问题描述
我正在自学数据结构,我正在尝试测量将 append 方法实现到数组数据结构的有效和低效方法之间的时间复杂度差异。也就是说,根据我在纸上做的一些数学计算,效率低下的方法应该是 O(n) = n^2,而有效的方法应该是 O(n) = n。
问题是,当我运行模拟并在图形上绘制两种情况时,低效方法按预期执行,但有效方法执行 O(n) = 1。我做错了什么吗?
import datetime
import time
import random
import matplotlib.pyplot as plt
import numpy as np
# Inefficient append
class PyListInef:
def __init__(self):
self.items = []
def append(self,item):
# Inefficient append -> appending n items to the list causes a O(n) = n^2,since for each i for i in 1,2,3...n
# we need i * k operations in order to append every element to the new list. Then,by weak induction we prove
# that the number of required operations is n(n+1)/2 which implies O(n) = n^2
self.items = self.items + [item]
# Using magic method for our PyList to be an iterable object.
def __iter__(self):
for c in self.items:
yield c
# Efficient append:
class PyList:
def __init__(self):
self.items = []
def append(self,item):
self.items.append(item)
def __iter__(self):
for c in self.items:
yield c
# The inefficient append running time
lst = PyListInef()
time_dict_inef = dict()
time_dict_ef = dict()
series = np.linspace(1,301,300)
time.sleep(2)
for i in range(300):
starttime = time.time()
for j in range(i):
lst.append(series[j])
elapsed_time = time.time() - starttime
time_dict_inef[i] = elapsed_time * 100000
# The efficient append running time
lst = PyList()
time.sleep(2)
for i in range(300):
starttime = time.time()
for j in range(i):
lst.append(series[j])
elapsed_time = time.time() - starttime
time_dict_ef[i] = elapsed_time * 100000
plt.figure(figsize = (14,7))
plt.plot(time_dict_inef.keys(),time_dict_inef.values())
plt.plot(time_dict_ef.keys(),time_dict_ef.values())
plt.xlabel('Number of elements to append')
plt.ylabel('Elapsed time (microseconds)')
plt.title('Comparison between efficient appending vs inefficient appending in a list data structure')
plt.show()
你能帮我指出我做错了什么吗?
解决方法
time.time()
的分辨率有限。您的“高效追加”时间足够快,它们通常会在 time.time()
分辨率的一个滴答声之前完成。请注意黄色图表中准确显示的两个时间:0 个刻度和 1 个刻度。图表右侧的 1-tick 时间更频繁,因为即使时间比单个 tick 短,时间越长也意味着在运行时发生该滴答的概率更高。如果您使用更大的输入运行,您最终会看到 2-tick 时间甚至更多。 (另请注意,100000 中没有足够的 0,因此您的时间差了 10 倍。)