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个组合。
我已经找到了一些代码(通过Googling),这些代码显然可以满足我的需求,但是我发现代码相当不透明并且对使用它很谨慎。另外,我觉得必须有一个更优雅的解决方案。
对我而言,唯一发生的就是循环遍历十进制整数1–32768,并将其转换为二进制,然后使用二进制表示形式作为筛选器来选择适当的数字。
有谁知道更好的方法吗?使用map()
,也许?
product
,等等)
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个组合。
我已经找到了一些代码(通过Googling),这些代码显然可以满足我的需求,但是我发现代码相当不透明并且对使用它很谨慎。另外,我觉得必须有一个更优雅的解决方案。
对我而言,唯一发生的就是循环遍历十进制整数1–32768,并将其转换为二进制,然后使用二进制表示形式作为筛选器来选择适当的数字。
有谁知道更好的方法吗?使用map()
,也许?
product
,等等)
Answers:
itertools.combinations(iterable, r)
从可迭代的输入中返回元素的r长度子序列。
组合按字典顺序排序。因此,如果将输入的iterable排序,则将按排序顺序生成组合元组。
从2.6开始,包括电池!
list(itertools.combinations(iterable, r))
r
,即元素的任何长度子序列的组合。
这个答案错过了一个方面:OP要求所有组合……而不仅仅是长度“ r”的组合。
因此,您要么必须遍历所有长度“ L”:
import itertools
stuff = [1, 2, 3]
for L in range(0, len(stuff)+1):
for subset in itertools.combinations(stuff, L):
print(subset)
或者-如果您想变得眼花or乱(或者想让以后阅读代码的人都屈服)-您可以生成“ combinations()”生成器链,然后进行迭代:
from itertools import chain, combinations
def all_subsets(ss):
return chain(*map(lambda x: combinations(ss, x), range(0, len(ss)+1)))
for subset in all_subsets(stuff):
print(subset)
itertools.combinations
将项目顺序保留在它产生的列表中。因此,如果按词法对输入进行排序,那么每个输出也将被分类。
itertools.combinations
按字典顺序生成n个中的k个组合,但不是n个中直到 k个的所有组合。据我所知,powerset
生成的所有组合最多为 k,但不按字典顺序排列:powerset([1,2])-> [(),(1,),(2,),(1,2)] 。它不应该是:[(),(1,),(1,2),(2,)]吗?
这是一个懒惰的单行代码,也使用itertools:
from itertools import compress, product
def combinations(items):
return ( set(compress(items,mask)) for mask in product(*[[0,1]]*len(items)) )
# alternative: ...in product([0,1], repeat=len(items)) )
答案背后的主要思想是:有2 ^ N个组合-与长度为N的二进制字符串的数目相同。对于每个二进制字符串,请选择与“ 1”相对应的所有元素。
items=abc * mask=###
|
V
000 ->
001 -> c
010 -> b
011 -> bc
100 -> a
101 -> a c
110 -> ab
111 -> abc
注意事项:
len(...)
的items
(解决方法:如果items
是这样的迭代就像一台发电机,与第一把它变成一个列表items=list(_itemsArg)
)items
不是随机的(解决方法:不要疯了){2,2,1}
并{2,1,1}
都将崩溃{2,1}
(解决方法:使用collections.Counter
作为一个下拉更换set
;它基本上是一个多集...尽管你可能需要以后使用tuple(sorted(Counter(...).elements()))
,如果你需要它是可哈希)演示版
>>> list(combinations(range(4)))
[set(), {3}, {2}, {2, 3}, {1}, {1, 3}, {1, 2}, {1, 2, 3}, {0}, {0, 3}, {0, 2}, {0, 2, 3}, {0, 1}, {0, 1, 3}, {0, 1, 2}, {0, 1, 2, 3}]
>>> list(combinations('abcd'))
[set(), {'d'}, {'c'}, {'c', 'd'}, {'b'}, {'b', 'd'}, {'c', 'b'}, {'c', 'b', 'd'}, {'a'}, {'a', 'd'}, {'a', 'c'}, {'a', 'c', 'd'}, {'a', 'b'}, {'a', 'b', 'd'}, {'a', 'c', 'b'}, {'a', 'c', 'b', 'd'}]
在@Dan H 极力支持的答案的评论powerset()
中,itertools
文档中提到了该配方—包括Dan本人的配方。但是,到目前为止,还没有人将其发布为答案。由于它可能是解决问题的最佳方法,即使不是最好的方法之一,并且在另一位评论者的鼓励下,它显示如下。该函数产生的所有列表中的元件的独特组合每长度可能的(包括那些含有零和所有的元素)。
注意:如果,微妙的不同,目标是获得唯一的独特元素的组合,改线s = list(iterable)
,以s = list(set(iterable))
消除任何重复的元素。无论如何,iterable
最终会变成的事实list
意味着它将与生成器一起使用(与其他几个答案不同)。
from itertools import chain, combinations
def powerset(iterable):
"powerset([1,2,3]) --> () (1,) (2,) (3,) (1,2) (1,3) (2,3) (1,2,3)"
s = list(iterable) # allows duplicate elements
return chain.from_iterable(combinations(s, r) for r in range(len(s)+1))
stuff = [1, 2, 3]
for i, combo in enumerate(powerset(stuff), 1):
print('combo #{}: {}'.format(i, combo))
输出:
combo #1: ()
combo #2: (1,)
combo #3: (2,)
combo #4: (3,)
combo #5: (1, 2)
combo #6: (1, 3)
combo #7: (2, 3)
combo #8: (1, 2, 3)
list()
首先转换是什么?
这是使用递归的一种:
>>> import copy
>>> def combinations(target,data):
... for i in range(len(data)):
... new_target = copy.copy(target)
... new_data = copy.copy(data)
... new_target.append(data[i])
... new_data = data[i+1:]
... print new_target
... combinations(new_target,
... new_data)
...
...
>>> target = []
>>> data = ['a','b','c','d']
>>>
>>> combinations(target,data)
['a']
['a', 'b']
['a', 'b', 'c']
['a', 'b', 'c', 'd']
['a', 'b', 'd']
['a', 'c']
['a', 'c', 'd']
['a', 'd']
['b']
['b', 'c']
['b', 'c', 'd']
['b', 'd']
['c']
['c', 'd']
['d']
new_data = copy.copy(data)
-此行是多余的,因为据我看到的,它不会对任何影响
这种单行代码为您提供了所有组合(如果原始列表/集合包含不同的元素,则在0
和n
项之间n
),并使用本机方法itertools.combinations
:
from itertools import combinations
input = ['a', 'b', 'c', 'd']
output = sum([map(list, combinations(input, i)) for i in range(len(input) + 1)], [])
from itertools import combinations
input = ['a', 'b', 'c', 'd']
output = sum([list(map(list, combinations(input, i))) for i in range(len(input) + 1)], [])
输出将是:
[[],
['a'],
['b'],
['c'],
['d'],
['a', 'b'],
['a', 'c'],
['a', 'd'],
['b', 'c'],
['b', 'd'],
['c', 'd'],
['a', 'b', 'c'],
['a', 'b', 'd'],
['a', 'c', 'd'],
['b', 'c', 'd'],
['a', 'b', 'c', 'd']]
在线尝试:
['b', 'a']
。
TypeError: can only concatenate list (not "map") to list
我同意Dan H的观点,Ben确实要求所有组合。itertools.combinations()
没有给出所有组合。
另一个问题是,如果可迭代的输入很大,则最好返回一个生成器而不是列表中的所有内容:
iterable = range(10)
for s in xrange(len(iterable)+1):
for comb in itertools.combinations(iterable, s):
yield comb
您可以使用以下简单代码在python中生成列表的所有组合
import itertools
a = [1,2,3,4]
for i in xrange(0,len(a)+1):
print list(itertools.combinations(a,i))
结果将是:
[()]
[(1,), (2,), (3,), (4,)]
[(1, 2), (1, 3), (1, 4), (2, 3), (2, 4), (3, 4)]
[(1, 2, 3), (1, 2, 4), (1, 3, 4), (2, 3, 4)]
[(1, 2, 3, 4)]
我认为我可以为那些寻求答案的人添加此功能,而无需导入itertools或任何其他额外的库。
def powerSet(items):
"""
Power set generator: get all possible combinations of a list’s elements
Input:
items is a list
Output:
returns 2**n combination lists one at a time using a generator
Reference: edx.org 6.00.2x Lecture 2 - Decision Trees and dynamic programming
"""
N = len(items)
# enumerate the 2**N possible combinations
for i in range(2**N):
combo = []
for j in range(N):
# test bit jth of integer i
if (i >> j) % 2 == 1:
combo.append(items[j])
yield combo
简单良率生成器用法:
for i in powerSet([1,2,3,4]):
print (i, ", ", end="")
上面用法示例的输出:
[],[1],[2],[1,2],[3],[1,3],[2,3],[1,2,3],[4],[1,4] ,[2、4],[1、2、4],[3、4],[1、3、4],[2、3、4],[1、2、3、4],
这是涉及使用itertools.combinations
函数的另一种解决方案(单行),但是这里我们使用了双重列表理解(与for循环或求和相对):
def combs(x):
return [c for i in range(len(x)+1) for c in combinations(x,i)]
演示:
>>> combs([1,2,3,4])
[(),
(1,), (2,), (3,), (4,),
(1, 2), (1, 3), (1, 4), (2, 3), (2, 4), (3, 4),
(1, 2, 3), (1, 2, 4), (1, 3, 4), (2, 3, 4),
(1, 2, 3, 4)]
下面是一个“标准递归答案”,类似于其他类似的答案https://stackoverflow.com/a/23743696/711085。(我们实际上不必担心会耗尽堆栈空间,因为我们不可能处理所有N!个置换。)
它依次访问每个元素,然后接受或离开它(我们可以从该算法直接看到2 ^ N基数)。
def combs(xs, i=0):
if i==len(xs):
yield ()
return
for c in combs(xs,i+1):
yield c
yield c+(xs[i],)
演示:
>>> list( combs(range(5)) )
[(), (0,), (1,), (1, 0), (2,), (2, 0), (2, 1), (2, 1, 0), (3,), (3, 0), (3, 1), (3, 1, 0), (3, 2), (3, 2, 0), (3, 2, 1), (3, 2, 1, 0), (4,), (4, 0), (4, 1), (4, 1, 0), (4, 2), (4, 2, 0), (4, 2, 1), (4, 2, 1, 0), (4, 3), (4, 3, 0), (4, 3, 1), (4, 3, 1, 0), (4, 3, 2), (4, 3, 2, 0), (4, 3, 2, 1), (4, 3, 2, 1, 0)]
>>> list(sorted( combs(range(5)), key=len))
[(),
(0,), (1,), (2,), (3,), (4,),
(1, 0), (2, 0), (2, 1), (3, 0), (3, 1), (3, 2), (4, 0), (4, 1), (4, 2), (4, 3),
(2, 1, 0), (3, 1, 0), (3, 2, 0), (3, 2, 1), (4, 1, 0), (4, 2, 0), (4, 2, 1), (4, 3, 0), (4, 3, 1), (4, 3, 2),
(3, 2, 1, 0), (4, 2, 1, 0), (4, 3, 1, 0), (4, 3, 2, 0), (4, 3, 2, 1),
(4, 3, 2, 1, 0)]
>>> len(set(combs(range(5))))
32
使用列表理解:
def selfCombine( list2Combine, length ):
listCombined = str( ['list2Combine[i' + str( i ) + ']' for i in range( length )] ).replace( "'", '' ) \
+ 'for i0 in range(len( list2Combine ) )'
if length > 1:
listCombined += str( [' for i' + str( i ) + ' in range( i' + str( i - 1 ) + ', len( list2Combine ) )' for i in range( 1, length )] )\
.replace( "', '", ' ' )\
.replace( "['", '' )\
.replace( "']", '' )
listCombined = '[' + listCombined + ']'
listCombined = eval( listCombined )
return listCombined
list2Combine = ['A', 'B', 'C']
listCombined = selfCombine( list2Combine, 2 )
输出为:
['A', 'A']
['A', 'B']
['A', 'C']
['B', 'B']
['B', 'C']
['C', 'C']
该代码采用带有嵌套列表的简单算法...
# FUNCTION getCombos: To generate all combos of an input list, consider the following sets of nested lists...
#
# [ [ [] ] ]
# [ [ [] ], [ [A] ] ]
# [ [ [] ], [ [A],[B] ], [ [A,B] ] ]
# [ [ [] ], [ [A],[B],[C] ], [ [A,B],[A,C],[B,C] ], [ [A,B,C] ] ]
# [ [ [] ], [ [A],[B],[C],[D] ], [ [A,B],[A,C],[B,C],[A,D],[B,D],[C,D] ], [ [A,B,C],[A,B,D],[A,C,D],[B,C,D] ], [ [A,B,C,D] ] ]
#
# There is a set of lists for each number of items that will occur in a combo (including an empty set).
# For each additional item, begin at the back of the list by adding an empty list, then taking the set of
# lists in the previous column (e.g., in the last list, for sets of 3 items you take the existing set of
# 3-item lists and append to it additional lists created by appending the item (4) to the lists in the
# next smallest item count set. In this case, for the three sets of 2-items in the previous list. Repeat
# for each set of lists back to the initial list containing just the empty list.
#
def getCombos(listIn = ['A','B','C','D','E','F'] ):
listCombos = [ [ [] ] ] # list of lists of combos, seeded with a list containing only the empty list
listSimple = [] # list to contain the final returned list of items (e.g., characters)
for item in listIn:
listCombos.append([]) # append an emtpy list to the end for each new item added
for index in xrange(len(listCombos)-1, 0, -1): # set the index range to work through the list
for listPrev in listCombos[index-1]: # retrieve the lists from the previous column
listCur = listPrev[:] # create a new temporary list object to update
listCur.append(item) # add the item to the previous list to make it current
listCombos[index].append(listCur) # list length and append it to the current list
itemCombo = '' # Create a str to concatenate list items into a str
for item in listCur: # concatenate the members of the lists to create
itemCombo += item # create a string of items
listSimple.append(itemCombo) # add to the final output list
return [listSimple, listCombos]
# END getCombos()
""
)。
我知道这是更为实际使用itertools得到所有的组合,但你可以想代码实现这个部分,只有列表中理解,如果你这么碰巧的愿望,给予了很多
对于两对组合:
lambda l: [(a, b) for i, a in enumerate(l) for b in l[i+1:]]
而且,对于三对组合,就这么简单:
lambda l: [(a, b, c) for i, a in enumerate(l) for ii, b in enumerate(l[i+1:]) for c in l[i+ii+2:]]
import itertools
combs_3 = lambda l: [
(a, b, c) for i, a in enumerate(l)
for ii, b in enumerate(l[i+1:])
for c in l[i+ii+2:]
]
data = ((1, 2), 5, "a", None)
print("A:", list(itertools.combinations(data, 3)))
print("B:", combs_3(data))
# A: [((1, 2), 5, 'a'), ((1, 2), 5, None), ((1, 2), 'a', None), (5, 'a', None)]
# B: [((1, 2), 5, 'a'), ((1, 2), 5, None), ((1, 2), 'a', None), (5, 'a', None)]
不使用itertools:
def combine(inp):
return combine_helper(inp, [], [])
def combine_helper(inp, temp, ans):
for i in range(len(inp)):
current = inp[i]
remaining = inp[i + 1:]
temp.append(current)
ans.append(tuple(temp))
combine_helper(remaining, temp, ans)
temp.pop()
return ans
print(combine(['a', 'b', 'c', 'd']))
这是两个实现 itertools.combinations
返回列表的一个
def combinations(lst, depth, start=0, items=[]):
if depth <= 0:
return [items]
out = []
for i in range(start, len(lst)):
out += combinations(lst, depth - 1, i + 1, items + [lst[i]])
return out
一个还发电机
def combinations(lst, depth, start=0, prepend=[]):
if depth <= 0:
yield prepend
else:
for i in range(start, len(lst)):
for c in combinations(lst, depth - 1, i + 1, prepend + [lst[i]]):
yield c
请注意,建议为这些函数提供帮助函数,因为prepend参数是静态的,并且不会在每次调用时更改
print([c for c in combinations([1, 2, 3, 4], 3)])
# [[1, 2, 3], [1, 2, 4], [1, 3, 4], [2, 3, 4]]
# get a hold of prepend
prepend = [c for c in combinations([], -1)][0]
prepend.append(None)
print([c for c in combinations([1, 2, 3, 4], 3)])
# [[None, 1, 2, 3], [None, 1, 2, 4], [None, 1, 3, 4], [None, 2, 3, 4]]
这是一个非常肤浅的案例,但总比后悔要安全
怎么样..使用字符串而不是列表,但是同样的事情..字符串可以像Python中的列表一样对待:
def comb(s, res):
if not s: return
res.add(s)
for i in range(0, len(s)):
t = s[0:i] + s[i + 1:]
comb(t, res)
res = set()
comb('game', res)
print(res)
没有 itertools
Python 3,您可以执行以下操作:
def combinations(arr, carry):
for i in range(len(arr)):
yield carry + arr[i]
yield from combinations(arr[i + 1:], carry + arr[i])
最初在哪里 carry = "".
3个功能:
import sys
def permutations(a):
return combinations(a, len(a))
def combinations(a, n):
if n == 1:
for x in a:
yield [x]
else:
for i in range(len(a)):
for x in combinations(a[:i] + a[i+1:], n-1):
yield [a[i]] + x
def combinationsNoOrder(a, n):
if n == 1:
for x in a:
yield [x]
else:
for i in range(len(a)):
for x in combinationsNoOrder(a[:i], n-1):
yield [a[i]] + x
if __name__ == "__main__":
for s in combinations(list(map(int, sys.argv[2:])), int(sys.argv[1])):
print(s)
这是我的实现
def get_combinations(list_of_things):
"""gets every combination of things in a list returned as a list of lists
Should be read : add all combinations of a certain size to the end of a list for every possible size in the
the list_of_things.
"""
list_of_combinations = [list(combinations_of_a_certain_size)
for possible_size_of_combinations in range(1, len(list_of_things))
for combinations_of_a_certain_size in itertools.combinations(list_of_things,
possible_size_of_combinations)]
return list_of_combinations
您也可以使用优质包装中的Powerset功能more_itertools
。
from more_itertools import powerset
l = [1,2,3]
list(powerset(l))
# [(), (1,), (2,), (3,), (1, 2), (1, 3), (2, 3), (1, 2, 3)]
我们还可以验证它是否符合OP的要求
from more_itertools import ilen
assert ilen(powerset(range(15))) == 32_768
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 = 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
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')