结论:L.append()方法在追加元素的多次测试中表现速度均更快。
思考:L.append(x)后面的参数直接是给元素x,少了生成新列表[x]的过程
运行结果:
代码:
import datetime
import matplotlib.pyplot as mp
def get_time(fn):
def f(*args):
t0 = datetime.datetime.now()
print('\nstart', t0)
fn(*args)
t1 = datetime.datetime.now()
print(' end', t1, '\n运行耗时', t1-t0)
Ltime.append((t1-t0).seconds)
return f
@get_time
def f1(n):
L = []
for x in range(n):
L += [n]
@get_time
def f2(n):
L = []
for x in range(n):
L.extend([n])
@get_time
def f3(n):
L = []
for x in range(n):
L.append(n)
X, y1, y2, y3 = [], [], [], []
x = 10000000
for x in range(x, 20*x, 4*x):
Ltime = []
f1(x)
f2(x)
f3(x)
X.append(x)
y1.append(Ltime[0])
y2.append(Ltime[1])
y3.append(Ltime[2])
mp.plot(X, y1, linestyle='-', label="+=", color='red')
mp.plot(X, y2, linestyle='-', label="L.extend", color='blue')
mp.plot(X, y3, linestyle='-', label="L.append", color='orange')
for y, colors in zip([y1, y2, y3], ['red', 'blue', 'orange']):
mp.scatter(X, y,
marker='o', # 点型 ~ matplotlib.markers
s=60, # 大小
edgecolor=colors, # 边缘色
facecolor='white', # 填充色
zorder=3 # 绘制图层编号 (编号越大,图层越靠上)
)
mp.legend()
mp.show()结论:对于需要涉及大量in查找操作的一批数据,最好使用字典或集合。
运行结果:
import datetime
import matplotlib.pyplot as mp
import numpy as np
import hashlib
def hex_sha1(STR):
s1 = hashlib.sha1()
s1.update(STR.encode('utf8'))
return s1.hexdigest().upper()
# print(hex_sha1('E5AC7D06E7C9A0F5'))
# E18937A9639FE98F8A24348ED0BBE3061A5465A8
def f(xx, iters):
t0 = datetime.datetime.now()
print('查找开始', t0)
print('类型', type(iters), xx, len(iters))
print(xx in iters)
t1 = datetime.datetime.now()
print('查找结束', t1)
print('耗时', (t1-t0).seconds, (t1-t0).microseconds)
return (t1-t0).microseconds
sed = 10**7 # 储存md5的数
# 1) 先准备数据
print(datetime.datetime.now())
L_all = np.array([hex_sha1(str(x))*2 for x in range(sed+1)])
print(datetime.datetime.now())
X, y = [], [] # X数量, y是各个函数时间[[y1,y2,y3],[...],...]
for x in list(range(0, sed+1, int(sed/10))):
X.append(x)
Ly = [] # y1,y2,y3运行时间
print()
for iters in ('L_all[:x+1]',
'list(L_all[:x+1])',
'dict.fromkeys(L_all[:x+1], None)',
'set(L_all[:x+1])'):
iters = eval(iters)
yn = f(hex_sha1(str(x))*2, iters)
Ly.append(yn)
y.append(Ly)
y1, y2, y3, y4 = (np.array(y)/1000).T
mp.plot(X, y1, linestyle='-', label="array", color='green')
mp.plot(X, y2, linestyle='-', label="list", color='red')
mp.plot(X, y3, linestyle='-', label="dict", color='blue')
mp.plot(X, y4, linestyle='-', label="set", color='orange')
mp.title('speed test', fontsize=20)
mp.xlabel('Num', fontsize=12)
mp.ylabel('Time(ms)', fontsize=12)
for y, colors in zip([y1, y2, y3, y4], ['green', 'red', 'blue', 'orange']):
mp.scatter(X, y,
marker='o', # 点型 ~ matplotlib.markers
s=60, # 大小
edgecolor=colors, # 边缘色
facecolor='white', # 填充色
zorder=3 # 绘制图层编号 (编号越大,图层越靠上)
)
mp.legend()
mp.show()上一个实例的对比不是很明显,主要是用单个元素在一个大空间内查找,太浪费资源了。
在这基础上,思考了一下,为何不用事先定义好的每一个元素就地查找呢。
改进版:
结论:集合和字典查找最快,其次为np.array,最慢的是list和tuple
运行结果:

代码:
# 对比某个元素在"列表/元组/字典/集合"的查找速度
# 结论:对于x in X的操作,集合字典在查找速度上有明显优势
import datetime
import numpy as np
import matplotlib.pyplot as mp
from collections import OrderedDict
def f(xx):
t0 = datetime.datetime.now()
for i in L:
if i in xx:
continue
t1 = datetime.datetime.now()
print('-->', type(xx), '遍历结束,耗时%s秒' % ((t1 - t0).total_seconds()))
return (t1 - t0).total_seconds()
X, y = [], []
n = 5*10**4
L_all = np.arange(0, n)
for x in range(0, n+1, 10000):
print('\nx', x)
L = L_all[:x+1]
L_result = []
L_result.append(f(L))
L_result.append(f(list(L)))
L_result.append(f(tuple(L)))
L_result.append(f(set(L)))
L_result.append(f(dict(zip(L, [None] * len(L)))))
L_result.append(f(OrderedDict(zip(L, [None] * len(L)))))
X.append(x)
y.append(L_result)
mp.title('speed test', fontsize=20)
mp.xlabel('Num', fontsize=12)
mp.ylabel('Time(s)', fontsize=12)
for y, lable, colors in zip(
(np.array(y)).T,
['np.array', 'list', 'tuple', 'set', 'dict', 'Orderdict'],
['green', 'red', 'blue', 'orange', 'yellow', 'pink']):
mp.plot(X, y, linestyle='-', label=lable, color=colors)
mp.scatter(X, y,
marker='o', # 点型 ~ matplotlib.markers
s=60, # 大小
edgecolor=colors, # 边缘色
facecolor='white', # 填充色
zorder=3 # 绘制图层编号 (编号越大,图层越靠上)
)
mp.legend()
mp.show()
