Say I have some time-series data in the form of a simple array.
X1 = np.array[(1, 2, 3, 4]
The Hankel matrix can be obtained by using scipy.linalg.hankel, which would look something like this:
hankel(X1)
array([[1, 2, 3, 4],
[2, 3, 4, 0],
[3, 4, 0, 0],
[4, 0, 0, 0]])
Now assume I had a larger array in the form of
X2 = np.array([1, 2, 3, 4, 5, 6, 7])
What I want to do is fill in the zeros in this matrix with the numbers that are next in the index (specific to each row). Taking the same Hankel matrix earlier by using the first four values in the array X2, I'd like to see the following output:
hankel(X2[:4])
array([[1, 2, 3, 4],
[2, 3, 4, 5],
[3, 4, 5, 6],
[4, 5, 6, 7]])
How would I do this? I'd ideally like to use this for larger data.
Appreciate any tips or pointers given. Thanks!
If you have a matrix with the appropriate index values into your dataset, you can use integer array indexing directly into your dataset.
To create the index matrix, you can simply use the upper-left quadrant of a double-sized Hankel array. There are likely simpler ways to create the index matrix, but this does the trick.
>>> X = np.array([9, 8, 7, 6, 5, 4, 3])
>>> N = 4 # the size of the "window"
>>> indices = scipy.linalg.hankel(np.arange(N*2))[:N, :N]
>>> indices
array([[0, 1, 2, 3],
[1, 2, 3, 4],
[2, 3, 4, 5],
[3, 4, 5, 6]])
>>> X[indices]
array([[9, 8, 7, 6],
[8, 7, 6, 5],
[7, 6, 5, 4],
[6, 5, 4, 3]])
Related
Is it possible to shuffle two 2D tensors in PyTorch by their rows, but maintain the same order for both? I know you can shuffle a 2D tensor by rows with the following code:
a=a[torch.randperm(a.size()[0])]
To elaborate:
If I had 2 tensors
a = torch.tensor([[1, 1, 1, 1, 1],
[2, 2, 2, 2, 2],
[3, 3, 3, 3, 3]])
b = torch.tensor([[4, 4, 4, 4, 4],
[5, 5, 5, 5, 5],
[6, 6, 6, 6, 6]])
And ran them through some function/block of code to shuffle randomly but maintain correlation and produce something like the following
a = torch.tensor([[2, 2, 2, 2, 2],
[1, 1, 1, 1, 1],
[3, 3, 3, 3, 3]])
b = torch.tensor([[5, 5, 5, 5, 5],
[4, 4, 4, 4, 4],
[6, 6, 6, 6, 6]])
My current solution is converting to a list, using the random.shuffle() function like below.
a_list = a.tolist()
b_list = b.tolist()
temp_list = list(zip(a_list , b_list ))
random.shuffle(temp_list) # Shuffle
a_temp, b_temp = zip(*temp_list)
a_list, b_list = list(a_temp), list(b_temp)
# Convert back to tensors
a = torch.tensor(a_list)
b = torch.tensor(b_list)
This takes quite a while and was wondering if there is a better way.
You mean
indices = torch.randperm(a.size()[0])
a=a[indices]
b=b[indices]
?
Doing np.roll(a, 1, axis = 1) on:
a = np.array([
[6, 3, 9, 2, 3],
[1, 7, 8, 1, 2],
[5, 4, 2, 2, 4],
[3, 9, 7, 6, 5],
])
results in the correct:
array([
[3, 6, 3, 9, 2],
[2, 1, 7, 8, 1],
[4, 5, 4, 2, 2],
[5, 3, 9, 7, 6]
])
The documentation says:
If a tuple, then axis must be a tuple of the same size, and each of the given axes is shifted by the corresponding number.
Now I like to roll rows of a by different values, like [1,2,1,3] meaning, first row will be rolled by 1, second by 2, third by 1 and forth by 3. But np.roll(a, [1,2,1,3], axis=(1,1,1,1)) doesn't seem to do it. What would be the correct interpretation of the sentence in the docs?
By specifying a tuple in np.roll you can roll an array along various axes. For example, np.roll(a, (3,2), axis=(0,1)) will shift each element of a by 3 places along axis 0, and it will also shift each element by 2 places along axis 1. np.roll does not have an option to roll each row by a different amount. You can do it though for example as follows:
import numpy as np
a = np.array([
[6, 3, 9, 2, 3],
[1, 7, 8, 1, 2],
[5, 4, 2, 2, 4],
[3, 9, 7, 6, 5],
])
shifts = np.c_[[1,2,1,3]]
a[np.c_[:a.shape[0]], (np.r_[:a.shape[1]] - shifts) % a.shape[1]]
It gives:
array([[3, 6, 3, 9, 2],
[1, 2, 1, 7, 8],
[4, 5, 4, 2, 2],
[7, 6, 5, 3, 9]])
I have a 2D Numpy ndarray, x, that I need to split in square subregions of size s. For each subregion, I want to get the greatest element (which I do), and its position within that subregion (which I can't figure out).
Here is a minimal example:
>>> x = np.random.randint(0, 10, (6,8))
>>> x
array([[9, 4, 8, 9, 5, 7, 3, 3],
[3, 1, 8, 0, 7, 7, 5, 1],
[7, 7, 3, 6, 0, 2, 1, 0],
[7, 3, 9, 8, 1, 6, 7, 7],
[1, 6, 0, 7, 5, 1, 2, 0],
[8, 7, 9, 5, 8, 3, 6, 0]])
>>> h, w = x.shape
>>> s = 2
>>> f = x.reshape(h//s, s, w//s, s)
>>> mx = np.max(f, axis=(1, 3))
>>> mx
array([[9, 9, 7, 5],
[7, 9, 6, 7],
[8, 9, 8, 6]])
For example, the 8 in the lower left corner of mx is the greatest element from subregion [[1,6], [8, 7]] in the lower left corner of x.
What I want is to get an array similar to mx, that keeps the indices of the largest elements, like this:
[[0, 1, 1, 2],
[0, 2, 3, 2],
[2, 2, 2, 2]]
where, for example, the 2 in the lower left corner is the index of 8 in the linear representation of [[1, 6], [8, 7]].
I could do it like this: np.argmax(f[i, :, j, :]) and iterate over i and j, but the speed difference is enormous for large amounts of computation. To give you an idea, I'm trying to use (only) Numpy for max pooling. Basically, I'm asking if there is a faster alternative than what I'm using.
Here's one approach -
# Get shape of output array
m,n = np.array(x.shape)//s
# Reshape and permute axes to bring the block as rows
x1 = x.reshape(h//s, s, w//s, s).swapaxes(1,2).reshape(-1,s**2)
# Use argmax along each row and reshape to output shape
out = x1.argmax(1).reshape(m,n)
Sample input, output -
In [362]: x
Out[362]:
array([[9, 4, 8, 9, 5, 7, 3, 3],
[3, 1, 8, 0, 7, 7, 5, 1],
[7, 7, 3, 6, 0, 2, 1, 0],
[7, 3, 9, 8, 1, 6, 7, 7],
[1, 6, 0, 7, 5, 1, 2, 0],
[8, 7, 9, 5, 8, 3, 6, 0]])
In [363]: out
Out[363]:
array([[0, 1, 1, 2],
[0, 2, 3, 2],
[2, 2, 2, 2]])
Alternatively, to simplify things, we could use scikit-image that does the heavy work of reshaping and permuting axes for us -
In [372]: from skimage.util import view_as_blocks as viewB
In [373]: viewB(x, (s,s)).reshape(-1,s**2).argmax(1).reshape(m,n)
Out[373]:
array([[0, 1, 1, 2],
[0, 2, 3, 2],
[2, 2, 2, 2]])
I need to create a matrix in Python containing a list of unknown arrays r that have this form:
r_i=[r1,r2,r3,r4,th_12,th_13]
I am running for statements with a couple of if conditions that will give me in output a number of r_i arrays that I don't know from the beginning.
I am looking for a function like append that I normally use to create a vector with all the solutions that I generate, but this time each solution is not a single value but an array of 6 values and I am not able to generate what I want.
I need to create a matrix like this one, where every r_1 has the form of the code that I wrote above.
EDIT: I would like to generate a numpy array (R_tot should be a numpy array).
You can generate the array normally as I explained in my comment:
r_tot = []
for r_i in however_many_rs_there_are: # each r_i contains an array of 6 values
r_tot.append(r_i)
You can then convert r_tot into a numpy array like so:
import numpy
np_array = numpy.array(r_tot)
Here's a very simple proof of concept:
>>> import random, numpy
>>> r_tot = []
>>> for i in range(0,random.randint(1,20)): # append an arbitrary number of arrays
r_i = [1,2,3,4,5,6] # all of size six
r_tot.append(r_i) # to r_tot
>>> np_array = numpy.array(r_tot) # then convert to numpy array!
>>> np_array # did it work?
array([[1, 2, 3, 4, 5, 6],
[1, 2, 3, 4, 5, 6],
[1, 2, 3, 4, 5, 6],
[1, 2, 3, 4, 5, 6],
[1, 2, 3, 4, 5, 6],
[1, 2, 3, 4, 5, 6], # yeaaaah
[1, 2, 3, 4, 5, 6],
[1, 2, 3, 4, 5, 6],
[1, 2, 3, 4, 5, 6],
[1, 2, 3, 4, 5, 6],
[1, 2, 3, 4, 5, 6],
[1, 2, 3, 4, 5, 6],
[1, 2, 3, 4, 5, 6]])
I want to extract the second and the 3rd to the fifth columns of the NumPy array, how would I go about it?
A = array([[0, 1, 2, 3, 4, 5, 6], [4, 5, 6, 7, 4, 5, 6]])
A[:, [1, 4:6]]
This obviously doesn't work.
Assuming I've understood you -- it's usually a good idea to explicitly specify the output you want, because it's not obvious -- you could use numpy.r_:
In [27]: A
Out[27]:
array([[0, 1, 2, 3, 4, 5, 6],
[4, 5, 6, 7, 4, 5, 6]])
In [28]: A[:, [1,3,4,5]]
Out[28]:
array([[1, 3, 4, 5],
[5, 7, 4, 5]])
In [29]: A[:, r_[1, 3:6]]
Out[29]:
array([[1, 3, 4, 5],
[5, 7, 4, 5]])
In [37]: A[1:, r_[1, 3:6]]
Out[37]: array([[5, 7, 4, 5]])
which you can then flatten or reshape as you like. r_ is basically a convenience function to generate the right indices, e.g.
In [30]: r_[1, 3:6]
Out[30]: array([1, 3, 4, 5])
Perhaps you are looking for this?
In [10]: A[1:, [1]+range(3,6)]
Out[10]: array([[5, 7, 4, 5]])
Note this gives you the second, fourth, fifth and six columns of all rows but the first.
The second element is A[:,1]. Elements 3-5 (I'm assuming you want inclusive) are A[:,2:5]. You won't be able to extract them with a single call. To get them as an array, you could do
import numpy as np
A = np.array([[0, 1, 2, 3, 4, 5, 6], [4, 5, 6, 7, 4, 5, 6]])
my_cols = np.hstack((A[:,1][...,np.newaxis], A[:,2:5]))
The np.newaxis stuff is just to make A[:,1] a 2D array, consistent with A[:,2:5].
Hope this helps.