I have used interp2 in Matlab, such as the following code, that is part of #rayryeng's answer in: Three dimensional (3D) matrix interpolation in Matlab:
d = size(volume_image)
[X,Y] = meshgrid(1:1/scaleCoeff(2):d(2), 1:1/scaleCoeff(1):d(1));
for ind = z
%Interpolate each slice via interp2
M2D(:,:,ind) = interp2(volume_image(:,:,ind), X, Y);
end
Example of Dimensions:
The image size is 512x512 and the number of slices is 133. So:
volume_image(rows, columns, slices in 3D dimenson) : 512x512x133 in 3D dimenson
X: 288x288
Y: 288x288
scaleCoeff(2): 0.5625
scaleCoeff(1): 0.5625
z = 1 up to 133 ,hence z: 1x133
ind: 1 up to 133
M2D(:,:,ind) finally is 288x288x133 in 3D dimenson
Aslo, Matlabs syntax for size: (rows, columns, slices in 3rd dimenson) and Python syntax for size: (slices in 3rd dim, rows, columns).
However, after convert the Matlab code to Python code occurred an error, ValueError: Invalid length for input z for non rectangular grid:
for ind in range(0, len(z)+1):
M2D[ind, :, :] = interpolate.interp2d(X, Y, volume_image[ind, :, :]) # ValueError: Invalid length for input z for non rectangular grid
What is wrong? Thank you so much.
In MATLAB, interp2 has as arguments:
result = interp2(input_x, input_y, input_z, output_x, output_y)
You are using only the latter 3 arguments, the first two are assumed to be input_x = 1:size(input_z,2) and input_y = 1:size(input_z,1).
In Python, scipy.interpolate.interp2 is quite different: it takes the first 3 input arguments of the MATLAB function, and returns an object that you can call to get interpolated values:
f = scipy.interpolate.interp2(input_x, input_y, input_z)
result = f(output_x, output_y)
Following the example from the documentation, I get to something like this:
from scipy import interpolate
x = np.arange(0, volume_image.shape[2])
y = np.arange(0, volume_image.shape[1])
f = interpolate.interp2d(x, y, volume_image[ind, :, :])
xnew = np.arange(0, volume_image.shape[2], 1/scaleCoeff[0])
ynew = np.arange(0, volume_image.shape[1], 1/scaleCoeff[1])
M2D[ind, :, :] = f(xnew, ynew)
[Code not tested, please let me know if there are errors.]
You might be interested in scipy.ndimage.zoom. If you are interpolating from one regular grid to another, it is much faster and easier to use than scipy.interpolate.interp2d.
See this answer for an example:
https://stackoverflow.com/a/16984081/1295595
You'd probably want something like:
import scipy.ndimage as ndimage
M2D = ndimage.zoom(volume_image, (1, scaleCoeff[0], scaleCoeff[1])
Related
I have big binary 3D data and I want to re-arrange the data such as it is a sequence of values in order achieved by parsing the original data as sub-arrays of size (4x4x4).
For example, if the data is 2D and I want to re-arrange the data from 2x2 sub-arrays
example image
I used simple loops for this but just iterating over the loops took way more times, I am trying to to use some numpy functions to do so but I am new to SciPy
My code looks like this
x,y,z = 1200,800,400
data = np.fromfile(file_name, dtype=np.float32)
data.shape = (z,y,x)
new_data = np.empty(shape=x*y*z, dtype = np.float32)
index = 0
for zz in range(0,z,4):
for yy in range(0,y,4):
for xx in range(0,x,4):
for zShift in range(4):
for yShift in range(4):
for xShift in range(4):
new_data[index] = data[zz+zShift][yy+yShift][xx+xShift]
index+=1
new_data.tofile(output)
However, this takes a lot of time, any better implementation ideas?
As I said, the code works as intended, however, I need a smarter, pythonic way to achieve my output
Thank you!
x,y,z = 1200,800,400
data = np.empty([x,y,z])
# numpy calculates the shape of -1
out = data.reshape(-1, 4, 4, 4)
out.shape
>>> (6000000, 4, 4, 4)
Perform the following test, for smaller data and block size:
x, y, z = 4, 4, 4 # Dimensions
stp = 2 # Block size (in each dimension)
# Create the test array
arr = np.arange(x * y * z).reshape((x, y, z))
And to create a list of "blocks", run:
new_data = []
for xx in range(0, x, stp):
for yy in range(0, y, stp):
for zz in range(0, z, stp):
print('Index:', xx, yy, zz)
obj = arr[xx:xx+stp, yy:yy+stp, zz:zz+stp].copy()
print(obj)
new_data.append(obj)
In the target version of your code:
restore original values of x, y and z,
read the array from your source,
change stp back to 4,
drop test printouts.
Note also that your code adds individual elements to new_data,
only iterating over blocks of size 4 * 4 * 4,
whereas you wrote that you want a sequence of smaller arrays
(i.e. slices) of size 4 * 4 * 4, what my code does.
So if you need a list of slices (smaller arrays), not a single
4-D array, use my code.
I found an article which is about epipolar geometry.
I calculated the fundamental matrix. Now Iam trying to find the line on which a corresponding point lays as described in the article:
I calculated the line which is in homogeneous coordinates. How could I plot this line into the picture like in the example? I thought about transforming the line from homogeneous to inhomogeneous coordinates. I think this can be achieved by dividing x and y by z
For example, homogeneous:
x=0.0295
y=0.9996
z=-265.1531
to inhomogeneous:
x=0.0295/-265.1531
y=0.9996/-265.1531
so:
x=-0.0001112564778612809
y=0.0037698974667842843
Those numbers seem wrong to me, because theyre so small. Is this the correct approach?
How could I plot my result into an image?
the x, y and z you have are the parameters of the "Epipolar Lines" equation that appear under the "line in the image" formula in the slides, but labelled a, b and c respectively, i.e:
au + bv + c = 0
solutions to this are points on the line. e.g. in Python I'd define a as some points on the picture's x-axis, and solve for b:
import numpy as np
F = np.array([
[-0.00310695, -0.0025646, 2.96584],
[-0.028094, -0.00771621, 56.3813],
[13.1905, -29.2007, -9999.79],
])
p_l = np.array([
[343.53],
[221.70],
[ 1.0],
])
lt = F # p_l
# if you want to normalise
lt /= np.sqrt(sum(lt[:2] ** 2))
# should give your values [0.0295, 0.9996, -265.2]
print(lt)
a, b, c = lt.ravel()
x = np.array([0, 400])
y = -(x*a + c) / b
and then just draw a line between these points
I have a 2D Numpy array that represents an image, and I want to create a surface plot of image intensity using matplotlib.surface_plot. For this function, I need to convert the 2D array A[x,y] => z into three arrays: [x0,...,xN], [y0,...,yN] and [z0,...,zN]. I can see how to do this conversion element-by-element:
X = []
Y = []
Z = []
for x in range( A.shape[ 0 ] ):
for y in range( A.shape[ 1 ] ):
X.append( x )
Y.append( y )
Z.append( A[x,y] )
but I'm wondering whether there is a more Pythonic way to do this?
a very simple way to do this could be to basically use the code shown in the matplotlib example. assuming x and y representing the sizes of the two dims in your image array A, you could do
import numpy as np
from mpl_toolkits.mplot3d import Axes3D
import matplotlib.pyplot as plt
# generate some input data that looks nice on a color map:
A = np.mgrid[0:10:0.1,0:10:0.1][0]
X = np.arange(0, A.shape[0], 1)
Y = np.arange(0, A.shape[1], 1)
X, Y = np.meshgrid(X, Y)
fig = plt.figure()
ax = fig.gca(projection='3d')
surf = ax.plot_surface(X, Y, A, cmap='viridis',
linewidth=0, antialiased=False)
gives
You possibly don't need to construct the actual grid, because some pyplot functions accept 1d arrays for x and y, implying that a grid is to be constructed. It seems that Axes3D.plot_surface (which I presume you meant) does need 2d arrays as input, though.
So to get your grid the easiest way is using np.indices to get the indices corresponding to your array:
>>> import numpy as np
...
... # dummy data
... A = np.random.random((3,4)) # randoms of shape (3,4)
...
... # get indices
... x,y = np.indices(A.shape) # both arrays have shape (3,4)
...
... # prove that the indices correspond to the values of A
... print(all(A[i,j] == A[x[i,j], y[i,j]] for i in x.ravel() for j in y.ravel()))
True
The resulting arrays all have the same shape as A, which should be correct for most use cases. If for any reason you really need a flattened 1d array, you should use x.ravel() etc. to get a flattened view of the same 2d array.
I should note though that the standard way to visualize images (due to the short-wavelength variation of the data) is pyplot.imshow or pyplot.pcolormesh which can give you pixel-perfect visualization, albeit in two dimensions.
We agree X, Y and Z have different sizes (N for X and Y and N^2 for Z) ? If yes:
X looks not correct (you add several times the same values)
something like:
X = list(range(A.shape[0])
Y = list(range(A.shape[1])
Z = [A[x,y] for x in X for y in Y]
I want to use the dendogram of scipy.
I have the following data:
I have a list with seven different means. For example:
Y = [71.407452200146807, 0, 33.700136456196823, 1112.3757110973756, 31.594949722819372, 34.823881975554166, 28.36368420190157]
Each mean is calculate for a different user. For example:
X = ["user1", "user2", "user3", "user4", "user5", "user6", "user7"]
My aim is to display the data described above with the help of a dendorgram.
I tried the following:
Y = [71.407452200146807, 0, 33.700136456196823, 1112.3757110973756, 31.594949722819372, 34.823881975554166, 28.36368420190157]
X = ["user1", "user2", "user3", "user4", "user5", "user6", "user7"]
# Attempt with matrix
#X = np.concatenate((X, Y),)
#Z = linkage(X)
Z = linkage(Y)
# Plot the dendogram with the results above
dendrogram(Z, leaf_rotation=45., leaf_font_size=12. , show_contracted=True)
plt.style.use("seaborn-whitegrid")
plt.title("Dendogram to find clusters")
plt.ylabel("Distance")
plt.show()
But it says:
ValueError: Length n of condensed distance matrix 'y' must be a binomial coefficient, i.e.there must be a k such that (k \choose 2)=n)!
I already tried to convert my data into a matrix. With:
# Attempt with matrix
#X = np.concatenate((X, Y),)
#Z = linkage(X)
But that doesn´t work too!
Are there any suggestions?
Thanks :-)
The first argument of linkage is either an n x m array, representing n points in m-dimensional space, or a one-dimensional array containing the condensed distance matrix. These are two very different meanings! The first is the raw data, i.e. the observations. The second format assumes that you have already computed all the distances between your observations, and you are providing these distances to linkage, not the original points.
It looks like you want the first case (raw data), with m = 1. So you must reshape the input to have shape (n, 1).
Replace this:
Z = linkage(Y)
with:
Z = linkage(np.reshape(Y, (len(Y), 1)))
So you are using 7 observations in Y len(Y) = 7.
But as per documentation of Linkage, the number of observations len(Y) should be such that.
{n \choose 2} = len(Y)
which means
1/2 * (n -1) * n = len(Y)
so length of Y should be such that n is a valid integer.
I've read the masked array documentation several times now, searched everywhere and feel thoroughly stupid. I can't figure out for the life in me how to apply a mask from one array to another.
Example:
import numpy as np
y = np.array([2,1,5,2]) # y axis
x = np.array([1,2,3,4]) # x axis
m = np.ma.masked_where(y>2, y) # filter out values larger than 5
print m
[2 1 -- 2]
print np.ma.compressed(m)
[2 1 2]
So this works fine.... but to plot this y axis, I need a matching x axis. How do I apply the mask from the y array to the x array? Something like this would make sense, but produces rubbish:
new_x = x[m.mask].copy()
new_x
array([5])
So, how on earth is that done (note the new x array needs to be a new array).
Edit:
Well, it seems one way to do this works like this:
>>> import numpy as np
>>> x = np.array([1,2,3,4])
>>> y = np.array([2,1,5,2])
>>> m = np.ma.masked_where(y>2, y)
>>> new_x = np.ma.masked_array(x, m.mask)
>>> print np.ma.compressed(new_x)
[1 2 4]
But that's incredibly messy! I'm trying to find a solution as elegant as IDL...
I had a similar issue, but involving loads more masking commands and more arrays to apply them. My solution is that I do all the masking on one array and then use the finally masked array as the condition in the mask_where command.
For example:
y = np.array([2,1,5,2]) # y axis
x = np.array([1,2,3,4]) # x axis
m = np.ma.masked_where(y>5, y) # filter out values larger than 5
new_x = np.ma.masked_where(np.ma.getmask(m), x) # applies the mask of m on x
The nice thing is you can now apply this mask to many more arrays without going through the masking process for each of them.
Why not simply
import numpy as np
y = np.array([2,1,5,2]) # y axis
x = np.array([1,2,3,4]) # x axis
m = np.ma.masked_where(y>2, y) # filter out values larger than 5
print list(m)
print np.ma.compressed(m)
# mask x the same way
m_ = np.ma.masked_where(y>2, x) # filter out values larger than 5
# print here the list
print list(m_)
print np.ma.compressed(m_)
code is for Python 2.x
Also, as proposed by joris, this do the work new_x = x[~m.mask].copy() giving an array
>>> new_x
array([1, 2, 4])
This may not bee 100% what OP wanted to know,
but it's a cute little piece of code I use all the time -
if you want to mask several arrays the same way, you can use this generalized function to mask a dynamic number of numpy arrays at once:
def apply_mask_to_all(mask, *arrays):
assert all([arr.shape == mask.shape for arr in arrays]), "All Arrays need to have the same shape as the mask"
return tuple([arr[mask] for arr in arrays])
See this example usage:
# init 4 equally shaped arrays
x1 = np.random.rand(3,4)
x2 = np.random.rand(3,4)
x3 = np.random.rand(3,4)
x4 = np.random.rand(3,4)
# create a mask
mask = x1 > 0.8
# apply the mask to all arrays at once
x1, x2, x3, x4 = apply_mask_to_all(m, x1, x2, x3, x4)