在Python中映射数组的好方法是什么？

在此代码中,数据数组由某种算法计算.我简化了它：假设我们有10个元素可以继续(在实际应用中它通常是10e 6而不是10)：

number_of_elements = 10
element_id_1 = [0,1,2,3,0] # size = number_of_elements
element_id_2 = [0,2]                      # size = max(element_id_1)

然后按如下方式使用这些数组：

my_element = one_of_my_10_elements # does not matter where it comes from
my_element_position = elt_position_in_element_id_1 # does not matter how
id_1 = element_id_1[my_element_position]
if id_1 == 0:
   id_2 = None
else:
   id_2 = element_id_2[id_1-1]
   modify(my_element,some_other_data[id_2])

什么是Pythonic / numpy管理这种关系的方式,即获得给定元素的id_2？

我已经看过掩盖的阵列,但我还没有找到一种方法来使用它们进行这种配置.为元素实现一个类,它将在计算后存储id_2并稍后提供它,这让我想到与数组操作相比非常差的计算时间.我错了吗？

UPD.遗留代码中当前完成的更大范例：

import numpy as np
number_of_elements = 10

elements = np.arange(number_of_elements,dtype=int)  # my elements IDs
# elements data
# where element_x[7] provides X value for element 7
#   and element_n[7] provides N value for element 7
element_x = np.arange(number_of_elements,dtype=np.float)
element_n = np.arange(number_of_elements,dtype=np.int32)

# array defining subsets of elements
# where
# element_id_1[1] = element_id_1[3] = element_id_1[4] means elements 1,3 and 4 have something in common
# and
# element_id_1[9] = 0 means element 9 does not belong to any group
element_id_1 = np.array([0,0])  # size = number_of_elements

# array defining other data for each group of elements
# element_id_2[0] means elements of group 1 (elements 1,3 and 4) have no data associated
# element_id_2[1] = 1 means elements of group 2 (elements 2 and 5) have data associated: other_x[element_id_2[1]-1] = 7.
# element_id_2[2] = 2 means elements of group 3 (elements 6 and 8) have data associated: other_x[element_id_2[1]-1] = 5.
element_id_2 = np.array([0,2])  # size = max(element_id_1)
other_x = np.array([7.,5.]) # size = max(element_id_2)

# work with elements
for my_element_position in elements:
    id_1 = element_id_1[my_element_position]

    if id_1 == 0:
        print 'element %d,skipping'%(my_element_position)
        continue

    id_2 = element_id_2[id_1-1]

    if id_2 > 0:
        # use element_x[my_element_position],element_n[my_element_position] and other_x[id_2] to compute more data
        print 'element %d,using other_x[%d] = %f'%(my_element_position,id_2,other_x[id_2-1])
    else:
        # use element_x[my_element_position] and element_n[my_element_position] to compute more data
        print 'element %d,not using other_x'%(my_element_position)

我知道切片一个numpy数组应该比迭代它更快,我已经得到了以下切片：

elements_to_skip = np.where(element_id_1[:] == 0)[0]
for my_element_position in elements_to_skip:
    print 'element %d,skipping'%(my_element_position)

elements_with_id1 = np.where(element_id_1[:] > 0)[0]
array1 = element_id_1[elements_with_id1]
array1 = element_id_2[array1-1]
array1 = np.where(array1[:] > 0)[0]
elements_with_data = elements_with_id1[array1]
id_2_array = element_id_2[element_id_1[elements_with_data]-1]
for my_element_position,id_2 in zip(elements_with_data,id_2_array):
    print 'element %d,other_x[id_2-1])

elements_without_data = np.delete(elements,np.concatenate((elements_to_skip,elements_with_data)))
for my_element_position in elements_without_data:
    print 'element %d,not using other_x'%(my_element_position)

这与上面的代码片段给出了相同的结果.你有没有办法让这个难以理解的代码变得更好？这种方法比以前的代码片段更值得推荐吗？

最佳答案

如果我有类似的问题,我会使用hashMaps. python中的dict几乎与大多数语言中的hashMap相同.
详细信息检查：Python dictionary implementation
所以类似于：

id2_dict = {}
my_element = one_of_my_10_elements # does not matter where it comes from
my_element_position = elt_position_in_element_id_1 # does not matter how
id_1 = element_id_1[my_element_position]
if id_1 == 0:
   id2_dict[id_1] = None
else:
   id2_dict[id_1] = id2_dict[id_1-1]

考虑到数据的性质(整数),您可能希望使用列表,但如果您的id_1值很稀疏,那么您将浪费大量空间并采用较少的pythonic方法.但是,如果您的id_1值跨越整数范围,并且在某些范围内密集,那么请使用列表并相应地处理索引. list会为你节省散列部分,但会减少pythonic并且难以维护.
tl; dr：如果id_1s密集且几乎跨越范围,则使用list和id_1作为索引(带有一些索引移位),否则使用(hashmap)dict和id_1作为键.

在Python中映射数组的好方法是什么？

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