查找提供的Sum值的组合 [英] Finding Combinations to the provided Sum value
问题描述
我有一系列这样的数字
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),其总和等于给定数字 m
。
我试图用下面的代码解决这个问题:
def sum_count (m):##其中m是itertools导入组合所需的
的总和
myseq = []
对于范围内的i(1,len(myvar )):
mycomb = list(组合(mass,i)); #获得长度的组合i
mycomb = [list(j)for mycomb in mycomb];
在范围内(len(mycomb)-1,-1,-1):
如果sum(mycomb [j])== m:
myseq.append(mycomb [j ])
return(myseq)
当我把 m = 270
(例如)它给了我:
[[114,156] ,[57,99,114]]
但从 myvar
表示还有其他组合的总和等于270.我在哪里无法理解。
TL
href =http://chat.stackoverflow.com/users/1903116/thefourtheye> thefourtheye : def subsets_with_sum(lst,target,with_replacement = False):
x = 0如果with_replacement else 1
def _a(idx,l,r,t):
if t == sum(l) :r.append(l)
eli f t < sum(l):在范围(idx,len(lst))中返回
:
_a(u + x,l + [lst [u]],r,t)
return r
return _a(0,[],[],target)
: : : :<好吧 - 用一些逻辑快速和简单地应用你的数据得出结论:你有正确的答案: p>
#data
vals = [57,71,87,97,109,101,103,113,114,115 ,128,129,131,137,147,156,163,186]
target = 270
>>>来自itertools导入组合
>>> (梳子)==目标]
[(114,156),(57,99,114)] $ b [comb for i in range(1,20)for comb in combinations(vals,i) $ b
然而,也许你想使用 combinations_with_replacement
,它允许从初始列表中多次使用值,而不是一次。
使用 itertools.combinations_with_replacement
:
>> > from itertools import combinations_with_replacement
>>> [comb for i in range(1,20)for comb in combination_with_replacement(vals,i)if sum(comb)== target]
>>> #结果需要太长时间...
您可以使它成为一个健壮的函数:
pre $ def $ subset_with_sum(lst,target,subset_lengths = range(1,20),method ='组合'):
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,它是多更快,并且不需要导入:
def a(lst,target,with_replacement = False) :
def _a(idx,l,r,t,w):
如果t == sum(l):r.append(l)
elif t < sum(l):返回
在范围内(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 = [57,71,87,97,109,101,103,113,114,115,128,129,131,137,147,156,163,186]
>>> 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]]
$ b def _a(idx,l,r,t,w):
if t == sum(l):r.append(l)
elif t < sum(l):返回
在范围内(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,如果sum(梳子)和comb(梳子)在
中,则getattr(itertools,method)(lst,i) ==目标]
vals = [57,71,87,97,109,101,103,113,114,115,128,129,131,137,147,156,163,186 ]
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 (b(vals,270,method ='combinations_with_replacement'),from __main__ import vals,270,True),from __main__ import vals,a,number = 10)
print timeit b,数字= 10)
计时输出:
没有替换
0.0273933852733
0.683039054001
替换
0.0177899423427
...等了很长时间...没有结果...
结论:
新方法(a)至少快20倍。
I have series of numbers like this
myvar = [57, 71, 87, 97, 99, 101, 103, 113, 114, 115, 128, 129, 131, 137, 147, 156, 163, 186]
Now I want to calculate all such possible combinations (of length 1 to 20) whose sum is equal to a given number m
.
I tried to solve with following code as :
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)
When I put m = 270
(for example) it gives me :
[[114, 156], [57, 99, 114]]
But is quite evident from the myvar
that there are still other combinations which have a sum equal to 270. Where am I failing to comprehend.
TL;DR:
Discuss different methods, best method is listed here for ease of access and was originally written by thefourtheye:
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)
note: the above method is modified with improvements from the original version below
Original Post:
Well - A quick and simple application of your data with some logic concludes that you have the correct answer:
# data
vals = [57, 71, 87, 97, 99, 101, 103, 113, 114, 115, 128, 129, 131, 137, 147, 156, 163, 186]
target = 270
Using 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)]
However, maybe you wanted to use combinations_with_replacement
which lets values be used multiple times from the initial list as opposed to only once.
Using 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 ...
You can make it into a robust function:
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)]
Another method, provided by thefourtheye , it is much faster, and requires no imports:
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]]
Timing:
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)
Timing Output:
no replacement
0.0273933852733
0.683039054001
with replacement
0.0177899423427
... waited a long time ... no results ...
conclusion:
The new method (a) is at least 20 times faster.
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