python 超好用的迭代兵器库itertools，十八般兵器哪18般？

Posted 2021-09-16 Hann Yang

知识点

在古典小说和传统评话中，常说武艺高强的人是“十八般武艺样样精通”，这十八般武艺是指使用“十八般兵器”的功夫和技能。哪十八般呢？

十八般兵器在武术界中最普遍的说法是：刀、枪、剑、戟、斧、钺、钩、叉、鞭、锏、锤、抓、镗、棍、槊、棒、拐、流星。

汉武于元封四年（公元前107），经过严格的挑选和整理，筛选出18种类型的兵器：矛、镗、刀、戈、槊、鞭、锏、剑、锤、抓、戟、弓、钺、斧、牌、棍、枪、叉。
三国时代，著名的兵器鉴别家吕虔，根据兵器的特点，对汉武帝钦定的“十八般兵器”重新排列为九长九短。九长：戈、矛、戟、槊、镗、钺、棍、枪、叉；九短：斧、戈、牌、箭、鞭、剑、锏、锤、抓。
明代《五杂俎》和清代《坚集》两书所载，“十八般兵器”为弓、弩、枪、刀、剑、矛、盾、斧、钺、戟、黄、锏、挝、殳（棍）、叉、耙头、锦绳套索、白打（拳术）。后人称其为“小十八般”。

迭代器

也叫生成器，它最大的优势就是延迟计算按需使用，节省内存空间、提高运行效率。

迭代工具库 itertools 中共有18个函数，恰好似“迭代界”的十八般兵器，掌握了这些功夫和技能也可以说是“十八般武艺样样精通”！：

>>> import itertools
>>> tools = [func for func in dir(itertools) if func[0]>='a']
>>> len(tools)
18
>>> tools
['accumulate', 'chain', 'combinations', 'combinations_with_replacement', 'compress', 
'count', 'cycle', 'dropwhile', 'filterfalse', 'groupby', 'islice', 'permutations', 
'product', 'repeat', 'starmap', 'takewhile', 'tee', 'zip_longest']

1. 累加器 accumulate 

>>> import itertools as it
>>> it.accumulate(range(11))
<itertools.accumulate object at 0x0A0C9988>
>>> list(it.accumulate(range(11)))
[0, 1, 3, 6, 10, 15, 21, 28, 36, 45, 55]
>>> 

1乘2乘3...一直乘到n有阶乘运算 n! ，但1加2加3...一直加到n，一般都没有定义“累和”运算，还需循环来计算。现在有了这个函数可以代替用用的，比如1加到100：

>>> list(it.accumulate(range(1+100)))[-1]
5050
>>> 

2. 连接器 chain

连接多个迭代器，或其它可迭代对象

>>> import itertools as it
>>> it.chain(range(3),[3,4,5],{6,7},(i for i in range(8,11)))
<itertools.chain object at 0x0A0BF3B8>
>>> list(it.chain(range(4),[4,5],{6,7},(i for i in range(8,11))))
[0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10]
>>> 

3. 组合器 combinations

from itertools import combinations as comb
>>> comb1 = comb(range(4), 3)
>>> list(comb1)
[(0, 1, 2), (0, 1, 3), (0, 2, 3), (1, 2, 3)]
>>> comb2 = comb(range(1,6), 3)
>>> list(comb2)
[(1, 2, 3), (1, 2, 4), (1, 2, 5), (1, 3, 4), (1, 3, 5),
 (1, 4, 5), (2, 3, 4), (2, 3, 5), (2, 4, 5), (3, 4, 5)]
>>> comb3 = comb(range(1,6), 4)
>>> list(comb3)
[(1, 2, 3, 4), (1, 2, 3, 5), (1, 2, 4, 5), (1, 3, 4, 5), (2, 3, 4, 5)]
>>> 

4. 可重复组合器 combinations_with_replacement

>>> from itertools import combinations_with_replacement as Comb2
>>> comb1 = Comb2(range(4), 3)
>>> list(comb1)
[(0, 0, 0), (0, 0, 1), (0, 0, 2), (0, 0, 3), (0, 1, 1), (0, 1, 2), (0, 1, 3),
 (0, 2, 2), (0, 2, 3), (0, 3, 3), (1, 1, 1), (1, 1, 2), (1, 1, 3), (1, 2, 2),
 (1, 2, 3), (1, 3, 3), (2, 2, 2), (2, 2, 3), (2, 3, 3), (3, 3, 3)]
>>> comb2 = Comb2(range(1,6), 3)
>>> list(comb2)
[(1, 1, 1), (1, 1, 2), (1, 1, 3), (1, 1, 4), (1, 1, 5), (1, 2, 2), (1, 2, 3),
 (1, 2, 4), (1, 2, 5), (1, 3, 3), (1, 3, 4), (1, 3, 5), (1, 4, 4), (1, 4, 5),
 (1, 5, 5), (2, 2, 2), (2, 2, 3), (2, 2, 4), (2, 2, 5), (2, 3, 3), (2, 3, 4),
 (2, 3, 5), (2, 4, 4), (2, 4, 5), (2, 5, 5), (3, 3, 3), (3, 3, 4), (3, 3, 5),
 (3, 4, 4), (3, 4, 5), (3, 5, 5), (4, 4, 4), (4, 4, 5), (4, 5, 5), (5, 5, 5)]
>>> 

5. 排列器 permutations

>>> import itertools as it
>>> list(it.permutations([1,2,3]))
[(1, 2, 3), (1, 3, 2), (2, 1, 3), (2, 3, 1), (3, 1, 2), (3, 2, 1)]
>>> # 数字1、2、3能组成哪些三位数？
>>> [i[0]*100+i[1]*10+i[2] for i in it.permutations([1,2,3])]
[123, 132, 213, 231, 312, 321]
>>> 

6. 压缩器 compress

按照真值表来精简迭代器，筛选出部分值

>>> import itertools as it
>>> i = it.compress(range(6), (1,1,0,0,1,0))
>>> list(i)
[0, 1, 4]
>>> 

7. 切片器 islice

>>> import itertools as it
>>> islice = it.islice(range(100),0,9,2)
>>> list(islice)
[0, 2, 4, 6, 8]
>>> iSlice = it.islice(range(1,100),0,9,2)
>>> list(iSlice)
[1, 3, 5, 7, 9]
>>> # 可以不指定起始和步长，直接指定个数
>>> list(it.islice(range(1,100),10))
[1, 11, 21, 31, 41, 51, 61, 71, 81, 91]
>>> 

8. 计数器 count

因为生成器只提供说法不是数据集，直接用 list(count1)会死循环的，可以用islice()指定一下个数。

>>> import itertools as it
>>> count1 = it.count(start=0,step=3)
>>> list(it.islice(count1,12))
[0, 3, 6, 9, 12, 15, 18, 21, 24, 27, 30, 33]
>>> count2 = it.count(start=100,step=-2)
>>> list(it.islice(count2,10))
[100, 98, 96, 94, 92, 90, 88, 86, 84, 82]
>>> 

9. 循环器 cycle

>>> import itertools as it
>>> list(it.islice(it.cycle('ABC'),10))
['A', 'B', 'C', 'A', 'B', 'C', 'A', 'B', 'C', 'A']
>>> list(it.islice(it.cycle([1,2,3,4]),10))
[1, 2, 3, 4, 1, 2, 3, 4, 1, 2]
>>> 

10. 重复器 repeat

>>> import itertools as it
>>> list(it.repeat(5,10))
[5, 5, 5, 5, 5, 5, 5, 5, 5, 5]
>>> list(it.repeat([1,2],5))
[[1, 2], [1, 2], [1, 2], [1, 2], [1, 2]]
>>> 

11. 舍真器 dropwhile

舍弃不满足条件的元素，但当条件不满足即停止筛选

>>> import itertools as it
>>> lst = [1,3,5,2,4,6,10,11,7,8,12,15]
>>> list(it.dropwhile(lambda i:i<9,lst))
[10, 11, 7, 8, 12, 15]
>>> list(it.dropwhile(lambda i:i%2,lst))
[2, 4, 6, 10, 11, 7, 8, 12, 15]
>>> 

12. 留真器 takewhile

留下满足条件的元素，但当条件不满足即停止筛选

>>> import itertools as it
>>> list(it.takewhile(lambda i:i<6, range(10)))
[0, 1, 2, 3, 4, 5]
>>> lst = [1,3,5,2,4,6,10,11,7,8,12,15]
>>> list(it.takewhile(lambda i:i<11,lst))
[1, 3, 5, 2, 4, 6, 10]
>>> list(it.takewhile(lambda i:i%6,lst))
[1, 3, 5, 2, 4]
>>> 

13. 筛假器 filterfalse

舍弃满足条件的所有元素，留下所有不满足条件的

>>> import itertools as it
>>> lst = [1,3,5,2,4,6,10,11,7,8,12,15]
>>> list(it.filterfalse(lambda i:i<9,lst))
[10, 11, 12, 15]
>>> list(it.filterfalse(lambda i:i%2,lst))
[2, 4, 6, 10, 8, 12]
>>> 

14. 分组器 groupby

>>> import itertools as it
>>> group = it.groupby(range(20), lambda i:not 8<i<16)
>>> for i,j in group: print(i,list(j))

True [0, 1, 2, 3, 4, 5, 6, 7, 8]
False [9, 10, 11, 12, 13, 14, 15]
True [16, 17, 18, 19]
>>> 

15. 乘积器 product

>>> import itertools as it
>>> list(it.product('ABC',(1,2)))
[('A', 1), ('A', 2), ('B', 1), ('B', 2), ('C', 1), ('C', 2)]

16. 映射器 starmap

>>> import itertools as it
>>> list(it.starmap(str.isupper, 'AbCDefgH'))
[True, False, True, True, False, False, False, True]
>>> list(it.starmap(lambda a,b,c:a+b+c,([1,2,3],[4,5,6],[7,8,9])))
[6, 15, 24]
>>> list(it.starmap(lambda *a:sum(a),[range(5),range(10),range(101)]))
[10, 45, 5050]
>>> 

17. 元组器 tee

返回多个迭代器的元组

>>> import itertools as it
>>> [list(i) for i in it.tee([1,2,3],3)]
[[1, 2, 3], [1, 2, 3], [1, 2, 3]]
>>> it.tee([1,2,3],3)
(<itertools._tee object at 0x030711A8>,
 <itertools._tee object at 0x03078228>,
 <itertools._tee object at 0x0131FFE8>)
>>> 

18. 打包器 zip_longest

与内置函数zip()类似，但元素个数以最长的迭代器为准

>>> import itertools as it
>>> list(it.zip_longest('ABCDE',range(1,4)))
[('A', 1), ('B', 2), ('C', 3), ('D', None), ('E', None)]
>>> list(zip('ABCDE',range(1,4)))
[('A', 1), ('B', 2), ('C', 3)]
>>> list(it.zip_longest('ABCDE',range(1,4),[1,2,3,4]))
[('A', 1, 1), ('B', 2, 2), ('C', 3, 3), ('D', None, 4), ('E', None, None)]
>>> 

名字我随便起的，形像就好。看下来如何？十八兵器，样样精通了吗？其实掌握个几样“称手的”即可，何必面面俱到呢 ^_^