On 4/23/07, Charles R Harris <[EMAIL PROTECTED]> wrote:
On 4/23/07, Travis Oliphant <[EMAIL PROTECTED]> wrote:
>
> Christian Marquardt wrote:
> > Hello,
> >
> > The following is what I expected...
> >
> >>>> y = 1234
> >>>> x = array([1], dtype = "uint64")
> >>>> print x + y, (x +
On 4/23/07, Travis Oliphant <[EMAIL PROTECTED]> wrote:
Christian Marquardt wrote:
> Hello,
>
> The following is what I expected...
>
>>>> y = 1234
>>>> x = array([1], dtype = "uint64")
>>>> print x + y, (x + y).dtype.type
>[1235]
>
>
This is "what you expect" only because y is
But even C89 required that x == (x/y)*y + (x%y), and that's not the case
here.
w
On Mon, 23 Apr 2007, David M. Cooke wrote:
> On Apr 23, 2007, at 16:41 , Christian Marquardt wrote:
> > On Mon, April 23, 2007 22:29, Christian Marquardt wrote:
> >> Actually,
> >>
> >> it happens for normal integer
Christian Marquardt wrote:
> Hello,
>
> The following is what I expected...
>
>>>> y = 1234
>>>> x = array([1], dtype = "uint64")
>>>> print x + y, (x + y).dtype.type
>[1235]
>
>
This is "what you expect" only because y is a scalar and cannot
determine the "kind" of the output
On Apr 23, 2007, at 16:41 , Christian Marquardt wrote:
On Mon, April 23, 2007 22:29, Christian Marquardt wrote:
Actually,
it happens for normal integers as well:
n = np.array([-5, -100, -150])
n // 100
array([ 0, -1, -1])
-5//100, -100//100, -150//100
(-1, -1, -2)
and finally:
n
Christopher Barker wrote:
> if numpy.reshape() is delegating to ndarray.reshape() couldn't they
> share docstrings somehow?
Yes, of course. I got bored halfway through the conversions and didn't get to
doing the methods. If you want to speed up the process, please submit a patch.
--
Robert Ker
Keith Goodman wrote:
> At the moment numpy.reshape and array.reshape have different doc
> strings (I'm using numpy 1.0.2.dev3546). The one I pasted is from
> numpy.reshape.
And I see from there:
:See also:
numpy.ndarray.reshape() is the equivalent method.
so it looks like they are th
Hmmm,
On Mon, April 23, 2007 22:29, Christian Marquardt wrote:
> Actually,
>
> it happens for normal integers as well:
>
>>>> n = np.array([-5, -100, -150])
>>>> n // 100
>array([ 0, -1, -1])
>>>> -5//100, -100//100, -150//100
>(-1, -1, -2)
and finally:
>>> n % 100
arra
Actually,
it happens for normal integers as well:
>>> n = np.array([-5, -100, -150])
>>> n // 100
array([ 0, -1, -1])
>>> -5//100, -100//100, -150//100
(-1, -1, -2)
On Mon, April 23, 2007 22:20, Christian Marquardt wrote:
> Dear all,
>
> this is odd:
>
>>>> import numpy as np
Dear all,
this is odd:
>>> import numpy as np
>>> fact = 2825L * 86400L
>>> nn = np.array([-20905000L])
>>> nn
array([-20905000], dtype=int64)
>>> nn[0] // fact
0
But:
>>> long(nn[0]) // fact
-1L
Is this a bug in numpy, or in python's implementation of longs? I w
On 4/23/07, Christopher Barker <[EMAIL PROTECTED]> wrote:
> Charles R Harris wrote:
> > Here's a better doc string that explains "This will be a new view
> > object if possible; otherwise, it will return a copy."
>
> Does this exist somewhere, or are you contributing it now?
At the moment
Charles R Harris wrote:
> Here's a better doc string that explains "This will be a new view
> object if possible; otherwise, it will return a copy."
Does this exist somewhere, or are you contributing it now?
> I think that it should raise an error, or warn, if it needs to make a
> copy,
On 4/23/07, Keith Goodman <[EMAIL PROTECTED]> wrote:
On 4/23/07, Christopher Barker <[EMAIL PROTECTED]> wrote:
> reshape(...)
> a.reshape(d1, d2, ..., dn, order='c')
>
> Return a new array from this one. The new array must have the same
>
> number of elements as self. Also alway
El dl 23 de 04 del 2007 a les 12:47 -0400, en/na Anne Archibald va
escriure:
> On 23/04/07, Pierre GM <[EMAIL PROTECTED]> wrote:
>
> > Note that in addition of the bitwise operators, you can use the "logical_"
> > functions. OK, you'll still end up w/ temporaries, but I wonder whether
> > there
>
On 4/23/07, Christopher Barker <[EMAIL PROTECTED]> wrote:
> reshape(...)
> a.reshape(d1, d2, ..., dn, order='c')
>
> Return a new array from this one. The new array must have the same
>
> number of elements as self. Also always returns a view or raises a
> ValueError if that i
On 4/23/07, Christopher Barker <[EMAIL PROTECTED]> wrote:
Gael Varoquaux wrote:
> Unless I miss something obvious "a.reshape()" doesn't modify a, which is
> somewhat missleading, IMHO.
quite correct. .reshape() creates a new array that shared data with the
original:
Sometimes it do, sometime
On Monday 23 April 2007 13:36:26 Christopher Barker wrote:
> Gael Varoquaux wrote:
> > Unless I miss something obvious "a.reshape()" doesn't modify a, which is
> > somewhat missleading, IMHO.
>
> quite correct. .reshape() creates a new array that shared data with the
> original:
Mmh. My understand
Gael Varoquaux wrote:
> Unless I miss something obvious "a.reshape()" doesn't modify a, which is
> somewhat missleading, IMHO.
quite correct. .reshape() creates a new array that shared data with the
original:
>>> import numpy
>>> a = numpy.zeros((2,3))
>>> help(a.reshape)
Help on built-in fu
Gael Varoquaux wrote:
> On Mon, Apr 23, 2007 at 12:20:51PM -0500, Robert Kern wrote:
>> .reshape()
>
> Unless I miss something obvious "a.reshape()" doesn't modify a, which is
> somewhat missleading, IMHO.
Nope, you caught me in a moment of stupidity.
--
Robert Kern
"I have come to believe tha
On Mon, Apr 23, 2007 at 10:20:43AM -0700, Timothy Hochberg wrote:
>Just set the shape of the array:
>somearray.shape = newshape
Of course ! Thanks
On Mon, Apr 23, 2007 at 12:20:51PM -0500, Robert Kern wrote:
> .reshape()
Unless I miss something obvious "a.reshape()" doesn't modify a
Gael Varoquaux wrote:
> Hi,
>
> I thought I remembered there was a way to reshape in-place an array, but
> neither google, nor greping my mailbox brings anything out.
> Am I wrong, or is there indeed a way to reshape in-place an array ?
.reshape()
--
Robert Kern
"I have come to believe that th
On 4/23/07, Gael Varoquaux <[EMAIL PROTECTED]> wrote:
Hi,
I thought I remembered there was a way to reshape in-place an array, but
neither google, nor greping my mailbox brings anything out.
Am I wrong, or is there indeed a way to reshape in-place an array ?
Just set the shape of the array:
On 23/04/07, Gael Varoquaux <[EMAIL PROTECTED]> wrote:
> Hi,
>
> I thought I remembered there was a way to reshape in-place an array, but
> neither google, nor greping my mailbox brings anything out.
> Am I wrong, or is there indeed a way to reshape in-place an array ?
Sometimes it's just impossib
Hi,
I thought I remembered there was a way to reshape in-place an array, but
neither google, nor greping my mailbox brings anything out.
Am I wrong, or is there indeed a way to reshape in-place an array ?
Cheers,
Gaël
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On 23/04/07, Pierre GM <[EMAIL PROTECTED]> wrote:
> Note that in addition of the bitwise operators, you can use the "logical_"
> functions. OK, you'll still end up w/ temporaries, but I wonder whether there
> couldn't be some tricks to bypass that...
If you're really determined not to make many t
Robert Kern wrote:
> Certainly. How about this?
>
> mask = (a<0)
> a[mask] = numpy.random.normal(0, 1, size=mask.sum())
>
That's slick. I believe it's precisely what I'm after.
Appreciate it,
-Mark
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> When you say "no python temps" I guess you mean, no temporary
> *variables*? If I understand correctly, this allocates a temporary
> boolean array to hold the result of "a<0".
Indeed, hence my precising "no *python* temps". There still be a tmp created
at one point or another (if I'm not mista
> > Have you tried nonzero() ?
>
> Nonzero isn't quite what I'm after, as the tests are more complicated
> than what I illustrated in my example.
Tests such as (a<0)&(b>1) will give you arrays of booleans. The nonzero give
you where the two conditions are met (viz, where the results is True, or 1
On 23/04/07, Pierre GM <[EMAIL PROTECTED]> wrote:
> Have you tried nonzero() ?
>
> a[a<0] = numpy.random.normal(0,1)
>
> will put a random number from the normal distribution where your initial a is
> negative. No Python loops needed, no Python temps.
When you say "no python temps" I guess you me
Mark.Miller wrote:
> Pierre GM wrote:
>> a[a<0] = numpy.random.normal(0,1)
>
> This is a neat construct that I didn't realize was possible. However,
> it has the undesirable (in my case) effect of placing a single new
> random number in each locations where a<0. While this could work, I
> id
Excellent suggestions...just a few comments:
Pierre GM wrote:
> On Monday 23 April 2007 10:37:57 Mark.Miller wrote:
>> Greetings:
>>
>> In some of my code, I need to use large matrix of random numbers that
>> meet specific criteria (i.e., some random numbers need to be removed and
>> replaces with
Oh, I pressed "send" too early.
Just an addition:
numpy.where creates a new array from some condition. If you only want to
change elements of an existing array that satisfies a given condition,
indexing is far more efficient: no temporary is created. Hence the suggestion
of
a[a<0]
_
> I have two arrays:
>
> a = numpy.array([0,1,2,3,4,5,6,7,8,9])
> b = numpy.array([0,0,1,1,2,2,0,1,2,3])
>
> I would like to get the part of a that corresponds
> to where b is equal to i.
>
> For example:
>
> i = 0 => ([0,1,6])
> i = 1 => ([2,3,7])
a[b == 1]
and
a[b == 0]
work too, btw.
-ste
On Mon, Apr 23, 2007 at 11:44:08AM -0400, Steve Lianoglou wrote:
> > I have two arrays:
> > a = numpy.array([0,1,2,3,4,5,6,7,8,9])
> > b = numpy.array([0,0,1,1,2,2,0,1,2,3])
> > I would like to get the part of a that corresponds
> > to where b is equal to i.
> > For example:
> > i = 0 => ([0,1,
On Monday 23 April 2007 10:37:57 Mark.Miller wrote:
> Greetings:
>
> In some of my code, I need to use large matrix of random numbers that
> meet specific criteria (i.e., some random numbers need to be removed and
> replaces with new ones).
>
> I have been working with .any() and .where() to facili
Hi,
On Apr 23, 2007, at 11:30 AM, Tommy Grav wrote:
> I have two arrays:
>
> a = numpy.array([0,1,2,3,4,5,6,7,8,9])
> b = numpy.array([0,0,1,1,2,2,0,1,2,3])
>
> I would like to get the part of a that corresponds
> to where b is equal to i.
>
> For example:
>
> i = 0 => ([0,1,6])
> i = 1 => ([2,3,
I have two arrays:
a = numpy.array([0,1,2,3,4,5,6,7,8,9])
b = numpy.array([0,0,1,1,2,2,0,1,2,3])
I would like to get the part of a that corresponds
to where b is equal to i.
For example:
i = 0 => ([0,1,6])
i = 1 => ([2,3,7])
Cheers
Tommy
___
Num
Greetings:
In some of my code, I need to use large matrix of random numbers that
meet specific criteria (i.e., some random numbers need to be removed and
replaces with new ones).
I have been working with .any() and .where() to facilitate this process.
In the code below, .any() is used in a w
On Mon, April 23, 2007 01:28, Charles R Harris wrote:
> Looks like a bug to me:
>
> In [5]: x = array([1],dtype=uint64)
>
> In [6]: type(x[0])
> Out[6]:
>
> In [7]: type(x[0]+1)
> Out[7]:
>
> Chuck
Yeah. Especially as it works apparently fine for uint32 and int64.
Christian.
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