Re: Very Strange Problem
On Wed, Jul 29, 2009 at 3:10 PM, Omer Khalidomer.kha...@cern.ch wrote: Hi Dave, Thanks for your reply. I actually didn't cut and paste my code as it was dispersed in different places, i typed the logic behind my code in the email (and obiviously made some typos, indentations is some thing else) but my Please, do not do that. It's very difficult to debug code that hasn't been seen. The code you posted has numerous problems (that likely have nothing to do with your actual problem.) If you're going to post code, try to recreate the issue with a small runnable script. If you can't do that, as it sounds like it would be difficult in this case as your code is dispersed in different places, at least post the relevant portions of the actual code. Don't re-type. real code does not have these problems as my application runs fine with out errors... Except that the line where i want to update the value doesn't get updated and no exception is thrown. What's surprising for me is that i am doing the same thing in hundreds of places in my 3k+ line code but by some reason this part doesn't work... As far as the global variables are concerned, i am using them in other places too and didn't see any problems. I think some thing else is going on here as the statement above and below my modified lines get executed. If the statements above and below that line(s) are executed, then that line is certainly being executed as well. Try introducing some additional print statements to verify your mental model of what's happening: # set it to 1 print jobs print jobs[index] print jobs[index]['v'] jobs[index]['v'] = 1 print jobs print jobs[index] print jobs[index]['v'] print Set to 1 Is there a way in Python to debug memory address or to see where in memory this object is stored, and is there a lock on it or else? You are barking up the wrong tree. Somewhere in your code you're doing something silly that's causing your issue. -- http://mail.python.org/mailman/listinfo/python-list
Re: Very Strange Problem
Omer Khalid wrote:
Hi,
I am having a very strange problem with modifying a variable in a list
in my program. Here is the code:
# a list that contains dictionary objects
jobs = []
index=5
for each in range(index):
jobs.append({'v':0})
some_function(index):
if jobs[index]['v'] == 0:
# set it to 1
jobs[index]['v'] = 1
print Set to 1
else:
print Already set to 1
loop():
index=0
for each in range(len(jobs)):
some_function(index)
index +=1
Apparently, the jobs[index]['v'] never get updated in the some_function
but the print statement afterwards get printed...
What's really surprising is that there are no errors or exceptions and
my my program runs in a single thread...so i have been unable to explain
this behavior.
Any insight would be much appreciated!
Well, when I insert the missing 'def's in the function definitions, it
works for me.
--
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Re: Very Strange Problem
Ricardo Aráoz wrote:
MRAB wrote:
Omer Khalid wrote:
Hi,
I am having a very strange problem with modifying a variable in a
list in my program. Here is the code:
# a list that contains dictionary objects
jobs = []
index=5
for each in range(index):
jobs.append({'v':0})
some_function(index):
if jobs[index]['v'] == 0:
# set it to 1
jobs[index]['v'] = 1
print Set to 1
else:
print Already set to 1
loop():
index=0
for each in range(len(jobs)):
some_function(index)
index +=1
Apparently, the jobs[index]['v'] never get updated in the
some_function but the print statement afterwards get printed...
What's really surprising is that there are no errors or exceptions
and my my program runs in a single thread...so i have been unable to
explain this behavior.
Any insight would be much appreciated!
Well, when I insert the missing 'def's in the function definitions, it
works for me.
Hi Omer, what he is trying to convey in his rude manner is that you are
missing def in your function definitions. It is probably a beginners
mistake.
That is :
from some_function(index):
to def some_function(index):
from loop():
to def loop():
I have not tried your code so you should believe him when he states he
has actually run the code.
HTH
Omer says the print statement afterwards get printed, but the code
provided would have raised a SyntaxError, so omitting the 'def's can't
be the cause of the actual problem reported.
--
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Re: Very Strange Problem
Omer Khalid wrote:
Hi,
I am having a very strange problem with modifying a variable in a list in my
program. Here is the code:
# a list that contains dictionary objects
jobs = []
index=5
for each in range(index):
jobs.append({'v':0})
some_function(index):
if jobs[index]['v'] == 0:
# set it to 1
jobs[index]['v'] = 1
print Set to 1
else:
print Already set to 1
loop():
index=0
for each in range(len(jobs)):
some_function(index)
index +=1
Apparently, the jobs[index]['v'] never get updated in the some_function but
the print statement afterwards get printed...
What's really surprising is that there are no errors or exceptions and my my
program runs in a single thread...so i have been unable to explain this
behavior.
Any insight would be much appreciated!
Cheers
Omer
There are four things to fix before the program does anything much at
all. Two places you're missing the def, indentation is inconsistent,
and you never actually call either of the functions. The first three
are syntax errors, so presumably your cut/paste in your computer is broken.
Once I make those four corrections, I get the following output:
Set to 1
Set to 1
Set to 1
Set to 1
Set to 1
But you never said what you got, nor what you expected. That's
certainly what I'd expect. And if you make a second call to loop() in
your outer code, you get five copies of Already set to 1
BTW, there are a number of things that could be done better. The main
one I'll point out is that you shouldn't re-use a global variable
'index' as a local with different meaning. As someone else pointed out,
since the global is a constant, making it all uppercase is the convention.
DaveA
--
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Re: Very Strange Problem
Hi Dave,
Thanks for your reply. I actually didn't cut and paste my code as it was
dispersed in different places, i typed the logic behind my code in the email
(and obiviously made some typos, indentations is some thing else) but my
real code does not have these problems as my application runs fine with out
errors...
Except that the line where i want to update the value doesn't get updated
and no exception is thrown. What's surprising for me is that i am doing the
same thing in hundreds of places in my 3k+ line code but by some reason this
part doesn't work...
As far as the global variables are concerned, i am using them in other
places too and didn't see any problems.
I think some thing else is going on here as the statement above and below my
modified lines get executed.
Is there a way in Python to debug memory address or to see where in memory
this object is stored, and is there a lock on it or else?
Thanks,
Omer
**
On Wed, Jul 29, 2009 at 8:56 PM, Dave Angel da...@ieee.org wrote:
Omer Khalid wrote:
Hi,
I am having a very strange problem with modifying a variable in a list in
my
program. Here is the code:
# a list that contains dictionary objects
jobs = []
index=5
for each in range(index):
jobs.append({'v':0})
some_function(index):
if jobs[index]['v'] == 0:
# set it to 1
jobs[index]['v'] = 1
print Set to 1
else:
print Already set to 1
loop():
index=0
for each in range(len(jobs)):
some_function(index)
index +=1
Apparently, the jobs[index]['v'] never get updated in the some_function
but
the print statement afterwards get printed...
What's really surprising is that there are no errors or exceptions and my
my
program runs in a single thread...so i have been unable to explain this
behavior.
Any insight would be much appreciated!
Cheers
Omer
There are four things to fix before the program does anything much at all.
Two places you're missing the def, indentation is inconsistent, and you
never actually call either of the functions. The first three are syntax
errors, so presumably your cut/paste in your computer is broken.
Once I make those four corrections, I get the following output:
Set to 1
Set to 1
Set to 1
Set to 1
Set to 1
But you never said what you got, nor what you expected. That's certainly
what I'd expect. And if you make a second call to loop() in your outer
code, you get five copies of Already set to 1
BTW, there are a number of things that could be done better. The main one
I'll point out is that you shouldn't re-use a global variable 'index' as a
local with different meaning. As someone else pointed out, since the global
is a constant, making it all uppercase is the convention.
DaveA
--
http://mail.python.org/mailman/listinfo/python-list
Re: Very Strange Problem
Omer Khalid wrote: Hi, I am having a very strange problem with modifying a variable in a list in my program. Here is the code: To me, this sentence clearly implies that the code that follows is the code that had the problem. Since the posted code cannot run, it clearly is not. People should test code to be posted before posting unless they clearly label it as 'untested'. Original posters, of course, should run the code first. tjr -- http://mail.python.org/mailman/listinfo/python-list
Re: Very Strange Problem
Omer Khalid wrote:
Hi Dave,
Thanks for your reply. I actually didn't cut and paste my code as it was
dispersed in different places, i typed the logic behind my code in the email
(and obiviously made some typos, indentations is some thing else) but my
real code does not have these problems as my application runs fine with out
errors...
Except that the line where i want to update the value doesn't get updated
and no exception is thrown. What's surprising for me is that i am doing the
same thing in hundreds of places in my 3k+ line code but by some reason this
part doesn't work...
As far as the global variables are concerned, i am using them in other
places too and didn't see any problems.
I think some thing else is going on here as the statement above and below my
modified lines get executed.
Is there a way in Python to debug memory address or to see where in memory
this object is stored, and is there a lock on it or else?
Thanks,
Omer
**
On Wed, Jul 29, 2009 at 8:56 PM, Dave Angel da...@ieee.org wrote:
Omer Khalid wrote:
Hi,
I am having a very strange problem with modifying a variable in a list in
my
program. Here is the code:
# a list that contains dictionary objects
jobs = []
index=5
for each in range(index):
jobs.append({'v':0})
some_function(index):
if jobs[index]['v'] == 0:
# set it to 1
jobs[index]['v'] = 1
print Set to 1
else:
print Already set to 1
loop():
index=0
for each in range(len(jobs)):
some_function(index)
index +=1
Apparently, the jobs[index]['v'] never get updated in the some_function
but
the print statement afterwards get printed...
What's really surprising is that there are no errors or exceptions and my
my
program runs in a single thread...so i have been unable to explain this
behavior.
Any insight would be much appreciated!
Cheers
Omer
There are four things to fix before the program does anything much at all.
Two places you're missing the def, indentation is inconsistent, and you
never actually call either of the functions. The first three are syntax
errors, so presumably your cut/paste in your computer is broken.
Once I make those four corrections, I get the following output:
Set to 1
Set to 1
Set to 1
Set to 1
Set to 1
But you never said what you got, nor what you expected. That's certainly
what I'd expect. And if you make a second call to loop() in your outer
code, you get five copies of Already set to 1
BTW, there are a number of things that could be done better. The main one
I'll point out is that you shouldn't re-use a global variable 'index' as a
local with different meaning. As someone else pointed out, since the global
is a constant, making it all uppercase is the convention.
DaveA
(You top-posted, so your ,message is out of sequence. More and more
people are doing that in this list.)
...Except that the line where i want to update the value doesn't
get updated...
And what makes you think that? You never answered my question. What did you
expect for output, and what did you get? I got exactly what I expected, when I
ran it.
... Is there a way in Python to debug memory address or
to see where in memory this object is stored, and
is there a lock on it or else?
If there really were a bug in the language, you might need such a tool.
I use Komodo IDE as a debugger, but there was no need in this case.
Adding a few print statements might clear up your confusion, but since
you haven't spelled out what it is, I can't suggest where. How about if
you just add aprint jobsat the beginning of some_function() ?
Then you could see things getting updated perfectly.
DaveA
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Re: very strange problem in 2.4
Ok, so I found out that even though mylist[] and all objects in it were fine ie id(mylist[i]) != id(mylist[all others]) what was happening is that during a reproduction function a shallow copies were being made making all offspring (genetic algorithm) have different id(mylist[0..n]), however the actual attributes were referenced for offspring on the same parent(s). Fixed it with a deepcopy and used properties insead of java like get sets. thanks for all the help, Conor -- http://mail.python.org/mailman/listinfo/python-list
Re: very strange problem in 2.4
Ok, so I found out that even though mylist[] and all objects in it were fine ie id(mylist[i]) != id(mylist[all others]) what was happening is that during a reproduction function a shallow copies were being made making all offspring (genetic algorithm) have different id(mylist[0..n]), however the actual attributes were referenced for offspring on the same parent(s). Fixed it with a deepcopy and used properties insead of java like get sets. thanks for all the help, Conor -- http://mail.python.org/mailman/listinfo/python-list
Re: very strange problem in 2.4
Ok, so I found out that even though mylist[] and all objects in it were fine ie id(mylist[i]) != id(mylist[all others]) what was happening is that during a reproduction function a shallow copies were being made making all offspring (genetic algorithm) have different id(mylist[0..n]), however the actual attributes were referenced for offspring on the same parent(s). Fixed it with a deepcopy and used properties insead of java like get sets. thanks for all the help, Conor -- http://mail.python.org/mailman/listinfo/python-list
Re: very strange problem in 2.4
Ok, so I found out that even though mylist[] and all objects in it were fine ie id(mylist[i]) != id(mylist[all others]) what was happening is that during a reproduction function a shallow copies were being made making all offspring (genetic algorithm) have different id(mylist[0..n]), however the actual attributes were referenced for offspring on the same parent(s). Fixed it with a deepcopy and used properties insead of java like get sets. thanks for all the help, Conor -- http://mail.python.org/mailman/listinfo/python-list
Re: very strange problem in 2.4
Ok, so I found out that even though mylist[] and all objects in it were fine ie id(mylist[i]) != id(mylist[all others]) what was happening is that during a reproduction function a shallow copies were being made making all offspring (genetic algorithm) have different id(mylist[0..n]), however the actual attributes were referenced for offspring on the same parent(s). Fixed it with a deepcopy and used properties insead of java like get sets. thanks for all the help, Conor -- http://mail.python.org/mailman/listinfo/python-list
Re: very strange problem in 2.4
John Zenger wrote: Your list probably contains several references to the same object, instead of several different objects. This happens often when you use a technique like: list = [ object ] * 100 ..because although this does make copies when object is an integer, it just makes references in other cases. it always creates new references. the only thing that distinguishes immutable objects (like integers) from mutable objects (like lists) is that integers don't have any methods that let you modify their contents. there's no this object is mutable flag inside the object, and there's no code in the list multiply operation, or anywhere else, that looks for such a flag. /F -- http://mail.python.org/mailman/listinfo/python-list
Re: very strange problem in 2.4
[EMAIL PROTECTED] a écrit : The Problem (very basic, but strange): I have a list holding a population of objects, each object has 5 vars and appropriate funtions to get or modify the vars. Which are probably not necessary: http://dirtsimple.org/2004/12/python-is-not-java.html (in short: Python as a mechanism named properties that allow you to gateway attribute access thru hiddens getter and setter). When objects in the list have identical vars (like all = 5 for var a and all = 10 for var b across all vars and objects) and i change self.mylist[i].change_var_a(5) to a new value, in this case var a in object i to 5, now all vars of type a in all objects in my list are changed to 5 instead of just var a in object mylist[i], which is my goal. (snip) What is python doing? Am I missing something? Any ideas at all would be wonderful? It would have helped if you had posted your code. Anyway, at least two things could lead to the behaviour you describe: 1/ you have class attributes instead of instance attributes, ie: class MyClass(object): # this attribute belongs to the class class_attrib = 42 def __init__(self, instance_attrib): # this attribute belongs to the instance self.instance_attrib = instance_attrib 2/ you in fact have in your list multiple references to the same instance. My 2 cents -- http://mail.python.org/mailman/listinfo/python-list
very strange problem in 2.4
The Problem (very basic, but strange): I have a list holding a population of objects, each object has 5 vars and appropriate funtions to get or modify the vars. When objects in the list have identical vars (like all = 5 for var a and all = 10 for var b across all vars and objects) and i change self.mylist[i].change_var_a(5) to a new value, in this case var a in object i to 5, now all vars of type a in all objects in my list are changed to 5 instead of just var a in object mylist[i], which is my goal. if i print self.mylist[i].return_var_a() right after I change var a in object i, I get the correct change, however when i print out the whole list, the last change to var a in the last object modified takes over for all objects in the list. note: all the vars not being modified must be the same across all objects in the list, the var being modified need not be the same as the one before it in the list (but will be once just one of the identical object are changed). The value changed in the last object var modified takes over for all object vars making them exactly identical. If, for example, half the list has objects with random vars init. and the other half is identical, as above, and I perform the same operation, as above, to one of the identical var objects self.mylist[i].change_var_a(5) (to an object that has identicals in the list) all the identicals are changed in the same way as above, however the objects that have different var values are unchanged. What is python doing? Am I missing something? Any ideas at all would be wonderful? -- http://mail.python.org/mailman/listinfo/python-list
Re: very strange problem in 2.4
Your list probably contains several references to the same object, instead of several different objects. This happens often when you use a technique like: list = [ object ] * 100 ..because although this does make copies when object is an integer, it just makes references in other cases. [EMAIL PROTECTED] wrote: The Problem (very basic, but strange): I have a list holding a population of objects, each object has 5 vars and appropriate funtions to get or modify the vars. When objects in the list have identical vars (like all = 5 for var a and all = 10 for var b across all vars and objects) and i change self.mylist[i].change_var_a(5) to a new value, in this case var a in object i to 5, now all vars of type a in all objects in my list are changed to 5 instead of just var a in object mylist[i], which is my goal. if i print self.mylist[i].return_var_a() right after I change var a in object i, I get the correct change, however when i print out the whole list, the last change to var a in the last object modified takes over for all objects in the list. note: all the vars not being modified must be the same across all objects in the list, the var being modified need not be the same as the one before it in the list (but will be once just one of the identical object are changed). The value changed in the last object var modified takes over for all object vars making them exactly identical. If, for example, half the list has objects with random vars init. and the other half is identical, as above, and I perform the same operation, as above, to one of the identical var objects self.mylist[i].change_var_a(5) (to an object that has identicals in the list) all the identicals are changed in the same way as above, however the objects that have different var values are unchanged. What is python doing? Am I missing something? Any ideas at all would be wonderful? -- http://mail.python.org/mailman/listinfo/python-list
Re: very strange problem in 2.4
On Fri, 07 Apr 2006 21:18:12 -0400, John Zenger wrote: Your list probably contains several references to the same object, instead of several different objects. This happens often when you use a technique like: list = [ object ] * 100 ..because although this does make copies when object is an integer, it just makes references in other cases. Wrong. It always makes references. L = [1]*3 id(L[0]), id(L[1]), id(L[2]) (155972920, 155972920, 155972920) This isn't a caching issue either, it also happens for objects which aren't cached: x = 4591034.56472 y = 4591034.56472 x == y True x is y False L = [x]*3 x is L[0] is L[1] is L[2] True y is L[0] False -- Steven. -- http://mail.python.org/mailman/listinfo/python-list
Re: very strange problem in 2.4
John Zenger wrote: Your list probably contains several references to the same object, instead of several different objects. This happens often when you use a technique like: list = [ object ] * 100 This is most likely what's going on. To the OP: please post the relevant code, including how you create mylist and the definitions of change_var_a and return_var_a. I suspect you're doing something like this: \ class C(object): def __init__(self, x): self.x = x def __repr__(self): return 'C(%r)' % self.x mylist = [C(0)]*3 + [C(1)]*3 mylist [C(0), C(0), C(0), C(1), C(1), C(1)] mylist[0].x = 2 [C(2), C(2), C(2), C(1), C(1), C(1)] When you should do something like: mylist = [C(0) for i in range(3)] + [C(1) for i in range(3)] [C(0), C(0), C(0), C(1), C(1), C(1)] mylist[0].x = 2 [C(2), C(0), C(0), C(1), C(1), C(1)] --Ben -- http://mail.python.org/mailman/listinfo/python-list
