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Math 第n个组合

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Math 第n个组合,math,statistics,probability,combinatorics,Math,Statistics,Probability,Combinatorics,是否有一种直接的方法可以获得nCr所有组合的有序集合的第n个组合 示例：我有四个元素：[6,4,2,1]。一次取三个的所有可能组合为： [6,4,2]，[6,4,1]，[6,2,1]，[4,2,1] 是否有一种算法可以在有序结果集中给出例如第三个答案[6，2，1]，而不枚举所有以前的答案？只是一个粗略的草图： 将数字排列到元组的上三角矩阵中： A(n-1,n-1) Aij = [i+1, j-1] 若首先遍历矩阵行，将得到两个元素按递增顺序的组合。要概括为三个元素，请将矩阵行视为另一个三

是否有一种直接的方法可以获得nCr所有组合的有序集合的第n个组合

示例：我有四个元素：[6,4,2,1]。一次取三个的所有可能组合为： [6,4,2]，[6,4,1]，[6,2,1]，[4,2,1]

是否有一种算法可以在有序结果集中给出例如第三个答案[6，2，1]，而不枚举所有以前的答案？

只是一个粗略的草图： 将数字排列到元组的上三角矩阵中：

A(n-1,n-1)   
Aij = [i+1, j-1]
若首先遍历矩阵行，将得到两个元素按递增顺序的组合。要概括为三个元素，请将矩阵行视为另一个三角形矩阵，而不是向量。它创建了一个立方体的角

至少我会这样处理这个问题

让我澄清一下，您不必存储矩阵，您需要计算索引。 让我做一个维度的例子，原则上你们可以扩展到20个维度（簿记可能很糟糕）

这个二维示例适用于任何数字n，并且您不必将排列列表化。

一种方法是使用位的属性。这仍然需要一些枚举，但您不必枚举每个集合

例如，您的集合中有4个数字。因此，如果您正在生成4个数字的所有可能组合，您可以按如下方式枚举它们：

{6, 4, 2, 1} 0000 - {(no numbers in set)} 0001 - {1} 0010 - {2} 0011 - {2, 1} ... 1111 - {6, 4, 2, 1} {6, 4, 2, 1} 0000-{（集合中没有数字）} 0001 - {1} 0010 - {2} 0011 - {2, 1} ... 1111 - {6, 4, 2, 1} 查看每个“位”如何对应于“该数字是否在您的集合中”？我们在这里看到有16种可能性（2^4）

所以现在我们可以遍历并找到所有只有3位开启的可能性。这将告诉我们存在的“3”的所有组合：

0111 - {4, 2, 1} 1011 - {6, 2, 1} 1101 - {6, 4, 1} 1110 - {6, 4, 2} 0111 - {4, 2, 1} 1011 - {6, 2, 1} 1101 - {6, 4, 1} 1110 - {6, 4, 2} 让我们将每个二进制值重写为十进制值：

0111 = 7 1011 = 11 1101 = 13 1110 = 14 0111 = 7 1011 = 11 1101 = 13 1110 = 14 现在我们已经这样做了-好吧，你说你想要“第三次”枚举。让我们看看第三大数字：11。它具有位模式1011。对应于。。。{6,2,1}

酷

基本上，您可以对任何集合使用相同的概念。现在我们所做的就是将问题从“枚举所有集合”转化为“枚举所有整数”。对于您的问题来说，这可能要容易得多。

注意，您可以通过递归生成包含第一个元素的所有组合，然后生成不包含第一个元素的所有组合来生成序列。在这两种递归情况下，删除第一个元素以获得n-1个元素的所有组合。在Python中：

def combination(l, r):
    if r == 0:
        yield []
    elif len(l) == r:
        yield l
    else:
        for c in (combination(l[1:], r-1)):
            yield l[0:1]+c
        for c in (combination(l[1:], r)):
            yield c
每当您通过这样的选择生成序列时，您都可以通过计算一个选择生成的元素数并将计数与k进行比较来递归地生成第k个元素。如果k小于计数，则您做出该选择。否则，减去计数，然后在该点重复您可能做出的其他选择。如果总是有
b
选项，您可以将其视为在base
b
中生成一个数字。如果选择的数量不同，该技术仍然有效。在伪代码中（当所有选项始终可用时）：

如果你颠倒了选择的顺序，你就颠倒了顺序

def reverseKthCombination(k, l, r):
    if r == 0:
        return []
    elif len(l) == r:
        return l
    else:
        i=nCr(len(l)-1, r)
        if k < i:
            return reverseKthCombination(k, l[1:], r)
        else:
            return l[0:1] + reverseKthCombination(k-i, l[1:], r-1)
  • TLDR？只需滚动到最底部，即可找到我的最终解决方案
当我在寻找获取索引的方法时，我偶然发现了这个问题，如果指定的组合位于字典排序的列表中，那么它将位于，反之亦然，从一些潜在的非常大的对象集中选择对象，但在后者中找不到太多（问题的反面并不那么难以捉摸）

因为我也解决了（我认为是）你的确切问题，然后我想我会把我的解决方案发布到这里

**
编辑：我的需求也是您的需求——我看到了答案，认为递归很好。现在，经过六年的漫长岁月，你终于拥有了它；只需向下滚动。
**

对于您在问题中提出的要求（我认为是这样），这将很好地完成工作：

def iterCombinations(n, k):
if k==1:
    for i in range(n):
        yield [i]
    return
result = []
for a in range(k-1, n):
    for e in iterCombinations(n, k-1):
        if e[-1] == a:
            break
        yield e + [a]
然后，您可以在按降序排列的集合中查找该项（或使用一些等效的比较方法），因此对于所讨论的情况：

>>> itemsDescending = [6,4,2,1]
>>> for c in iterCombinations(4, 3):
...     [itemsDescending[i] for i in c]
...
[6, 4, 2]
[6, 4, 1]
[6, 2, 1]
[4, 2, 1]
但是，在Python中，这也可能是开箱即用的：

>>> import itertools
>>> for c in itertools.combinations(itemsDescending, 3):
...     c
...
(6, 4, 2)
(6, 4, 1)
(6, 2, 1)
(4, 2, 1)
以下是我为满足非递归算法的需求所做的工作，该算法不创建或遍历任意方向的有序列表，而是使用简单但有效的nCr非递归实现，选择（n，k）：

要获取排序列表中某个（从零开始）索引的组合，请执行以下操作：

def iterCombination(index, n, k):
    '''Yields the items of the single combination that would be at the provided
    (0-based) index in a lexicographically sorted list of combinations of choices
    of k items from n items [0,n), given the combinations were sorted in 
    descending order. Yields in descending order.
    '''
    if index < 0 or index >= choose(n, k):
        return
    n -= 1
    for i in range(k):
        while choose(n, k) > index:
            n -= 1
        yield n
        index -= choose(n, k)
        n -= 1
        k -= 1
这对于你的例子来说太过分了（但我现在意识到这只是一个例子）；这是每个索引的顺序：

def exampleUseCase(itemsDescending=[6,4,2,1], k=3):
    n = len(itemsDescending)
    print("index -> combination -> and back again:")
    for i in range(choose(n, k)):
        c = [itemsDescending[j] for j in iterCombination(i, n, k)][-1::-1]
        index = indexOfCombination([itemsDescending.index(v) for v in c])
        print("{0} -> {1} -> {2}".format(i, c, index))

>>> exampleUseCase()
index -> combination -> and back again:
0 -> [6, 4, 2] -> 0
1 -> [6, 4, 1] -> 1
2 -> [6, 2, 1] -> 2
3 -> [4, 2, 1] -> 3
这可以找到给定长列表的索引，或在眨眼间返回某个天文索引的组合，例如：

>>> choose(2016, 37)
9617597205504126094112265433349923026485628526002095715212972063686138242753600
>>> list(iterCombination(_-1, 2016, 37))
[2015, 2014, 2013, 2012, 2011, 2010, 2009, 2008, 2007, 2006, 2005, 2004, 2003,
2002, 2001, 2000, 1999, 1998, 1997, 1996, 1995, 1994, 1993, 1992, 1991, 1990, 1989,
1988, 1987, 1986, 1985, 1984, 1983, 1982, 1981, 1980, 1979]

或者，因为这是最后一个，并且由于选择（n，k）中的反射，所以可以很快，这里有一个从右边到中间，看起来也一样快…

>>> choose(2016, 37)//2
4808798602752063047056132716674961513242814263001047857606486031843069121376800
>>> list(iterCombination(_, 2016, 37))
[1978, 1973, 1921, 1908, 1825, 1775, 1747, 1635, 1613, 1598, 1529, 1528, 1521,
1445, 1393, 1251, 1247, 1229, 1204, 1198, 922, 901, 794, 699, 685, 633, 619, 598,
469, 456, 374, 368, 357, 219, 149, 93, 71]
最后一个例子暂停思考了一秒钟，但你不会吗

>>> import random
>>> rSet = set(random.randint(0, 10000000) for i in range(900))
>>> len(rSet)
900
>>> rList = sorted(rSet, reverse=True)
>>> combinations.indexOfCombination(rList)
61536587905102303838316048492163850175478325236595592744487336325506086930974887
88085020093159925576117511028315621934208381981476407812702689774826510322023536
58905845549371069786639595263444239118366962232872361362581506476113967993096033
00541202874946853699568596881200225925266331936183173583581021914595163799417151
30442624813775945054888304722079206982972852037480516813527237183254850056012217
59834465303543702263588008387352235149083914737690225710105023486226582087736870
38383323140972279867697434315252036074490127510158752080225274972225311906715033
86851377357968649982293794242170046400174118714525559851836064661141086690326842
25236658978135989907667078625869419802333512020715700514133380517628637151215549
05922388534567108671308819960483147825031620798631811671493891643972220604919591
22785587505280326638477135315176731640100473359830821781905546117103137944239120
34912084544221250309244925308316352643060056100719194985568284049903555621750881
39419639825279398618630525081169688672242833238889454445237928356800414839702024
66807635358129606994342005075585962080795273287472139515994244684088406544976674
84183671032002497594936116837768233617073949894918741875863985858049825755901232
89317507965160689287607868119414903299382093412911433254998227245783454244894604
83654290108678890682359278892580855226717964180806265176337132759167920384512456
91624558534942279041452960272707049107641475225516294235268581475735143470692000
78400891862852130481822509803019636619427631175355448729708451565341764545325720
79277290914349746541071731127111532099038538549697091038496002102703737347343739
96398832832674081286904287066696046621691978697914823322322650123025472624927566
99891468668052668317066769517155581261265629289158798073055495539590686279250097
27295943276536772955923599217742543093669565147228386873469711200278811335649924
13587219640724942441913695193417732608127949738209466313175361161142601108707568
19470026889319648128790363676253707359290547393198350533094409863254710237344552
47692325209744353688541868412075798500629908908768438513508959321262250985142709
19794478379412756202638771417821781240327337108495689300616872374578607430951230
96908870723878513999404242546015617238957825116802801618973562178005776911079790
22026655573872019955677676783191505879571719659770550759779880002320421606755826
75809722478174545846409923210824885805972611279030267270741509747224602604003738
30411365119180944456819762167312738395140461035991994771968906979578667047734952
21981545694935313345331923300019842406900689401417602004228459137311983483386802
30352489602769346000257761959413965109940729263098747702427952104316612809425394
85037536245288888254374135695390839718978818689595231708490351927063849922772653
26064826999661128817511630298712833048667406916285156973335575847429111697259113
53969532522640227276562651123634766230804871160471143157687290382053412295542343
14022687833967461351170188107671919648640149202504369991478703293224727284508796
06843631262345918398240286430644564444566815901074110609701319038586170760771099
41252989796265436701638358088345892387619172572763571929093224171759199798290520
71975442996399826830220944004118266689537930602427572308646745061258472912222347
18088442198837834539211242627770833874751143136048704550494404981971932449150098
52555927020553995188323691320225317096340687798498057634440618188905647503384292
79493920419695886724506109053220167190536026635080266763647744881063220423654648
36855624855494077960732944499038847158715263413026604773216510801253044020991845
89652657529729792772055725210165026891724511953666038764273616212464901231675592
46950937136633665320781952510620087284589083139308516989522633786063418913473703
96532777760440118656525488729217328376766171004246127636983612583177565603918697
15557602015171235214344399010185766876727226408494760175957535995025356361689144
85181975631986409708533731043231896096597038345028523539733981468056497208027899
6245509252811753667386001506195
然而，从该索引返回到它在前面的实现中表示的900-choose-10000000的组合将非常缓慢（因为它只是在每次迭代时从n中减去一）

对于这样大的组合列表，我们可以对空间进行二进制搜索，而我们添加的开销意味着对于小的组合列表，搜索速度只会稍微慢一点：

def iterCombination(index, n, k):
    '''Yields the items of the single combination that would be at the provided
    (0-based) index in a lexicographically sorted list of combinations of choices
    of k items from n items [0,n), given the combinations were sorted in 
    descending order. Yields in descending order.
    '''
    if index < 0 or n < k or n < 1 or k < 1 or choose(n, k) <= index:
        return
    for i in range(k, 0, -1):
        d = (n - i) // 2 or 1
        n -= d
        while 1:
            nCi = choose(n, i)
            while nCi > index:
                d = d // 2 or 1
                n -= d
                nCi = choose(n, i)
            if d == 1:
                break
            n += d
            d //= 2
            n -= d
        yield n
        index -= nCi
最后提醒您，对于问题的用例，您将执行以下操作：

def combinationAt(index, itemsDescending, k):
    return [itemsDescending[i] for i in
            list(iterCombination(index, len(itemsDescending), k))[-1::-1]]

>>> itemsDescending = [6,4,2,1]
>>> numberOfItemsBeingChosen = 3
>>> zeroBasedIndexWanted = 1
>>> combinationAt(zeroBasedIndexWanted, itemsDescending, numberOfItemsBeingChosen)
[6, 4, 1]
Fr 
def exampleUseCase(itemsDescending=[6,4,2,1], k=3):
    n = len(itemsDescending)
    print("index -> combination -> and back again:")
    for i in range(choose(n, k)):
        c = [itemsDescending[j] for j in iterCombination(i, n, k)][-1::-1]
        index = indexOfCombination([itemsDescending.index(v) for v in c])
        print("{0} -> {1} -> {2}".format(i, c, index))

>>> exampleUseCase()
index -> combination -> and back again:
0 -> [6, 4, 2] -> 0
1 -> [6, 4, 1] -> 1
2 -> [6, 2, 1] -> 2
3 -> [4, 2, 1] -> 3

>>> choose(2016, 37)
9617597205504126094112265433349923026485628526002095715212972063686138242753600
>>> list(iterCombination(_-1, 2016, 37))
[2015, 2014, 2013, 2012, 2011, 2010, 2009, 2008, 2007, 2006, 2005, 2004, 2003,
2002, 2001, 2000, 1999, 1998, 1997, 1996, 1995, 1994, 1993, 1992, 1991, 1990, 1989,
1988, 1987, 1986, 1985, 1984, 1983, 1982, 1981, 1980, 1979]

>>> choose(2016, 37)//2
4808798602752063047056132716674961513242814263001047857606486031843069121376800
>>> list(iterCombination(_, 2016, 37))
[1978, 1973, 1921, 1908, 1825, 1775, 1747, 1635, 1613, 1598, 1529, 1528, 1521,
1445, 1393, 1251, 1247, 1229, 1204, 1198, 922, 901, 794, 699, 685, 633, 619, 598,
469, 456, 374, 368, 357, 219, 149, 93, 71]

>>> import random
>>> rSet = set(random.randint(0, 10000000) for i in range(900))
>>> len(rSet)
900
>>> rList = sorted(rSet, reverse=True)
>>> combinations.indexOfCombination(rList)
61536587905102303838316048492163850175478325236595592744487336325506086930974887
88085020093159925576117511028315621934208381981476407812702689774826510322023536
58905845549371069786639595263444239118366962232872361362581506476113967993096033
00541202874946853699568596881200225925266331936183173583581021914595163799417151
30442624813775945054888304722079206982972852037480516813527237183254850056012217
59834465303543702263588008387352235149083914737690225710105023486226582087736870
38383323140972279867697434315252036074490127510158752080225274972225311906715033
86851377357968649982293794242170046400174118714525559851836064661141086690326842
25236658978135989907667078625869419802333512020715700514133380517628637151215549
05922388534567108671308819960483147825031620798631811671493891643972220604919591
22785587505280326638477135315176731640100473359830821781905546117103137944239120
34912084544221250309244925308316352643060056100719194985568284049903555621750881
39419639825279398618630525081169688672242833238889454445237928356800414839702024
66807635358129606994342005075585962080795273287472139515994244684088406544976674
84183671032002497594936116837768233617073949894918741875863985858049825755901232
89317507965160689287607868119414903299382093412911433254998227245783454244894604
83654290108678890682359278892580855226717964180806265176337132759167920384512456
91624558534942279041452960272707049107641475225516294235268581475735143470692000
78400891862852130481822509803019636619427631175355448729708451565341764545325720
79277290914349746541071731127111532099038538549697091038496002102703737347343739
96398832832674081286904287066696046621691978697914823322322650123025472624927566
99891468668052668317066769517155581261265629289158798073055495539590686279250097
27295943276536772955923599217742543093669565147228386873469711200278811335649924
13587219640724942441913695193417732608127949738209466313175361161142601108707568
19470026889319648128790363676253707359290547393198350533094409863254710237344552
47692325209744353688541868412075798500629908908768438513508959321262250985142709
19794478379412756202638771417821781240327337108495689300616872374578607430951230
96908870723878513999404242546015617238957825116802801618973562178005776911079790
22026655573872019955677676783191505879571719659770550759779880002320421606755826
75809722478174545846409923210824885805972611279030267270741509747224602604003738
30411365119180944456819762167312738395140461035991994771968906979578667047734952
21981545694935313345331923300019842406900689401417602004228459137311983483386802
30352489602769346000257761959413965109940729263098747702427952104316612809425394
85037536245288888254374135695390839718978818689595231708490351927063849922772653
26064826999661128817511630298712833048667406916285156973335575847429111697259113
53969532522640227276562651123634766230804871160471143157687290382053412295542343
14022687833967461351170188107671919648640149202504369991478703293224727284508796
06843631262345918398240286430644564444566815901074110609701319038586170760771099
41252989796265436701638358088345892387619172572763571929093224171759199798290520
71975442996399826830220944004118266689537930602427572308646745061258472912222347
18088442198837834539211242627770833874751143136048704550494404981971932449150098
52555927020553995188323691320225317096340687798498057634440618188905647503384292
79493920419695886724506109053220167190536026635080266763647744881063220423654648
36855624855494077960732944499038847158715263413026604773216510801253044020991845
89652657529729792772055725210165026891724511953666038764273616212464901231675592
46950937136633665320781952510620087284589083139308516989522633786063418913473703
96532777760440118656525488729217328376766171004246127636983612583177565603918697
15557602015171235214344399010185766876727226408494760175957535995025356361689144
85181975631986409708533731043231896096597038345028523539733981468056497208027899
6245509252811753667386001506195

def iterCombination(index, n, k):
    '''Yields the items of the single combination that would be at the provided
    (0-based) index in a lexicographically sorted list of combinations of choices
    of k items from n items [0,n), given the combinations were sorted in 
    descending order. Yields in descending order.
    '''
    if index < 0 or n < k or n < 1 or k < 1 or choose(n, k) <= index:
        return
    for i in range(k, 0, -1):
        d = (n - i) // 2 or 1
        n -= d
        while 1:
            nCi = choose(n, i)
            while nCi > index:
                d = d // 2 or 1
                n -= d
                nCi = choose(n, i)
            if d == 1:
                break
            n += d
            d //= 2
            n -= d
        yield n
        index -= nCi

def iterCombination(index, n, k):
    '''Yields the items of the single combination that would be at the provided
    (0-based) index in a lexicographically sorted list of combinations of choices
    of k items from n items [0,n), given the combinations were sorted in 
    descending order. Yields in descending order.
    '''
    nCk = 1
    for nMinusI, iPlus1 in zip(range(n, n - k, -1), range(1, k + 1)):
        nCk *= nMinusI
        nCk //= iPlus1
    curIndex = nCk
    for k in range(k, 0, -1):
        nCk *= k
        nCk //= n
        while curIndex - nCk > index:
            curIndex -= nCk
            nCk *= (n - k)
            nCk -= nCk % k
            n -= 1
            nCk //= n
        n -= 1
        yield n

def combinationAt(index, itemsDescending, k):
    return [itemsDescending[i] for i in
            list(iterCombination(index, len(itemsDescending), k))[-1::-1]]

>>> itemsDescending = [6,4,2,1]
>>> numberOfItemsBeingChosen = 3
>>> zeroBasedIndexWanted = 1
>>> combinationAt(zeroBasedIndexWanted, itemsDescending, numberOfItemsBeingChosen)
[6, 4, 1]

def nth_combination(iterable, r, index):
    'Equivalent to list(combinations(iterable, r))[index]'
    pool = tuple(iterable)
    n = len(pool)
    if r < 0 or r > n:
        raise ValueError
    c = 1
    k = min(r, n-r)
    for i in range(1, k+1):
        c = c * (n - k + i) // i
    if index < 0:
        index += c
    if index < 0 or index >= c:
        raise IndexError
    result = []
    while r:
        c, n, r = c*r//n, n-1, r-1
        while index >= c:
            index -= c
            c, n = c*(n-r)//n, n-1
        result.append(pool[-1-n])
    return tuple(result)

iterable, r, index = [6, 4, 2, 1], 3, 2

nth_combination(iterable, r, index)
# (6, 2, 1)

import itertools as it


list(it.combinations(iterable, r))[index]
# (6, 2, 1)

> pip install more_itertools