fill Numpy array with axisymmetric values

Question

I'm trying to find a fast way to fill a Numpy array with rotation symmetric values. Imagine an array of zeros containing a cone shaped area. I have a 1D array of values and want to rotate it 360° around the center of the array. There is no 2D function like z=f(x,y), so I can't calculate the 2D values explicitly. I have something that works, but the for-loop is too slow for big arrays. This should make a circle:

values = np.ones(100)
x = np.arange(values.size)-values.size/2+0.5
y = values.size/2-0.5-np.arange(values.size)
x,y = np.meshgrid(x,y)
grid = np.rint(np.sqrt(x**2+y**2))
arr = np.zeros_like(grid)
for i in np.arange(values.size/2):
    arr[grid==i] = values[i+values.size/2]

My 1D array is of course not as simple. Can someone think of a way to get rid of the for-loop?

Update: I want to make a circular filter for convolutional blurring. Before I used np.outer(values,values) which gave me a rectangular filter. David's hint allows me to create a circular filter very fast. See below:

square filter with np.outer()

circular filter with David's answer

Answer 1

You can use fancy indexing to achieve this:

values = np.ones(100)
x = np.arange(values.size)-values.size/2+0.5
y = values.size/2-0.5-np.arange(values.size)
x,y = np.meshgrid(x,y)
grid = np.rint(np.sqrt(x**2+y**2)).astype(np.int)

arr = np.zeros_like(grid)
size_half = values.size // 2
inside = (grid < size_half)
arr[inside] = values[grid[inside] + size_half]

Here, inside select the indices that lie inside the circle, since only these items can be derived from values.

Answer 2

You can do something like that:

x=y=np.arange(-500,501)
r=np.random.randint(0,256,len(x)/np.sqrt(2)+1)
X,Y=np.meshgrid(x,y)
im=(X*X+Y*Y)**(1/2)
circles=r.take(np.int64(im))
plt.imshow(circles)