A matrix problem. Please help!

[Carl]

Peter Otten wrote:

> Carl wrote:
> 
>> 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(random.uniform(0, 2)))
>> ru = N.reshape(ru, (5, 5))
>> 
>>>>> ru
>> 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:
>> 
>>>>> ru
>> 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
> 
> (Disclaimer: Numeric newbie code)
> 
>>>> import random
>>>> import Numeric
>>>> import itertools
>>>> def oneone():
> ...     while 1:
> ...             r = int(random.uniform(0, 2))
> ...             yield r
> ...             if r: break
> ...     for r in itertools.repeat(0):
> ...             yield r
> ...
>>>> ru = Numeric.array([[x for x,y in itertools.izip(oneone(), range(5))]
> for z in range(5)])
>>>> ru
> 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]])
>>>>
> 
> Or, more seriously, i. e. assuming that you have no influence on the
> initial matrix:
> 
>>>> r
> array([[1, 1, 1],
>        [0, 1, 1],
>        [0, 0, 1]])
>>>> for i in r:
> ...     for k in Numeric.nonzero(i)[1:]:
> ...             i[k] = 0
> ...
>>>> r
> array([[1, 0, 0],
>        [0, 1, 0],
>        [0, 0, 1]])
> 
> Peter


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