admin 管理员组文章数量: 887021
‘h’
**(2)choices(seq,weights=None, k)——对序列类型进行k次重复采样,可设置权重**
choices([‘win’, ‘lose’, ‘draw’], k=5)
[‘draw’, ‘lose’, ‘draw’, ‘draw’, ‘draw’]
choices([‘win’, ‘lose’, ‘draw’], [4,4,2], k=10)
[‘lose’, ‘draw’, ‘lose’, ‘win’, ‘draw’, ‘lose’, ‘draw’, ‘win’, ‘win’, ‘lose’]
中间的就是权重
**(3)shuffle(seq)——将序列类型中元素随机排列,返回打乱后的序列**
numbers = [“one”, “two”, “three”, “four”]
shuffle(numbers)
numbers
[‘four’, ‘one’, ‘three’, ‘two’]
**(4)sample(pop, k)——从pop类型中随机选取k个元素,以列表类型返回**
sample([10, 20, 30, 40, 50], k=3)
[20, 30, 10]
**5、概率分布——以高斯分布为例**
**gauss(mean, std)——生产一个符合高斯分布的随机数**
number = gauss(0, 1)
number
0.6331522345532208
**多生成几个**
import matplotlib.pyplot as plt
res = [gauss(0, 1) for i in range(100000)]
plt.hist(res, bins=1000)
plt.show()
![image-20220929194045846](https://img-blog.csdnimg/img_convert/753c0746373f205936e06eb51593a89d.png)
【例1】用random库实现简单的微信红包分配
import random
def red_packet(total, num):
for i in range(1, num):
per = random.uniform(0.01, total/(num-i+1)*2) # 保证每个人获得红包的期望是total/num
total = total - per
print(“第{}位红包金额: {:.2f}元”.format(i, per))
else:
print(“第{}位红包金额: {:.2f}元”.format(num, total))
red_packet(10, 5)
第1位红包金额: 1.85元
第2位红包金额: 3.90元
第3位红包金额: 0.41元
第4位红包金额: 3.30元
第5位红包金额: 0.54元
import random
import numpy as np
def red_packet(total, num):
ls = []
for i in range(1, num):
per = round(random.uniform(0.01, total/(num-i+1)*2), 2) # 保证每个人获得红包的期望是total/num
ls.append(per)
total = total - per
else:
ls.append(total)
return ls
重复发十万次红包,统计每个位置的平均值(约等于期望)
res = []
for i in range(100000):
ls = red_packet(10,5)
res.append(ls)
res = np.array(res)
print(res[:10])
np.mean(res, axis=0)
[[1.71 1.57 0.36 1.25 5.11]
[1.96 0.85 1.46 3.29 2.44]
[3.34 0.27 1.9 0.64 3.85]
[1.99 1.08 3.86 1.69 1.38]
[1.56 1.47 0.66 4.09 2.22]
[0.57 0.44 1.87 5.81 1.31]
[0.47 1.41 3.97 1.28 2.87]
[2.65 1.82 1.22 2.02 2.29]
[3.16 1.2 0.3 3.66 1.68]
[2.43 0.16 0.11 0.79 6.51]]
array([1.9991849, 2.0055725, 2.0018144, 2.0022472, 1.991181 ])
【例2】生产4位由数字和英文字母构成的验证码
import random
import string
print(string.digits)
print(string.ascii_letters)
s=string.digits + string.ascii_letters
v=random.sample(s,4)
print(v)
print(‘’.join(v))
0123456789
abcdefghijklmnopqrstuvwxyzABCDEFGHIJKLMNOPQRSTUVWXYZ
[‘n’, ‘Q’, ‘4’, ‘7’]
nQ47
### 9.3 collections库——容器数据类型
![image-20220929194627479](https://img-blog.csdnimg/img_convert/f7b7f67ed03ab1c398b348620d83eb54.png)
import collections
#### 9.3.1 namedtuple——具名元组
* 点的坐标,仅看数据,很难知道表达的是一个点的坐标
p = (1, 2)
* 构建一个新的元组子类
定义方法如下:typename 是元组名字,field\_names 是域名
collections.namedtuple(typename, field_names, *, rename=False, defaults=None, module=None)
Point = collections.namedtuple(“Point”, [“x”, “y”])
p = Point(1, y=2)
p
Point(x=1, y=2)
* 可以调用属性
print(p.x)
print(p.y)
1
2
* 有元组的性质
print(p[0])
print(p[1])
x, y = p
print(x)
print(y)
1
2
1
2
* 确实是元组的子类
print(isinstance(p, tuple))
True
【例】模拟扑克牌
Card = collections.namedtuple(“Card”, [“rank”, “suit”])
ranks = [str(n) for n in range(2, 11)] + list(“JQKA”)
suits = “spades diamonds clubs hearts”.split()
print(“ranks”, ranks)
print(“suits”, suits)
cards = [Card(rank, suit) for rank in ranks
for suit in suits]
cards
ranks [‘2’, ‘3’, ‘4’, ‘5’, ‘6’, ‘7’, ‘8’, ‘9’, ‘10’, ‘J’, ‘Q’, ‘K’, ‘A’]
suits [‘spades’, ‘diamonds’, ‘clubs’, ‘hearts’]
[Card(rank=‘2’, suit=‘spades’),
Card(rank=‘2’, suit=‘diamonds’),
Card(rank=‘2’, suit=‘clubs’),
Card(rank=‘2’, suit=‘hearts’),
Card(rank=‘3’, suit=‘spades’),
Card(rank=‘3’, suit=‘diamonds’),
Card(rank=‘3’, suit=‘clubs’),
Card(rank=‘3’, suit=‘hearts’),
Card(rank=‘4’, suit=‘spades’),
Card(rank=‘4’, suit=‘diamonds’),
Card(rank=‘4’, suit=‘clubs’),
Card(rank=‘4’, suit=‘hearts’),
Card(rank=‘5’, suit=‘spades’),
Card(rank=‘5’, suit=‘diamonds’),
Card(rank=‘5’, suit=‘clubs’),
Card(rank=‘5’, suit=‘hearts’),
Card(rank=‘6’, suit=‘spades’),
Card(rank=‘6’, suit=‘diamonds’),
Card(rank=‘6’, suit=‘clubs’),
Card(rank=‘6’, suit=‘hearts’),
Card(rank=‘7’, suit=‘spades’),
Card(rank=‘7’, suit=‘diamonds’),
Card(rank=‘7’, suit=‘clubs’),
Card(rank=‘7’, suit=‘hearts’),
Card(rank=‘8’, suit=‘spades’),
Card(rank=‘8’, suit=‘diamonds’),
Card(rank=‘8’, suit=‘clubs’),
Card(rank=‘8’, suit=‘hearts’),
Card(rank=‘9’, suit=‘spades’),
Card(rank=‘9’, suit=‘diamonds’),
Card(rank=‘9’, suit=‘clubs’),
Card(rank=‘9’, suit=‘hearts’),
Card(rank=‘10’, suit=‘spades’),
Card(rank=‘10’, suit=‘diamonds’),
Card(rank=‘10’, suit=‘clubs’),
Card(rank=‘10’, suit=‘hearts’),
Card(rank=‘J’, suit=‘spades’),
Card(rank=‘J’, suit=‘diamonds’),
Card(rank=‘J’, suit=‘clubs’),
Card(rank=‘J’, suit=‘hearts’),
Card(rank=‘Q’, suit=‘spades’),
Card(rank=‘Q’, suit=‘diamonds’),
Card(rank=‘Q’, suit=‘clubs’),
Card(rank=‘Q’, suit=‘hearts’),
Card(rank=‘K’, suit=‘spades’),
Card(rank=‘K’, suit=‘diamonds’),
Card(rank=‘K’, suit=‘clubs’),
Card(rank=‘K’, suit=‘hearts’),
Card(rank=‘A’, suit=‘spades’),
Card(rank=‘A’, suit=‘diamonds’),
Card(rank=‘A’, suit=‘clubs’),
Card(rank=‘A’, suit=‘hearts’)]
from random import *
洗牌
shuffle(cards)
cards
[Card(rank=‘J’, suit=‘hearts’),
Card(rank=‘A’, suit=‘hearts’),
Card(rank=‘3’, suit=‘hearts’),
Card(rank=‘8’, suit=‘hearts’),
Card(rank=‘K’, suit=‘hearts’),
Card(rank=‘7’, suit=‘spades’),
Card(rank=‘5’, suit=‘hearts’),
Card(rank=‘A’, suit=‘spades’),
Card(rank=‘10’, suit=‘spades’),
Card(rank=‘J’, suit=‘diamonds’),
Card(rank=‘K’, suit=‘clubs’),
Card(rank=‘4’, suit=‘spades’),
Card(rank=‘2’, suit=‘diamonds’),
Card(rank=‘Q’, suit=‘spades’),
Card(rank=‘A’, suit=‘clubs’),
Card(rank=‘A’, suit=‘diamonds’),
Card(rank=‘6’, suit=‘hearts’),
Card(rank=‘7’, suit=‘diamonds’),
Card(rank=‘5’, suit=‘diamonds’),
Card(rank=‘10’, suit=‘clubs’),
Card(rank=‘8’, suit=‘clubs’),
Card(rank=‘9’, suit=‘clubs’),
Card(rank=‘6’, suit=‘clubs’),
Card(rank=‘6’, suit=‘diamonds’),
Card(rank=‘5’, suit=‘clubs’),
Card(rank=‘3’, suit=‘diamonds’),
Card(rank=‘4’, suit=‘hearts’),
Card(rank=‘3’, suit=‘clubs’),
Card(rank=‘7’, suit=‘hearts’),
Card(rank=‘2’, suit=‘spades’),
Card(rank=‘J’, suit=‘clubs’),
Card(rank=‘9’, suit=‘spades’),
Card(rank=‘J’, suit=‘spades’),
Card(rank=‘10’, suit=‘hearts’),
Card(rank=‘2’, suit=‘clubs’),
Card(rank=‘8’, suit=‘diamonds’),
Card(rank=‘6’, suit=‘spades’),
Card(rank=‘10’, suit=‘diamonds’),
Card(rank=‘9’, suit=‘hearts’),
Card(rank=‘3’, suit=‘spades’),
Card(rank=‘8’, suit=‘spades’),
Card(rank=‘Q’, suit=‘clubs’),
Card(rank=‘Q’, suit=‘hearts’),
Card(rank=‘5’, suit=‘spades’),
Card(rank=‘7’, suit=‘clubs’),
Card(rank=‘4’, suit=‘clubs’),
Card(rank=‘2’, suit=‘hearts’),
Card(rank=‘K’, suit=‘diamonds’),
Card(rank=‘K’, suit=‘spades’),
Card(rank=‘Q’, suit=‘diamonds’),
Card(rank=‘4’, suit=‘diamonds’),
Card(rank=‘9’, suit=‘diamonds’)]
随机抽一张牌
choice(cards)
Card(rank=‘4’, suit=‘hearts’)
随机抽多张牌
sample(cards, k=5)
[Card(rank=‘4’, suit=‘hearts’),
Card(rank=‘2’, suit=‘clubs’),
Card(rank=‘Q’, suit=‘diamonds’),
Card(rank=‘9’, suit=‘spades’),
Card(rank=‘10’, suit=‘hearts’)]
#### 9.3.2 Counter——计数器工具
from collections import Counter
s = “牛奶奶找刘奶奶买牛奶”
colors = [‘red’, ‘blue’, ‘red’, ‘green’, ‘blue’, ‘blue’]
cnt_str = Counter(s)
cnt_color = Counter(colors)
print(cnt_str)
print(cnt_color)
Counter({‘奶’: 5, ‘牛’: 2, ‘找’: 1, ‘刘’: 1, ‘买’: 1})
Counter({‘blue’: 3, ‘red’: 2, ‘green’: 1})
* 是字典的一个子类
print(isinstance(Counter(), dict))
True
* 最常见的统计——most\_commom(n)
提供 n 个频率最高的元素和计数
cnt_color.most_common(2)
[(‘blue’, 3), (‘red’, 2)]
* 元素展开——elements()
list(cnt_str.elements())
[‘牛’, ‘牛’, ‘奶’, ‘奶’, ‘奶’, ‘奶’, ‘奶’, ‘找’, ‘刘’, ‘买’]
* 其他一些加减操作
c = Counter(a=3, b=1)
d = Counter(a=1, b=2)
c+d
Counter({‘a’: 4, ‘b’: 3})
【例】从一副牌中抽取10张,大于10的比例有多少
cards = collections.Counter(tens=16, low_cards=36)
seen = sample(list(cards.elements()), k=10)
print(seen)
[‘tens’, ‘low_cards’, ‘low_cards’, ‘low_cards’, ‘tens’, ‘tens’, ‘low_cards’, ‘low_cards’, ‘low_cards’, ‘low_cards’]
seen.count(‘tens’) / 10
0.3
#### 9.3.3 deque——双向队列
列表访问数据非常快速
插入和删除操作非常慢——通过移动元素位置来实现
特别是 insert(0, v) 和 pop(0),在列表开始进行的插入和删除操作
**双向队列可以方便的在队列两边高效、快速的增加和删除元素**
from collections import deque
d = deque(‘cde’)
d
deque([‘c’, ‘d’, ‘e’])
d.append(“f”) # 右端增加
d.append(“g”)
d.appendleft(“b”) # 左端增加
d.appendleft(“a”)
d
deque([‘a’, ‘b’, ‘c’, ‘d’, ‘e’, ‘f’, ‘g’])
d.pop() # 右端删除
d.popleft() # 左端删除
d
deque([‘b’, ‘c’, ‘d’, ‘e’, ‘f’])
deque 其他用法可参考官方文档
### 9.4 itertools库——迭代器
![image-20220929194634540](https://img-blog.csdnimg/img_convert/3f777b75d1b6c3f2882904078d6e46f8.png)
#### 9.4.1 排列组合迭代器
(1)product——笛卡尔积
import itertools
for i in itertools.product(‘ABC’, ‘01’):
print(i)
(‘A’, ‘0’)
(‘A’, ‘1’)
(‘B’, ‘0’)
(‘B’, ‘1’)
(‘C’, ‘0’)
(‘C’, ‘1’)
for i in itertools.product(‘ABC’, repeat=3): # 相当于3组ABC的笛卡尔积
print(i)
(‘A’, ‘A’, ‘A’)
(‘A’, ‘A’, ‘B’)
(‘A’, ‘A’, ‘C’)
(‘A’, ‘B’, ‘A’)
(‘A’, ‘B’, ‘B’)
(‘A’, ‘B’, ‘C’)
(‘A’, ‘C’, ‘A’)
(‘A’, ‘C’, ‘B’)
(‘A’, ‘C’, ‘C’)
(‘B’, ‘A’, ‘A’)
(‘B’, ‘A’, ‘B’)
(‘B’, ‘A’, ‘C’)
(‘B’, ‘B’, ‘A’)
(‘B’, ‘B’, ‘B’)
(‘B’, ‘B’, ‘C’)
(‘B’, ‘C’, ‘A’)
(‘B’, ‘C’, ‘B’)
(‘B’, ‘C’, ‘C’)
(‘C’, ‘A’, ‘A’)
(‘C’, ‘A’, ‘B’)
(‘C’, ‘A’, ‘C’)
(‘C’, ‘B’, ‘A’)
(‘C’, ‘B’, ‘B’)
(‘C’, ‘B’, ‘C’)
(‘C’, ‘C’, ‘A’)
(‘C’, ‘C’, ‘B’)
(‘C’, ‘C’, ‘C’)
(2) permutations——排列
for i in itertools.permutations(‘ABCD’, 3): # 3 是排列的长度
print(i)
(‘A’, ‘B’, ‘C’)
(‘A’, ‘B’, ‘D’)
(‘A’, ‘C’, ‘B’)
(‘A’, ‘C’, ‘D’)
(‘A’, ‘D’, ‘B’)
(‘A’, ‘D’, ‘C’)
(‘B’, ‘A’, ‘C’)
(‘B’, ‘A’, ‘D’)
(‘B’, ‘C’, ‘A’)
(‘B’, ‘C’, ‘D’)
(‘B’, ‘D’, ‘A’)
(‘B’, ‘D’, ‘C’)
(‘C’, ‘A’, ‘B’)
(‘C’, ‘A’, ‘D’)
(‘C’, ‘B’, ‘A’)
(‘C’, ‘B’, ‘D’)
(‘C’, ‘D’, ‘A’)
(‘C’, ‘D’, ‘B’)
(‘D’, ‘A’, ‘B’)
(‘D’, ‘A’, ‘C’)
(‘D’, ‘B’, ‘A’)
(‘D’, ‘B’, ‘C’)
(‘D’, ‘C’, ‘A’)
(‘D’, ‘C’, ‘B’)
for i in itertools.permutations(range(3)):
print(i)
(0, 1, 2)
(0, 2, 1)
(1, 0, 2)
(1, 2, 0)
(2, 0, 1)
(2, 1, 0)
(3)combinations——组合 其结果元素不能重复
for i in itertoolsbinations(‘ABCD’, 2): # 2是组合的长度
print(i)
(‘A’, ‘B’)
(‘A’, ‘C’)
(‘A’, ‘D’)
(‘B’, ‘C’)
(‘B’, ‘D’)
(‘C’, ‘D’)
for i in itertoolsbinations(range(4), 3):
print(i)
(0, 1, 2)
(0, 1, 3)
(0, 2, 3)
(1, 2, 3)
(4)combinations\_with\_replacement——元素可重复组合
for i in itertoolsbinations_with_replacement(‘ABC’, 2): # 2是组合的长度
print(i)
(‘A’, ‘A’)
(‘A’, ‘B’)
(‘A’, ‘C’)
(‘B’, ‘B’)
(‘B’, ‘C’)
(‘C’, ‘C’)
for i in itertools.product(‘ABC’,repeat=2):
print(i)
(‘A’, ‘A’)
(‘A’, ‘B’)
(‘A’, ‘C’)
(‘B’, ‘A’)
(‘B’, ‘B’)
(‘B’, ‘C’)
(‘C’, ‘A’)
(‘C’, ‘B’)
(‘C’, ‘C’)
#### 9.4.2 拉链
(1)zip——短拉链
把相同位置上的元素组合在一起
for i in zip(“ABC”, “012”, “xyz”):
print(i)
(‘A’, ‘0’, ‘x’)
(‘B’, ‘1’, ‘y’)
(‘C’, ‘2’, ‘z’)
长度不一时,执行到最短的对象处,就停止
for i in zip(“ABC”, [0, 1, 2, 3, 4, 5]): # 注意zip是内置的,不需要加itertools
print(i)
(‘A’, 0)
(‘B’, 1)
(‘C’, 2)
文末有福利领取哦~
👉一、Python所有方向的学习路线
Python所有方向的技术点做的整理,形成各个领域的知识点汇总,它的用处就在于,你可以按照上面的知识点去找对应的学习资源,保证自己学得较为全面。
👉二、Python必备开发工具
👉三、Python视频合集
观看零基础学习视频,看视频学习是最快捷也是最有效果的方式,跟着视频中老师的思路,从基础到深入,还是很容易入门的。
👉 四、实战案例
光学理论是没用的,要学会跟着一起敲,要动手实操,才能将自己的所学运用到实际当中去,这时候可以搞点实战案例来学习。(文末领读者福利)
👉五、Python练习题
检查学习结果。
👉六、面试资料
我们学习Python必然是为了找到高薪的工作,下面这些面试题是来自阿里、腾讯、字节等一线互联网大厂最新的面试资料,并且有阿里大佬给出了权威的解答,刷完这一套面试资料相信大家都能找到满意的工作。
👉因篇幅有限,仅展示部分资料,这份完整版的Python全套学习资料已经上传
网上学习资料一大堆,但如果学到的知识不成体系,遇到问题时只是浅尝辄止,不再深入研究,那么很难做到真正的技术提升。
需要这份系统化学习资料的朋友,可以戳这里无偿获取
一个人可以走的很快,但一群人才能走的更远!不论你是正从事IT行业的老鸟或是对IT行业感兴趣的新人,都欢迎加入我们的的圈子(技术交流、学习资源、职场吐槽、大厂内推、面试辅导),让我们一起学习成长!
版权声明:本文标题:Python基础(九) time random collections itertools标准库详解(1) 内容由网友自发贡献,该文观点仅代表作者本人, 转载请联系作者并注明出处:http://www.freenas.com.cn/jishu/1726377424h948208.html, 本站仅提供信息存储空间服务,不拥有所有权,不承担相关法律责任。如发现本站有涉嫌抄袭侵权/违法违规的内容,一经查实,本站将立刻删除。
发表评论