我有一个包含15个数字的列表,我需要编写一些代码来生成这些数字的所有32,768种组合。
我已经找到了一些代码(通过谷歌搜索),显然做了我正在寻找的,但我发现代码相当不透明,并谨慎使用它。而且我觉得肯定有更优雅的解决方案。
我唯一想到的就是循环十进制整数1-32768并将其转换为二进制,并使用二进制表示作为筛选器来挑选适当的数字。
有人知道更好的办法吗?使用map(),也许?
我有一个包含15个数字的列表,我需要编写一些代码来生成这些数字的所有32,768种组合。
我已经找到了一些代码(通过谷歌搜索),显然做了我正在寻找的,但我发现代码相当不透明,并谨慎使用它。而且我觉得肯定有更优雅的解决方案。
我唯一想到的就是循环十进制整数1-32768并将其转换为二进制,并使用二进制表示作为筛选器来挑选适当的数字。
有人知道更好的办法吗?使用map(),也许?
当前回答
我来晚了,但我想分享我找到的解决这个问题的方法: 具体来说,我想要做顺序组合,所以对于“STAR”,我想要“STAR”,“TA”,“AR”,但不是“SR”。
lst = [S, T, A, R]
lstCombos = []
for Length in range(0,len(lst)+1):
for i in lst:
lstCombos.append(lst[lst.index(i):lst.index(i)+Length])
可以通过在最后一行之前添加额外的if来过滤重复:
lst = [S, T, A, R]
lstCombos = []
for Length in range(0,len(lst)+1):
for i in lst:
if not lst[lst.index(i):lst.index(i)+Length]) in lstCombos:
lstCombos.append(lst[lst.index(i):lst.index(i)+Length])
如果由于某种原因,这将在输出中返回空白列表,这发生在我身上,我添加:
for subList in lstCombos:
if subList = '':
lstCombos.remove(subList)
其他回答
我在这个话题上有点晚了,但我想我可以帮助别人。
你可以使用itertools中的product:
from itertools import product
n = [1, 2, 3]
result = product(n, repeat=3) # You can change the repeat more then n length
print(list(result))
输出:
[(1, 1, 1), (1, 1, 2), (1, 1, 3), (1, 2, 1), (1, 2, 2), (1, 2, 3), (1, 3, 1),
(1, 3, 2), (1, 3, 3), (2, 1, 1), (2, 1, 2), (2, 1, 3), (2, 2, 1), (2, 2, 2),
(2, 2, 3), (2, 3, 1), (2, 3, 2), (2, 3, 3), (3, 1, 1), (3, 1, 2), (3, 1, 3),
(3, 2, 1), (3, 2, 2), (3, 2, 3), (3, 3, 1), (3, 3, 2), (3, 3, 3)]
另一个例子,但是改变了repeat参数:
from itertools import product
n = [1, 2, 3]
result = product(n, repeat=4) # Changing repeat to 4
print(list(result))
输出:
(1, 1, 2, 3), (1, 1, 3, 1), (1, 1, 3, 2), (1, 1, 3, 3), (1, 2, 1, 1),
(1, 2, 1, 2), (1, 2, 1, 3), (1, 2, 2, 1), (1, 2, 2, 2), (1, 2, 2, 3),
(1, 2, 3, 1), (1, 2, 3, 2), (1, 2, 3, 3), (1, 3, 1, 1), (1, 3, 1, 2),
(1, 3, 1, 3), (1, 3, 2, 1), (1, 3, 2, 2), (1, 3, 2, 3), (1, 3, 3, 1),
(1, 3, 3, 2), (1, 3, 3, 3), (2, 1, 1, 1), (2, 1, 1, 2), (2, 1, 1, 3),
(2, 1, 2, 1), (2, 1, 2, 2), (2, 1, 2, 3), (2, 1, 3, 1), (2, 1, 3, 2),
(2, 1, 3, 3), (2, 2, 1, 1), (2, 2, 1, 2), (2, 2, 1, 3), (2, 2, 2, 1),
(2, 2, 2, 2), (2, 2, 2, 3), (2, 2, 3, 1), (2, 2, 3, 2), (2, 2, 3, 3),
(2, 3, 1, 1), (2, 3, 1, 2), (2, 3, 1, 3), (2, 3, 2, 1), (2, 3, 2, 2),
(2, 3, 2, 3), (2, 3, 3, 1), (2, 3, 3, 2), (2, 3, 3, 3), (3, 1, 1, 1),
(3, 1, 1, 2), (3, 1, 1, 3), (3, 1, 2, 1), (3, 1, 2, 2), (3, 1, 2, 3),
(3, 1, 3, 1), (3, 1, 3, 2), (3, 1, 3, 3), (3, 2, 1, 1), (3, 2, 1, 2),
(3, 2, 1, 3), (3, 2, 2, 1), (3, 2, 2, 2), (3, 2, 2, 3), (3, 2, 3, 1),
(3, 2, 3, 2), (3, 2, 3, 3), (3, 3, 1, 1), (3, 3, 1, 2), (3, 3, 1, 3),
(3, 3, 2, 1), (3, 3, 2, 2), (3, 3, 2, 3), (3, 3, 3, 1), (3, 3, 3, 2),
(3, 3, 3, 3)]```
如文档中所述
def combinations(iterable, r):
# combinations('ABCD', 2) --> AB AC AD BC BD CD
# combinations(range(4), 3) --> 012 013 023 123
pool = tuple(iterable)
n = len(pool)
if r > n:
return
indices = list(range(r))
yield tuple(pool[i] for i in indices)
while True:
for i in reversed(range(r)):
if indices[i] != i + n - r:
break
else:
return
indices[i] += 1
for j in range(i+1, r):
indices[j] = indices[j-1] + 1
yield tuple(pool[i] for i in indices)
x = [2, 3, 4, 5, 1, 6, 4, 7, 8, 3, 9]
for i in combinations(x, 2):
print i
看看itertools.combination:
itertools.combinations (iterable, r) 返回元素的r长度子序列 输入迭代对象。 组合是按字典排序顺序发出的。那么,如果 Input iterable已排序,则 组合元组将在 排序顺序。
从2.6开始,电池包括在内!
flag = 0
requiredCals =12
from itertools import chain, combinations
def powerset(iterable):
s = list(iterable) # allows duplicate elements
return chain.from_iterable(combinations(s, r) for r in range(len(s)+1))
stuff = [2,9,5,1,6]
for i, combo in enumerate(powerset(stuff), 1):
if(len(combo)>0):
#print(combo , sum(combo))
if(sum(combo)== requiredCals):
flag = 1
break
if(flag==1):
print('True')
else:
print('else')
这种方法可以很容易地移植到所有支持递归的编程语言中(没有itertools,没有yield,没有列表理解):
def combs(a):
if len(a) == 0:
return [[]]
cs = []
for c in combs(a[1:]):
cs += [c, c+[a[0]]]
return cs
>>> combs([1,2,3,4,5])
[[], [1], [2], [2, 1], [3], [3, 1], [3, 2], ..., [5, 4, 3, 2, 1]]