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 3 1137
parinfo = [{'value':0., 'fixed':0, 'limited':[0,0], 'limits':[0.,0.]}]*6
With this, you are creating a list with 6 references to the same list.
Note that the left operand of '*' is evaluated only once before
"multiplying" it six times.
Regards,
Tito
Jean_Francois Moulin wrote in
news:ma***************************************@pyt hon.org in
comp.lang.python:
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...
This should help:
<url: http://www.python.org/doc/faq/progra...do-i-create-a-
multidimensional-list>
As a TinyUrl: http://tinyurl.com/s8akj
Rob.
-- http://www.victim-prime.dsl.pipex.com/
Jean_Francois Moulin wrote:
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
The first line builds a list of six dictionaries with initialised keys.
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
xvec = [{'value':0}]*6
xids = [id(x) for x in xvec]
print xids
Should print a list of six integers, all equal.
So all elements in your list are the same.
Another way to construct the desired list:
yvec = [{'value':0} for i in range(6)]
yids = [id(y) for y in yvec]
print yids
The elements in this list should be all different. This thread has been closed and replies have been disabled. Please start a new discussion. Similar topics
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