我基本上是在寻找组合的 python版本List<List<int>>
给定一个列表列表,我需要一个新列表,该列表给出列表之间所有可能的项目组合。
[[1,2,3],[4,5,6],[7,8,9,10]] -> [[1,4,7],[1,4,8],...,[3,6,10]]
列表的数量是未知的,因此我需要适用于所有情况的东西。优雅奖励积分!
我基本上是在寻找组合的 python版本List<List<int>>
给定一个列表列表,我需要一个新列表,该列表给出列表之间所有可能的项目组合。
[[1,2,3],[4,5,6],[7,8,9,10]] -> [[1,4,7],[1,4,8],...,[3,6,10]]
列表的数量是未知的,因此我需要适用于所有情况的东西。优雅奖励积分!
Answers:
>>> import itertools
>>> a = [[1,2,3],[4,5,6],[7,8,9,10]]
>>> list(itertools.product(*a))
[(1, 4, 7), (1, 4, 8), (1, 4, 9), (1, 4, 10), (1, 5, 7), (1, 5, 8), (1, 5, 9), (1, 5, 10), (1, 6, 7), (1, 6, 8), (1, 6, 9), (1, 6, 10), (2, 4, 7), (2, 4, 8), (2, 4, 9), (2, 4, 10), (2, 5, 7), (2, 5, 8), (2, 5, 9), (2, 5, 10), (2, 6, 7), (2, 6, 8), (2, 6, 9), (2, 6, 10), (3, 4, 7), (3, 4, 8), (3, 4, 9), (3, 4, 10), (3, 5, 7), (3, 5, 8), (3, 5, 9), (3, 5, 10), (3, 6, 7), (3, 6, 8), (3, 6, 9), (3, 6, 10)]
a
输入作为输入,遍历itertools.product(*a)
和处理yield
由产生的元组itertools
和反向版本(例如创建一个列表,reverse()
然后将其转换回元组)。最好问一个新问题。
最优雅的解决方案是在python 2.6中使用itertools.product。
如果您不使用Python 2.6,itertools.product的文档实际上会显示等效的功能,以“手动”方式完成产品:
def product(*args, **kwds):
# product('ABCD', 'xy') --> Ax Ay Bx By Cx Cy Dx Dy
# product(range(2), repeat=3) --> 000 001 010 011 100 101 110 111
pools = map(tuple, args) * kwds.get('repeat', 1)
result = [[]]
for pool in pools:
result = [x+[y] for x in result for y in pool]
for prod in result:
yield tuple(prod)
listOLists = [[1,2,3],[4,5,6],[7,8,9,10]]
for list in itertools.product(*listOLists):
print list;
希望您能像我初次接触时一样优雅。
Numpy可以做到:
>>> import numpy
>>> a = [[1,2,3],[4,5,6],[7,8,9,10]]
>>> [list(x) for x in numpy.array(numpy.meshgrid(*a)).T.reshape(-1,len(a))]
[[ 1, 4, 7], [1, 5, 7], [1, 6, 7], ....]
直接执行此任务的递归没有什么问题,如果您需要使用字符串的版本,则可以满足您的需求:
combinations = []
def combine(terms, accum):
last = (len(terms) == 1)
n = len(terms[0])
for i in range(n):
item = accum + terms[0][i]
if last:
combinations.append(item)
else:
combine(terms[1:], item)
>>> a = [['ab','cd','ef'],['12','34','56']]
>>> combine(a, '')
>>> print(combinations)
['ab12', 'ab34', 'ab56', 'cd12', 'cd34', 'cd56', 'ef12', 'ef34', 'ef56']
一个人可以为此使用基本的python。该代码需要一个函数来拉平列表的列表:
def flatten(B): # function needed for code below;
A = []
for i in B:
if type(i) == list: A.extend(i)
else: A.append(i)
return A
然后可以运行:
L = [[1,2,3],[4,5,6],[7,8,9,10]]
outlist =[]; templist =[[]]
for sublist in L:
outlist = templist; templist = [[]]
for sitem in sublist:
for oitem in outlist:
newitem = [oitem]
if newitem == [[]]: newitem = [sitem]
else: newitem = [newitem[0], sitem]
templist.append(flatten(newitem))
outlist = list(filter(lambda x: len(x)==len(L), templist)) # remove some partial lists that also creep in;
print(outlist)
输出:
[[1, 4, 7], [2, 4, 7], [3, 4, 7],
[1, 5, 7], [2, 5, 7], [3, 5, 7],
[1, 6, 7], [2, 6, 7], [3, 6, 7],
[1, 4, 8], [2, 4, 8], [3, 4, 8],
[1, 5, 8], [2, 5, 8], [3, 5, 8],
[1, 6, 8], [2, 6, 8], [3, 6, 8],
[1, 4, 9], [2, 4, 9], [3, 4, 9],
[1, 5, 9], [2, 5, 9], [3, 5, 9],
[1, 6, 9], [2, 6, 9], [3, 6, 9],
[1, 4, 10], [2, 4, 10], [3, 4, 10],
[1, 5, 10], [2, 5, 10], [3, 5, 10],
[1, 6, 10], [2, 6, 10], [3, 6, 10]]
from itertools import product
list_vals = [['Brand Acronym:CBIQ', 'Brand Acronym :KMEFIC'],['Brand Country:DXB','Brand Country:BH']]
list(product(*list_vals))
输出:
[(''品牌缩写:CBIQ','品牌国家:DXB'),
('品牌缩写:CBIQ','品牌国家:BH'),
('品牌缩写:KMEFIC','品牌国家:DXB'),
( '品牌缩写:KMEFIC','品牌国家:BH')]
*a
吗?