I have been able to improve the leak considerably with your suggestion
and also attaching the dataFrame instead of passing it as an argument
of lm every time. As an added benefit this reduces the number of copy,
which increases performance. It is still leaking, though....
Laurent
2008/11/21 Laurent Gautier <[EMAIL PROTECTED]>:
> laurent oget wrote:
>>
>> is there something i can do to prevent the following script from leaking
>> memory?
>
>
> There seems to be something happening.
> (more below)
>
>> Laurent
>>
>> from rpy2.robjects import r,RVector,RDataFrame
>> from rpy2.rlike.container import TaggedList
>> from math import sin
>> import array
>>
>> size=100
>> xx=[]
>> for x in range(size):
>> xx.append(RVector(array.array('f',[sin(0.1*(y+x))for y in
>> range(1000)])))
>>
>
>
> (side note: the line above could write easier
>
> from rpy2.robjects import FloatVector
>
> for x in range(size):
> xx.append(FloatVector([sin(0.1 * (x+y) for y in range(1000)]))
> )
>
>
>>
>> formula='x%s~%s'%(0,'+'.join(['x%d'%(i+1) for i in range(size-1)]))
>>
>> t=['x%d'%i for i in range(size)]
>> df=RDataFrame(TaggedList(xx,tags=t))
>>
>> done=False
>> i=0
>> while not done:
>> print i
>> i=i+1
>> model=r.lm(formula,df)
>
> I thought that the following would solve the problem... but it does not.
> The increase in running speed suggests that this is because there is less
> copying (therefore the problem about copying and protecting R objects and
> never unprotecting them). The growth of the process is slower, but still
> there...
>
>
> from rpy2.robjects import globalEnv
>
> globalEnv["dataf"] = df
>
> done=False
> i=0
> while not done:
> print i
> i=i+1
> model=r('lm("%s", data=dataf)' %formula)
>
>
> The problem will likely require a closer look.
>
>
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