On 06.02.2012 22:27, Sturla Molden wrote:
# Make a 4D view of this data, such that b[i,j]
# is a 2D block with shape (4,4) (e.g. b[0,0] is
# the same as a[:4, :4]).
b = as_strided(a, shape=(a.shape[0]/4, a.shape[1]/4, 4, 4),
strides=(4*a.strides[0], 4*a.strides[1],
Hi
This is elegant and very fast as well!
On Tue, Feb 7, 2012 at 2:57 PM, Sturla Molden stu...@molden.no wrote:
On 06.02.2012 22:27, Sturla Molden wrote:
# Make a 4D view of this data, such that b[i,j]
# is a 2D block with shape (4,4) (e.g. b[0,0] is
# the same as a[:4, :4]).
b =
On 07.02.2012 15:27, eat wrote:
This is elegant and very fast as well!
Just be aware that it depends on C ordered input. So:
m,n = data.shape
cond = lamda x : (x = t1) (x = t2)
x = cond(np.ascontiguousarray(data)).reshape((m//4, 4, n//4, 4))
found = np.any(np.any(x, axis=1),
for i in range(m):
for j in range(n):
found[i//4,j//4] = cond(x[i,j])
Blah, that should be
found[i//4,j//4] |= cond(x[i,j])
Sturla
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Short answer: Create 16 view arrays, each with a stride of 4 in both
dimensions. Test them against the conditions and combine the tests with an |=
operator. Thus you replace the nested loop with one that has only 16
iterations. Or reshape to 3 dimensions, the last with length 4, and you can do
On Mon, Feb 6, 2012 at 2:57 PM, Sturla Molden stu...@molden.no wrote:
Short answer: Create 16 view arrays, each with a stride of 4 in both
dimensions. Test them against the conditions and combine the tests with an
|= operator. Thus you replace the nested loop with one that has only 16
Something like this:
m,n = data.shape
x = data.reshape((m,n//4,4))
z = (x[0::4,...] = t1) (x[0::4,...] = t1)
z |= (x[1::4,...] = t1) (x[1::4,...] = t1)
z |= (x[2::4,...] = t1) (x[2::4,...] = t1)
z |= (x[3::4,...] = t1) (x[3::4,...] = t1)
found = np.any(z, axis=2)
Sturla
Sendt fra min iPad
The last t1 on each lineis of course t2. Sorry for the typo. Hard to code on an
ipad ;-)
Sturla
Sendt fra min iPad
Den 6. feb. 2012 kl. 22:12 skrev Sturla Molden stu...@molden.no:
Something like this:
m,n = data.shape
x = data.reshape((m,n//4,4))
z = (x[0::4,...] = t1) (x[0::4,...]
# Make a 4D view of this data, such that b[i,j]
# is a 2D block with shape (4,4) (e.g. b[0,0] is
# the same as a[:4, :4]).
b = as_strided(a, shape=(a.shape[0]/4, a.shape[1]/4, 4, 4),
strides=(4*a.strides[0], 4*a.strides[1], a.strides[0],
a.strides[1]))
Yes :-) Being
Hi,
On Mon, Feb 6, 2012 at 9:16 PM, Moroney, Catherine M (388D)
catherine.m.moro...@jpl.nasa.gov wrote:
Hello,
I have to write a code to downsample an array in a specific way, and I am
hoping that
somebody can tell me how to do this without the nested do-loops. Here is
the problem
Hi,
Sorry for my latest post, hands way too quick ;(
On Mon, Feb 6, 2012 at 9:16 PM, Moroney, Catherine M (388D)
catherine.m.moro...@jpl.nasa.gov wrote:
Hello,
I have to write a code to downsample an array in a specific way, and I am
hoping that
somebody can tell me how to do this without
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