Dear Robert,
So how should I do this? I tried
>>>> d1 = {1: array([2, 3, 4]).copy(), 2: ''}
>>>> l1 = []
>>>> for i in range(3): l1.append(d1.copy())
and it still doesn't work.
I think my problem is: how to create a new array every time I append
it to the list...
Thanks in advance,
- ZMY
On Apr 17, 12:36 pm, Robert Kern <[EMAIL PROTECTED]> wrote:
> ZMYwrote:
> > I am new to Numpy/Pylab, and I am trying to construct a list of
> > dictionaries with arrays as the items, for example:
>
> >>>> dict = {1: array([2, 3, 4]), 2: ''}
> >>>> list1 = []
> >>>> for i in range(3): list1.append(dict.copy())
> > ...
> >>>> list1
> > [{1: array([2, 3, 4]), 2: ''}, {1: array([2, 3, 4]), 2: ''}, {1:
> > array([2, 3, 4]), 2: ''}]
> >>>> list1[0][1][1]=100
> >>>> list1
> > [{1: array([ 2, 100, 4]), 2: ''}, {1: array([ 2, 100, 4]), 2:
> > ''}, {1: array([ 2, 100, 4]), 2: ''}]
>
> >>>> list1[0][2]='Jack'
> >>>> list1
> > [{1: array([ 2, 100, 4]), 2: 'Jack'}, {1: array([ 2, 100, 4]),
> > 2: ''}, {1: array([ 2, 100, 4]), 2: ''}]
>
> > So the strings can be assigned seperately but arrays can not. What is
> > the problem here?
>
> When you call dict.copy(), all it does is make a copy of the dictionary; each
> copy of the dictionary still refers to the same objects.
>
> list1[i][1] refers to the same array for all i, so when you modify it in-place
> like you did, you will see the modifications in every reference to that
> object.
>
> list[i][2] also referred to the same string for all i until you *replaced* the
> entry in the 0'th dictionary.
>
> --
> Robert Kern
>
> "I have come to believe that the whole world is an enigma, a harmless enigma
> that is made terrible by our own mad attempt to interpret it as though it had
> an underlying truth."
> -- Umberto Eco
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