how to find position of dictionary values

[lee]

On Sep 1, 1:45 pm, Bruno Desthuilliers <bruno.
42.desthuilli... at websiteburo.invalid> wrote:
> lee a écrit :
>
> > hi,
> > i have a dictionary as follows :
> > kev :  {'phno': ['dgsd', 'gsdg', 'dfsdf', 'g'], 'email': ['dg',
> > 'sgsd', 'sdfsdf', 'gdf'], 'name': ['ds', 'dsg', 'dsfds', 'fgdf'],
> > 'address': ['sdg', 'dsgsdg', 'sdf', 'dfg']}
>
> > if user is enters the 3rd item of key phno,
> > ie "dfsdf" in my dict,
> > how can i find it is the third  item in the internal list of phno of
> > that dictionary?
>
> It's quite simple (hint : read the FineManual(tm) for dict.items() and
> list.index()), but  1/totally inefficient and 2/not garanteed to yield a
> single value (what if 'dfsdf' happens to be also the 4th item of the
> list bound to key 'address'   ?).
>
> May I suggest you rethink your data structure instead ? What you have
> here is obviously a collection of 'phno/email/name/address'records.
> These records shouldn't be split across different objects. Assuming
> 'phno' is a unique identifier for each record, a better data structure
> would be:
>
> records =  {
>     'dgsd' : {'email': 'dg', 'name' : 'ds', 'address' : 'sdg'},
>     'gsdg' : {'email': 'sgsd', 'name':'ds', 'address' : 'dsgsdg'},
>     # etc
>
> }
>
> This way, the lookup is as simple and efficient as possible.
>
> My 2 cents....

hi,
 i agree with u, my data strusture is not efficient. but all the
records,viz...name,phno, email,address are all generated at runtime ,
when the user enters them. so how can i design my datastructure in
that case?