how to find position of dictionary values

[lee]

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? thanks you.

[lee]

On Sep 1, 2:37 pm, "Diez B. Roggisch" <de...@nospam.w eb.dewrote:

lee wrote:

On Sep 1, 1:45 pm, Bruno Desthuilliers <bruno.
42.desthuilli.. .@websiteburo.i nvalidwrote:

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 yielda
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...n ame,phno, email,address are all generated at runtime ,
when the user enters them. so how can i design my datastructure in
that case?

Are "u" short on keystrokes? You are not textmessaging here...

Regarding the actual question: there is no difference in building your or
the other structure. It's only a question of which key you use first.
Instead of first looking up the type of the record ("phno" or some such),
do that with the name of the user. If no record exists, create one. Then
populate the record with the user's values. Like this:

user = "dsdf"
phonenumber = "123"

record = records.setdefa ult(user, {})
record["phno"] = phonenumber

Diez

i am soory for that keystrokes. can anyone tell me how can i change
the value of key.
suppose i have a dictionary

kev = {'kabir': ['k****@kabir.co m', '1234', 'missuri'], 'shri':
['s***@shri.com' , '23423', 'india'], 'marsa': ['m****@marsa.co m',
'2345', 'brazil'], 'sandeep': ['s******@sandee p.com', '007',
'canada']}
how can i change the key to something like 'sabir' and how can i
change the values of kabir?

[Wojtek Walczak]

On Mon, 1 Sep 2008 03:51:13 -0700 (PDT), lee wrote:

i am soory for that keystrokes. can anyone tell me how can i change
the value of key.
suppose i have a dictionary

kev = {'kabir': ['k****@kabir.co m', '1234', 'missuri'], 'shri':
['s***@shri.com' , '23423', 'india'], 'marsa': ['m****@marsa.co m',
'2345', 'brazil'], 'sandeep': ['s******@sandee p.com', '007',
'canada']}

how can i change the key to something like 'sabir' and

kev['sabir'] = kev['kabir']
del kev['kabir']

how can i
change the values of kabir?

kev['sabir'][0] = 's****@sabir.co m'

*untested*

--
Regards,
Wojtek Walczak,
http://tosh.pl/gminick/

[lee]

On Sep 1, 3:59 pm, Wojtek Walczak <[e-mail] twrote:

On Mon, 1 Sep 2008 03:51:13 -0700 (PDT), lee wrote:

i am soory for that keystrokes. can anyone tell me how can i change
the value of key.
suppose i have a dictionary

kev = {'kabir': ['[e-mail] m', '1234', 'missuri'], 'shri':
['[e-mail]' , '23423', 'india'], 'marsa': ['[e-mail] m',
'2345', 'brazil'], 'sandeep': ['sand...@sandee p.com', '007',
'canada']}
how can i change the key to something like 'sabir' and

kev['sabir'] = kev['kabir']
del kev['kabir']

how can i
change the values of kabir?

kev['sabir'][0] = '[e-mail] m'

*untested*

--
Regards,
Wojtek Walczak,http://tosh.pl/gminick/

thanks wojtek, it worked :)

[Bruno Desthuilliers]

lee a écrit :

On Sep 1, 2:37 pm, "Diez B. Roggisch" <de...@nospam.w eb.dewrote:

>lee wrote:

(snip)

>> 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?

<ot>

>Are "u" short on keystrokes? You are not textmessaging here...

i am soory for that keystrokes

Is your cap key broken ?-)

</ot>

(snip)

>. can anyone tell me how can i change
the value of key.
suppose i have a dictionary

kev = {'kabir': ['k****@kabir.co m', '1234', 'missuri'], 'shri':
['s***@shri.com' , '23423', 'india'], 'marsa': ['m****@marsa.co m',
'2345', 'brazil'], 'sandeep': ['s******@sandee p.com', '007',
'canada']}
how can i change the key to something like 'sabir' and how can i
change the values of kabir?

I don't mean to be harsh, but did you read the FineManual(tm) ? All this
is very basic (no pun) stuff, mostly documented, and easy to try out
using the interactive interpreter. This ng is very newbie-friendly, but
this doesn't mean you can hope to get by without doing the minimal
effort of going thru the tutorial and trying a couple things by yourself
before asking.

Hope you'll understand the above remark intends to be an helpful advice...

[Terry Reedy]

Alexandru Palade wrote:

lookfor = 'dfsdf'
for item, value in kev.items():
if lookfor in value:
print item
print value.index(loo kfor)
break # assuming you only want one result

slight variation:

lookfor = 'dfsdf'
for item, value in kev.items():
for i, val in enumerate(value ):
if val == lookfor:
print item, i
break # assuming you only want one result
else:
print lookfor, 'not found'

This is what for-else is meant for.

If you want 0 to many lookfor occurences,

lookfor = 'dfsdf'
hits = []
for item, value in kev.items():
for i, val in enumerate(value ):
if val == lookfor:
hits.append((it em, i))

print hits

One-liner fanatics would, of course, rewrite this as

hits = [(item, i) for item, value in kev.items() for i, val in
enumerate(value ) if val == lookfor]

Terry Jan Reedy

tjr