Implement merge_sort with multiprocessing solution
I tried to write a merge sort with multiprocessing solution
from heapq import merge
from multiprocessing import Process
def merge_sort1(m):
if len(m) < 2:
return m
middle = len(m) // 2
left = Process(target=merge_sort1, args=(m[:middle],))
left.start()
right = Process(target=merge_sort1, args=(m[middle:],))
right.start()
for p in (left, right):
p.join()
result = list(merge(left, right))
return result
Test it with arr
In [47]: arr = list(range(9))
In [48]: random.shuffle(arr)
It repost error:
In [49]: merge_sort1(arr)
TypeError: 'Process' object is not iterable
What's the problem with my code?
python
asked Nov 26 '18 at 3:52
JawSawJawSaw
4,74911938
add a comment |
I tried to write a merge sort with multiprocessing solution
from heapq import merge
from multiprocessing import Process
def merge_sort1(m):
if len(m) < 2:
return m
middle = len(m) // 2
left = Process(target=merge_sort1, args=(m[:middle],))
left.start()
right = Process(target=merge_sort1, args=(m[middle:],))
right.start()
for p in (left, right):
p.join()
result = list(merge(left, right))
return result
Test it with arr
In [47]: arr = list(range(9))
In [48]: random.shuffle(arr)
It repost error:
In [49]: merge_sort1(arr)
TypeError: 'Process' object is not iterable
What's the problem with my code?
python
asked Nov 26 '18 at 3:52
JawSawJawSaw
4,74911938
add a comment |
I tried to write a merge sort with multiprocessing solution
from heapq import merge
from multiprocessing import Process
def merge_sort1(m):
if len(m) < 2:
return m
middle = len(m) // 2
left = Process(target=merge_sort1, args=(m[:middle],))
left.start()
right = Process(target=merge_sort1, args=(m[middle:],))
right.start()
for p in (left, right):
p.join()
result = list(merge(left, right))
return result
Test it with arr
In [47]: arr = list(range(9))
In [48]: random.shuffle(arr)
It repost error:
In [49]: merge_sort1(arr)
TypeError: 'Process' object is not iterable
What's the problem with my code?
python
asked Nov 26 '18 at 3:52
JawSawJawSaw
4,74911938
I tried to write a merge sort with multiprocessing solution
from heapq import merge
from multiprocessing import Process
def merge_sort1(m):
if len(m) < 2:
return m
middle = len(m) // 2
left = Process(target=merge_sort1, args=(m[:middle],))
left.start()
right = Process(target=merge_sort1, args=(m[middle:],))
right.start()
for p in (left, right):
p.join()
result = list(merge(left, right))
return result
Test it with arr
In [47]: arr = list(range(9))
In [48]: random.shuffle(arr)
It repost error:
In [49]: merge_sort1(arr)
TypeError: 'Process' object is not iterable
What's the problem with my code?
python
python
asked Nov 26 '18 at 3:52
JawSawJawSaw
4,74911938
asked Nov 26 '18 at 3:52
JawSawJawSaw
4,74911938
asked Nov 26 '18 at 3:52
JawSawJawSaw
4,74911938
asked Nov 26 '18 at 3:52
JawSawJawSaw
4,74911938
asked Nov 26 '18 at 3:52
JawSawJawSaw
4,74911938
4,74911938
add a comment |
add a comment |
1 Answer
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oldest
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merge(left, right) tries to merge two processes, whereas you presumably want to merge the two lists that resulted from each process. Note that return value of the function passed to Process is lost; it is a different process, not just a different thread, and you can't very easily shuffle data back to parent, so Python doesn't do that, by default. You need to be explicit and code such a channel yourself. Fortunately, there are multiprocessing datatypes to help you; for example, multiprocessing.Pipe:
from heapq import merge
import random
import multiprocessing
def merge_sort1(m, send_end=None):
if len(m) < 2:
result = m
else:
middle = len(m) // 2
inputs = [m[:middle], m[middle:]]
pipes = [multiprocessing.Pipe(False) for _ in inputs]
processes = [multiprocessing.Process(target=merge_sort1, args=(input, send_end))
for input, (recv_end, send_end) in zip(inputs, pipes)]
for process in processes: process.start()
for process in processes: process.join()
results = [recv_end.recv() for recv_end, send_end in pipes]
result = list(merge(*results))
if send_end:
send_end.send(result)
else:
return result
arr = list(range(9))
random.shuffle(arr)
print(merge_sort1(arr))
answered Nov 26 '18 at 4:10
AmadanAmadan
133k13146197
add a comment |
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1 Answer
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1 Answer
1
active
oldest
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active
oldest
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active
oldest
votes
merge(left, right) tries to merge two processes, whereas you presumably want to merge the two lists that resulted from each process. Note that return value of the function passed to Process is lost; it is a different process, not just a different thread, and you can't very easily shuffle data back to parent, so Python doesn't do that, by default. You need to be explicit and code such a channel yourself. Fortunately, there are multiprocessing datatypes to help you; for example, multiprocessing.Pipe:
from heapq import merge
import random
import multiprocessing
def merge_sort1(m, send_end=None):
if len(m) < 2:
result = m
else:
middle = len(m) // 2
inputs = [m[:middle], m[middle:]]
pipes = [multiprocessing.Pipe(False) for _ in inputs]
processes = [multiprocessing.Process(target=merge_sort1, args=(input, send_end))
for input, (recv_end, send_end) in zip(inputs, pipes)]
for process in processes: process.start()
for process in processes: process.join()
results = [recv_end.recv() for recv_end, send_end in pipes]
result = list(merge(*results))
if send_end:
send_end.send(result)
else:
return result
arr = list(range(9))
random.shuffle(arr)
print(merge_sort1(arr))
answered Nov 26 '18 at 4:10
AmadanAmadan
133k13146197
add a comment |
merge(left, right) tries to merge two processes, whereas you presumably want to merge the two lists that resulted from each process. Note that return value of the function passed to Process is lost; it is a different process, not just a different thread, and you can't very easily shuffle data back to parent, so Python doesn't do that, by default. You need to be explicit and code such a channel yourself. Fortunately, there are multiprocessing datatypes to help you; for example, multiprocessing.Pipe:
from heapq import merge
import random
import multiprocessing
def merge_sort1(m, send_end=None):
if len(m) < 2:
result = m
else:
middle = len(m) // 2
inputs = [m[:middle], m[middle:]]
pipes = [multiprocessing.Pipe(False) for _ in inputs]
processes = [multiprocessing.Process(target=merge_sort1, args=(input, send_end))
for input, (recv_end, send_end) in zip(inputs, pipes)]
for process in processes: process.start()
for process in processes: process.join()
results = [recv_end.recv() for recv_end, send_end in pipes]
result = list(merge(*results))
if send_end:
send_end.send(result)
else:
return result
arr = list(range(9))
random.shuffle(arr)
print(merge_sort1(arr))
answered Nov 26 '18 at 4:10
AmadanAmadan
133k13146197
add a comment |
merge(left, right) tries to merge two processes, whereas you presumably want to merge the two lists that resulted from each process. Note that return value of the function passed to Process is lost; it is a different process, not just a different thread, and you can't very easily shuffle data back to parent, so Python doesn't do that, by default. You need to be explicit and code such a channel yourself. Fortunately, there are multiprocessing datatypes to help you; for example, multiprocessing.Pipe:
from heapq import merge
import random
import multiprocessing
def merge_sort1(m, send_end=None):
if len(m) < 2:
result = m
else:
middle = len(m) // 2
inputs = [m[:middle], m[middle:]]
pipes = [multiprocessing.Pipe(False) for _ in inputs]
processes = [multiprocessing.Process(target=merge_sort1, args=(input, send_end))
for input, (recv_end, send_end) in zip(inputs, pipes)]
for process in processes: process.start()
for process in processes: process.join()
results = [recv_end.recv() for recv_end, send_end in pipes]
result = list(merge(*results))
if send_end:
send_end.send(result)
else:
return result
arr = list(range(9))
random.shuffle(arr)
print(merge_sort1(arr))
answered Nov 26 '18 at 4:10
AmadanAmadan
133k13146197
merge(left, right) tries to merge two processes, whereas you presumably want to merge the two lists that resulted from each process. Note that return value of the function passed to Process is lost; it is a different process, not just a different thread, and you can't very easily shuffle data back to parent, so Python doesn't do that, by default. You need to be explicit and code such a channel yourself. Fortunately, there are multiprocessing datatypes to help you; for example, multiprocessing.Pipe:
from heapq import merge
import random
import multiprocessing
def merge_sort1(m, send_end=None):
if len(m) < 2:
result = m
else:
middle = len(m) // 2
inputs = [m[:middle], m[middle:]]
pipes = [multiprocessing.Pipe(False) for _ in inputs]
processes = [multiprocessing.Process(target=merge_sort1, args=(input, send_end))
for input, (recv_end, send_end) in zip(inputs, pipes)]
for process in processes: process.start()
for process in processes: process.join()
results = [recv_end.recv() for recv_end, send_end in pipes]
result = list(merge(*results))
if send_end:
send_end.send(result)
else:
return result
arr = list(range(9))
random.shuffle(arr)
print(merge_sort1(arr))
answered Nov 26 '18 at 4:10
AmadanAmadan
133k13146197
edited Nov 26 '18 at 4:28
answered Nov 26 '18 at 4:10
AmadanAmadan
133k13146197
answered Nov 26 '18 at 4:10
AmadanAmadan
133k13146197
answered Nov 26 '18 at 4:10
AmadanAmadan
133k13146197
133k13146197
add a comment |
add a comment |
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