python中的映射_在Python中映射数组的好方法是什么？

我有一个旧的遗留Fortran代码,将从Python调用.

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？