Python: parallel code running slower than sequential version
$begingroup$
I have a sequential code where I am counting unique events occurring at a timestamp given the data on time intervals. The sequential code I have prepared is:
a=list of timestamps of size 100.
number=
for i in range(100):
indices=numpy.argwhere((a[i] >= dataset[:,0]) & (a[i] <= dataset[:,1]))[:,0]
number.append(len(set(dataset[indices,2])))
Since the actual size of a is large, it is expected to take large number of days to complete. Therefore, I created a parallel version of the code:
num_cores = multiprocessing.cpu_count()
inputs = range(100)
def processInput(i):
indices=numpy.argwhere((a[i] >= dataset[:,0]) & (a[i] <= dataset[:,1]))[:,0]
return(len(set(dataset[indices,2])))
results = Parallel(n_jobs=num_cores)(delayed(processInput)(i) for i in inputs)
Surprisingly, the sequential version on 100 elements is taking 2 minutes to complete and the parallel version takes about 9 minutes. Why
python performance
asked 2 mins ago
shaifali Guptashaifali Gupta
12
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shaifali Gupta is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
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add a comment |
$begingroup$
I have a sequential code where I am counting unique events occurring at a timestamp given the data on time intervals. The sequential code I have prepared is:
a=list of timestamps of size 100.
number=
for i in range(100):
indices=numpy.argwhere((a[i] >= dataset[:,0]) & (a[i] <= dataset[:,1]))[:,0]
number.append(len(set(dataset[indices,2])))
Since the actual size of a is large, it is expected to take large number of days to complete. Therefore, I created a parallel version of the code:
num_cores = multiprocessing.cpu_count()
inputs = range(100)
def processInput(i):
indices=numpy.argwhere((a[i] >= dataset[:,0]) & (a[i] <= dataset[:,1]))[:,0]
return(len(set(dataset[indices,2])))
results = Parallel(n_jobs=num_cores)(delayed(processInput)(i) for i in inputs)
Surprisingly, the sequential version on 100 elements is taking 2 minutes to complete and the parallel version takes about 9 minutes. Why
python performance
asked 2 mins ago
shaifali Guptashaifali Gupta
12
New contributor
shaifali Gupta is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
$endgroup$
add a comment |
$begingroup$
I have a sequential code where I am counting unique events occurring at a timestamp given the data on time intervals. The sequential code I have prepared is:
a=list of timestamps of size 100.
number=
for i in range(100):
indices=numpy.argwhere((a[i] >= dataset[:,0]) & (a[i] <= dataset[:,1]))[:,0]
number.append(len(set(dataset[indices,2])))
Since the actual size of a is large, it is expected to take large number of days to complete. Therefore, I created a parallel version of the code:
num_cores = multiprocessing.cpu_count()
inputs = range(100)
def processInput(i):
indices=numpy.argwhere((a[i] >= dataset[:,0]) & (a[i] <= dataset[:,1]))[:,0]
return(len(set(dataset[indices,2])))
results = Parallel(n_jobs=num_cores)(delayed(processInput)(i) for i in inputs)
Surprisingly, the sequential version on 100 elements is taking 2 minutes to complete and the parallel version takes about 9 minutes. Why
python performance
asked 2 mins ago
shaifali Guptashaifali Gupta
12
New contributor
shaifali Gupta is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
$endgroup$
I have a sequential code where I am counting unique events occurring at a timestamp given the data on time intervals. The sequential code I have prepared is:
a=list of timestamps of size 100.
number=
for i in range(100):
indices=numpy.argwhere((a[i] >= dataset[:,0]) & (a[i] <= dataset[:,1]))[:,0]
number.append(len(set(dataset[indices,2])))
Since the actual size of a is large, it is expected to take large number of days to complete. Therefore, I created a parallel version of the code:
num_cores = multiprocessing.cpu_count()
inputs = range(100)
def processInput(i):
indices=numpy.argwhere((a[i] >= dataset[:,0]) & (a[i] <= dataset[:,1]))[:,0]
return(len(set(dataset[indices,2])))
results = Parallel(n_jobs=num_cores)(delayed(processInput)(i) for i in inputs)
Surprisingly, the sequential version on 100 elements is taking 2 minutes to complete and the parallel version takes about 9 minutes. Why
python performance
python performance
asked 2 mins ago
shaifali Guptashaifali Gupta
12
New contributor
shaifali Gupta is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
asked 2 mins ago
shaifali Guptashaifali Gupta
12
New contributor
shaifali Gupta is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
asked 2 mins ago
shaifali Guptashaifali Gupta
12
New contributor
shaifali Gupta is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
asked 2 mins ago
shaifali Guptashaifali Gupta
12
asked 2 mins ago
shaifali Guptashaifali Gupta
12
12
New contributor
shaifali Gupta is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
New contributor
shaifali Gupta is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
shaifali Gupta is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
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