J
Jean_Francois Moulin
Hi all,
I tried this piece of code (FWIW, it was taken as is from a help section of mpfit, a mathematical routine for least square fitting):
parinfo = [{'value':0., 'fixed':0, 'limited':[0,0], 'limits':[0.,0.]}]*6
parinfo[0]['fixed'] = 1
parinfo[4]['limited'][0] = 1
parinfo[4]['limits'][0] = 50.
The first line builds a list of six dictionaries with initialised keys.
I expected that the last three lines would only affect the corresponding keys of the corresponding dictionnary and that I would end up with a fully initialised list where only the 'fixed' key of the first dict would be 1, and the first values of limited and limits for dict number 4 would be 1 and 50. respectively....
This is not so!
I end up with all dictionaries being identical and having their 'fixed' key set to 1, and limited[0]==1 and limits[0]==50.
I do not understand this behaviour...
Thanks for helping a newbie.
JF
I tried this piece of code (FWIW, it was taken as is from a help section of mpfit, a mathematical routine for least square fitting):
parinfo = [{'value':0., 'fixed':0, 'limited':[0,0], 'limits':[0.,0.]}]*6
parinfo[0]['fixed'] = 1
parinfo[4]['limited'][0] = 1
parinfo[4]['limits'][0] = 50.
The first line builds a list of six dictionaries with initialised keys.
I expected that the last three lines would only affect the corresponding keys of the corresponding dictionnary and that I would end up with a fully initialised list where only the 'fixed' key of the first dict would be 1, and the first values of limited and limits for dict number 4 would be 1 and 50. respectively....
This is not so!
I end up with all dictionaries being identical and having their 'fixed' key set to 1, and limited[0]==1 and limits[0]==50.
I do not understand this behaviour...
Thanks for helping a newbie.
JF