from itertools import count
# 生成從2開(kāi)始的偶數(shù)序列（2,4,6,...）
even_nums = count(start=2, step=2)
#使用next函數(shù)獲取元素
print(next(even_nums))
print(next(even_nums))
#for循環(huán)
for i in even_nums:
    print(i)
    if i == 10000000:
        break
# 取前5個(gè)元素
print(list(next(even_nums) for_ in range(5)))  # [2, 4, 6, 8, 10]

from itertools import cycle
status = cycle(['running', 'paused', 'stopped'])
# 模擬3次狀態(tài)切換
for _ in range(30):
    print(next(status))  #running,paused,stopped,running...

from itertools import repeat
# 生成5個(gè)0組成的列表
zeros = list(repeat(0, 5))  # [0, 0, 0, 0, 0]
# 與map結(jié)合：計(jì)算多個(gè)數(shù)的平方
squares = list(map(lambda x: x**2, repeat(3, 4)))  # [9, 9, 9, 9]

from itertools import chain
list1 = [1, 2, 'a']
list2 = ('a', 'b')
list3 = {True, False}
list4 = {'a':1, 'v':2}
# 合并為一個(gè)迭代器
merged = chain(list1,list2,list3,list4)
print(list(merged))  
#輸出為：
[1, 2, 'a', 'a', 'b', False, True, 'a', 'v']

from itertools import groupby
words = ['apple', 'banana', 'orange', 'avocado', 'berry']
# 務(wù)必先使用sorted排序 然后按首字母分組
for key, group in groupby(sorted(words), key=lambda x: x[0]):
    print(f"{key}: {list(group)}")
輸出：
a: ['apple', 'avocado']
b: ['banana', 'berry']
o: ['orange']
#如果不排序，直接使用groupby函數(shù)，會(huì)生成重復(fù)的key，與預(yù)期不符
for key, group in groupby(words, key=lambda x: x[0]):
    print(f"{key}: {list(group)}")
輸出：
a: ['apple']
b: ['banana']
o: ['orange']
a: ['avocado']
b: ['berry']

from itertools import zip_longest
names = ['Alice', 'Bob']
ages = [25, 30, 35]
# 填充N(xiāo)one
result = list(zip_longest(names, ages, fillvalue='-'))
print(result)  # [('Alice', 25), ('Bob', 30), ('-', 35)]

from itertools import permutations
# 生成'ABC'的2元素排列
perms = permutations('ABC', 2)
print(list(perms))  
#輸出： [('A', 'B'), ('A', 'C'), ('B', 'A'), ('B', 'C'), ('C', 'A'), ('C', 'B')]

from itertools import combinations
# 生成'ABC'的2元素組合
combs = combinations('ABC', 2)
print(list(combs))  # [('A', 'B'), ('A', 'C'), ('B', 'C')]

from itertools import product
colors = ['red', 'blue']
sizes = ['S', 'M', 'L']
# 生成所有可能的組合
products = product(colors, sizes)
print(list(products)) 
#輸出：
[('red', 'S'), ('red', 'M'), ('red', 'L'), 
 ('blue', 'S'), ('blue', 'M'), ('blue', 'L')]
 

fromitertoolsimportfilterfalse
numbers=[1,2,3,4,5,6]
# 保留奇數(shù)
odds=filterfalse(lambdax:x%2==0,numbers)
print(list(odds))
#輸出： [1, 3, 5]

from itertools import filterfalse
numbers = [1, 2, 3, 4, 5, 0,'',False,True]
# 保留奇數(shù)
odds = filterfalse(None,numbers)
print(list(odds))
#輸出：[0, '', False]

from itertools import islice, count # 生成從1開(kāi)始的無(wú)限序列 nums = count(1) # 取第3到第7個(gè)并且步長(zhǎng)為2的元素，相當(dāng)于取第3個(gè)，第5個(gè)，第7個(gè)元素（索引從0開(kāi)始） sliced = islice(nums, 2, 7, 2) print(list(sliced))   # 輸出：[3, 5, 7] 

fromitertoolsimporttakewhile
numbers=[1,3,5,2,4,6]
# 取小于5的數(shù)
result=takewhile(lambdax:x<5,numbers)
print(list(result))
#輸出： [1, 3]

from itertools import permutations, islice
import random
# 生成1-10的隨機(jī)排列
def random_permutation(n):
    numbers = list(range(1, n+1))
    # 打亂順序后生成排列
    random.shuffle(numbers)
    return list(islice(permutations(numbers), 1))[0]
print("隨機(jī)排列：", random_permutation(5)) 
#輸出 例如: (3, 1, 5, 2, 4)

from itertools import groupby
from collections import Counter
text = "apple banana apple orange banana apple"
words = text.split()
# 按單詞分組并統(tǒng)計(jì)數(shù)量
word_groups = groupby(sorted(words))
word_count = {word: len(list(group)) for word, group in word_groups}
print("單詞頻率：", word_count)  # {'apple': 3, 'banana': 2, 'orange': 1}
# 更簡(jiǎn)潔方式：結(jié)合Counter
print("Counter統(tǒng)計(jì)：", Counter(words))  
#輸出：
單詞頻率： {'apple': 3, 'banana': 2, 'orange': 1}
Counter統(tǒng)計(jì)： Counter({'apple': 3, 'banana': 2, 'orange': 1})