求和任意数量数组的所有可能组合，并应用限制和返回索引 [英] summing all possible combinations of an arbitrary number of arrays and applying limits and returning indices
问题描述
这是对的修改>这个问题，除了元素本身之外，我还想返回数组元素的索引.我已经成功地修改了arraysums()，arraysums_recursive()，但是我却在为arraysums_recursive_anyvals()苦苦挣扎.详细信息如下:
This is a modification of this question in which I would like to return the indices of the arrays elements in addition to the elements themselves. I've successfully modified arraysums(), arraysums_recursive(), but I'm struggling with arraysums_recursive_anyvals(). Here are the details:
我修改了arraysums():
def arraysums(arrays,lower,upper):
products = itertools.product(*arrays)
result = list()
indices = itertools.product(*[np.arange(len(arr)) for arr in arrays])
index = list()
for n,k in zip(products,indices):
s = sum(n)
if lower <= s <= upper:
result.append(n)
index.append(k)
return result,index
现在它返回元素和元素的索引:
It now returns the elements and the indices of the elements:
N = 8
a = np.arange(N)
b = np.arange(N)-N/2
arraysums((a,b),lower=5,upper=6)
([(2, 3),
(3, 2),
(3, 3),
(4, 1),
(4, 2),
(5, 0),
(5, 1),
(6, -1),
(6, 0),
(7, -2),
(7, -1)],
[(2, 7),
(3, 6),
(3, 7),
(4, 5),
(4, 6),
(5, 4),
(5, 5),
(6, 3),
(6, 4),
(7, 2),
(7, 3)])
我还修改了@unutbu的递归解决方案，该解决方案也返回与arraysums()相同的结果:
I also modified @unutbu's recursive solution which also returns the same result as arraysums():
def arraysums_recursive(arrays, lower, upper):
if len(arrays) <= 1:
result = [(item,) for item in arrays[0] if lower <= item <= upper]
index = [] # this needs to be fixed
else:
result = []
index = []
for item in arrays[0]:
subarrays = [[item2 for item2 in arr if item2 <= upper-item]
for arr in arrays[1:]]
result.extend(
[(item,)+tup for tup in arraysums(
subarrays, lower-item, upper-item)[0]])
index.extend(
[(item,)+tup for tup in arraysums(
subarrays, lower-item, upper-item)[1]])
return result,index
最后，我修改了arraysums_recursive_anyvals()，但是我似乎无法理解为什么它不返回索引:
Finally, I modified arraysums_recursive_anyvals(), but I can't seem to understand why it does not return the indices:
def arraysums_recursive_anyvals(arrays, lower, upper):
if len(arrays) <= 1:
result = [(item,) for item in arrays[0] if lower <= item <= upper]
index = [] # this needs to be fixed
else:
minval = min(item for arr in arrays for item in arr)
# Subtract minval from arrays to guarantee all the values are positive
arrays = [[item-minval for item in arr] for arr in arrays]
# Adjust the lower and upper bounds accordingly
lower -= minval*len(arrays)
upper -= minval*len(arrays)
result = []
index = []
for item in arrays[0]:
subarrays = [[item2 for item2 in arr if item2 <= upper-item]
for arr in arrays[1:]]
if min(len(arr) for arr in subarrays) == 0:
continue
result.extend(
[(item,)+tup for tup in arraysums_recursive(
subarrays, lower-item, upper-item)[0]])
index.extend(
[(item,)+tup for tup in arraysums_recursive(
subarrays, lower-item, upper-item)[1]])
# Readjust the result by adding back minval
result = [tuple([item+minval for item in tup]) for tup in result]
return result,index
结果:
arraysums_recursive_anyvals((a,b),lower=5,upper=6)
([(2, 3),
(3, 2),
(3, 3),
(4, 1),
(4, 2),
(5, 0),
(5, 1),
(6, -1),
(6, 0),
(7, -2),
(7, -1)],
[])
推荐答案
arraysums_recursive的主要功能是它抛出了可能对结果无用的值:
A key feature of arraysums_recursive is that it throws out values which can not possibly contribute to the result:
subarrays = [[item2 for item2 in arr if item2 <= upper-item]
for arr in arrays[1:]]
尽管扔东西使索引的记录变得复杂，但这并不难.
首先，在arraysums_recursive中，展开arrays以包括索引以及项目值:
While throwing things out complicates the recording of indices, it's not too hard.
First, in arraysums_recursive expand arrays to include the index as well as the item value:
def arraysums_recursive(arrays, lower, upper):
arrays = [[(i, item) for i, item in enumerate(arr)] for arr in arrays]
...
index, result = zip(*arraysums_recursive_all_positive(arrays, lower, upper))
return result, index
现在重写arraysums_recursive_all_positive来处理arrays，它由(index, item) tuples 的列表组成.
Now rewrite arraysums_recursive_all_positive to handle arrays which consist of a list of lists of (index, item) tuples.
def arraysums_recursive(arrays, lower, upper):
arrays = [[(i, item) for i, item in enumerate(arr)] for arr in arrays]
minval = min(item for arr in arrays for i, item in arr)
# Subtract minval from arrays to guarantee all the values are positive
arrays = [[(i, item-minval) for i, item in arr] for arr in arrays]
# Adjust the lower and upper bounds accordingly
lower -= minval*len(arrays)
upper -= minval*len(arrays)
index, result = zip(*arraysums_recursive_all_positive(arrays, lower, upper))
# Readjust the result by adding back minval
result = [tuple([item+minval for item in tup]) for tup in result]
return result, index
def arraysums_recursive_all_positive(arrays, lower, upper):
# Assumes all values in arrays are positive
if len(arrays) <= 1:
result = [((i,), (item,)) for i, item in arrays[0] if lower <= item <= upper]
else:
result = []
for i, item in arrays[0]:
subarrays = [[(i, item2) for i, item2 in arr if item2 <= upper-item]
for arr in arrays[1:]]
if min(len(arr) for arr in subarrays) == 0:
continue
result.extend(
[((i,)+i_tup, (item,)+item_tup) for i_tup, item_tup in
arraysums_recursive_all_positive(subarrays, lower-item, upper-item)])
return result
import numpy as np
N = 8
a = np.arange(N)
b = np.arange(N)-N/2
result, index = arraysums_recursive((a,b),lower=5,upper=6)
产量result:
[(2.0, 3.0),
(3.0, 2.0),
(3.0, 3.0),
(4.0, 1.0),
(4.0, 2.0),
(5.0, 0.0),
(5.0, 1.0),
(6.0, -1.0),
(6.0, 0.0),
(7.0, -2.0),
(7.0, -1.0)]
和index:
((2, 7),
(3, 6),
(3, 7),
(4, 5),
(4, 6),
(5, 4),
(5, 5),
(6, 3),
(6, 4),
(7, 2),
(7, 3))
这篇关于求和任意数量数组的所有可能组合，并应用限制和返回索引的文章就介绍到这了，希望我们推荐的答案对大家有所帮助，也希望大家多多支持IT屋！
