Parallel arithmetic?

[Terrance N. Phillip]

Given a and b, two equal length lists of integers, I want c to be
[a1-b1, a2-b2, ... , an-bn]. I can do something like:

c = [0] * len(a)
for ndx, item in enumerate(a):
c[ndx] = item - b[ndx]

But I'm wondering if there's a better way, perhaps that avoids a loop?

Nick.

(I seem to recall from my distant past that this sort of thing was dead
easy with APL... c = a-b, more or less.)

N

[Paul Rubin]

Re: Parallel arithmetic?

"Terrance N. Phillip" <mediocre_perso n@hotmail.com> writes:[color=blue]
> Given a and b, two equal length lists of integers, I want c to be
> [a1-b1, a2-b2, ... , an-bn].[/color]

c = [a[i] - b[i] for i in xrange(len(a))]

[Michael Hoffman]

Re: Parallel arithmetic?

Terrance N. Phillip wrote:[color=blue]
> Given a and b, two equal length lists of integers, I want c to be
> [a1-b1, a2-b2, ... , an-bn]. I can do something like:
>
> c = [0] * len(a)
> for ndx, item in enumerate(a):
> c[ndx] = item - b[ndx]
>
> But I'm wondering if there's a better way, perhaps that avoids a loop?[/color]

Here's one way:

c = [a_item - b_item for a_item, b_item in zip(a, b)]

And another:

import operator
c = map(operator.su b, a, b)
--
Michael Hoffman

[Lonnie Princehouse]

Re: Parallel arithmetic?

There are many ways to do this. None of them avoids looping,
technically, although you can easily avoid the "for" syntax.

-- Simple but wastes some memory
c = [i-j for i,j in zip(a,b)]

-- Using itertools.izip (python 2.3)
c = [i-j for i,j in itertools.izip( a,b) ]

-- Generator expression (python 2.4)
c = ( i-j for i,j in itertools.izip( a,b) )

[Robert Kern]

Re: Parallel arithmetic?

Terrance N. Phillip wrote:[color=blue]
> Given a and b, two equal length lists of integers, I want c to be
> [a1-b1, a2-b2, ... , an-bn]. I can do something like:
>
> c = [0] * len(a)
> for ndx, item in enumerate(a):
> c[ndx] = item - b[ndx]
>
> But I'm wondering if there's a better way, perhaps that avoids a loop?
>
> Nick.
>
> (I seem to recall from my distant past that this sort of thing was dead
> easy with APL... c = a-b, more or less.)[/color]

If you're doing this kind of thing often, look into using Numeric.

http://numeric.scipy.org

In [21]: from Numeric import array

In [22]: a = array(range(10) )

In [23]: b = array(range(10, 20))

In [24]: c = a - b

In [25]: c
Out[25]: [-10,-10,-10,-10,-10,-10,-10,-10,-10,-10,]

--
Robert Kern
rkern@ucsd.edu

"In the fields of hell where the grass grows high
Are the graves of dreams allowed to die."
-- Richard Harter

[jepler@unpythonic.net]

Re: Parallel arithmetic?

If you use numarray, you *can* write
c = a-b
[color=blue][color=green][color=darkred]
>>> import numarray
>>> a = numarray.array([1,2,3])
>>> b = numarray.array([5,0,2])
>>> c = a-b
>>> c[/color][/color][/color]
array([-4, 2, 1])

numarray is packaged separately from Python.

http://www.stsci.edu/resources/software_hardware/numarray

Jeff

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[Terry Reedy]

Re: Parallel arithmetic?

[color=blue]
>"#"map" will be removed from the next versions of python.[/color]

The next version will be 2.5. Map will not go away then.
In 3.0, in the indefinite future, it might go away, it might just be moved.

tjr

[Terrance N. Phillip]

Re: Parallel arithmetic?

Thank-you very much for all the excellent replies. I'm thinking of using
this to determine if a sequence is a "run" (as in a card game). If I've
got a sorted hand [3, 4, 5, 6, 7], then I know I've got a 5-card run
because [4, 5, 6, 7] - [3, 4, 5, 6] == [1, 1, 1, 1]. I want to avoid
something like
if h[0] == h[1]-1 and h[1] == h[2]-1 ...

Nick.

[Robert Kern]

Re: Parallel arithmetic?

Terrance N. Phillip wrote:[color=blue]
> Thank-you very much for all the excellent replies. I'm thinking of using
> this to determine if a sequence is a "run" (as in a card game). If I've
> got a sorted hand [3, 4, 5, 6, 7], then I know I've got a 5-card run
> because [4, 5, 6, 7] - [3, 4, 5, 6] == [1, 1, 1, 1]. I want to avoid
> something like
> if h[0] == h[1]-1 and h[1] == h[2]-1 ...[/color]

In that case:

http://aspn.activestate.com/ASPN/Cookbook/Python/Recipe/415504

--
Robert Kern
rkern@ucsd.edu

"In the fields of hell where the grass grows high
Are the graves of dreams allowed to die."
-- Richard Harter

[Dennis Lee Bieber]

Re: Parallel arithmetic?

On Thu, 04 Aug 2005 20:59:33 -0500, "Terrance N. Phillip"
<mediocre_perso n@hotmail.com> declaimed the following in
comp.lang.pytho n:
[color=blue]
> Thank-you very much for all the excellent replies. I'm thinking of using
> this to determine if a sequence is a "run" (as in a card game). If I've
> got a sorted hand [3, 4, 5, 6, 7], then I know I've got a 5-card run[/color]

A sorted list?

if (hand[-1] - hand[0]) == (len(hand) - 1)

would seem to do it for your example.

Actually, if you KNOW the list is only 5 entries long

if (hand[4] - hand[0]) == 4

would do it.

Or any equivalent... (hand[0] + 4) == hand[4]


--[color=blue]
> =============== =============== =============== =============== == <
> wlfraed@ix.netc om.com | Wulfraed Dennis Lee Bieber KD6MOG <
> wulfraed@dm.net | Bestiaria Support Staff <
> =============== =============== =============== =============== == <
> Home Page: <http://www.dm.net/~wulfraed/> <
> Overflow Page: <http://wlfraed.home.ne tcom.com/> <[/color]

[jburgy]

Re: Parallel arithmetic?

Dennis Lee Bieber wrote:[color=blue]
> On Thu, 04 Aug 2005 20:59:33 -0500, "Terrance N. Phillip"
> <mediocre_perso n@hotmail.com> declaimed the following in
> comp.lang.pytho n:
>[color=green]
> > Thank-you very much for all the excellent replies. I'm thinking of using
> > this to determine if a sequence is a "run" (as in a card game). If I've
> > got a sorted hand [3, 4, 5, 6, 7], then I know I've got a 5-card run[/color]
>
> A sorted list?
>
> if (hand[-1] - hand[0]) == (len(hand) - 1)
>
> would seem to do it for your example.
>
> Actually, if you KNOW the list is only 5 entries long
>
> if (hand[4] - hand[0]) == 4
>[/color]

It's cute but wrong! How 'bout hand = [ 0, 0, 0, 4 ]? It's sorted,
passes your test and does not meet the OP's requirement :(
[color=blue]
> would do it.
>
> Or any equivalent... (hand[0] + 4) == hand[4]
>
>
> --[color=green]
> > =============== =============== =============== =============== == <
> > wlfraed@ix.netc om.com | Wulfraed Dennis Lee Bieber KD6MOG <
> > wulfraed@dm.net | Bestiaria Support Staff <
> > =============== =============== =============== =============== == <
> > Home Page: <http://www.dm.net/~wulfraed/> <
> > Overflow Page: <http://wlfraed.home.ne tcom.com/> <[/color][/color]