itertools 模块和 collections 模块的使用

• itertools 模块的使用

• 这两个模块极其重要，极其重要，在科学运算，代数运算中的作用极大
• 这两个模块是非计科专业对抗计科专业最重要的武器之一
  • 甚至就是最重要的那个（在内存足够大的情况下）

itertools模块

count(start, step)

• 生成从start开始的无限迭代器，步长为step

from itertools import count

for num in count(3,-2):
    print(num)

cycle(iterable)

• 对一个可迭代对象进行无限循环

from itertools import cycle

ITER1 = [1,2,3,4,5]
ITER2 = (1,2,3,4,5)

cycle1 = cycle(ITER1)
cycle2 = cycle(ITER2)

for _ in range(5):
    print(next(cycle1))
    print(next(cycle2))

repeat(element, times) 

• 将element重复生成times次

>>> repeat(10, 3)
repeat(10, 3)
>>> repeat(10, 3).__str__

>>> print(repeat(10, 3))
repeat(10, 3)
>>> type(repeat(10, 3))

 

from itertools import repeat

for num in repeat(10, 3):
    print(num)

 

•  然而我们比较一下

from itertools import repeat
from itertools import count
from itertools import cycle

print(type(repeat(10, 3)))
print(type(count(10, 3)))
print(type(cycle([10, ])))

chain(*iterables)

• 将多个可迭代对象连接起来，返回一个迭代器。
  • 可迭代对象
    • 列表
    • 字符串
    • 元组
    • 字典

from itertools import chain

iter1 = [1, 2, 3]
iter2 = "string"
iter3 = (111,222,333)
iter4 = {"1111":1111,"2222":2222,"3333":3333}

combined = chain(iter1,iter2,iter3,iter4)

for item in combined:
    print(item)

compress(data, selectors)

• 根据selectors的真假值对data进行筛选，返回一个迭代器
  • 有用，但不多，因为可以用mao替代

from itertools import compress

data = [1, 2, 3, 4, 5, 6, 7, 8, 9, 0]
selectors = [True, True, True, False, True, True, False, True, False, True]

filtered = compress(data, selectors)

for item in filtered:
    print(item)

dropwhile(predicate, iterable)

• 从iterable中删掉predicate为真的元素，返回剩余的元素组成的迭代器 

from itertools import dropwhile

data = [1, 3, 5, 7, 9, 2, 4, 6, 8, 10]

def fun(x):
    return (x+1)**x < 300

droped = dropwhile(fun, data)

for iter_term in droped:
    print(iter_term)

takewhile(predicate, iterable)

• 从iterable中取出predicate为真的元素，直到遇到第一个为假的元素为止，组成一个迭代器

from itertools import takewhile

data = [1, 1.3, 1.5, 7, 2.9, 1.2, 1.4, 3.6, 1.8, 1.0]

def fun(x):
    return (x+1)**x < 300

taked = takewhile(fun, data)

for iter_term in taked:
    print(iter_term)

•  那么我们抛出一个问题，怎么样才能取出全部为真的元素呢？
  • dropwhile取反即可

permutations(iterable, r)

• 返回iterable中所有长度为r的排列（元组）组成的列表

from itertools import permutations

data = [1,2,3,4,5]
result = permutations(data,2)


for iter_term in result:
    print(iter_term)

combinations(iterable, r)

• 返回iterable中所有长度为r的组合（元组）组成的列表

from itertools import combinations

data = [1,2,3,4,5]
result = combinations(data,2)


for iter_term in result:
    print(iter_term)

product(*iterables, repeat)

• 返回iterable中所有长度为r的笛卡尔积（元组）组成的列表

from itertools import product

numbers = [1, 2, 3]
capital = ['A', 'B', 'C']
lowercase = ['a', 'b', 'c']

result = product(numbers, capital, lowercase)

for iter_term in result:
    print(iter_term)

• 只传入单个iterable，返回所有长度为r的可重复的排列

>>> k = itertools.product
>>> lk = k(["1","2","3","4"],3)
Traceback (most recent call last):
  File "", line 1, in 
    lk = k(["1","2","3","4"],3)
TypeError: 'int' object is not iterable
>>> lk = k(["1","2","3","4"],repeat=3)
>>> lk

>>> list(lk)
[('1', '1', '1'), ('1', '1', '2'), ('1', '1', '3'), ('1', '1', '4'), ('1', '2', '1'), ('1', '2', '2'), ('1', '2', '3'), ('1', '2', '4'), ('1', '3', '1'), ('1', '3', '2'), ('1', '3', '3'), ('1', '3', '4'), ('1', '4', '1'), ('1', '4', '2'), ('1', '4', '3'), ('1', '4', '4'), ('2', '1', '1'), ('2', '1', '2'), ('2', '1', '3'), ('2', '1', '4'), ('2', '2', '1'), ('2', '2', '2'), ('2', '2', '3'), ('2', '2', '4'), ('2', '3', '1'), ('2', '3', '2'), ('2', '3', '3'), ('2', '3', '4'), ('2', '4', '1'), ('2', '4', '2'), ('2', '4', '3'), ('2', '4', '4'), ('3', '1', '1'), ('3', '1', '2'), ('3', '1', '3'), ('3', '1', '4'), ('3', '2', '1'), ('3', '2', '2'), ('3', '2', '3'), ('3', '2', '4'), ('3', '3', '1'), ('3', '3', '2'), ('3', '3', '3'), ('3', '3', '4'), ('3', '4', '1'), ('3', '4', '2'), ('3', '4', '3'), ('3', '4', '4'), ('4', '1', '1'), ('4', '1', '2'), ('4', '1', '3'), ('4', '1', '4'), ('4', '2', '1'), ('4', '2', '2'), ('4', '2', '3'), ('4', '2', '4'), ('4', '3', '1'), ('4', '3', '2'), ('4', '3', '3'), ('4', '3', '4'), ('4', '4', '1'), ('4', '4', '2'), ('4', '4', '3'), ('4', '4', '4')]
>>> 

combinations_with_replacement

>>> f = itertools.combinations_with_replacement
>>> f(["1","2","3","4"],3)

>>> lf = f(["1","2","3","4"],3)
>>> list(lf)
[('1', '1', '1'), ('1', '1', '2'), ('1', '1', '3'), ('1', '1', '4'), ('1', '2', '2'), ('1', '2', '3'), ('1', '2', '4'), ('1', '3', '3'), ('1', '3', '4'), ('1', '4', '4'), ('2', '2', '2'), ('2', '2', '3'), ('2', '2', '4'), ('2', '3', '3'), ('2', '3', '4'), ('2', '4', '4'), ('3', '3', '3'), ('3', '3', '4'), ('3', '4', '4'), ('4', '4', '4')]

collections 模块

Counter 类

Counter 的__str__ 方法

>>> counter1
Counter({1: 1, 2: 1, 3: 1, 4: 1, 5: 1, 6: 1})
>>> counter2
Counter({6: 3, 5: 2, 2: 1, 7: 1, 8: 1})
 

Counter 的交并差运算

from collections import Counter

set1  = set([1, 2, 3, 4, 4, 5, 6, 6])
list2 =     [2, 5, 5, 6, 6, 6, 7, 8]

counter1 = Counter(set1)
counter2 = Counter(list2)

intersection = counter1 & counter2
print("Intersection:", intersection.keys())

union = counter1 | counter2
print("Union:", union.keys())

difference = counter1 - counter2
print("Difference1:", difference.keys())

difference = counter2 - counter1
print("Difference2:", difference.keys())

most_common()

• 按照元素出现的次数进行排序

>>> counter2.most_common

>>> counter2.most_common[0]
Traceback (most recent call last):
  File "", line 1, in
    counter2.most_common[0]
TypeError: 'method' object is not subscriptable
>>> 
 

subtract()

• 在这里我们顺便再看另一种counter的定义方式
  • 这是直接减的
  • 区别于Counter 的直接 - 运算

from collections import Counter

counter1 = Counter(a=3, b=2, c=1)
counter2 = Counter(a=1, b=2, c=3)

counter1.subtract(counter2)
print(counter1)
print(counter1 - counter2)

Counter({'a': 2, 'b': 0, 'c': -2})
Counter({'a': 1})
 

• 添加一串代码

from collections import Counter

counter1 = Counter(a=3, b=2, c=1)
counter3 = Counter(a=1, d=2, c=3)

counter1.subtract(counter3)
print(counter1)
print(counter1 - counter3)

• "可哈希"（hashable）是指一个对象可以作为字典的键或集合的元素