I'm pleased to announce the release of PyOpenCL 0.90, a complete, functional Python wrapper around OpenCL.

Its home page is at http://mathema.tician.de/software/pyopencl.

PyOpenCL has complete documentation and a Wiki available.

Download PyOpenCL here from its page in the Python package index.

Since there are a few competing efforts, what sets PyOpenCL apart?

  • Object cleanup tied to lifetime of objects. This idiom, often called RAII in C++, makes it much easier to write correct, leak- and crash-free code.[/*:m:20t6iq6e]
  • Completeness. PyOpenCL puts the full power of OpenCL’s API at your disposal, if you wish. Every obscure get_info() query and all CL calls are accessible.[/*:m:20t6iq6e]
  • Automatic Error Checking. All errors are automatically translated into Python exceptions.[/*:m:20t6iq6e]
  • Speed. PyOpenCL’s base layer is written in C++, so all the niceties above are virtually free.[/*:m:20t6iq6e]
  • Helpful, complete Documentation[/*:m:20t6iq6e]
  • Liberal license. PyOpenCL is open-source and free for commercial, academic, and private use under the MIT/X11 license.[/*:m:20t6iq6e]


Here's some sample code to give you an impression:

Code :
import pyopencl as cl
import numpy
import numpy.linalg as la
 
a = numpy.random.rand(50000).astype(numpy.float32)
b = numpy.random.rand(50000).astype(numpy.float32)
 
ctx = cl.create_context_from_type(cl.device_type.ALL)
queue = cl.CommandQueue(ctx)
 
mf = cl.mem_flags
a_buf = cl.create_host_buffer(
        ctx, mf.READ_ONLY | mf.COPY_HOST_PTR, a)
b_buf = cl.create_host_buffer(
        ctx, mf.READ_ONLY | mf.COPY_HOST_PTR, b)
dest_buf = cl.create_buffer(ctx, mf.WRITE_ONLY, b.nbytes)
 
prg = cl.create_program_with_source(ctx, """
    __kernel void sum(__global const float *a,
    __global const float *b, __global float *c)
    {
      int gid = get_global_id(0);
      c[gid] = a[gid] + b[gid];
    }
    """).build()
 
prg.sum(queue, a.shape, a_buf, b_buf, dest_buf)
 
a_plus_b = numpy.empty_like(a)
cl.enqueue_read_buffer(queue, dest_buf, a_plus_b).wait()
 
print la.norm(a_plus_b - (a+b))