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))>>> 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)))>>> 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>> 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)(, , )>>> 

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 ^_^