My code for a LocalBinaryPatern in python is running very slow. Any tips?
No else after return
You have code like
if x:
    return 0
else:
    ...

Just drop the else, because you previously already returned. Use else branches only if the code converges again after the two branches, not if you exit the function (raise, return or exit()).
Double Initialization
You have code creating two lists:
        xshift, yshift = [0 for _ in range(8)], [0 for _ in range(8)]

        for i in range(8):
            xshift[i] = - radius * sin(i*abase)
            yshift[i] = radius * cos(i*abase)

You can combine those:
xshift = [-radius * sin(i * abase) for i in range(8)]
xshift = [ radius * cos(i * abase) for i in range(8)]

Further Dead Code

Initialization of h, w at the beginning can be moved down, after checking the radius.
lbp = [0 for _ in range(8)] -- you're doing that once before the loop and then in every iteration.

C-isms in Loop Iteration
It looks like you're used to writing in C (or a similar language) where you can only access an array by index. It shows e.g. in this code:
sum_lbp = 0
for i in range(len(feat)):
    sum_lbp += feat[i]

In Python, you would write this loop like this instead:
sum_lbp = 0
for value in feat:
    sum_lbp += value

An additional minor improvement specific to this case is to write it in one line:
sum_lbp = sum(feat)

Get familiar with the tools Python gives you at hand, they can make your life easier. In this case it's just the sum() function, but also you're not making use of e.g. dictionaries, which could make this code cleaner.
Redundant Parentheses
Here are two redundant pairs of partheses:
for x in (range(radius, (w-radius))):

Same code without them:
for x in range(radius, w - radius):

Redundant Type Conversions
Looking at int(lbp[i]*base[i]), I wonder why you int() the product. Both only contain integers, as far as I can tell.
Off-by-one Error
This code looks suspicious:
feat = [0 for _ in range(255)]
...
bits = 0
for i in range(8):
    bits += int(lbp[i]*base[i])

if bits &lt; 255:
    feat[bits] += 1

Since an octet can take 256 different values, I'd expect feat to have 256 values as well. That way you could eliminate the bits &lt; 255 check. Another way to simplify this is to reuse the sum() function again. Also, base[i] is 1 &lt;&lt; i and lbp[i] is either zero or one, so you can also write this lbp[i] &lt;&lt; i eliminating base entirely.
Further Redundant Code
Look at the inner loop:
# init `lbp` with zeroes
lbp = [0 for _ in range(8)]
# actually fill `lbp`
for i in range(8):
    ...
    if image[ry][rx] &gt; 0:
        lbp[i] = 1
# compute values from `lbp`                
bits = 0
for i in range(8):
    bits += int(lbp[i]*base[i])

This uses too many temporaries. Instead, try this approach:
bits = 0
for i in range(8):
    ...
    if image[ry][rx] &gt; 0:
        bits += 1 &lt;&lt; i