Python基础（九） time random collections itertools标准库详解(1)

choice("python")

'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()

【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

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)

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

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 itertools.combinations('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 itertools.combinations(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 itertools.combinations_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)

（2）zip_longest——长拉链

长度不一时，执行到最长的对象处，就停止，缺省元素用None或指定字符替代

for i in itertools.zip_longest("ABC", "012345"):
    print(i)

('A', '0')
('B', '1')
('C', '2')
(None, '3')
(None, '4')
(None, '5')

for i in itertools.zip_longest("ABC", "012345", fillvalue = "?"):
    print(i)

('A', '0')
('B', '1')
('C', '2')
('?', '3')
('?', '4')
('?', '5')

9.4.3 无穷迭代器

（1）count(start=0, step=1)——计数

创建一个迭代器，它从 start 值开始，返回均匀间隔的值

itertools.count(10)
10
11
12
.
.
.

（2）cycle(iterable)——循环

创建一个迭代器，返回 iterable 中所有元素，无限重复

itertools.cycle("ABC")
A
B
C
A
B
C
.
.
.

（3）repeat(object [, times])——重复

创建一个迭代器，不断重复 object 。除非设定参数 times ，否则将无限重复

for i in itertools.repeat(10, 3):
    print(i)

10
10
10

9.4.4 其他

（1）chain(iterables)——锁链

把一组迭代对象串联起来，形成一个更大的迭代器

for i in itertools.chain('ABC', [1, 2, 3]):
    print(i)

A
B
C
1
2
3

（2）enumerate(iterable, start=0)——枚举（Python内置）

产出由两个元素组成的元组，结构是（index, item）,其中index 从start开始，item从iterable中取


![img](https://p6-xtjj-sign.byteimg.com/tos-cn-i-73owjymdk6/f86e8ad0e3964a34b78ad8670c0bf7d9~tplv-73owjymdk6-jj-mark-v1:0:0:0:0:5o6Y6YeR5oqA5pyv56S-5Yy6IEAg5py65Zmo5a2m5Lmg5LmL5b-DQUk=:q75.awebp?rk3s=f64ab15b&x-expires=1771251770&x-signature=NKILcIyP%2FqLdzWawTftiDH94vAs%3D)
![img](https://p6-xtjj-sign.byteimg.com/tos-cn-i-73owjymdk6/2100fdf4ec4e49daa00602d90389bea0~tplv-73owjymdk6-jj-mark-v1:0:0:0:0:5o6Y6YeR5oqA5pyv56S-5Yy6IEAg5py65Zmo5a2m5Lmg5LmL5b-DQUk=:q75.awebp?rk3s=f64ab15b&x-expires=1771251770&x-signature=6zdhtpxU%2FhL0RO0zq9GEs2bMOMw%3D)

**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。**

**[需要这份系统化的资料的朋友，可以添加戳这里获取](https://gitee.com/vip204888)**


**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！**