A matrix problem. Please help!

[Carl]

I have the following matrix that I want to transform:

import random
import Numeric as N
ru = []
for i in range(25):
ru.append(int(r andom.uniform(0 , 2)))
ru = N.reshape(ru, (5, 5))
[color=blue][color=green][color=darkred]
>>> ru[/color][/color][/color]
array([[1, 0, 1, 1, 1],
[0, 1, 1, 1, 0],
[1, 1, 0, 0, 1],
[0, 1, 0, 0, 1],
[1, 1, 0, 1, 1]])

Trailing numbers (ie, from left to right) after the first encounter of a "1"
for all rows should equal "0".

Thus, I want this:
[color=blue][color=green][color=darkred]
>>> ru[/color][/color][/color]
array([[1, 0, 0, 0, 0],
[0, 1, 0, 0, 0],
[1, 0, 0, 0, 0],
[0, 1, 0, 0, 0],
[1, 0, 0, 0, 0]])

Does anyone have a suggestion of a fast and easy way to accomplish the
above?

Carl

[Peter Otten]

Re: A matrix problem. Please help!

Carl wrote:
[color=blue]
> I have the following matrix that I want to transform:
>
> import random
> import Numeric as N
> ru = []
> for i in range(25):
> ru.append(int(r andom.uniform(0 , 2)))
> ru = N.reshape(ru, (5, 5))
>[color=green][color=darkred]
>>>> ru[/color][/color]
> array([[1, 0, 1, 1, 1],
> [0, 1, 1, 1, 0],
> [1, 1, 0, 0, 1],
> [0, 1, 0, 0, 1],
> [1, 1, 0, 1, 1]])
>
> Trailing numbers (ie, from left to right) after the first encounter of a
> "1" for all rows should equal "0".
>
> Thus, I want this:
>[color=green][color=darkred]
>>>> ru[/color][/color]
> array([[1, 0, 0, 0, 0],
> [0, 1, 0, 0, 0],
> [1, 0, 0, 0, 0],
> [0, 1, 0, 0, 0],
> [1, 0, 0, 0, 0]])
>
> Does anyone have a suggestion of a fast and easy way to accomplish the
> above?
>
> Carl[/color]

(Disclaimer: Numeric newbie code)
[color=blue][color=green][color=darkred]
>>> import random
>>> import Numeric
>>> import itertools
>>> def oneone():[/color][/color][/color]
.... while 1:
.... r = int(random.unif orm(0, 2))
.... yield r
.... if r: break
.... for r in itertools.repea t(0):
.... yield r
....[color=blue][color=green][color=darkred]
>>> ru = Numeric.array([[x for x,y in itertools.izip( oneone(), range(5))][/color][/color][/color]
for z in range(5)])[color=blue][color=green][color=darkred]
>>> ru[/color][/color][/color]
array([[0, 0, 1, 0, 0],
[0, 0, 0, 1, 0],
[1, 0, 0, 0, 0],
[0, 0, 0, 1, 0],
[0, 0, 1, 0, 0]])[color=blue][color=green][color=darkred]
>>>[/color][/color][/color]

Or, more seriously, i. e. assuming that you have no influence on the initial
matrix:
[color=blue][color=green][color=darkred]
>>> r[/color][/color][/color]
array([[1, 1, 1],
[0, 1, 1],
[0, 0, 1]])[color=blue][color=green][color=darkred]
>>> for i in r:[/color][/color][/color]
.... for k in Numeric.nonzero (i)[1:]:
.... i[k] = 0
....[color=blue][color=green][color=darkred]
>>> r[/color][/color][/color]
array([[1, 0, 0],
[0, 1, 0],
[0, 0, 1]])

Peter

[Carl]

Re: A matrix problem. Please help!

Peter Otten wrote:
[color=blue]
> Carl wrote:
>[color=green]
>> I have the following matrix that I want to transform:
>>
>> import random
>> import Numeric as N
>> ru = []
>> for i in range(25):
>> ru.append(int(r andom.uniform(0 , 2)))
>> ru = N.reshape(ru, (5, 5))
>>[color=darkred]
>>>>> ru[/color]
>> array([[1, 0, 1, 1, 1],
>> [0, 1, 1, 1, 0],
>> [1, 1, 0, 0, 1],
>> [0, 1, 0, 0, 1],
>> [1, 1, 0, 1, 1]])
>>
>> Trailing numbers (ie, from left to right) after the first encounter of a
>> "1" for all rows should equal "0".
>>
>> Thus, I want this:
>>[color=darkred]
>>>>> ru[/color]
>> array([[1, 0, 0, 0, 0],
>> [0, 1, 0, 0, 0],
>> [1, 0, 0, 0, 0],
>> [0, 1, 0, 0, 0],
>> [1, 0, 0, 0, 0]])
>>
>> Does anyone have a suggestion of a fast and easy way to accomplish the
>> above?
>>
>> Carl[/color]
>
> (Disclaimer: Numeric newbie code)
>[color=green][color=darkred]
>>>> import random
>>>> import Numeric
>>>> import itertools
>>>> def oneone():[/color][/color]
> ... while 1:
> ... r = int(random.unif orm(0, 2))
> ... yield r
> ... if r: break
> ... for r in itertools.repea t(0):
> ... yield r
> ...[color=green][color=darkred]
>>>> ru = Numeric.array([[x for x,y in itertools.izip( oneone(), range(5))][/color][/color]
> for z in range(5)])[color=green][color=darkred]
>>>> ru[/color][/color]
> array([[0, 0, 1, 0, 0],
> [0, 0, 0, 1, 0],
> [1, 0, 0, 0, 0],
> [0, 0, 0, 1, 0],
> [0, 0, 1, 0, 0]])[color=green][color=darkred]
>>>>[/color][/color]
>
> Or, more seriously, i. e. assuming that you have no influence on the
> initial matrix:
>[color=green][color=darkred]
>>>> r[/color][/color]
> array([[1, 1, 1],
> [0, 1, 1],
> [0, 0, 1]])[color=green][color=darkred]
>>>> for i in r:[/color][/color]
> ... for k in Numeric.nonzero (i)[1:]:
> ... i[k] = 0
> ...[color=green][color=darkred]
>>>> r[/color][/color]
> array([[1, 0, 0],
> [0, 1, 0],
> [0, 0, 1]])
>
> Peter[/color]


Thanks Peter,

This:

for i in ru:
for k in N.nonzero(i)[1:]:
i[k] = 0

works very nice, but I'm a little worried about performance problem, since
the matrix ru is typically of size 10,000 to 100,000 rows by 10 to 60
columns.

I have tried to figure out how to do it without loops, but have not managed
to do that yet.

Carl

[Jeff Epler]

Re: A matrix problem. Please help!

Well, my strategy is to get a new array with 1 in the spots where you
want a 1 in the output, and other values elsewhere. Then I can use
comparison to get a (boolean) array with 1s in the right place

I didn't find a satisfactory way to do that, but here's one that seemed
to work:[color=blue][color=green][color=darkred]
>>> ru[/color][/color][/color]
array([[0, 0, 1, 0, 1],
[1, 0, 0, 1, 1],
[0, 0, 0, 1, 1],
[1, 1, 1, 0, 0],
[1, 1, 0, 1, 0]])[color=blue][color=green][color=darkred]
>>> N.add.accumulat e(N.add.accumul ate(ru, 1), 1)[/color][/color][/color]
array([[ 0, 0, 1, 2, 4],
[ 1, 2, 3, 5, 8],
[ 0, 0, 0, 1, 3],
[ 1, 3, 6, 9, 12],
[ 1, 3, 5, 8, 11]])[color=blue][color=green][color=darkred]
>>> _ == 1[/color][/color][/color]
array([[0, 0, 1, 0, 0],
[1, 0, 0, 0, 0],
[0, 0, 0, 1, 0],
[1, 0, 0, 0, 0],
[1, 0, 0, 0, 0]], type=Bool)

The first accumulate makes everything to the left of a 1 >0, and the
second accumulate makes it >1.

Jeff

[Carl]

Re: A matrix problem. Please help!

Jeff Epler wrote:
[color=blue]
> Well, my strategy is to get a new array with 1 in the spots where you
> want a 1 in the output, and other values elsewhere. Then I can use
> comparison to get a (boolean) array with 1s in the right place
>
> I didn't find a satisfactory way to do that, but here's one that seemed
> to work:[color=green][color=darkred]
>>>> ru[/color][/color]
> array([[0, 0, 1, 0, 1],
> [1, 0, 0, 1, 1],
> [0, 0, 0, 1, 1],
> [1, 1, 1, 0, 0],
> [1, 1, 0, 1, 0]])[color=green][color=darkred]
>>>> N.add.accumulat e(N.add.accumul ate(ru, 1), 1)[/color][/color]
> array([[ 0, 0, 1, 2, 4],
> [ 1, 2, 3, 5, 8],
> [ 0, 0, 0, 1, 3],
> [ 1, 3, 6, 9, 12],
> [ 1, 3, 5, 8, 11]])[color=green][color=darkred]
>>>> _ == 1[/color][/color]
> array([[0, 0, 1, 0, 0],
> [1, 0, 0, 0, 0],
> [0, 0, 0, 1, 0],
> [1, 0, 0, 0, 0],
> [1, 0, 0, 0, 0]], type=Bool)
>
> The first accumulate makes everything to the left of a 1 >0, and the
> second accumulate makes it >1.
>
> Jeff[/color]

Jeff,

Your solution works very well. I tried to do something similar, but with
N.cumsum instead of N.add.accumulua te. However, I never did the second
step, so I ended up with two "1:s" in each row and was, therefore, never
able to do the booelan comparison.

Thanks for yopur help!

Carl