I know this has been answered before, but it seems that executing the script directly \"python filename.py\" does not work. I have Python 2.6.2 on SuSE Linux.
Code:<
Restructure your code so that the f()
function is defined before you create instance of Pool. Otherwise the worker cannot see your function.
#!/usr/bin/python
# -*- coding: utf-8 -*-
from multiprocessing import Pool
def f(x):
return x*x
p = Pool(1)
p.map(f, [1, 2, 3])
The problem I had was solved by using if __name__ == "__main__"
as pointed out by Tamás; in Eclipse for Windows the examples do not work under the interpreter.
This is explained in
http://docs.python.org/2/library/multiprocessing
One possibility is that your python file has the same name as a module:
in pickle.py, you have the error coming from:
def find_class(self, module, name):
# Subclasses may override this
__import__(module)
mod = sys.modules[module] # <- here mod will reference your test/__init__.py
klass = getattr(mod, name)
return klass
This comes from the fact that with p = Pool(1)
the main process forks processes (threads vs processes) before it creates the function f. As stated in Bartosz answer the spawned processes do not have access to the new function.
def f1(x):
...
p = Pool(1) # p is spawned and is now an independent process, knows f1
def f(x): # p doesn't not share this object
...
This one works:
#!/usr/bin/python
# -*- coding: utf-8 -*-
from multiprocessing import Pool
def f(x):
return x*x
if __name__ == "__main__":
p = Pool(1)
p.map(f, [1, 2, 3])
I'm not 100% sure why your code does not work, but I guess the reason is that child processes launched by the multiprocessing
module try to import the main module (to have access to the methods you defined), and the if __name__ == "__main__"
stanza is required not to execute the initialization code where you set up your pool.