注意:您可以通过递归地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

_th 元素,并将计数与k进行比较。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 选项中,您可以将其视为生成基数中的数字 B

kth(k, choicePoints)
    if choicePoints is empty
        return empty list
    for each choice in head of choicePoints:
        if k < size of choice
            return choice and kth(k, tail of choicePoints)
        else
            k -= size of choice
    signal exception: k is out-of-bounds

这将为您提供一个基于0的索引。如果要基于1,请将比较更改为 k <= size of choice .

棘手的部分(伪代码中未指定的部分)是选择的大小取决于之前的选择。请注意,伪代码可用于解决比问题更一般的情况。

对于这个特定的问题,有两种选择( =2)和第一选择的尺寸(即包括1 ^圣 元素)由 C _r-1 . 这里有一个实现(需要合适的 nCr ):

def kthCombination(k, l, r):
    if r == 0:
        return []
    elif len(l) == r:
        return l
    else:
        i=nCr(len(l)-1, r-1)
        if k < i:
            return l[0:1] + kthCombination(k, l[1:], r-1)
        else:
            return kthCombination(k-i, l[1:], r)

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)

使用它:

>>> l = [6, 4, 2, 1]
>>> [kthCombination(k, [6, 4, 2, 1], 3) for k in range(nCr(len(l), 3)) ]
[[6, 4, 2], [6, 4, 1], [6, 2, 1], [4, 2, 1]]
>>> powOf2s=[2**i for i in range(4,-1,-1)]
>>> [sum(kthCombination(k, powOf2s, 3)) for k in range(nCr(len(powOf2s), 3))]
[28, 26, 25, 22, 21, 19, 14, 13, 11, 7]
>>> [sum(reverseKthCombination(k, powOf2s, 3)) for k in range(nCr(len(powOf2s), 3))]
[7, 11, 13, 14, 19, 21, 22, 25, 26, 28]

• 太长,读不下去了只需滚动到最底部,即可查看我的 最终解决方案

当我在寻找同时获得两个结果的方法时,我偶然发现了这个问题 特定的 结合 如果它位于 列表和 大对象集 在后者上找不到太多(你的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:我的要求 是 你的要求也是-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。
**

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)

以下是我为满足我的要求所做的(实际上也是为了满足你的要求!)属于 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非递归实现 _N C _R ,选择(n,k):

def choose(n, k):
    '''Returns the number of ways to choose k items from n items'''
    reflect = n - k
    if k > reflect:
        if k > n:
            return 0
        k = reflect
    if k == 0:
        return 1
    for nMinusIPlus1, i in zip(range(n - 1, n - k, -1), range(2, k + 1)):
        n = n * nMinusIPlus1 // i
    return n

以获取某个索引(从零开始)中的组合

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

颠倒 有序列表:

def indexOfCombination(combination):
    '''Returns the (0-based) index the given combination would have if it were in
    a reverse-lexicographically sorted list of combinations of choices of
    len(combination) items from any possible number of items (given the
    combination's length and maximum value)
   - combination must already be in descending order,
     and it's items drawn from the set [0,n).
    '''
    result = 0
    for i, a in enumerate(combination):
        result += choose(a, i + 1)
    return result

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

这可以找到给定某个长列表的索引,或者返回列表中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,并且可能是由于选择(n,k)中的反射而快,这里有一个从右边在中间,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

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

从这里你可能会注意到 choose 有取消的条款,如果我们取消所有的东西,我们最终会得到更快的实现,我认为。。。

该问题的最优函数

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

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]

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。

{6, 4, 2, 1}

0000 - {(no numbers in set)}
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}

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

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。它有位模式1011。它对应于{6, 2, 1}

酷!

来自Python3.6 itertools recipes :

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)

实际上:

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

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

或者,如文档字符串中所述:

import itertools as it


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

另见 more_itertools -实现的第三方库 this recipe 为你。通过以下方式安装:

> pip install more_itertools

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]

至少我会这样处理这个问题

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)。

ij = (i*i + i)/2 + j // ij is also the combination number
(i,j) = decompose(ij) // from ij one can recover i,j components
I = i // actual first index
J = j + 1 // actual second index

这个二维的例子适用于任何数字n,你不必把排列列成表格。

是的,有没有一种直接的方法可以得到nCr所有组合的有序集合的第n个组合?假设您需要生成第0、第3、第6个。。给定集合的组合。您可以直接生成它,而无需使用JNuberTools在两者之间生成组合。您甚至可以生成下一个第十亿个组合(如果设置的大小较大) 下面是代码示例:

JNumberTools.combinationsOf(list)
        .uniqueNth(8,1000_000_000) //skip to billionth combination of size 8
        .forEach(System.out::println);

<dependency>
    <groupId>io.github.deepeshpatel</groupId>
    <artifactId>jnumbertools</artifactId>
    <version>1.0.0</version>
</dependency>