numpy-discussion

[Numpy-discussion] stumped numpy user seeks help

[Mathew Y. <my...@jp...>]

My head is about to explode.

I have an M by N array of floats. Associated with the columns are 
character labels
['a','b','b','c','d','e','e','e']  note: already sorted so duplicates 
are contiguous

I want to replace the 2 'b' columns with the sum of the 2 columns. 
Similarly, replace the 3 'e' columns with the sum of the 3 'e' columns.

The resulting array still has M rows but less than N columns. Anyone? 
Could be any harder than Sudoku.

Mathew

Re: [Numpy-discussion] stumped numpy user seeks help

[Keith G. <kwg...@gm...>]

On 8/29/06, Mathew Yeates <my...@jp...> wrote:

> I have an M by N array of floats. Associated with the columns are
> character labels
> ['a','b','b','c','d','e','e','e']  note: already sorted so duplicates
> are contiguous
>
> I want to replace the 2 'b' columns with the sum of the 2 columns.
> Similarly, replace the 3 'e' columns with the sum of the 3 'e' columns.

Make a cumsum of the array. Find the index of the last 'a', last 'b',
etc and make the reduced array from that. Then take the diff of the
columns.

I know that's vague, but so is my understanding of python/numpy.

Or even more vague: make a function that does what you want.

Re: [Numpy-discussion] stumped numpy user seeks help

[Charles R H. <cha...@gm...>]

On 8/29/06, Keith Goodman <kwg...@gm...> wrote:
>
> On 8/29/06, Mathew Yeates <my...@jp...> wrote:
>
> > I have an M by N array of floats. Associated with the columns are
> > character labels
> > ['a','b','b','c','d','e','e','e']  note: already sorted so duplicates
> > are contiguous
> >
> > I want to replace the 2 'b' columns with the sum of the 2 columns.
> > Similarly, replace the 3 'e' columns with the sum of the 3 'e' columns.
>
> Make a cumsum of the array. Find the index of the last 'a', last 'b',
> etc and make the reduced array from that. Then take the diff of the
> columns.
>
> I know that's vague, but so is my understanding of python/numpy.
>
> Or even more vague: make a function that does what you want.


Or you could use searchsorted on the labels to get a sequence of ranges.
What you have is a sort of binning applied to columns instead of values in a
vector. Or, if the overhead isn't to much, use a dictionary of with (keys:
array) entries. Index thru the columns adding keys, when the key is new
insert a column copy, when it is already present add the new column to the
old one.

Chuck

Re: [Numpy-discussion] stumped numpy user seeks help

[Sven S. <sve...@gm...>]

Mathew Yeates schrieb:
> My head is about to explode.
> 
> I have an M by N array of floats. Associated with the columns are 
> character labels
> ['a','b','b','c','d','e','e','e']  note: already sorted so duplicates 
> are contiguous
> 
> I want to replace the 2 'b' columns with the sum of the 2 columns. 
> Similarly, replace the 3 'e' columns with the sum of the 3 'e' columns.
> 
> The resulting array still has M rows but less than N columns. Anyone? 
> Could be any harder than Sudoku.
> 


Hi,
I don't have time for this ;-) , but I learnt something useful along the
way...

import numpy as n
m = n.ones([2,6])
a = ['b', 'c', 'c', 'd', 'd', 'd']

startindices = set([a.index(x) for x in a])
out = n.empty([m.shape[0], 0])
for i in startindices:
    temp = n.mat(m[:, i : i + a.count(a[i])]).sum(axis = 1)
    out = n.hstack([out, temp])

print out

Not sure if axis = 1 is needed, but until the defaults have settled a
bit it can't hurt. You need python 2.4 for the built-in <set>, and <out>
will be a numpy matrix, use <asarray> if you don't like that. But here
it's really nice to work with matrices, because otherwise .sum() will
give you a 1-d array sometimes, and that will suddenly look like a row
to <hstack> (instead of a nice column vector) and wouldn't work --
that's why matrices are so great and everybody should be using them ;-)

hth,
sven

Re: [Numpy-discussion] stumped numpy user seeks help

[Bill B. <wb...@gm...>]

On 8/30/06, Sven Schreiber <sve...@gm...> wrote:
> Mathew Yeates schrieb:
> will be a numpy matrix, use <asarray> if you don't like that. But here
> it's really nice to work with matrices, because otherwise .sum() will
> give you a 1-d array sometimes, and that will suddenly look like a row
> to <hstack> (instead of a nice column vector) and wouldn't work --
> that's why matrices are so great and everybody should be using them ;-)

column_stack would work perfectly in place of hstack there if it only
didn't have the silly behavior of transposing arguments that already
are 2-d.

For reminders, here's the replacement implementation of column_stack I
proposed on July 21:

def column_stack(tup):
   def transpose_1d(array):
        if array.ndim<2: return _nx.transpose(atleast_2d(array))
        else: return array
   arrays = map(transpose_1d,map(atleast_1d,tup))
   return _nx.concatenate(arrays,1)

This was in a big ticket I submitted about overhauling r_,c_,etc,
which was largely ignored.  Maybe I should resubmit this by itself...

--bb

Re: [Numpy-discussion] stumped numpy user seeks help

[Stefan v. d. W. <st...@su...>]

On Tue, Aug 29, 2006 at 03:46:45PM -0700, Mathew Yeates wrote:
> My head is about to explode.
>=20
> I have an M by N array of floats. Associated with the columns are=20
> character labels
> ['a','b','b','c','d','e','e','e']  note: already sorted so duplicates=20
> are contiguous
>=20
> I want to replace the 2 'b' columns with the sum of the 2 columns.=20
> Similarly, replace the 3 'e' columns with the sum of the 3 'e' columns.
>=20
> The resulting array still has M rows but less than N columns. Anyone?=20
> Could be any harder than Sudoku.

I attach one possible solution (allowing for the same column name
occurring in different places, i.e. ['a','b','b','a']).  I'd be glad
for any suggestions on how to clean up the code.

Regards
St=E9fan