OpenCL bincount - opencl

Question

I am trying to implement a bincount operation in OpenCL which allocates an output buffer and uses indices from x to accumulate some weights at the same index (assume that num_bins == max(x)). This is equivalent to the following python code:
out = np.zeros_like(num_bins)
for i in range(len(x)):
out[x[i]] += weight[i]
return out
What I have is the following:
import pyopencl as cl
import numpy as np
ctx = cl.create_some_context()
queue = cl.CommandQueue(ctx)
prg = cl.Program(ctx, """
__kernel void bincount(__global int *res_g, __global const int* x_g, __global const int* weight_g)
{
int gid = get_global_id(0);
res_g[x_g[gid]] += weight_g[gid];
}
""").build()
# test
x = np.arange(5, dtype=np.int32).repeat(2) # [0, 0, 1, 1, 2, 2, 3, 3, 4, 4]
x_g = cl.Buffer(ctx, cl.mem_flags.READ_WRITE | cl.mem_flags.COPY_HOST_PTR, hostbuf=x)
weight = np.arange(10, dtype=np.int32) # [0, 1, 2, 3, 4, 5, 6, 7, 8, 9]
weight_g = cl.Buffer(ctx, cl.mem_flags.READ_WRITE | cl.mem_flags.COPY_HOST_PTR, hostbuf=weight)
res_g = cl.Buffer(ctx, cl.mem_flags.READ_WRITE, 4 * 5)
prg.bincount(queue, [10], None, res_g, x_g, weight_g)
# transfer back to cpu
res_np = np.empty(5).astype(np.int32)
cl.enqueue_copy(queue, res_np, res_g)
Output in res_np:
array([1, 3, 5, 7, 9], dtype=int32)
Expected output:
array([1, 5, 9, 13, 17], dtype=int32)
How do I accumulate the elements that are indexed more than once?
EDIT
The above is a contrived example, in my real-world application x will be indices from a sliding window algorithm:
x = np.array([ 0, 1, 2, 4, 5, 6, 8, 9, 10, 1, 2, 3, 5, 6, 7, 9, 10,
11, 4, 5, 6, 8, 9, 10, 12, 13, 14, 5, 6, 7, 9, 10, 11, 13,
14, 15, 8, 9, 10, 12, 13, 14, 16, 17, 18, 9, 10, 11, 13, 14, 15,
17, 18, 19, 20, 21, 22, 24, 25, 26, 28, 29, 30, 21, 22, 23, 25, 26,
27, 29, 30, 31, 24, 25, 26, 28, 29, 30, 32, 33, 34, 25, 26, 27, 29,
30, 31, 33, 34, 35, 28, 29, 30, 32, 33, 34, 36, 37, 38, 29, 30, 31,
33, 34, 35, 37, 38, 39], dtype=np.int32)
weight = np.array([1, 0, 0, 1, 0, 0, 1, 0, 0, 1, 0, 0, 1, 0, 0, 1, 0, 0, 1, 0, 0, 1,
0, 0, 1, 0, 0, 1, 0, 0, 1, 0, 0, 1, 0, 0, 1, 0, 0, 1, 0, 0, 1, 0,
0, 1, 0, 0, 1, 0, 0, 1, 0, 0, 1, 0, 0, 1, 0, 0, 1, 0, 0, 1, 0, 0,
1, 0, 0, 1, 0, 0, 1, 0, 0, 1, 0, 0, 1, 0, 0, 1, 0, 0, 1, 0, 0, 1,
0, 0, 1, 0, 0, 1, 0, 0, 1, 0, 0, 1, 0, 0, 1, 0, 0, 1, 0, 0], dtype=np.int32)
There is a pattern which becomes more apparent when reshaping x to (2,3,2,3,3). But I am having a hard time figuring out how the approach given by #doqtor can be used here and especially if it is easy enough to generalize.
The expected output is:
array([1, 1, 0, 0, 2, 2, 0, 0, 3, 3, 0, 0, 2, 2, 0, 0, 1, 1, 0, 0, 1, 1,
0, 0, 2, 2, 0, 0, 3, 3, 0, 0, 2, 2, 0, 0, 1, 1, 0, 0], dtype=int32)

Answer 1

The problem is that OpenCL buffer to which weights are accumulated is not initialized (zeroed). Fixing that:
res_np = np.zeros(5).astype(np.int32)
res_g = cl.Buffer(ctx, cl.mem_flags.WRITE_ONLY | cl.mem_flags.COPY_HOST_PTR, hostbuf=res_np)
prg.bincount(queue, [10], None, res_g, x_g, weight_g)
# transfer back to cpu
cl.enqueue_copy(queue, res_np, res_g)
Returns correct results: [ 1 5 9 13 17]
====== Update ==========
As #Kevin noticed there is race condition here too. If there is any pattern it could be addressed this way without using synchronization, for example processing every 2 elements by 1 work item:
__kernel void bincount(__global int *res_g, __global const int* x_g, __global const int* weight_g)
{
int gid = get_global_id(0);
for(int x = gid*2; x < gid*2+2; ++x)
res_g[x_g[x]] += weight_g[x];
}
Then schedule 5 work items:
prg.bincount(queue, [5], None, res_g, x_g, weight_g)