Select as dictionary...

[Besturk.Net Admin]

Hi..
I am using python with postgresql.
And i have a query :

aia.execute("SELECT id, w from list")
links=aia.fetchall()
print links

and result
[(1, 5), (2,5).......] (2 million result)

I want to see this result directly as a dictionary:

{1: 5, 2: 5 .....}

How do i select in this format ?

I'm sorry my bad english.

King Regards...

[Bruno Desthuilliers]

Besturk.Net Admin a écrit :

Hi..
I am using python with postgresql.
And i have a query :

aia.execute("SELECT id, w from list")
links=aia.fetchall()
print links

and result
[(1, 5), (2,5).......] (2 million result)

I want to see this result directly as a dictionary:

{1: 5, 2: 5 .....}

How do i select in this format ?

IIRC, postgres' db-api connector (well, at least one of them - I don't
know which one you're using) has a DictCursor. You should find all you
want to know in the relevant doc.

FWIW:
http://www.initd.org/tracker/psycopg/wiki/PsycopgTwoFaq

HTH

[J. Clifford Dyer]

On Mon, Oct 01, 2007 at 06:32:07AM -0700, Besturk.Net Admin wrote regarding Select as dictionary...:

>
aia.execute("SELECT id, w from list")
links=aia.fetchall()
print links

and result
[(1, 5), (2,5).......] (2 million result)

I want to see this result directly as a dictionary:

{1: 5, 2: 5 .....}

How do i select in this format ?

Try this:

aia.execute("SELECT id, w from list")
links=aia.fetchall()
linkdict = dict()
for k,v in links:
linkdict[k] = v
print linkdict
Cheers,
Cliff

[Stargaming]

On Mon, 01 Oct 2007 09:57:46 -0400, J. Clifford Dyer wrote:

On Mon, Oct 01, 2007 at 06:32:07AM -0700, Besturk.Net Admin wrote
regarding Select as dictionary...:

>>
aia.execute("SELECT id, w from list") links=aia.fetchall()
print links

and result
[(1, 5), (2,5).......] (2 million result)

I want to see this result directly as a dictionary:

{1: 5, 2: 5 .....}

How do i select in this format ?

Try this:

aia.execute("SELECT id, w from list")
links=aia.fetchall()
linkdict = dict()
for k,v in links:
linkdict[k] = v
print linkdict

Besides using the already pointed out DB adapters, you can easily
transform a list of items into a dictionary with the `dict` constructor::

>>links = [(1, 5), (2, 5), (3, 10)]
dict(links)

{1: 5, 2: 5, 3: 10}

Stargaming

[M.-A. Lemburg]

On 2007-10-01 15:32, Besturk.Net Admin wrote:

Hi..
I am using python with postgresql.
And i have a query :

aia.execute("SELECT id, w from list")
links=aia.fetchall()
print links

and result
[(1, 5), (2,5).......] (2 million result)

I want to see this result directly as a dictionary:

{1: 5, 2: 5 .....}

How do i select in this format ?

You can easily convert the above list to a dictionary:

d = dict(links)

--
Marc-Andre Lemburg
eGenix.com

Professional Python Services directly from the Source (#1, Oct 01 2007)

>>Python/Zope Consulting and Support ... http://www.egenix.com/
mxODBC.Zope.Database.Adapter ... http://zope.egenix.com/
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Registered at Amtsgericht Duesseldorf: HRB 46611

[Duncan Booth]

"J. Clifford Dyer" <jc*@sdf.lonestar.orgwrote:

aia.execute("SELECT id, w from list")
links=aia.fetchall()
linkdict = dict()
for k,v in links:
linkdict[k] = v
print linkdict

Wouldn't it be simpler just to do:

aia.execute("SELECT id, w from list")
linkdict=dict(aia.fetchall())

even better would be to use an iterator to avoid fetching the entire
resultset as a list.

[Carsten Haese]

On Mon, 2007-10-01 at 09:57 -0400, J. Clifford Dyer wrote:

On Mon, Oct 01, 2007 at 06:32:07AM -0700, Besturk.Net Admin wrote regarding Select as dictionary...:

aia.execute("SELECT id, w from list")
links=aia.fetchall()
print links

and result
[(1, 5), (2,5).......] (2 million result)

I want to see this result directly as a dictionary:

{1: 5, 2: 5 .....}

How do i select in this format ?

Try this:

aia.execute("SELECT id, w from list")
links=aia.fetchall()
linkdict = dict()
for k,v in links:
linkdict[k] = v
print linkdict

Improvement 1: Use the fact that dict can be initialized from a sequence
of key/value pairs:

aia.execute("SELECT id, w from list")
linkdict = dict(aia.fetchall())

Improvement 2: Use an iterator instead of reading all rows into memory:

aia.execute("SELECT id, w from list")
linkdict = dict(iter(aia.fetchone,None))

--
Carsten Haese
http://informixdb.sourceforge.net

[Carsten Haese]

On Mon, 2007-10-01 at 15:50 +0200, Bruno Desthuilliers wrote:

Besturk.Net Admin a écrit :

I want to see this result directly as a dictionary:

{1: 5, 2: 5 .....}

How do i select in this format ?

IIRC, postgres' db-api connector (well, at least one of them - I don't
know which one you're using) has a DictCursor.

That would return the results as a list of dictionaries like this:

[{'id':1, 'w':5}, {'id':2, 'w':5}, ...]

--
Carsten Haese
http://informixdb.sourceforge.net

[Carsten Haese]

linkdict = dict(iter(aia.fetchone,None))

And by the way, that line can be shortened to "linkdict=dict(aia)" if
the cursor object supports the iterator protocol, but that's not a
mandatory feature in DB-API v2. The longer form is guaranteed to work
with any DB-API v2 compliant implementation.

--
Carsten Haese
http://informixdb.sourceforge.net

[J. Clifford Dyer]

On Mon, Oct 01, 2007 at 10:12:04AM -0400, Carsten Haese wrote regarding Re: Select as dictionary...:

>
On Mon, 2007-10-01 at 09:57 -0400, J. Clifford Dyer wrote:

>

Try this:

aia.execute("SELECT id, w from list")
links=aia.fetchall()
linkdict = dict()
for k,v in links:
linkdict[k] = v
print linkdict

Improvement 1: Use the fact that dict can be initialized from a sequence
of key/value pairs:

aia.execute("SELECT id, w from list")
linkdict = dict(aia.fetchall())

This is only an improvement if the SQL query remains a list of 2-tuples. If the OP wants to add more values, the more verbose version allows for easier extensibility.

for k,v1,v2 in links:
linkdict[k] = (v1, v2)

Of course even better would be not to have to add variables, so maybe:

for link in links:
linkdict[link[0]] = link[1:],

which changes the output format somewhat, but keeps the access by dict keyed to the DB id.

Improvement 2: Use an iterator instead of reading all rows into memory:

aia.execute("SELECT id, w from list")
linkdict = dict(iter(aia.fetchone,None))

Agreed. A much better solution.

Cheers,
Cliff

[Bruno Desthuilliers]

J. Clifford Dyer a écrit :

On Mon, Oct 01, 2007 at 03:50:59PM +0200, Bruno Desthuilliers wrote regarding Re: Select as dictionary...:

>IIRC, postgres' db-api connector (well, at least one of them - I don't
know which one you're using) has a DictCursor. You should find all you
want to know in the relevant doc.

Correct me if I'm wrong, but I think the DictCursor would output the following:

[{'id': 1, 'value': 5}, {'id': 2, 'value': 5}, ...]

But the OP wanted one dict with IDs as keys:

{1: 5, 2: 5, ...}

Hmmm... I did answer too fast, indeed. Thanks for pointing that out.

[Abandoned]

On 1 Ekim, 18:13, Bruno Desthuilliers <bruno.
[e-mail]:

J. Clifford Dyer a écrit :

On Mon, Oct 01, 2007 at 03:50:59PM +0200, Bruno Desthuilliers wrote regarding Re: Select as dictionary...:

IIRC, postgres' db-api connector (well, at least one of them - I don't
know which one you're using) has a DictCursor. You should find all you
want to know in the relevant doc.

Correct me if I'm wrong, but I think the DictCursor would output the following:

[{'id': 1, 'value': 5}, {'id': 2, 'value': 5}, ...]

But the OP wanted one dict with IDs as keys:

{1: 5, 2: 5, ...}

Hmmm... I did answer too fast, indeed. Thanks for pointing that out.

Very very very thanks to everybody.
I'm a new user in this group and you are wonderfull persons :)

[Abandoned]

Also if i need a list id what can i do ?

aia.execute("SELECT id, w from list")
links=aia.fetchall()

I want to..

idlist=[1, 2, 3] ( I don't want to use FOR and APPEND because the
query have 2 million result and i want to speed)

[Diez B. Roggisch]

Abandoned wrote:

Also if i need a list id what can i do ?

aia.execute("SELECT id, w from list")
links=aia.fetchall()

I want to..

idlist=[1, 2, 3] ( I don't want to use FOR and APPEND because the
query have 2 million result and i want to speed)

It will always return a list of tuples, so even if you'd do
select id from list
you'd end up with

[(1,), ...]

But you shouldn't bother - a simple

idlist = [v[0] for v in links]

should do it. Because this program shows that it takes only a split-second
to process 2 million list-entries and create a new one:

import time

two_million = range(2000000)

start = time.time()
[i * 2 for i in two_million]

used = time.time() - start

print used
Need 0.8 seconds on my machine. So - don't bother.

Diez

[Carsten Haese]

On Mon, 2007-10-01 at 10:13 -0700, Abandoned wrote:

Also if i need a list id what can i do ?

aia.execute("SELECT id, w from list")
links=aia.fetchall()

I want to..

idlist=[1, 2, 3] ( I don't want to use FOR and APPEND because the
query have 2 million result and i want to speed)

The canonical replacement for a for/append loop is a list comprehension:

idlist = [x[0] for x in links]

HTH,

--
Carsten Haese
http://informixdb.sourceforge.net

[Robert Kern]

wang frank wrote:

>
Hi,

I need to help to install these two packages on debian. I want to know
what packages do I need to? I have installed fftw3,fftww2, sfftw2,
atlas. Did I miss anything? In what way I can install an optimized numpy
and scipy, since my project is very big and speed is important.

Be sure to install the -dev packages for each of the above as well as
python-dev. You will probably also want to install the CPU-specific versions of
ATLAS, too (like atlas3-sse2). If you *really* need speed, I recommend building
a more recent release of ATLAS yourself especially if you have a CPU released in
the last couple of years. The 3.7 series has seen a lot of improvements over the
3.6 released in Debian.

If you can live with the packaged versions of ATLAS and FFTW, then you might
also be able to live with the packaged python-numpy and python-scipy. They are
somewhat older than the current release, but that shouldn't be too bad.

If you want to build numpy and scipy yourself and you need more specific help,
please join us on scipy-user with your questions.

http://www.scipy.org/Mailing_Lists

--
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

[Duncan Booth]

Abandoned <be*****@gmail.comwrote:

( I don't want to use FOR and APPEND because the
query have 2 million result and i want to speed)

How many times do you plan on doing this? On my system I just timed adding
2 million elements to a dictionary: it took 0.35 seconds. Appending 2
million elements to a list took 0.43 seconds.

Unless you really do need to repeat this operation many many many times
then just write the obvious code instead of wasting your (and our) time
trying to micro-optimise.

[Steve Holden]

Abandoned wrote:

Also if i need a list id what can i do ?

aia.execute("SELECT id, w from list")
links=aia.fetchall()

I want to..

idlist=[1, 2, 3] ( I don't want to use FOR and APPEND because the
query have 2 million result and i want to speed)

It may not be practical for your algorithm, but I want to point out that
at this sort of query size it's much better (when you can) to repeatedly
call the execution cursor's fetchmany() method repeatedly. This avoids
having to allocate sufficient memory space to hold the whole query's output.

regards
Steve
--
Steve Holden [phone] [phone]
Holden Web LLC/Ltd http://www.holdenweb.com
Skype: holdenweb http://del.icio.us/steve.holden

Sorry, the dog ate my .sigline