数组递增的判断【python实现】
有时候需要对某一组数组的数据进行判断是否 递增 的场景，比如我在开发一些体育动作场景下，某些肢体动作是需要持续朝着垂直方向向上变化，那么z轴的值是会累增的。同理，逆向考虑，递减就是它的对立面。
下面是查找总结到的所有方式，如有补充可以评论区提出。
资料参考来源: Check if list is strictly increasing
1. zip() and all()
- Code:
test_list = [1, 4, 5, 7, 8, 10]
# Using zip() and all() to
# Check for strictly increasing list
res = all(i < j for i, j in zip(test_list, test_list[1:]))
print(f"Is list strictly increasing ? : {res}")
- Output:
Is list strictly increasing ? : True
时间复杂度: O(n), n是数组的长度。
2. reduce and lambda
- Code:
import functools
test_list = [1, 4, 5, 7, 8, 10]
res = bool((lambda list_demo: functools.reduce(lambda i, j: j if
i < j else 9999, list_demo) != 9999)(test_list))
print(f"Is list strictly increasing ? : {res}")
- Output:
Is list strictly increasing ? : True
时间复杂度: O(n), n是数组的长度。
3. itertools.starmap() + zip() + all()
- Code:
import itertools
test_list = [1, 4, 5, 7, 8, 10]
res = all(itertools.starmap(operator.le, zip(test_list, test_list[1:])))
print(f"Is list strictly increasing ? : {res}")
- Output:
Is list strictly increasing ? : True
时间复杂度: O(n), n是数组的长度。
4. sort() and extend()
- Code:
test_list = [1, 4, 5, 7, 8, 10]
res = False
new_list = []
new_list.extend(test_list)
test_list.sort()
if new_list == test_list:
res = True
print(f"Is list strictly increasing ? : {res}")
- Output:
Is list strictly increasing ? : True
时间复杂度: O(nlogn), 这里是sort()的时间复杂度
5. Use stacks
栈是一种后进先出的数据结构(Last in, first out)。
- Code:
def is_strictly_increasing(lst):
stack = []
for i in lst:
if stack and i <= stack[-1]:
return False
stack.append(i)
return True
test_list = [1, 4, 5, 7, 8, 10]
print(is_strictly_increasing(test_list)) # True
test_list = [1, 4, 5, 7, 7, 10]
print(is_strictly_increasing(test_list)) # False
时间复杂度: O(n)，原数组被遍历了一遍
空间复杂度: O(n)，栈可能要存储全部的n个原数组元素
6. numpy()
- Code:
import numpy as np
def is_increasing(lst):
# Converting input list to a numpy array
arr = np.array(lst)
# calculate the difference between adjacent elements of the array
diff = np.diff(arr)
# check if all differences are positive
# using the np.all() function
is_increasing = np.all(diff > 0)
# return the result
return is_increasing
# Input list
test_list = [1, 4, 5, 7, 8, 10]
# Printing original lists
print("Original list : " + str(test_list))
result = is_increasing(test_list)
print(result)
# True
时间复杂度: O(n)
7. itertools.pairwise() and all()
这里面就等于使用 pairwise() 替代了之前的 zip(list, list[1:]) 。
- Code:
from itertools import pairwise
# Function
def is_strictly_increasing(my_list):
# using pairwise method to iterate through the list and
# create pairs of adjacent elements.
# all() method checks if all pairs of adjacent elements
# satisfy the condition i < j, where i and j
# are the two elements in the pair.
if all(a < b for a, b in pairwise(my_list)):
return True
else:
return False
# Initializing list
test_list = [1, 4, 5, 7, 8, 10]
# Printing original lists
print("Original list : " + str(test_list))
# Checking for strictly increasing list
# using itertools pairwise() and all() method
res = is_strictly_increasing(test_list)
# Printing the result
print("Is list strictly increasing ? : " + str(res))
- Output:
Original list : [1, 4, 5, 7, 8, 10]
Is list strictly increasing ? : True
时间复杂度: O(n)