假设我有一个100元素的numpy数组。我对这个数组的一个子集执行一些计算 - 可能有20个元素满足某些条件。然后我在这个子集中选择一个索引，如何（有效地）恢复第一个数组中的索引？我不想对a中的所有值执行计算，因为它很昂贵，所以我只想在需要的地方执行它（满足条件的地方）。
这是一些伪代码来证明我的意思（这里的'条件'是列表理解）：
a = np.arange(100) # size = 100
b = some_function(a[[i for i in range(0,100,5)]]) # size = 20
Index = np.argmax(b)
# Index gives the index of the maximum value in b,
# but what I really want is the index of the element
# in a
修改
我不太清楚，所以我提供了一个更完整的例子。我希望这能让我更明确地了解自己的目标。我觉得有一些聪明而有效的方法可以做到这一点，没有一些循环或查找。
CODE：
import numpy as np
def some_function(arr):
return arr*2.0
a = np.arange(100)*2. # size = 100
b = some_function(a[[i for i in range(0,100,5)]]) # size = 20
Index = np.argmax(b)
print Index
# Index gives the index of the maximum value in b, but what I really want is
# the index of the element in a
# In this specific case, Index will be 19. So b[19] is the largest value
# in b. Now, what I REALLY want is the index in a. In this case, that would
# 95 because some_function(a[95]) is what made the largest value in b.
print b[Index]
print some_function(a[95])
# It is important to note that I do NOT want to change a. I will perform
# several calculations on SOME values of a, then return the indices of 'a' where
# all calculations meet some condition.
答案 0 :(得分：2)
我不确定我是否理解你的问题。所以，如果我错了，请纠正我。
假设您有类似
的内容a = np.arange(100)
condition = (a % 5 == 0) & (a % 7 == 0)
b = a[condition]
index = np.argmax(b)
# The following should do what you want
a[condition][index]
或者如果你不想使用面具：
a = np.arange(100)
b_indices = np.where(a % 5 == 0)
b = a[b_indices]
index = np.argmax(b)
# Get the value of 'a' corresponding to 'index'
a[b_indices][index]
这是你想要的吗？
答案 1 :(得分：0)
通常，在对数组进行任何更改之前，您将根据条件存储索引。您可以使用索引进行更改。
如果您的数组是a：
>>> a = np.random.random((10,5))
>>> a
array([[ 0.22481885, 0.80522855, 0.1081426 , 0.42528799, 0.64471832],
[ 0.28044374, 0.16202575, 0.4023426 , 0.25480368, 0.87047212],
[ 0.84764143, 0.30580141, 0.16324907, 0.20751965, 0.15903343],
[ 0.55861168, 0.64368466, 0.67676172, 0.67871825, 0.01849056],
[ 0.90980614, 0.95897292, 0.15649259, 0.39134528, 0.96317126],
[ 0.20172827, 0.9815932 , 0.85661944, 0.23273944, 0.86819205],
[ 0.98363954, 0.00219531, 0.91348196, 0.38197302, 0.16002007],
[ 0.48069675, 0.46057327, 0.67085243, 0.05212357, 0.44870942],
[ 0.7031601 , 0.50889065, 0.30199446, 0.8022497 , 0.82347358],
[ 0.57058441, 0.38748261, 0.76947605, 0.48145936, 0.26650583]])
b是您的子阵列：
>>> b = a[2:4,2:7]
>>> b
array([[ 0.16324907, 0.20751965, 0.15903343],
[ 0.67676172, 0.67871825, 0.01849056]])
可以证明a仍然拥有b中的数据：
>>> b.base
array([[ 0.22481885, 0.80522855, 0.1081426 , 0.42528799, 0.64471832],
[ 0.28044374, 0.16202575, 0.4023426 , 0.25480368, 0.87047212],
[ 0.84764143, 0.30580141, 0.16324907, 0.20751965, 0.15903343],
[ 0.55861168, 0.64368466, 0.67676172, 0.67871825, 0.01849056],
[ 0.90980614, 0.95897292, 0.15649259, 0.39134528, 0.96317126],
[ 0.20172827, 0.9815932 , 0.85661944, 0.23273944, 0.86819205],
[ 0.98363954, 0.00219531, 0.91348196, 0.38197302, 0.16002007],
[ 0.48069675, 0.46057327, 0.67085243, 0.05212357, 0.44870942],
[ 0.7031601 , 0.50889065, 0.30199446, 0.8022497 , 0.82347358],
[ 0.57058441, 0.38748261, 0.76947605, 0.48145936, 0.26650583]])
您可以通过两种方式对a和b进行更改：
>>> b+=1
>>> b
array([[ 1.16324907, 1.20751965, 1.15903343],
[ 1.67676172, 1.67871825, 1.01849056]])
>>> a
array([[ 0.22481885, 0.80522855, 0.1081426 , 0.42528799, 0.64471832],
[ 0.28044374, 0.16202575, 0.4023426 , 0.25480368, 0.87047212],
[ 0.84764143, 0.30580141, 1.16324907, 1.20751965, 1.15903343],
[ 0.55861168, 0.64368466, 1.67676172, 1.67871825, 1.01849056],
[ 0.90980614, 0.95897292, 0.15649259, 0.39134528, 0.96317126],
[ 0.20172827, 0.9815932 , 0.85661944, 0.23273944, 0.86819205],
[ 0.98363954, 0.00219531, 0.91348196, 0.38197302, 0.16002007],
[ 0.48069675, 0.46057327, 0.67085243, 0.05212357, 0.44870942],
[ 0.7031601 , 0.50889065, 0.30199446, 0.8022497 , 0.82347358],
[ 0.57058441, 0.38748261, 0.76947605, 0.48145936, 0.26650583]])
或者：
>>> a[2:4,2:7]+=1
>>> a
array([[ 0.22481885, 0.80522855, 0.1081426 , 0.42528799, 0.64471832],
[ 0.28044374, 0.16202575, 0.4023426 , 0.25480368, 0.87047212],
[ 0.84764143, 0.30580141, 1.16324907, 1.20751965, 1.15903343],
[ 0.55861168, 0.64368466, 1.67676172, 1.67871825, 1.01849056],
[ 0.90980614, 0.95897292, 0.15649259, 0.39134528, 0.96317126],
[ 0.20172827, 0.9815932 , 0.85661944, 0.23273944, 0.86819205],
[ 0.98363954, 0.00219531, 0.91348196, 0.38197302, 0.16002007],
[ 0.48069675, 0.46057327, 0.67085243, 0.05212357, 0.44870942],
[ 0.7031601 , 0.50889065, 0.30199446, 0.8022497 , 0.82347358],
[ 0.57058441, 0.38748261, 0.76947605, 0.48145936, 0.26650583]])
>>> b
array([[ 1.16324907, 1.20751965, 1.15903343],
[ 1.67676172, 1.67871825, 1.01849056]])
两者都是等价的，两者都不比另一个贵。因此，只要保留从b创建a的索引，就可以始终查看基础数组中已更改的数据。通常，在对切片执行操作时甚至不需要创建子数组。
修改
这假定some_func返回子数组中某些条件为真的索引。
我认为当函数返回索引并且您只想将该函数作为子数组提供时，您仍然需要存储该子数组的索引并使用它们来获取基本数组索引。例如：
>>> def some_func(a):
... return np.where(a>.8)
>>> a = np.random.random((10,4))
>>> a
array([[ 0.94495378, 0.55532342, 0.70112911, 0.4385163 ],
[ 0.12006191, 0.93091941, 0.85617421, 0.50429453],
[ 0.46246102, 0.89810859, 0.31841396, 0.56627419],
[ 0.79524739, 0.20768512, 0.39718061, 0.51593312],
[ 0.08526902, 0.56109783, 0.00560285, 0.18993636],
[ 0.77943988, 0.96168229, 0.10491335, 0.39681643],
[ 0.15817781, 0.17227806, 0.17493879, 0.93961027],
[ 0.05003535, 0.61873245, 0.55165992, 0.85543841],
[ 0.93542227, 0.68104872, 0.84750821, 0.34979704],
[ 0.06888627, 0.97947905, 0.08523711, 0.06184216]])
>>> i_off, j_off = 3,2
>>> b = a[i_off:,j_off:] #b
>>> i = some_func(b) #indicies in b
>>> i
(array([3, 4, 5]), array([1, 1, 0]))
>>> map(sum, zip(i,(i_off, j_off))) # indicies in a
[array([6, 7, 8]), array([3, 3, 2])]
修改2
这假定some_func返回子数组b的修改副本。
你的例子看起来像这样：
import numpy as np
def some_function(arr):
return arr*2.0
a = np.arange(100)*2. # size = 100
idx = np.array(range(0,100,5))
b = some_function(a[idx]) # size = 20
b_idx = np.argmax(b)
a_idx = idx[b_idx] # indices in a translated from indices in b
print b_idx, a_idx
print b[b_idx], a[a_idx]
assert b[b_idx] == 2* a[a_idx] #true!
答案 2 :(得分：0)
使用辅助数组a_index，它只是a元素的索引，因此a_index[3,5] = (3,5)。然后，您可以将原始索引设为a_index[condition == True][Index]。
如果您可以保证b是a的视图，则可以使用两个数组的memory layout信息来查找b和a的索引之间的转换。
答案 3 :(得分：0)
这样的事情有用吗？
mask = S == 1
ind_local = np.argmax(X[mask])
G = np.ravel_multi_index(np.where(mask), mask.shape)
ind_global = np.unravel_index(G[ind_local], mask.shape)
return ind_global
这将返回argmax的全局索引。