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QUESTION
I have source (src
) image(s) I wish to align to a destination (dst
) image using an Affine Transformation whilst retaining the full extent of both images during alignment (even the non-overlapping areas).
I am already able to calculate the Affine Transformation rotation and offset matrix, which I feed to scipy.ndimage.interpolate.affine_transform
to recover the dst
-aligned src
image.
The problem is that, when the images are not fuly overlapping, the resultant image is cropped to only the common footprint of the two images. What I need is the full extent of both images, placed on the same pixel coordinate system. This question is almost a duplicate of this one - and the excellent answer and repository there provides this functionality for OpenCV transformations. I unfortunately need this for scipy
's implementation.
Much too late, after repeatedly hitting a brick wall trying to translate the above question's answer to scipy
, I came across this issue and subsequently followed to this question. The latter question did give some insight into the wonderful world of scipy
's affine transformation, but I have as yet been unable to crack my particular needs.
The transformations from src
to dst
can have translations and rotation. I can get translations only working (an example is shown below) and I can get rotations only working (largely hacking around the below and taking inspiration from the use of the reshape
argument in scipy.ndimage.interpolation.rotate
). However, I am getting thoroughly lost combining the two. I have tried to calculate what should be the correct offset
(see this question's answers again), but I can't get it working in all scenarios.
Translation-only working example of padded affine transformation, which follows largely this repo, explained in this answer:
...ANSWER
Answered 2022-Mar-22 at 16:44If you have two images that are similar (or the same) and you want to align them, you can do it using both functions rotate and shift :
QUESTION
I have a function like e. g.:
...ANSWER
Answered 2022-Jan-17 at 21:14I might be misunderstanding your question, but this should do the trick:
QUESTION
I have tried pip install scipy
and everything appears fine, going through the path I opened the files and couldn't find any mention of the bootstrap library despite it being on their website: https://docs.scipy.org/doc/scipy/reference/generated/scipy.stats.bootstrap.html
After looking on Github https://github.com/scipy/scipy/blob/master/scipy/stats/_bootstrap.py I can see there was an update 5 days ago although I last ran the code three days ago with no issues
...ANSWER
Answered 2021-Dec-31 at 15:55I had this issue and solved it by re-installing the scipy package with pip install -U scipy
in order to upgrade to version 1.7
QUESTION
I have install Python 3.10 on Windows 10.
Then I installed numpy and matplotlib without problem.
But when I try to install scipy, I get a ton of errors.
The install sequence is below.
Is this related to needing MKL/BLAS libraries? If so, what should I install?
...ANSWER
Answered 2021-Oct-31 at 13:24In scipy's PyPI page, it looks like scipy doesn't support 3.10 as the meta says
QUESTION
Trying to solve a nonlinear program with inequality constraints using sequential quadratic programming. I've solved it with Python but I get inconsistent results in R.
The objective function takes a vector y
and a matrix X
and looks for weights W
that minimize the L2 norm. There are two constraints:
- each weight in the vector
W
is between 0 and 1 W
sums to 1
In Python I use scipy.optimize.fmin_slsqp
, which "implements the SLSQP Optimization subroutine originally implemented by Dieter Kraft":
ANSWER
Answered 2021-Oct-24 at 21:16This may be a rounding/precision issue. In the R implementation if you add a rounding option to the loss function:
QUESTION
Here's a basic example equation that I am trying to fit to an example data.
The goal is to find the best fit of k
for my data assuming the data follows the above equation. An obvious way to do so is to numerically integrate the equation and then use curve fitting methods to minimise the least squares and get k
.
Numerically integrating using odeint
and ivp_solve
and using them on curve_fit
produced rather drastic differences. The older odeint
produced a better fit compared to newer solve_ivp
. Best fit values of k
are very different too.
ANSWER
Answered 2021-Oct-10 at 09:00Check again what the input arguments of solve_ivp
are. The integration interval is given by the first two numbers in the t_span
argument, so in your application most values in sol.sol(t)
are obtained via wild extrapolation.
Correct that by giving the interval as [min(t),max(t)]
.
To get more compatible computations, explicitly set the error tolerances, as the default values need not be equal. For instance atol=1e-22, rtol=1e-9
so that only the relative tolerance has an effect.
It is curious how you use the args
mechanism. It was only recently introduced to solve_ivp
to be more compatible with odeint
. I would not use it in either case here, as the definition of the parameter and its use is contained in a 3-line block. It has its uses where proper encapsulation and isolation from other code is a real concern.
QUESTION
I am trying to undertake a 3D transformation using SciPy using a rotation vector that is in degrees.
I am using the following to test the setup:
...ANSWER
Answered 2021-Sep-30 at 04:36I suspect the problem is that you are running a different version of SciPy than the one the documents are describing.
QUESTION
In the r
programming language, the following
ANSWER
Answered 2021-Sep-12 at 17:31Turning a comment into an answer, BSpline.design_matrix
is constructing what you are after, in the csr sparse format. It'll be available from scipy 1.8 when it is released. Until then, you can either grab the master branch of scipy, or use a workaround suggested by the docs (https://scipy.github.io/devdocs/reference/generated/scipy.interpolate.BSpline.design_matrix.html#scipy.interpolate.BSpline.design_matrix) :
QUESTION
I have a function fun()
that accepts a NumPy ArrayLike and a "matrix", and returns a numpy array.
ANSWER
Answered 2021-Sep-02 at 13:56You can use typing.Protocol
to assert that the type implements __matmul__
.
QUESTION
I don't understand the return format of the ridge vertices for the function scipy.spatial.Voronoi. When using this function in 2D, the vertices are in pairs for one ridge, which is the format I expect, but in 3D, the number of vertices in ridges tends to have more than 2 points.
Why would a ridge need more than 2 points?
With some post-processing, can I simplify the format into 2 points per ridge?
Examples(The int in vor.ridge_vertices
refer to a point index in vor.vertices
)
ANSWER
Answered 2021-Aug-29 at 20:08In 2D, regions are separated by a single line segment, thus always 2 points per ridge. In 3D and up, regions separation "plane segments" are typically triangular, but they can have 4+ edges, too.
For sceletonization purposes, one approach would be to show the outline of the separation region, skipping virtual (-1) points. So, [3, 0, -1]
would translate to one line between points 3 and 0. [1, 0, 3, 2]
will generate segments 1-0, 0-3, 3-2, 2-1. As an additional improvement, ridges with 4+ points can be further split into trianges, so in case of [1, 0, 3, 2]
another segment would be 0-2 or 1-3.
I am still not sure if I got the question right, let me know if I didn't
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