知識點

在古典小說和傳統評話中，常說武藝高強的人是“十八般武藝樣樣精通”，這十八般武藝是指使用“十八般兵器”的功夫和技能。哪十八般呢？

十八般兵器在武術界中最普遍的說法是：刀、槍、劍、戟、斧、鉞、鉤、叉、鞭、锏、錘、抓、鏜、棍、槊、棒、拐、流星。

漢武于元封四年（公元前107），經過嚴格的挑選和整理，篩選出18種類型的兵器：矛、鏜、刀、戈、槊、鞭、锏、劍、錘、抓、戟、弓、鉞、斧、牌、棍、槍、叉。
三國時代，著名的兵器鑒別家呂虔，根據兵器的特點，對漢武帝欽定的“十八般兵器”重新排列為九長九短。九長：戈、矛、戟、槊、鏜、鉞、棍、槍、叉；九短：斧、戈、牌、箭、鞭、劍、锏、錘、抓。
明代《五雜俎》和清代《堅集》兩書所載，“十八般兵器”為弓、弩、槍、刀、劍、矛、盾、斧、鉞、戟、黃、锏、撾、殳（棍）、叉、耙頭、錦繩套索、白打（拳術）。后人稱其為“小十八般”。

迭代器

也叫生成器，它最大的優勢就是延遲計算按需使用，節省內存空間、提高運行效率。

迭代工具庫 itertools 中共有18個函數，恰好似“迭代界”的十八般兵器，掌握了這些功夫和技能也可以說是“十八般武藝樣樣精通”！：

>>> 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有階乘運算 n! ，但1加2加3...一直加到n，一般都沒有定義“累和”運算，還需循環來計算。現在有了這個函數可以代替用用的，比如1加到100：

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

2. 連接器 chain

連接多個迭代器，或其它可迭代對象

>>> 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. 組合器 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. 可重復組合器 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)]>>> # 數字1、2、3能組成哪些三位數？>>> [i[0]*100+i[1]*10+i[2] for i in it.permutations([1,2,3])][123, 132, 213, 231, 312, 321]>>> 

6. 壓縮器 compress

按照真值表來精簡迭代器，篩選出部分值

>>> 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長，直接指定個數>>> list(it.islice(range(1,100),10))[1, 11, 21, 31, 41, 51, 61, 71, 81, 91]>>> 

8. 計數器 count

因為生成器只提供說法不是數據集，直接用 list(count1)會死循環的，可以用islice()指定一下個數。

>>> 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. 循環器?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. 重復器 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

舍棄不滿足條件的元素，但當條件不滿足即停止篩選

>>> 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

留下滿足條件的元素，但當條件不滿足即停止篩選

>>> 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. 篩假器 filterfalse

舍棄滿足條件的所有元素，留下所有不滿足條件的

>>> 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. 分組器?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. 乘積器?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. 元組器 tee

返回多個迭代器的元組

>>> 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

與內置函數zip()類似，但元素個數以最長的迭代器為準

>>> 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)]>>> 

名字我隨便起的，形像就好。看下來如何？十八兵器，樣樣精通了嗎？其實掌握個幾樣“稱手的”即可，何必面面俱到呢 ^_^