Numpy 2D数组索引没有超出限制和剪切值

Question

I have indices array

a = np.array([
   [0, 0],
   [1, 1],
   [1, 9]
])

And 2D array

b = np.array([
   [0, 1, 2, 3],
   [5, 6, 7, 8]
])

I can do this one

b[a[:, 0], a[:, 1]]

But it'll be an exception ' out of bounds ', because 9 is out of range. I need a very fast way to make array slice by indices and it will be ideal if I can set a clip value, eg:

np.indexing_with_clipping(array=b, indices=a, clipping_value=0)
> array([0, 6, --> 0 = clipped value <--])

Answer 1

Here's an approach -

def indexing_with_clipping(arr, indices, clipping_value=0):
    idx = np.where(indices < arr.shape,indices,clipping_value)
    return arr[idx[:, 0], idx[:, 1]]

Sample runs -

In [266]: arr
Out[266]: 
array([[0, 1, 2, 3],
       [5, 6, 7, 8]])

In [267]: indices
Out[267]: 
array([[0, 0],
       [1, 1],
       [1, 9]])

In [268]: indexing_with_clipping(arr,indices,clipping_value=0)
Out[268]: array([0, 6, 5])

In [269]: indexing_with_clipping(arr,indices,clipping_value=1)
Out[269]: array([0, 6, 6])

In [270]: indexing_with_clipping(arr,indices,clipping_value=2)
Out[270]: array([0, 6, 7])

In [271]: indexing_with_clipping(arr,indices,clipping_value=3)
Out[271]: array([0, 6, 8])

---

With focus on memory and performance efficiency, here's an approach that modifies the indices within the function -

def indexing_with_clipping_v2(arr, indices, clipping_value=0):
    indices[indices >= arr.shape] = clipping_value
    return arr[indices[:, 0], indices[:, 1]]

Sample run -

In [307]: arr
Out[307]: 
array([[0, 1, 2, 3],
       [5, 6, 7, 8]])

In [308]: indices
Out[308]: 
array([[0, 0],
       [1, 1],
       [1, 9]])

In [309]: indexing_with_clipping_v2(arr,indices,clipping_value=2)
Out[309]: array([0, 6, 7])

Answer 2

You can use list comprehension:

b[
    [min(x,len(b[0])-1) for x in a[:,0]],
    [min(x,len(b[1])-1) for x in a[:,1]]
]

edit I used last array value as your clipping value, but you can replace the min() function with whatever you want (eg trenary operator)

edit2 OK, based on clarification in comments and all python-fu that I could put together, this snipped finally does what you need:

clipping_value = -1
tmp=np.append(b,[[clipping_value],[clipping_value]],axis=1)
tmp[zip(*[((x,y) if (x<b.shape[0] and y<b.shape[1]) else (0,b.shape[1])) for (x,y) in zip(a.transpose()[0],a.transpose()[1])])]

It is the same as above, just creates ndarray tmp , which is a copy of b but contains the clipping_value as its last element and then uses my previous solution to set indices so, that they point to the last element if either of the indices is bigger than dimensions of b .

I learned that there is reverse to the zip function and that the numpy arrays accept lists as indices. It was fun. Thanks.