I'm trying to the fill the offset diagonals of a matrix:
loss_matrix = np.zeros((125,125))
np.diagonal(loss_matrix, 3).fill(4)
ValueError: assignment destination is read-only
Two questions:
1) Without iterating over indexes, how can I set the offset diagonals of a numpy array?
2) Why is the result of np.diagonal read only? The documentation for numpy.diagonal reads: "In NumPy 1.10, it will return a read/write view and writing to the returned array will alter your original array."
np.__version__
'1.10.1'
Judging by the discussion on the NumPy issue tracker, it looks like the feature is stuck in limbo and they never got around to fixing the documentation to say it was delayed.
If you need writability, you can force it. This will only work on NumPy 1.9 and up, since np.diagonal makes a copy on lower versions:
diag = np.diagonal(loss_matrix, 3)
# It's not writable. MAKE it writable.
diag.setflags(write=True)
diag.fill(4)
In an older version, diagflat constructs an array from a diagonal.
In [180]: M=np.diagflat(np.ones(125-3)*4,3)
In [181]: M.shape
Out[181]: (125, 125)
In [182]: M.diagonal(3)
Out[182]:
array([ 4., 4., 4., 4., 4., 4., 4., 4., 4., 4., 4., 4., 4.,... 4.])
In [183]: np.__version__
Out[183]: '1.8.2'
Effectively it does this (working from its Python code)
res = np.zeros((125, 125))
i = np.arange(122)
fi = i+3+i*125
res.flat[fi] = 4
That is, it finds the flatten array equivalent indices of the diagonal.
I can also get fi with:
In [205]: i=np.arange(0,122)
In [206]: np.ravel_multi_index((i,i+3),(125,125))
Related
I have a sparse matrix A in python. I m going through code of a friend of mine and he uses A[:,i:i+1].toarray().flatten() in his code. As far as I m concerned, the program worked for him. However when I try to use it, I get:
from scipy import sparse
...
diagonals = [[2] * 3, [-1] * (3-1), [-1] * (3-1)]
offsets = [0, 1, -1]
B = sparse.diags(diagonals, offsets)
A = sparse.kronsum(B,B)
...
A[:,i:i+1].toarray().flatten()
Exception:
in __getitem__
raise NotImplementedError
NotImplementedError
My question, what to I need to implement or how could I access elements of a sparse matrix. Thanks for the help.
Most likely you have a bsr format matrix, while the code you have, was implemented using an older version of scipy and returns a csr or csc matrix. I don't know a good way of tracing this.
So if we run you code on scipy 1.7.2 :
type(A)
scipy.sparse.bsr.bsr_matrix
We can access the elements by:
A = sparse.kronsum(B,B,format = "csr")
A[:,i:i+1].toarray().flatten()
array([-1., 4., -1., 0., -1., 0., 0., 0., 0.])
Or
A = sparse.kronsum(B,B)
A.tocsr()[:,i:i+1].toarray().flatten()
I have a problem with a numpy array.
In particular, suppose to have a matrix
x = np.array([[1., 2., 3.], [4., 5., 6.]])
with shape (2,3), I want to convert the float numbers into list so to obtain the array [[[1.], [2.], [3.]], [[4.], [5.], [6.]]] with shape (2,3,1).
I tried to convert each float number to a list (i.e., x[0][0] = [x[0][0]]) but it does not work.
Can anyone help me? Thanks
What you want is adding another dimension to your numpy array. One way of doing it is using reshape:
x = x.reshape(2,3,1)
output:
[[[1.]
[2.]
[3.]]
[[4.]
[5.]
[6.]]]
There is a function in Numpy to perform exactly what #Valdi_Bo mentions. You can use np.expand_dims and add a new dimension along axis 2, as follows:
x = np.expand_dims(x, axis=2)
Refer:
np.expand_dims
Actually, you want to add a dimension (not level).
To do it, run:
result = x[...,np.newaxis]
Its shape is just (2, 3, 1).
Or save the result back under x.
You are trying to add a new dimension to the numpy array. There are multiple ways of doing this as other answers mentioned np.expand_dims, np.new_axis, np.reshape etc. But I usually use the following as I find it the most readable, especially when you are working with vectorizing multiple tensors and complex operations involving broadcasting (check this Bounty question that I solved with this method).
x[:,:,None].shape
(2,3,1)
x[None,:,None,:,None].shape
(1,2,1,3,1)
Well, maybe this is an overkill for the array you have, but definitely the most efficient solution is to use np.lib.stride_tricks.as_strided. This way no data is copied.
import numpy as np
x = np.array([[1., 2., 3.], [4., 5., 6.]])
newshape = x.shape[:-1] + (x.shape[-1], 1)
newstrides = x.strides + x.strides[-1:]
a = np.lib.stride_tricks.as_strided(x, shape=newshape, strides=newstrides)
results in:
array([[[1.],
[2.],
[3.]],
[[4.],
[5.],
[6.]]])
>>> a.shape
(2, 3, 1)
I have a numpy.array with a dimension dim_array. I'm looking forward to obtain a median filter like scipy.signal.medfilt(data, window_len).
This in fact doesn't work with numpy.array may be because the dimension is (dim_array, 1) and not (dim_array, ).
How to obtain such filter?
Next, another question, how can I obtain other filter, i.e., min, max, mean?
Based on this post, we could create sliding windows to get a 2D array of such windows being set as rows in it. These windows would merely be views into the data array, so no memory consumption and thus would be pretty efficient. Then, we would simply use those ufuncs along each row axis=1.
Thus, for example sliding-median` could be computed like so -
np.median(strided_app(data, window_len,1),axis=1)
For the other ufuncs, just use the respective ufunc names there : np.min, np.max & np.mean. Please note this is meant to give a generic solution to use ufunc supported functionality.
For the best performance, one must still look into specific functions that are built for those purposes. For the four requested functions, we have the builtins, like so -
Median : scipy.signal.medfilt.
Max : scipy.ndimage.filters.maximum_filter1d.
Min : scipy.ndimage.filters.minimum_filter1d.
Mean : scipy.ndimage.filters.uniform_filter1d
The fact that applying of a median filter with the window size 1 will not change the array gives us a freedom to apply the median filter row-wise or column-wise.
For example, this code
from scipy.ndimage import median_filter
import numpy as np
arr = np.array([[1., 2., 3.], [4., 5., 6.], [7., 8., 9.]])
median_filter(arr, size=3, cval=0, mode='constant')
#with cval=0, mode='constant' we set that input array is extended with zeros
#when window overlaps edges, just for visibility and ease of calculation
outputs an expected filtered with window (3, 3) array
array([[0., 2., 0.],
[2., 5., 3.],
[0., 5., 0.]])
because median_filter automatically extends the size to all dimensions, so the same effect we can get with:
median_filter(arr, size=(3, 3), cval=0, mode='constant')
Now, we can also apply median_filter row-wise with setting 1 to the first element of size
median_filter(arr, size=(1, 3), cval=0, mode='constant')
Output:
array([[1., 2., 2.],
[4., 5., 5.],
[7., 8., 8.]])
And column-wise with the same logic
median_filter(arr, size=(3, 1), cval=0, mode='constant')
Output:
array([[1., 2., 3.],
[4., 5., 6.],
[4., 5., 6.]])
I created a text file called 'column.txt' containing the following data:
1
2
3
4
9
8
Then I wrote the code below to transpose my data to a single-row text file.
import numpy as np
x=np.loadtxt('column.txt')
z=x.T
y=x.transpose()
np.savetxt('row.txt',y, fmt='%i')
I tried two different ways - using matrix multiplication (the commented line in my code) and using transpose command. The problem was the output was exactly the same as the input!
Afterwards, I added another column to the input file, ran the code and surprisingly this time the output was completely fine (The output contained two rows!)
So my question is:
Is there anyway to transpose a single column file to a single row one? If yes, could you please describe how?
You can use numpy.reshape to transpose data and change the shape of your array like the following:
>>> import numpy as np
>>> arr=np.loadtxt('column.txt')
>>> arr
array([ 1., 2., 3., 4., 9., 8.])
>>> arr.shape
(6,)
>>> arr=arr.reshape(6,1)
>>> arr
array([[ 1.],
[ 2.],
[ 3.],
[ 4.],
[ 9.],
[ 8.]])
or you can just give the number of an array dimension as an input to the numpy.loadtxt function
>>> np.loadtxt('column.txt', ndmin=2)
array([[ 1.],
[ 2.],
[ 3.],
[ 4.],
[ 9.],
[ 8.]])
But if you want to convert a single column to a single row and write it into a file just you need to do as following
>>> parr=arr.reshape(1,len(arr))
np.savetxt('column.txt',parr, fmt='%i')
If your input data only consists of a single column, np.loadtxt() will return an one-dimensional array. Transposing basically means to reverse the order of the axes. For a one-dimensional array with only a single axis, this is a no-op. You can convert the array into a two-dimensional array in many different ways, and transposing will work as expected for the two-dimensional array, e.g.
x = np.atleast_2d(np.loadtxt('column.txt'))
It is because the transpose of a 1D array is the same as itself, as there is no other dimension to transpose to.
You could try adding a 2nd dimension by doing this,
>>> import numpy as np
>>> x = np.array([[1], [2], [3], [4], [9], [8]])
>>> x.T
array([[1, 2, 3, 4, 9, 8]])
In numpy if you want to calculate the sinus of each entry of a matrix (elementise) then
a = numpy.arange(0,27,3).reshape(3,3)
numpy.sin(a)
will get the job done! If you want the power let's say to 2 of each entry
a**2
will do it.
But if you have a sparse matrix things seem more difficult. At least I haven't figured a way to do that besides iterating over each entry of a lil_matrix format and operate on it.
I've found this question on SO and tried to adapt this answer but I was not succesful.
The Goal is to calculate elementwise the squareroot (or the power to 1/2) of a scipy.sparse matrix of CSR format.
What would you suggest?
The following trick works for any operation which maps zero to zero, and only for those operations, because it only touches the non-zero elements. I.e., it will work for sin and sqrt but not for cos.
Let X be some CSR matrix...
>>> from scipy.sparse import csr_matrix
>>> X = csr_matrix(np.arange(10).reshape(2, 5), dtype=np.float)
>>> X.A
array([[ 0., 1., 2., 3., 4.],
[ 5., 6., 7., 8., 9.]])
The non-zero elements' values are X.data:
>>> X.data
array([ 1., 2., 3., 4., 5., 6., 7., 8., 9.])
which you can update in-place:
>>> X.data[:] = np.sqrt(X.data)
>>> X.A
array([[ 0. , 1. , 1.41421356, 1.73205081, 2. ],
[ 2.23606798, 2.44948974, 2.64575131, 2.82842712, 3. ]])
Update In recent versions of SciPy, you can do things like X.sqrt() where X is a sparse matrix to get a new copy with the square roots of elements in X.