Question

我有这样的一系列数字

myvar = [57, 71, 87, 97, 99, 101, 103, 113, 114, 115, 128, 129, 131, 137, 147, 156, 163, 186]

现在我想计算所有这些可能的组合（长度为1到20），其总和等于给定的数字m。

我尝试使用以下代码解决：

def sum_count(m):    ## Where m is the sum required

    from itertools import combinations

    myseq = []
    for i in range(1,len(myvar)):
        mycomb = list(combinations(mass,i));  # Getting combinations of length i
        mycomb = [list(j) for j in mycomb];
        for j in range(len(mycomb)-1,-1,-1):
            if sum(mycomb[j]) == m:
                myseq.append(mycomb[j])

    return(myseq)

当我放m = 270时（例如）它给了我：

[[114, 156], [57, 99, 114]]

但从myvar可以明显看出，还有其他组合的总和等于270.我在哪里都无法理解。

Answer 1

<强> TL; DR：

讨论不同的方法，此处列出了最佳方法以便于访问，最初由thefourtheye编写：

def subsets_with_sum(lst, target, with_replacement=False):
    x = 0 if with_replacement else 1
    def _a(idx, l, r, t):
        if t == sum(l): r.append(l)
        elif t < sum(l): return
        for u in range(idx, len(lst)):
            _a(u + x, l + [lst[u]], r, t)
        return r
    return _a(0, [], [], target)

注意：上述方法已根据以下原始版本的改进进行了修改

原帖：

嗯 - 通过一些逻辑快速简单地应用您的数据就可以得出正确的答案：

# data
vals = [57, 71, 87, 97, 99, 101, 103, 113, 114, 115, 128, 129, 131, 137, 147, 156, 163, 186]
target = 270

使用itertools.combinations：

>>> from itertools import combinations
>>> [comb for i in range(1, 20) for comb in combinations(vals, i) if sum(comb) == target]
[(114, 156), (57, 99, 114)]

但是，也许你想使用combinations_with_replacement，它允许从初始列表中多次使用值，而不是只使用一次。

使用itertools.combinations_with_replacement：

>>> from itertools import combinations_with_replacement
>>> [comb for i in range(1, 20) for comb in combinations_with_replacement(vals, i) if sum(comb) == target]
>>>  # result takes too long ...

你可以把它变成一个强大的功能：

def subsets_with_sum(lst, target, subset_lengths=range(1, 20), method='combinations'):   
    import itertools
    return [comb for i in subset_lengths for comb in
            getattr(itertools, method)(lst, i) if sum(comb) == target]

>>> subsets_with_sum(vals , 270)
[(114, 156), (57, 99, 114)]

另一种方法，由thefourtheye提供，远更快，并且不需要导入：

def a(lst, target, with_replacement=False):
    def _a(idx, l, r, t, w):
        if t == sum(l): r.append(l)
        elif t < sum(l): return
        for u in range(idx, len(lst)):
            _a(u if w else (u + 1), l + [lst[u]], r, t, w)
        return r
    return _a(0, [], [], target, with_replacement)


>>> s = [57, 71, 87, 97, 99, 101, 103, 113, 114, 115, 128, 129, 131, 137, 147, 156, 163, 186]
>>> a(s, 270)
[[57, 99, 114], [114, 156]]
>>> a(s, 270, True)
[[57, 57, 57, 99], [57, 57, 156], [57, 71, 71, 71], [57, 99, 114], [71, 71, 128], [114, 156]]

<强>定时：

def a(lst, target, with_replacement=False):
    def _a(idx, l, r, t, w):
        if t == sum(l): r.append(l)
        elif t < sum(l): return
        for u in range(idx, len(lst)):
            _a(u if w else (u + 1), l + [lst[u]], r, t, w)
        return r
    return _a(0, [], [], target, with_replacement)

def b(lst, target, subset_lengths=range(1, 21), method='combinations'):   
    import itertools
    return [comb for i in subset_lengths for comb in
            getattr(itertools, method)(lst, i) if sum(comb) == target]

vals = [57, 71, 87, 97, 99, 101, 103, 113, 114, 115, 128, 129, 131, 137, 147, 156, 163, 186]

from timeit import timeit
print 'no replacement'
print timeit("a(vals, 270)", "from __main__ import vals, a", number=10)
print timeit("b(vals, 270)", "from __main__ import vals, b", number=10)
print 'with replacement'
print timeit("a(vals, 270, True)", "from __main__ import vals, a", number=10)
print timeit("b(vals, 270, method='combinations_with_replacement')", "from __main__ import vals, b", number=10)

时间输出：

no replacement
0.0273933852733
0.683039054001
with replacement
0.0177899423427
... waited a long time ... no results ...

<强>结论

新方法（a）的速度至少快20倍。

Answer 2

如果您只是想要计算组合数，请使用：

aminoacid_masses = [57, 71, 87, 97, 99, 101, 103, 113, 114, 115, 128, 129, 131, 137, 147, 156, 163, 186]


def peptides(n, d):
    for m in aminoacid_masses:
        if n-m in d:
            d[n] = d.get(n,0)+d[n-m]
    return d


def pep_counter(M):
    dicc = {0:1}
    mn = min(aminoacid_masses)
    for i in range(M-mn+1):
        j = i+mn
        peptides(j,dicc)
    return dicc


# This line calls the routine and indexes the returned dict.  Both with the desired mass (the mass we want peptides to sum up to)
print(pep_counter(1024)[1024])

查找提供的Sum值的组合

2 个答案: