本文主要是介绍python的itertools库，希望对大家解决编程问题提供一定的参考价值，需要的开发者们随着小编来一起学习吧！
itertools常用的方法如下：
import itertools
1. 生成的列表累加，在生成新的列表x = itertools.accumulate(range(10))print(list(x))结果：[0, 1, 3, 6, 10, 15, 21, 28, 36, 45]
2. 连接多个列表或者迭代器x = itertools.chain(range(3), range(4), [3,2,1])print(list(x))结果：[0, 1, 2, 0, 1, 2, 3, 3, 2, 1]
3. 求列表或生成器中指定数目的元素不重复的所有组合x = itertools.combinations(range(4), 3)print(list(x))结果：[(0, 1, 2), (0, 1, 3), (0, 2, 3), (1, 2, 3)]
4. 允许重复元素的组合x = itertools.combinations_with_replacement( ABC , 2)print(list(x))结果：[( A , A ), ( A , B ), ( A , C ), ( B , B ), ( B , C ), ( C , C )]
5. 按照真值表筛选元素x = itertools.compress(range(5), (True, False, True, True, False))print(list(x))结果：[0, 2, 3]
6. 一个计数器,可以指定起始位置和步长x = itertools.count(start=20, step=-1) # X的类型为countprint(list(itertools.islice(x, 0, 10, 1)))结果：[20, 19, 18, 17, 16, 15, 14, 13, 12, 11]
7. 循环指定的列表和迭代器x = itertools.cycle( ABC )print(list(itertools.islice(x, 0, 10, 1)))结果：[ A , B , C , A , B , C , A , B , C , A ]
8. 按照真值函数丢弃掉列表和迭代器前面的元素x = itertools.dropwhile(lambda e: e < 5, range(10))print(list(x))结果：[5, 6, 7, 8, 9]
9. 保留对应真值为False的元素x = itertools.filterfalse(lambda e: e < 5, (1, 5, 3, 6, 9, 4))print(list(x))结果：[5, 6, 9]
10. 按照分组函数的值对元素进行分组x = itertools.groupby(range(10), lambda x: x < 5 or x > 8) for condition, numbers in x: print(condition, list(numbers)) True [0, 1, 2, 3, 4] False [5, 6, 7, 8] True [9]
11. 上文使用过的函数，对迭代器进行切片x = itertools.islice(range(10), 0, 9, 2)print(list(x))结果：[0, 2, 4, 6, 8]
12. 产生指定数目的元素的所有排列(顺序有关)x = itertools.permutations(range(4), 3)print(list(x))结果：[(0, 1, 2), (0, 1, 3), (0, 2, 1), (0, 2, 3), (0, 3, 1), (0, 3, 2), (1, 0, 2), (1, 0, 3), (1, 2, 0), (1, 2, 3), (1, 3, 0), (1, 3, 2), (2, 0, 1), (2, 0,3), (2, 1, 0), (2, 1, 3), (2, 3, 0), (2, 3, 1), (3, 0, 1), (3, 0, 2), (3, 1, 0), (3, 1, 2), (3, 2, 0), (3, 2, 1)]13. 产生多个列表和迭代器的(积)x = itertools.product( ABC , range(3))print(list(x))结果：[( A , 0), ( A , 1), ( A , 2), ( B , 0), ( B , 1), ( B , 2), ( C , 0), ( C , 1), ( C , 2)]14. 简单的生成一个拥有指定数目元素的迭代器x = itertools.repeat(0, 5)print(list(x))结果：[0, 0, 0, 0, 0]
15. 类似mapx = itertools.starmap(str.islower, aBCDefGhI )print(list(x))结果：[True, False, False, False, True, True, False, True, False]
16. 与dropwhile相反，保留元素直至真值函数值为假。x = itertools.takewhile(lambda e: e < 5, range(10))print(list(x))结果：[0, 1, 2, 3, 4]
17. 生成指定数目的迭代器x = itertools.tee(range(10), 2)for letters in x:print(list(letters))结果：[0, 1, 2, 3, 4, 5, 6, 7, 8, 9][0, 1, 2, 3, 4, 5, 6, 7, 8, 9]
18. 类似于zip，以较长的列表和迭代器的长度为准x = itertools.zip_longest(range(3), range(5))y = zip(range(3), range(5))>>> print(list(x))[(0, 0), (1, 1), (2, 2), (None, 3), (None, 4)]>>> print(list(y))[(0, 0), (1, 1), (2, 2)]这篇关于python的itertools库的文章就介绍到这儿，希望我们推荐的文章对编程师们有所帮助！
