这是对 this question 的修改除了元素本身之外,我还想返回数组元素的索引。我已经成功地修改了 arraysums()
、arraysums_recursive()
,但是我在使用 arraysums_recursive_anyvals()
时遇到了困难。以下是详细信息:
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
它现在返回元素和元素的索引:
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,它也返回与 arraysums()
相同的结果:
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()
,但我似乎无法理解为什么它不返回索引:
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)],
[])
最佳答案
arraysums_recursive
的一个关键特性是它会丢弃可能对结果没有贡献的值:
subarrays = [[item2 for item2 in arr if item2 <= upper-item]
for arr in arrays[1:]]
虽然把东西扔掉会使索引的记录变得复杂,但这并不难。
首先,在 arraysums_recursive
中展开 arrays
以包含索引和项目值:
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)
元组.
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)
产生结果
:
[(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)]
和索引
:
((2, 7),
(3, 6),
(3, 7),
(4, 5),
(4, 6),
(5, 4),
(5, 5),
(6, 3),
(6, 4),
(7, 2),
(7, 3))
关于python - 对任意数量的数组的所有可能组合求和并应用限制并返回索引,我们在Stack Overflow上找到一个类似的问题: https://stackoverflow.com/questions/40429466/