[Tutor] dictionary datatype

[Victor Bouffier]

On Tue, 2006-04-11 at 22:37 -0500, Jason Massey wrote:
> Works for me:
> 
> >>> dict1 = { 0x2018:u'k', 0x2019:u'd'}
> >>> n = 0x2018
> >>> print dict1[n]
> k
> >>> 
> 
> On 4/11/06, kakada <hokkakada at khmeros.info> wrote:
>         Hello all,
>         
>         For example, I have a dictionary:
>         dict1 = { 0x2018:u'k', 0x2019:u'd'}
>         
>         I assign:
>         n = 0x2018
>         print dict1[n]
>         
>         Then:
>         KeyError: '0x2018'
>         
>         But I can call directly:
>         print dict1[0x2018] 
>         
>         So, what is wrong with this? How can I solve it?
>         
>         Thx
>         
>         kakada
>         _______________________________________________
>         Tutor maillist  -  Tutor at python.org
>         http://mail.python.org/mailman/listinfo/tutor
> 
> _______________________________________________
> Tutor maillist  -  Tutor at python.org
> http://mail.python.org/mailman/listinfo/tutor

Look into the value of n after assignment:

In [285]: n = 0x2018

In [286]: n
Out[286]: 8216

n is taken as an integer containing the decimal number 8216, which in
hex is represented as 0x2018.

Try again to see if you did not mistakenly typed dict1['0x2018'] or else
defined n as such.

If you try dict1[n], dict1[0x2018], or dict1[8216] you get a correct
result, since the integer variable 'n' contains that value. Trying
dict1['0x2016'] gives you the error because the key does not exist.

define dict1 as:
dict1 = { 0x2018:u'k', 0x2019:u'd'}

and then display it whole:

In [299]: print dict1
{8216: u'k', 8217: u'd'}

Can you see your '0x2018' key anywhere?

HTH

Victor