，则我们将其标记为 True，否则标记为 False。

Python3

# Python3 code to demonstrate working of 
# K difference Consecutive elements
# Using list comprehension
  
# initializing list
test_list = [5, 6, 3, 2, 5, 3, 4]
  
# printing original list
print("The original list : " + str(test_list))
  
# initializing K 
K = 3
  
# using list comprehension and abs() to compute result
res = [True if abs(test_list[idx] - test_list[idx + 1]) == K else False
       for idx in range(len(test_list) - 1)]
      
# printing result 
print("The difference list result : " + str(res))

输出

The original list : [5, 6, 3, 2, 5, 3, 4]
The difference list result : [False, True, False, True, False, False]

方法 #2：使用 zip() + 列表理解

这是可以执行此任务的另一种方式。在这种情况下，连续列表使用 zip() 配对，计算通过列表理解运行。

Python3

# Python3 code to demonstrate working of 
# K difference Consecutive elements
# Using zip() + list comprehension
  
# initializing list
test_list = [5, 6, 3, 2, 5, 3, 4]
  
# printing original list
print("The original list : " + str(test_list))
  
# initializing K 
K = 3
  
# using list comprehension and abs() to compute result
# zip() used to pair Consecutive elements list
res = [abs(a - b) == K for a, b in zip(test_list, test_list[1:])]
      
# printing result 
print("The difference list result : " + str(res))

输出

The original list : [5, 6, 3, 2, 5, 3, 4]
The difference list result : [False, True, False, True, False, False]