Python使用itertools模块处理各类迭代对象

from itertools import count
# 生成从2开始的偶数序列（2,4,6,...）
even_nums = count(start=2, step=2)
#使用next函数获取元素
print(next(even_nums))
print(next(even_nums))
#for循环
for i in even_nums:
    print(i)
    if i == 10000000:
        break
# 取前5个元素
print(list(next(even_nums) for_ in range(5)))  # [2, 4, 6, 8, 10]

from itertools import cycle
status = cycle(['running', 'paused', 'stopped'])
# 模拟3次状态切换
for _ in range(30):
    print(next(status))  #running,paused,stopped,running...

from itertools import repeat
# 生成5个0组成的列表
zeros = list(repeat(0, 5))  # [0, 0, 0, 0, 0]
# 与map结合：计算多个数的平方
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}
# 合并为一个迭代器
merged = chain(list1,list2,list3,list4)
print(list(merged))  
#输出为：
[1, 2, 'a', 'a', 'b', False, True, 'a', 'v']

from itertools import groupby
words = ['apple', 'banana', 'orange', 'avocado', 'berry']
# 务必先使用sorted排序 然后按首字母分组
for key, group in groupby(sorted(words), key=lambda x: x[0]):
    print(f"{key}: {list(group)}")
输出：
a: ['apple', 'avocado']
b: ['banana', 'berry']
o: ['orange']
#如果不排序，直接使用groupby函数，会生成重复的key，与预期不符
for key, group in groupby(words, key=lambda x: x[0]):
    print(f"{key}: {list(group)}")
输出：
a: ['apple']
b: ['banana']
o: ['orange']
a: ['avocado']
b: ['berry']

from itertools import zip_longest
names = ['Alice', 'Bob']
ages = [25, 30, 35]
# 填充None
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))  
#输出： [('A', 'B'), ('A', 'C'), ('B', 'A'), ('B', 'C'), ('C', 'A'), ('C', 'B')]

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

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

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

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

from itertools import islice, count # 生成从1开始的无限序列 nums = count(1) # 取第3到第7个并且步长为2的元素，相当于取第3个，第5个，第7个元素（索引从0开始） sliced = islice(nums, 2, 7, 2) print(list(sliced))   # 输出：[3, 5, 7] 

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

from itertools import permutations, islice
import random
# 生成1-10的随机排列
def random_permutation(n):
    numbers = list(range(1, n+1))
    # 打乱顺序后生成排列
    random.shuffle(numbers)
    return list(islice(permutations(numbers), 1))[0]
print("随机排列：", random_permutation(5)) 
#输出 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()
# 按单词分组并统计数量
word_groups = groupby(sorted(words))
word_count = {word: len(list(group)) for word, group in word_groups}
print("单词频率：", word_count)  # {'apple': 3, 'banana': 2, 'orange': 1}
# 更简洁方式：结合Counter
print("Counter统计：", Counter(words))  
#输出：
单词频率： {'apple': 3, 'banana': 2, 'orange': 1}
Counter统计： Counter({'apple': 3, 'banana': 2, 'orange': 1})