Python3【时空复杂度】测试

空间复杂度

• sys.getsizeof

from sys import getsizeof
from numpy import array
from pandas import DataFrame
r = range(10)
sg = ''.join(str(i) for i in r)
ls = list(r)
st = set(r)
dt = {i: i for i in r}
t = tuple(r)
a = array(ls)
df = DataFrame(ls)
class C:pass
def f():pass
for i in ['r', 'sg', 'ls', 'st', 'dt', 't', 'a', 'df', 'C', 'f']:
    size = getsizeof(eval(i))
    print('\n%2s' % i, '\033[0;7m \033[0m' * (size // 14), size)
    # print(eval(i))

集合空间占用大于字典
元组空间占用小于列表

时间复杂度

Timer

from time import time
class T:
    def __init__(self):
        self.t = time()
    def __del__(self):
        t = time() - self.t
        print('\n', '\033[0;7m \033[0m' * int(t * 20), t)

for

from numpy import array
r = range(500000)
sg = ''.join(str(i) for i in r)
ls = list(r)
dt = {i: i for i in r}
st = {i for i in r}
tu = tuple(r)
a = array(r)

for j in ['r', 'sg', 'ls', 'dt', 'st', 'tu', 'a', 'a.tolist()']:
    t = T()
    for i in eval(j):pass
del t

结果	时间
生成器	0.09400558471679688
字符串	0.38802218437194824
列表	0.06100344657897949
字典	0.07000398635864258
集合	0.06800389289855957
元组	0.06100344657897949
numpy.array	0.1370079517364502
numpy.array.tolist()	0.128007173538208

元组≈列表>字典≈集合>生成器>>numpy.array.tolist()>numpy.array>>>字符串

in

r = range(10000)
sg = ''.join(str(i) for i in r)
ls = [str(i) for i in r]
dt = {str(i): i for i in r}
st = {str(i) for i in r}
t1 = tuple(r)
t2 = tuple(ls)

t = T()
for i in ls:
    if i in sg:pass  # 字符串
t = T()
for i in ls:
    if i in ls:pass  # 列表
t = T()
for i in ls:
    if i in dt:pass  # 字典
t = T()
for i in ls:
    if i in st:pass  # 集合
t = T()
for i in ls:
    if i in t1:pass  # 元组（int）
t = T()
for i in ls:
    if i in t2:pass  # 元组
del t

结果	时间
字符串	0.2870168685913086
列表	3.38619327545166
字典	0.005000591278076172
集合	0.004000186920166016
元组1	7.896451711654663
元组2	3.6772103309631348

集合>字典>>字符串>>列表>元组

列表、字典推导式

[]

r = range(100000)

t = T()
ls1 = [i+1 for i in r]
ls2 = [i+2 for i in r]

t = T()
ls1, ls2 = [], []
for i in r:
    ls1.append(i+1)
    ls2.append(i+2)

del t

列表推导式更快

{}

t = T()
dt = {i: i+1 for i in r}
t = T()
for i in r:
    dt[i] = i + 1
del t

0.012964963912963867
0.010970592498779297
字典推导式反而更慢

文件

from sys import getsizeof
from time import time
import os, pickle, json


class T:
    def __init__(self):
        self.t = time()

    def __del__(self):
        t = time() - self.t
        print('\n', '\033[0;7m \033[0m' * int(t * 40), t)


b = {i: i for i in range(300000)}

txt = 'a.txt'
py = 'a.py'
js = 'a.json'
pkl = 'a.pickle'

with open(txt, 'w', encoding='utf-8') as f:
    f.write(repr(b))
with open(py, 'w', encoding='utf-8') as f:
    f.write('b=' + repr(b))
with open(js, 'w', encoding='utf-8') as f:
    json.dump(b, f)
with open(pkl, 'wb') as f:
    pickle.dump(b, f)

print(getsizeof(b))
for f in [txt, py, js, pkl]:
    print(os.path.getsize(f))

t = T()
with open(txt, encoding='utf-8') as f:
    b = f.read()
t = T()
from a import b
t = T()
with open(js, encoding='utf-8') as f:
    b = json.load(f)
t = T()
with open(pkl, 'rb') as f:
    b = pickle.load(f)
del t

pickle占用空间最小，读取速度较快，力荐~~
from XX import xx 速度较慢

比较符：>!=<

几乎没什么差别

from time import time
n = 9999999
t0 = time()
[i >= 1000001 for i in range(n)]
t1 = time()
[i > 1000000 for i in range(n)]
t2 = time()
[i == 1000000 for i in range(n)]
t3 = time()
[i != 1000000 for i in range(n)]
t4 = time()
print(t1 - t0, t2 - t1, t3 - t2, t4 - t3)

0.7499642372131348 0.7569770812988281 0.7450189590454102 0.7629609107971191