How to get correct global variable with jupyter's timeit?
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0
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One first cell I have this:
from numba import cuda
@cuda.jit
def thread_counter_safe(global_counter):
cuda.atomic.add(global_counter, 0, 1) # Safely add 1 to offset 0 in global_counter array
On the next cell I have this:
global_counter = cuda.to_device(np.array([0], dtype=np.int32))
thread_counter_safe[64, 64](global_counter)
print('Should be %d:' % (64*64), global_counter.copy_to_host())
global_counter = cuda.to_device(np.array([0], dtype=np.int32))
%timeit thread_counter_safe[64, 64](global_counter)
print('Should be %d:' % (64*64), global_counter.copy_to_host())
Output of this second cell is something like this:
Should be 4096: [4096]
10000 loops, best of 3: 118 µs per loop
Should be 4096: [168390656]
Jupyter Notebook's timeit carries the global_counter accross its iteration test. How do it get it to give back global_counter correctly?
jupyter-notebook numba
edited yesterday
talonmies
58.7k17125192
asked yesterday
biocyberman
2,52922135
add a comment |
up vote
0
down vote
favorite
One first cell I have this:
from numba import cuda
@cuda.jit
def thread_counter_safe(global_counter):
cuda.atomic.add(global_counter, 0, 1) # Safely add 1 to offset 0 in global_counter array
On the next cell I have this:
global_counter = cuda.to_device(np.array([0], dtype=np.int32))
thread_counter_safe[64, 64](global_counter)
print('Should be %d:' % (64*64), global_counter.copy_to_host())
global_counter = cuda.to_device(np.array([0], dtype=np.int32))
%timeit thread_counter_safe[64, 64](global_counter)
print('Should be %d:' % (64*64), global_counter.copy_to_host())
Output of this second cell is something like this:
Should be 4096: [4096]
10000 loops, best of 3: 118 µs per loop
Should be 4096: [168390656]
Jupyter Notebook's timeit carries the global_counter accross its iteration test. How do it get it to give back global_counter correctly?
jupyter-notebook numba
edited yesterday
talonmies
58.7k17125192
asked yesterday
biocyberman
2,52922135
1
you would need to reset theglobal_counterto zero as part of the function that you are runningtimeiton. You can't safely/easily do that from thethread_counter_safecuda kernel, but you coulddefa new function that resets theglobal_counterto zero and then callsthread_counter_safe, and runtimeiton that.
– Robert Crovella
yesterday
Got it. A wrapper function would do, but still prefer some flag or parameter when calling timeit.
– biocyberman
yesterday
add a comment |
up vote
0
down vote
favorite
up vote
0
down vote
favorite
One first cell I have this:
from numba import cuda
@cuda.jit
def thread_counter_safe(global_counter):
cuda.atomic.add(global_counter, 0, 1) # Safely add 1 to offset 0 in global_counter array
On the next cell I have this:
global_counter = cuda.to_device(np.array([0], dtype=np.int32))
thread_counter_safe[64, 64](global_counter)
print('Should be %d:' % (64*64), global_counter.copy_to_host())
global_counter = cuda.to_device(np.array([0], dtype=np.int32))
%timeit thread_counter_safe[64, 64](global_counter)
print('Should be %d:' % (64*64), global_counter.copy_to_host())
Output of this second cell is something like this:
Should be 4096: [4096]
10000 loops, best of 3: 118 µs per loop
Should be 4096: [168390656]
Jupyter Notebook's timeit carries the global_counter accross its iteration test. How do it get it to give back global_counter correctly?
jupyter-notebook numba
edited yesterday
talonmies
58.7k17125192
asked yesterday
biocyberman
2,52922135
One first cell I have this:
from numba import cuda
@cuda.jit
def thread_counter_safe(global_counter):
cuda.atomic.add(global_counter, 0, 1) # Safely add 1 to offset 0 in global_counter array
On the next cell I have this:
global_counter = cuda.to_device(np.array([0], dtype=np.int32))
thread_counter_safe[64, 64](global_counter)
print('Should be %d:' % (64*64), global_counter.copy_to_host())
global_counter = cuda.to_device(np.array([0], dtype=np.int32))
%timeit thread_counter_safe[64, 64](global_counter)
print('Should be %d:' % (64*64), global_counter.copy_to_host())
Output of this second cell is something like this:
Should be 4096: [4096]
10000 loops, best of 3: 118 µs per loop
Should be 4096: [168390656]
Jupyter Notebook's timeit carries the global_counter accross its iteration test. How do it get it to give back global_counter correctly?
jupyter-notebook numba
jupyter-notebook numba
edited yesterday
talonmies
58.7k17125192
asked yesterday
biocyberman
2,52922135
edited yesterday
talonmies
58.7k17125192
asked yesterday
biocyberman
2,52922135
edited yesterday
talonmies
58.7k17125192
edited yesterday
talonmies
58.7k17125192
edited yesterday
talonmies
58.7k17125192
58.7k17125192
asked yesterday
biocyberman
2,52922135
asked yesterday
biocyberman
2,52922135
asked yesterday
biocyberman
2,52922135
2,52922135
1
you would need to reset theglobal_counterto zero as part of the function that you are runningtimeiton. You can't safely/easily do that from thethread_counter_safecuda kernel, but you coulddefa new function that resets theglobal_counterto zero and then callsthread_counter_safe, and runtimeiton that.
– Robert Crovella
yesterday
Got it. A wrapper function would do, but still prefer some flag or parameter when calling timeit.
– biocyberman
yesterday
add a comment |
1
you would need to reset theglobal_counterto zero as part of the function that you are runningtimeiton. You can't safely/easily do that from thethread_counter_safecuda kernel, but you coulddefa new function that resets theglobal_counterto zero and then callsthread_counter_safe, and runtimeiton that.
– Robert Crovella
yesterday
Got it. A wrapper function would do, but still prefer some flag or parameter when calling timeit.
– biocyberman
yesterday
1
1
you would need to reset the
global_counter to zero as part of the function that you are running timeiton. You can't safely/easily do that from the thread_counter_safe cuda kernel, but you could def a new function that resets the global_counter to zero and then calls thread_counter_safe, and run timeit on that.– Robert Crovella
yesterday
you would need to reset the
global_counter to zero as part of the function that you are running timeiton. You can't safely/easily do that from the thread_counter_safe cuda kernel, but you could def a new function that resets the global_counter to zero and then calls thread_counter_safe, and run timeit on that.– Robert Crovella
yesterday
Got it. A wrapper function would do, but still prefer some flag or parameter when calling timeit.
– biocyberman
yesterday
Got it. A wrapper function would do, but still prefer some flag or parameter when calling timeit.
– biocyberman
yesterday
add a comment |
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1
you would need to reset the
global_counterto zero as part of the function that you are runningtimeiton. You can't safely/easily do that from thethread_counter_safecuda kernel, but you coulddefa new function that resets theglobal_counterto zero and then callsthread_counter_safe, and runtimeiton that.– Robert Crovella
yesterday
Got it. A wrapper function would do, but still prefer some flag or parameter when calling timeit.
– biocyberman
yesterday