[EMAIL PROTECTED] wrote:
> Hi,
>
> I've reviewed the patch attached to bug #10022
> (https://gna.org/bugs/?10022), and have found an issue. The problem
> is with the use of two dictionaries for the previous and current run.
> The issue is that order in a dictionary is not guaranteed, and hence
> the comparison may not work.
Hi I have two thoughts here ( I haven't seen thr patches so I can't be
certain that I am on ther
1. if you compare two dictionaries with the same contents though the
absolute order of the contents are not guarunteed within the same python
interpreter they will have the same order and the dictionaries can be
compared e.g.
test = {'1':1,'2':2,'3':3}
test2={'1':1,'2':2,'3':3}
test==test2
True
test3={'1':1,'2':2,'3':4}
test3==test2
test4={'1':1,'2':2,'4':3}
test2==test4
False
whats more the way python is structured they would have forbidden
dictionary comparisons if they weren't possible
if you need to extract all they keys and compare the contents in order
then the thing to do is call keys on the two dictionaries and put the
results in a set and then lookup by the keys from the set while properly
guarding for missing keys
reading more on this the dictionaries you are looking at seem to be
keyed by residue (or is it an arbitary something else i.e. just and
index in which case ed comment is fair)
for i in xrange(len(self.relax.data.res['previous'])):
if hasattr(self.relax.data.res['previous'][i], 'model'):
and
for i in xrange(len(self.relax.data.res[run])):
if hasattr(self.relax.data.res[run][i], 'model'):
which you are looking up by an xrange call so the order of i by which
you are comparing seems to be well defined...
now from the look of the problem:
residue 1: prev=None curr=m2
residue 2: prev=None curr=m2
residue 3: prev=None curr=m2
residue 6: prev=m2 curr=m4
residue 7: prev=m2 curr=m1
residue 9: prev=m4 curr=m2
residue 11: prev=None curr=m2
residue 12: prev=m2 curr=m3
residue 13: prev=m2 curr=m3
residue 15: prev=m2 curr=m1
residue 16: prev=m3 curr=m2
residue 17: prev=m3 curr=m1
residue 18: prev=None curr=m3
surely we need to do the following to compare results excluding those which
have none from the previous round
lookup_keys = []
prev_models = ''
for i in xrange(len(self.relax.data.res['previous'])):
if hasattr(self.relax.data.res['previous'][i], 'model'):
if not self.relax.data.res['previous'][i].model == None:
prev_models = prev_models +
self.relax.data.res['previous'][i].model
lookup_keys.append(i)
# Create a string representation of the model-free models of the
current run.
curr_models = ''
for i in lookup_keys:
if hasattr(self.relax.data.res[run][i], 'model'):
if not self.relax.data.res[run][i].model == None:
curr_models = curr_models +
self.relax.data.res[run][i].model
# The test.
if prev_models == curr_models:
now I maybe missing a trick here because I came late to the party but
If not I hope it helps
regards
gary
> Maybe a simple test such as "if
> self.relax.data.res[run][i].model == None" or its negative after
> testing for the presence of the 'model' attribute, and removal of the
> dictionary would remove the problem. Significant simplifications to
> the code after the comment "# The test." could also be made.
>
> As for the code in the section "NOTE: the following code has not been
> extenstively tested", this does not directly address the bug itself
> and it will have problems with the dictionary ordering. I would
> prefer that this section not be present in the patch. If someone does
> have different residues in two different iterations of
> full_analysis.py then this error is much more severe and belongs
> elsewhere other than in the convergence tests.
>
> Cheers,
>
> Edward
>
>
>
>
>
>
--
-------------------------------------------------------------------
Dr Gary Thompson
Astbury Centre for Structural Molecular Biology,
University of Leeds, Astbury Building,
Leeds, LS2 9JT, West-Yorkshire, UK Tel. +44-113-3433024
email: [EMAIL PROTECTED] Fax +44-113-2331407
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