[英]Get the cartesian product of a series of lists?
如何从一组列表中获取笛卡尔积(每个可能的值组合)?
输入:
somelists = [
[1, 2, 3],
['a', 'b'],
[4, 5]
]
所需的 output:
[(1, 'a', 4), (1, 'a', 5), (1, 'b', 4), (1, 'b', 5), (2, 'a', 4), (2, 'a', 5) ...]
itertools.product
可从 Python 2.6 获得。
import itertools
somelists = [
[1, 2, 3],
['a', 'b'],
[4, 5]
]
for element in itertools.product(*somelists):
print(element)
这与,
for element in itertools.product([1, 2, 3], ['a', 'b'], [4, 5]):
print(element)
import itertools
>>> for i in itertools.product([1,2,3],['a','b'],[4,5]):
... print i
...
(1, 'a', 4)
(1, 'a', 5)
(1, 'b', 4)
(1, 'b', 5)
(2, 'a', 4)
(2, 'a', 5)
(2, 'b', 4)
(2, 'b', 5)
(3, 'a', 4)
(3, 'a', 5)
(3, 'b', 4)
(3, 'b', 5)
>>>
对于 Python 2.5 及更早版本:
>>> [(a, b, c) for a in [1,2,3] for b in ['a','b'] for c in [4,5]]
[(1, 'a', 4), (1, 'a', 5), (1, 'b', 4), (1, 'b', 5), (2, 'a', 4),
(2, 'a', 5), (2, 'b', 4), (2, 'b', 5), (3, 'a', 4), (3, 'a', 5),
(3, 'b', 4), (3, 'b', 5)]
这是product()
的递归版本(只是一个说明):
def product(*args):
if not args:
return iter(((),)) # yield tuple()
return (items + (item,)
for items in product(*args[:-1]) for item in args[-1])
例子:
>>> list(product([1,2,3], ['a','b'], [4,5]))
[(1, 'a', 4), (1, 'a', 5), (1, 'b', 4), (1, 'b', 5), (2, 'a', 4),
(2, 'a', 5), (2, 'b', 4), (2, 'b', 5), (3, 'a', 4), (3, 'a', 5),
(3, 'b', 4), (3, 'b', 5)]
>>> list(product([1,2,3]))
[(1,), (2,), (3,)]
>>> list(product([]))
[]
>>> list(product())
[()]
我会使用列表理解:
somelists = [
[1, 2, 3],
['a', 'b'],
[4, 5]
]
cart_prod = [(a,b,c) for a in somelists[0] for b in somelists[1] for c in somelists[2]]
import itertools
result = list(itertools.product(*somelists))
这是一个递归生成器,它不存储任何临时列表
def product(ar_list):
if not ar_list:
yield ()
else:
for a in ar_list[0]:
for prod in product(ar_list[1:]):
yield (a,)+prod
print list(product([[1,2],[3,4],[5,6]]))
输出:
[(1, 3, 5), (1, 3, 6), (1, 4, 5), (1, 4, 6), (2, 3, 5), (2, 3, 6), (2, 4, 5), (2, 4, 6)]
在 Python 2.6 及更高版本中,您可以使用“itertools.product”。 在旧版本的 Python 中,您可以使用文档中的以下(几乎 - 参见文档)等效代码,至少作为起点:
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)
两者的结果都是一个迭代器,所以如果你真的需要一个列表进行进一步处理,请使用list(result)
。
尽管已经有很多答案,但我想分享一些我的想法:
def cartesian_iterative(pools):
result = [[]]
for pool in pools:
result = [x+[y] for x in result for y in pool]
return result
def cartesian_recursive(pools):
if len(pools) > 2:
pools[0] = product(pools[0], pools[1])
del pools[1]
return cartesian_recursive(pools)
else:
pools[0] = product(pools[0], pools[1])
del pools[1]
return pools
def product(x, y):
return [xx + [yy] if isinstance(xx, list) else [xx] + [yy] for xx in x for yy in y]
def cartesian_reduct(pools):
return reduce(lambda x,y: product(x,y) , pools)
递归方法:
def rec_cart(start, array, partial, results):
if len(partial) == len(array):
results.append(partial)
return
for element in array[start]:
rec_cart(start+1, array, partial+[element], results)
rec_res = []
some_lists = [[1, 2, 3], ['a', 'b'], [4, 5]]
rec_cart(0, some_lists, [], rec_res)
print(rec_res)
迭代方法:
def itr_cart(array):
results = [[]]
for i in range(len(array)):
temp = []
for res in results:
for element in array[i]:
temp.append(res+[element])
results = temp
return results
some_lists = [[1, 2, 3], ['a', 'b'], [4, 5]]
itr_res = itr_cart(some_lists)
print(itr_res)
对上述可变变量风格的递归生成器解决方案的小修改:
def product_args(*args):
if args:
for a in args[0]:
for prod in product_args(*args[1:]) if args[1:] else ((),):
yield (a,) + prod
当然还有一个包装器,使它的工作方式与该解决方案完全相同:
def product2(ar_list):
"""
>>> list(product(()))
[()]
>>> list(product2(()))
[]
"""
return product_args(*ar_list)
有一个折衷:它检查递归是否应该在每个外循环上中断,并且一个收获:空调用时没有收益,例如product(())
,我认为这在语义上会更正确(参见 doctest )。
关于列表推导:数学定义适用于任意数量的参数,而列表推导只能处理已知数量的参数。
只是补充一点已经说过的内容:如果您使用 sympy,则可以使用符号而不是字符串,这使得它们在数学上很有用。
import itertools
import sympy
x, y = sympy.symbols('x y')
somelist = [[x,y], [1,2,3], [4,5]]
somelist2 = [[1,2], [1,2,3], [4,5]]
for element in itertools.product(*somelist):
print element
关于同情。
我相信这有效:
def cartesian_product(L):
if L:
return {(a,) + b for a in L[0]
for b in cartesian_product(L[1:])}
else:
return {()}
def my_product(pools: List[List[Any]], rules: Dict[Any, List[Any]], forbidden: List[Any]) -> Iterator[Tuple[Any]]:
"""
Compute the cartesian product except it rejects some combinations based on provided rules
:param pools: the values to calculate the Cartesian product on
:param rules: a dict specifying which values each value is incompatible with
:param forbidden: values that are never authorized in the combinations
:return: the cartesian product
"""
if not pools:
return
included = set()
# if an element has an entry of 0, it's acceptable, if greater than 0, it's rejected, cannot be negative
incompatibles = defaultdict(int)
for value in forbidden:
incompatibles[value] += 1
selections = [-1] * len(pools)
pool_idx = 0
def current_value():
return pools[pool_idx][selections[pool_idx]]
while True:
# Discard incompatibilities from value from previous iteration on same pool
if selections[pool_idx] >= 0:
for value in rules[current_value()]:
incompatibles[value] -= 1
included.discard(current_value())
# Try to get to next value of same pool
if selections[pool_idx] != len(pools[pool_idx]) - 1:
selections[pool_idx] += 1
# Get to previous pool if current is exhausted
elif pool_idx != 0:
selections[pool_idx] = - 1
pool_idx -= 1
continue
# Done if first pool is exhausted
else:
break
# Add incompatibilities of newly added value
for value in rules[current_value()]:
incompatibles[value] += 1
included.add(current_value())
# Skip value if incompatible
if incompatibles[current_value()] or \
any(intersection in included for intersection in rules[current_value()]):
continue
# Submit combination if we're at last pool
if pools[pool_idx] == pools[-1]:
yield tuple(pool[selection] for pool, selection in zip(pools, selections))
# Else get to next pool
else:
pool_idx += 1
我有一个案例,我必须获取一个非常大的笛卡尔积的第一个结果。 尽管我只想要一件物品,但这需要很长时间。 问题在于,由于结果的顺序,它必须遍历许多不需要的结果才能找到正确的结果。 因此,如果我有 10 个包含 50 个元素的列表,并且前两个列表的第一个元素不兼容,则它必须遍历最后 8 个列表的笛卡尔积,尽管它们都会被拒绝。
此实现允许在结果包含每个列表中的一项之前对其进行测试。 因此,当我检查某个元素是否与之前列表中已包含的元素不兼容时,我会立即转到当前列表的下一个元素,而不是遍历以下列表的所有产品。
以下代码是Using numpy to build an array of all combination of two arrays的 95% 副本,所有学分都在那里! 据说这要快得多,因为它仅在 numpy 中。
import numpy as np
def cartesian(arrays, dtype=None, out=None):
arrays = [np.asarray(x) for x in arrays]
if dtype is None:
dtype = arrays[0].dtype
n = np.prod([x.size for x in arrays])
if out is None:
out = np.zeros([n, len(arrays)], dtype=dtype)
m = int(n / arrays[0].size)
out[:,0] = np.repeat(arrays[0], m)
if arrays[1:]:
cartesian(arrays[1:], out=out[0:m, 1:])
for j in range(1, arrays[0].size):
out[j*m:(j+1)*m, 1:] = out[0:m, 1:]
return out
如果您不想从所有条目的第一个条目中获取 dtype,则需要将 dtype 定义为参数。 如果您有字母和数字作为项目,请使用 dtype = 'object'。 测试:
somelists = [
[1, 2, 3],
['a', 'b'],
[4, 5]
]
[tuple(x) for x in cartesian(somelists, 'object')]
出去:
[(1, 'a', 4),
(1, 'a', 5),
(1, 'b', 4),
(1, 'b', 5),
(2, 'a', 4),
(2, 'a', 5),
(2, 'b', 4),
(2, 'b', 5),
(3, 'a', 4),
(3, 'a', 5),
(3, 'b', 4),
(3, 'b', 5)]
这可以作为
[(x, y) for x in range(10) for y in range(10)]
另一个变量? 没问题:
[(x, y, z) for x in range(10) for y in range(10) for z in range(10)]
列表理解简单明了:
import itertools
somelists = [
[1, 2, 3],
['a', 'b'],
[4, 5]
]
lst = [i for i in itertools.product(*somelists)]
在 99% 的情况下,您应该使用itertools.product 。 它是用高效的 C 代码编写的,因此它可能比任何自定义实现都要好。
在 1% 的情况下,您需要纯 Python 算法,您可以使用以下算法。
def product(*args, repeat=1):
"""Find the Cartesian product of the arguments.
The interface is identical to itertools.product.
"""
# Initialize data structures and handle bad input
if len(args) == 0:
return []
gears = [tuple(arg) for arg in args] * repeat
for gear in gears:
if len(gear) == 0:
return []
tooth_numbers = [0] * len(gears)
result = [gear[0] for gear in gears]
# Rotate through all gears
finished = False
while not finished:
yield tuple(result)
# Get next result
gear_number = len(gears) - 1
while gear_number >= 0:
gear = gears[gear_number]
tooth_number = tooth_numbers[gear_number] + 1
if tooth_number < len(gear):
# No gear change is necessary, so exit the loop
result[gear_number] = gear[tooth_number]
tooth_numbers[gear_number] = tooth_number
break
result[gear_number] = gear[0]
tooth_numbers[gear_number] = 0
gear_number -= 1
else:
# We changed all the gears, so we are back at the beginning
finished = True
该接口与itertools.product的接口相同。 例如:
>>> list(product((1, 2), "ab"))
[(1, 'a'), (1, 'b'), (2, 'a'), (2, 'b')]
与此页面上的其他纯 Python 解决方案相比,此算法具有以下优点:
此代码基于PyPy 的 itertools.product 算法,该算法在 MIT 许可下发布。
巨石阵方法:
def giveAllLists(a, t):
if (t + 1 == len(a)):
x = []
for i in a[t]:
p = [i]
x.append(p)
return x
x = []
out = giveAllLists(a, t + 1)
for i in a[t]:
for j in range(len(out)):
p = [i]
for oz in out[j]:
p.append(oz)
x.append(p)
return x
xx= [[1,2,3],[22,34,'se'],['k']]
print(giveAllLists(xx, 0))
输出:
[[1, 22, 'k'], [1, 34, 'k'], [1, 'se', 'k'], [2, 22, 'k'], [2, 34, 'k'], [2, 'se', 'k'], [3, 22, 'k'], [3, 34, 'k'], [3, 'se', 'k']]
您可以使用标准库中的itertools.product
来获取笛卡尔积。 itertools
中其他很酷的相关实用程序包括permutations
、 combinations
和combinations_with_replacement
。 这是下面代码片段的 python codepen 的链接:
from itertools import product
somelists = [
[1, 2, 3],
['a', 'b'],
[4, 5]
]
result = list(product(*somelists))
print(result)
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