How to get correct global variable with jupyter's timeit?
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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 Nov 9 at 15:10
talonmies
59k17128195
asked Nov 9 at 10:58
biocyberman
2,55422135
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 Nov 9 at 15:10
talonmies
59k17128195
asked Nov 9 at 10:58
biocyberman
2,55422135
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
Nov 9 at 14:48
Got it. A wrapper function would do, but still prefer some flag or parameter when calling timeit.
– biocyberman
Nov 9 at 15:11
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 Nov 9 at 15:10
talonmies
59k17128195
asked Nov 9 at 10:58
biocyberman
2,55422135
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 Nov 9 at 15:10
talonmies
59k17128195
asked Nov 9 at 10:58
biocyberman
2,55422135
edited Nov 9 at 15:10
talonmies
59k17128195
asked Nov 9 at 10:58
biocyberman
2,55422135
edited Nov 9 at 15:10
talonmies
59k17128195
edited Nov 9 at 15:10
talonmies
59k17128195
edited Nov 9 at 15:10
talonmies
59k17128195
59k17128195
asked Nov 9 at 10:58
biocyberman
2,55422135
asked Nov 9 at 10:58
biocyberman
2,55422135
asked Nov 9 at 10:58
biocyberman
2,55422135
2,55422135
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
Nov 9 at 14:48
Got it. A wrapper function would do, but still prefer some flag or parameter when calling timeit.
– biocyberman
Nov 9 at 15:11
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
Nov 9 at 14:48
Got it. A wrapper function would do, but still prefer some flag or parameter when calling timeit.
– biocyberman
Nov 9 at 15:11
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
Nov 9 at 14:48
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
Nov 9 at 14:48
Got it. A wrapper function would do, but still prefer some flag or parameter when calling timeit.
– biocyberman
Nov 9 at 15:11
Got it. A wrapper function would do, but still prefer some flag or parameter when calling timeit.
– biocyberman
Nov 9 at 15:11
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
Nov 9 at 14:48
Got it. A wrapper function would do, but still prefer some flag or parameter when calling timeit.
– biocyberman
Nov 9 at 15:11