python 超好用的迭代兵器库itertools，十八般兵器哪18般？

她是你的谁

常识面

正在古典小道战传统说书中，常道技艺下强的人是“十八般技艺样样夺目”，那十八般技艺是教唆用“十八般刀兵”的工夫战妙技。哪十八般呢？
十八般刀兵正在技击界中最遍及的道法是：刀、枪、剑、戟、斧、钺、钩、叉、鞭、锏、锤、抓、镗、棍、槊、棒、拐、流星。

汉武于元启四年（公元前107），颠末严厉的选择战收拾整顿，挑选出18种规范的刀兵：盾、镗、刀、戈、槊、鞭、锏、剑、锤、抓、戟、弓、钺、斧、牌、棍、枪、叉。
三国时期，出名的刀兵辨别家吕虔，按照刀兵的特性，对汉武帝钦定的“十八般刀兵”从头布列为九少九短。九少：戈、盾、戟、槊、镗、钺、棍、枪、叉；九短：斧、戈、牌、箭、鞭、剑、锏、锤、抓。
明朝《五纯俎》战清朝《脆散》两书所载，“十八般刀兵”为弓、弩、枪、刀、剑、盾、盾、斧、钺、戟、黄、锏、挝、殳（棍）、叉、耙头、锦绳套索、利剑挨（拳术）。先人称其为“小十八般”。
迭代器

也叫天生器，它最年夜的劣势便是提早计较按需利用，节流内乱存空间、进步运转服从。
迭代东西库 itertools 中共有18个函数，刚好似“迭代界”的十八般刀兵，把握了那些工夫战妙技也能够道是“十八般技艺样样夺目”！：

>>> 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']

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1. 乏减器 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]
>>>

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1乘2乘3...不断乘到n有阶乘运算 n! ，但1减2减3...不断减到n，普通皆出有界说“乏战”运算，借需轮回去计较。如今有了那个函数能够代替用用的，比如1减到100：

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

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2. 毗连器 chain

毗连多个迭代器，或别的可迭代工具

>>> 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]
>>>

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3. 组开器 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)]
>>>

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4. 可反复组开器 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)]
>>>

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5. 布列器 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能构成哪些三位数？
>>> [i[0]*100+i[1]*10+i[2] for i in it.permutations([1,2,3])]
[123, 132, 213, 231, 312, 321]
>>>

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6. 紧缩器 compress

根据实值表去粗简迭代器，挑选出部门值

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

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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]
>>> # 能够没有指定肇端战步少，间接指定个数
>>> list(it.islice(range(1,100),10))
[1, 11, 21, 31, 41, 51, 61, 71, 81, 91]
>>>

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8. 计数器 count

由于天生器只供给道法没有是数据散，间接用 list(count1)会逝世轮回的，能够用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]
>>>

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9. 轮回器 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]
>>>

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10. 反复器 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]]
>>>

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11. 舍实器 dropwhile

舍弃没有满意前提的元素，但当前提没有满意即避免挑选

>>> 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]
>>>

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12. 留实器 takewhile

留下满意前提的元素，但当前提没有满意即避免挑选

>>> 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]
>>>

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13. 筛假器 filterfalse

舍弃满意前提的一切元素，留下一切没有满意前提的

>>> 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]
>>>

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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]
>>>

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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)]

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16. 映照器 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]
>>>

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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>)
>>>

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18. 挨包器 zip_longest

取内乱置函数zip()相似，但元素个数以最少的迭代器为准

>>> 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)]
>>>

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名字我随意起的，形像便好。看下去如何？十八刀兵，样样夺目了吗？实在把握个几样“称脚的”便可，何须八面玲珑呢 ^_^

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