Python - vectorizing a sliding window

Question

I'm trying to vectorize a sliding window operation. For the 1-d case a helpful example could go along the lines of: ``` x= vstack((np.array([range(10)]),np.array([range(10)]))) x[1,:]=np.where((x[0,:]<5)&(x[0,:]>0),x[1,x[0,:]+1],x[1,:]) ``` The n+1 value for each current value for indices <5. But I get this error: ``` x[1,:]=np.where((x[0,:]<2)&(x[0,:]>0),x[1,x[0,:]+1],x[1,:]) IndexError: index (10) out of range (0<=index<9) in dimension 1 ``` Curiously I wouldn't get this error for the n-1 value which would mean indices smaller than 0. It doesn't seem to mind: ``` x[1,:]=np.where((x[0,:]<5)&(x[0,:]>0),x[1,x[0,:]-1],x[1,:]) print(x) [[0 1 2 3 4 5 6 7 8 9] [0 0 1 2 3 5 6 7 8 9]] ``` Is there anyway around this? is my approach totally wrong? any comments would be appreciated. EDIT : This is what I would like to achieve, I flatten a matrix to an numpy array on which I want to calculate the mean of the 6x6 neighborhood of each cell: ``` matriz = np.array([[1,2,3,4,5], [6,5,4,3,2], [1,1,2,…

Accepted Answer

If I understand the problem correctly you would like to take the mean of all numbers 1 step around the index, neglecting the index.

I have patched your function to work, I believe you were going for something like this:

```
def original(matriz):

vector2 = np.ndarray.flatten(matriz)

nrows, ncols= matriz.shape
    vector = np.zeros(nrows*ncols,dtype='float64')

# Interior pixels
    for i in range(vector.shape[0]):
        if ( (i % ncols) != 0 and (i+1) % ncols != 0 and i>ncols and i<ncols*(nrows-1)):

vector[i] = np.mean(np.array([vector2[i-ncols-1],vector2[i-ncols],\
                        vector2[i-ncols+1],vector2[i-1],vector2[i+1],\
                        vector2[i+ncols-1],vector2[i+ncols],vector2[i+ncols+1]]))
```

I rewrote this using using slicing and views:

```
def mean_around(arr):
    arr=arr.astype(np.float64)

out= np.copy(arr[:-2,:-2])  #Top left corner
    out+= arr[:-2,2:]           #Top right corner
    out+= arr[:-2,1:-1]         #Top center
    out+= arr[2:,:-2]           #etc
    out+= arr[2:,2:]
    out+= arr[2:,1:-1]
    out+= arr[1:-1,2:]
    out+= arr[1:-1,:-2]

out/=8.0    #Divide by # of elements to obtain mean

cout=np.empty_like(arr)  #Create output array
    cout[1:-1,1:-1]=out      #Fill with out values
    cout[0,:]=0;cout[-1,:]=0;cout[:,0]=0;cout[:,-1]=0 #Set edges equal to zero

return  cout
```

Using `np.empty_like` and then filling the edges seemed slightly faster then `np.zeros_like`. First lets double check they give the same thing using your `matriz` array.

```
print np.allclose(mean_around(matriz),original(matriz))
True

print mean_around(matriz)
[[ 0.     0.     0.     0.     0.   ]
 [ 0.     2.5    2.75   3.125  0.   ]
 [ 0.     3.25   2.75   2.375  0.   ]
 [ 0.     1.875  2.     2.     0.   ]
 [ 0.     2.25   2.25   1.75   0.   ]
 [ 0.     0.     0.     0.     0.   ]]
```

Some timings:

```
a=np.random.rand(500,500)

print np.allclose(original(a),mean_around(a))
True

%timeit mean_around(a)
100 loops, best of 3: 4.4 ms per loop

%timeit original(a)
1 loops, best of 3: 6.6 s per loop
```

Roughly ~1500x speedup.

Looks like a good place to use numba:

```
def mean_numba(arr):
    out=np.zeros_like(arr)
    col,rows=arr.shape

for x in xrange(1,col-1):
        for y in xrange(1,rows-1):
            out[x,y]=(arr[x-1,y+1]+arr[x-1,y]+arr[x-1,y-1]+arr[x,y+1]+\
                      arr[x,y-1]+arr[x+1,y+1]+arr[x+1,y]+arr[x+1,y-1])/8.
    return out

nmean= autojit(mean_numba)
```

Now lets compare against all presented methods.

```
a=np.random.rand(5000,5000)

%timeit mean_around(a)
1 loops, best of 3: 729 ms per loop

%timeit nmean(a)
10 loops, best of 3: 169 ms per loop

#CT Zhu's answer
%timeit it_mean(a)
1 loops, best of 3: 36.7 s per loop

#Ali_m's answer
%timeit fast_local_mean(a,(3,3))
1 loops, best of 3: 4.7 s per loop

#lmjohns3's answer
%timeit scipy_conv(a)
1 loops, best of 3: 3.72 s per loop
```

A 4x speed with numba up is pretty nominal indicating that the numpy code is about as good as its going to get. I pulled the other codes as presented, although I did have to change @CTZhu's answer to include different array sizes.

Python - vectorizing a sliding window

Solution

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