[Tutor] Re: prepend variable name...

[don arnold]

----- Original Message -----
From: "kevin parks" <kp8 at mac.com>
To: "kevin parks" <kp8 at mac.com>
Cc: <tutor at python.org>
Sent: Tuesday, January 20, 2004 1:01 AM
Subject: [Tutor] Re: prepend variable name...


> I should add that, i think the answer lies in making a dictionary (with
> zip?), but what i am not sure of is how to then traverse a dictionary
> in order...
>

Yes, a dictionary is your best option (though zip doesn't enter the picture,
here). In fact, Danny just answered this same question a couple of days ago
with his 'fruitbasket' example. Instead of using a 'regular' variable to
hold your data, you use the variable name as a dictionary key and store your
data there:

myvars = {}
myvars['exI_2a'] = [11,5,7,2]
myvars['exI_2b'] = [0,10,5]
myvars['exI_2c'] = [7,6,9,1]
myvars['exI_2d'] = [4,7,2,7,11]

for key in myvars:
    print key
    print data
    print

[--output--]

exI_2d
[4, 7, 2, 7, 11]

exI_2c
[7, 6, 9, 1]

exI_2b
[0, 10, 5]

exI_2a
[11, 5, 7, 2]

You'll notice that the items aren't necessarily in key order. This is
because a dictionary is an unordered collection. To get around this, you can
use the dicionary's keys() method to generate a list of its keys, then sort
this list before iterating over it.

myvars = {}
myvars['exI_2a'] = [11,5,7,2]
myvars['exI_2b'] = [0,10,5]
myvars['exI_2c'] = [7,6,9,1]
myvars['exI_2d'] = [4,7,2,7,11]

keylist = myvars.keys()
keylist.sort()
for key in keylist:
    print key
    print myvars[key]
    print

[--output--]

exI_2a
[11, 5, 7, 2]

exI_2b
[0, 10, 5]

exI_2c
[7, 6, 9, 1]

exI_2d
[4, 7, 2, 7, 11]


> gosh... dictionaries... hmm... we didn't have those in C so i find that
> i forget how versatile they are. Anyway, i haven't really answered my
> own question yet... but i am i headed in the right direction?
>
> -kp--

Most definitely.

HTH,
Don