在计算机的世界中,同一个问题,使用不同的数据结构和算法实现,所使用的资源有很大差别
为了方便量化python中算法的资源消耗,对性能做测试非常有必要,这里针对stack做了python语言
下的性能分析。为后续算法分析做个基础。
代码:
import timeit from timeit import Timer class Stack: def __init__(self): self.items = [] def is_empty(self): return self.items == [] def push(self,item): self.items.append(item) def pop(self): return self.items.pop() def peek(self): return self.items[len(self.items) - 1] def size(self): return len(self.items) s = Stack() class StackA: def __init__(self): self.items = [] def is_empty(self): return self.items == [] def push(self,item): self.items.insert(0,item) def pop(self): return self.items.pop(0) def peek(self): return self.items[0] def size(self): return len(self.items) s1 = StackA() print("stack performing test:") def stack_push_test(): for i in range(1000): s.push('dog') def stack_pop_test(): for i in range(1000): s.pop() t1 = Timer("stack_push_test()","from __main__ import stack_push_test") print("stack_push_test ",t1.timeit(number=100),"milliseconds") t2 = Timer("stack_pop_test()","from __main__ import stack_pop_test") print("stack_pop_test ",t2.timeit(number=100),"milliseconds") print("stackA performing test:") def stacka_push_test(): for i in range(1000): s1.push('dog') def stacka_pop_test(): for i in range(1000): s1.pop() t3 = Timer("stacka_push_test()","from __main__ import stacka_push_test") print("stacka_push_test ",t3.timeit(number=100),"milliseconds") t4 = Timer("stacka_pop_test()","from __main__ import stacka_pop_test") print("stacka_pop_test ",t4.timeit(number=100),"milliseconds")
在linux上运行的主频为2.4G的系统上运行结果:
stack performing test: ('stack_push_test ', 0.020203113555908203, 'milliseconds') ('stack_pop_test ', 0.02147078514099121, 'milliseconds') stackA performing test: ('stacka_push_test ', 2.8804190158843994, 'milliseconds') ('stacka_pop_test ', 1.8630762100219727, 'milliseconds')
可以看出,不同的stack实现,对性能的影响是巨大的,显然是第一种stack实现的效率比较高。
具体原因是两者的算法复杂度是不一样的,第一种是O(1) 第二种是O(n).
作者:虚生 出处:https://www.cnblogs.com/dylancao/ 以音频和传感器算法为核心的智能可穿戴产品解决方案提供商 ,提供可穿戴智能软硬件解决方案的设计,开发和咨询服务。 勾搭热线:邮箱:1173496664@qq.com weixin:18019245820 市场技术对接群:347609188 |