I have an image stored as a numpy array:
image = np.array([
[[ 11, 12],
[ 13, 14],
[ 15, 16]],
[[ 21, 22],
[ 23, 24],
[ 25, 26]],
[[ 31, 32],
[ 33, 34],
[ 35, 36]]])
I print the shape and dimensions:
>>> print('image shape', image.shape, 'image dimensions', image.ndim)
image shape (3, 3, 2) image dimensions 3
I understand that the matrix is 3 of Height, 3 of Width and 2 of color channel(?)
But.... if the dimension of the matrix is 3 .. What is the difference between dimension and color channel?
The inherent properties of an array are shape and ndim. Shape is the size of an array in each dimension. image.ndim is just a shortcut for len(image.shape). Another related quantity is image.size, which is np.prod(image.shape). In numpy, dimensions are called "axes", and start with zero. So the first index in an array corresponds to axis 0, the second to axis 1, etc.
Which dimension is "channel" is entirely up to interpretation. Generally, it will be the last dimension, but I have seen it as the first as well. Even more generally, dimensions generally correspond to some useful coordinates in your array. Remember that not all arrays are images. Data can be totally arbitrary, and will be based on your use-case.
Your example is slightly confusing because image.ndim == image.shape[0] == image.shape[1]. Here is an example of an array containing multiple RGB images, all of the same size:
images = np.random.randint(255, size=(5, 7, 7, 3), dtype=np.uint8)
We can interpret axis 0 as the index of the image, axis 1 as the width, axis 2 as the height, and axis 3 as the color channel. In that case, we have five images of size 7x7, with three color channels each. If the channels are R, G, B, then images[3, 0, 0, 2] is the blue component of the upper-left pixel in the fourth image.
Related
I have a series of histograms that I plot over the top of each other using a for loop:
import matplotlib as plt
import proplot as pplt
cmap = colormap
fig = pplt.figure(figsize=(12, 10), dpi=300)
jj = [ 4, 3, 2, 1, 0]
for j in jj:
plt.fill_between(p[:,j], s[:, j], y2=0, alpha = 0.6, color = colormap[:,4-j], edgecolor=[0,0,0], linewidth=1.5)
The colormap in question is a manually specified list of RGB triplets (from Fabio Crameri's 'lajolla' map):
0.64566 0.823453 0.895061 0.924676 0.957142
0.277907 0.386042 0.526882 0.657688 0.803006
0.259453 0.301045 0.317257 0.331596 0.408285
Each color corresponds to data recorded under different conditions. I want the colorbar to have manually specified ticks corresponding to this variable (e.g. c = 30, 35, 40, 45, 50), but I can't seem to configure the colormap to not just pull the indices of the cmap matrix (0, 1, 2, 3, 4) as the values of the mapped variable. Trying to set the ticks outside of this range just result in them not being shown.
cbar = fig.colorbar(np.transpose(cmap))
cbar.set_ticks([30, 35, 40, 45, 50])
cbar.set_ticklabels([30, 35, 40, 45, 50])
Any idea how I can resolve this?
Tried shifting indices of colormap but this doesn't seem to work.
Trying to get the colorbar with ticks corresponding to the '30, 35, 40, 45, 50' values quoted above.
Let's say I have a mask (100, 100) of 1's and 0's:
mask = np.random.randint(0, 2, size=(100, 100))
How to convert this mask to get white-and-blue image in RGB:
1's from mask goes blue
0's from mask goes white
Easy as piece of cake. You need auxiliary array to do it:
img = np.zeros((100, 100, 3))
img[mask==0,:] = [1, 1, 1]
img[mask==1,:] = [0, 0, 1]
I have an n-dimensional numpy array of shape: (3, 3, 3, 64). I would like to increase the count along axis 2 by inserting zeros, so that the new shape is (3, 3, 4, 64).
How do I insert zeros to increase a given axis value of a numpy array?
Create a zeros arrays with same shape as the input, but the third axis being of length same as the padding length, which is 1 for our case and concatenate with the input array along the same axis (third axis). For concatenation, we can employ np.concatenate or np.dstack(because its third axis).
Thus, the implementation would be -
z = np.zeros((3, 3, 1, 64),dtype=a.dtype)
out = np.concatenate((a,z),axis=2) # Or np.dstack((a,z))
Sample run -
In [182]: a = np.random.randint(11,99,(3, 3, 3, 64)) # Array with all nonzeros
In [183]: z = np.zeros((3, 3, 1, 64),dtype=a.dtype)
In [184]: out = np.concatenate((a,z),axis=2)
In [185]: (out[:,:,-1,:]==0).all()
Out[185]: True
In [186]: out.shape
Out[186]: (3, 3, 4, 64)
# Another way to verify
In [187]: (out==0).sum()
Out[187]: 576
In [188]: 3*3*64
Out[188]: 576
I have two dense matrices, A [200000,10], B [10,100000]. I need to multiply them to get matrix C. I can't do that directly, since the resulting matrix won't fit into the memory. Moreover, I need only a few elements from the resulting matrix, like 1-2% of the total number of elements. I have a third matrix W [200000,100000] which is sparse and has non-zero elements on exactly those places which are interesting to me in the matrix C.
Is there a way to use W as a "mask" so that the resulting matrix C will be sparse and will contain only the needed elements?
Since a matrix multiplication is just a table of dot products, we can just perform the specific dot products we need, in a vectorized fashion.
import numpy as np
import scipy as sp
iX, iY = sp.nonzero(W)
values = np.sum(A[iX]*B[:, iY].T, axis=-1) #batched dot product
C = sp.sparse.coo_matrix(values, np.asarray([iX,iY]).T)
First, get the indexes of the non zero places in W, and then you can just get the (i,j) element of the result matrix by multiplying the i-th row in A with the j-th column in B, and save the result as a tuple (i,j,res) instead of saving it as a matrix (this is the right way to save sparse matrices).
Here's one approach using np.einsum for a vectorized solution -
from scipy import sparse
from scipy.sparse import coo_matrix
# Get row, col for the output array
r,c,_= sparse.find(W)
# Get the sum-reduction using valid rows and corresponding cols from A, B
out = np.einsum('ij,ji->i',A[r],B[:,c])
# Store as sparse matrix
out_sparse = coo_matrix((out, (r, c)), shape=W.shape)
Sample run -
1) Inputs :
In [168]: A
Out[168]:
array([[4, 6, 1, 1, 1],
[0, 8, 1, 3, 7],
[2, 8, 3, 2, 2],
[3, 4, 1, 6, 3]])
In [169]: B
Out[169]:
array([[5, 2, 4],
[2, 1, 3],
[7, 7, 2],
[5, 7, 5],
[8, 5, 0]])
In [176]: W
Out[176]:
<4x3 sparse matrix of type '<type 'numpy.bool_'>'
with 5 stored elements in Compressed Sparse Row format>
In [177]: W.toarray()
Out[177]:
array([[ True, False, False],
[False, False, False],
[ True, True, False],
[ True, False, True]], dtype=bool)
2) Using dense array to perform direct calculations and verify results later on :
In [171]: (A.dot(B))*W.toarray()
Out[171]:
array([[52, 0, 0],
[ 0, 0, 0],
[73, 57, 0],
[84, 0, 56]])
3) Use the proposed codes and get sparse matrix output :
In [172]: # Using proposed codes
...: r,c,_= sparse.find(W)
...: out = np.einsum('ij,ji->i',A[r],B[:,c])
...: out_sparse = coo_matrix((out, (r, c)), shape=W.shape)
...:
4) Finally verify results by converting to dense/array version and checking against direct version -
In [173]: out_sparse.toarray()
Out[173]:
array([[52, 0, 0],
[ 0, 0, 0],
[73, 57, 0],
[84, 0, 56]])
I am having an issue with Ipython - Numpy. I want to do the following operation:
x^T.x
with and x^T the transpose operation on vector x. x is extracted from a txt file with the instruction:
x = np.loadtxt('myfile.txt')
The problem is that if i use the transpose function
np.transpose(x)
and uses the shape function to know the size of x, I get the same dimensions for x and x^T. Numpy gives the size with a L uppercase indice after each dimensions. e.g.
print x.shape
print np.transpose(x).shape
(3L, 5L)
(3L, 5L)
Does anybody know how to solve this, and compute x^T.x as a matrix product?
Thank you!
What np.transpose does is reverse the shape tuple, i.e. you feed it an array of shape (m, n), it returns an array of shape (n, m), you feed it an array of shape (n,)... and it returns you the same array with shape(n,).
What you are implicitly expecting is for numpy to take your 1D vector as a 2D array of shape (1, n), that will get transposed into a (n, 1) vector. Numpy will not do that on its own, but you can tell it that's what you want, e.g.:
>>> a = np.arange(4)
>>> a
array([0, 1, 2, 3])
>>> a.T
array([0, 1, 2, 3])
>>> a[np.newaxis, :].T
array([[0],
[1],
[2],
[3]])
As explained by others, transposition won't "work" like you want it to for 1D arrays.
You might want to use np.atleast_2d to have a consistent scalar product definition:
def vprod(x):
y = np.atleast_2d(x)
return np.dot(y.T, y)
I had the same problem, I used numpy matrix to solve it:
# assuming x is a list or a numpy 1d-array
>>> x = [1,2,3,4,5]
# convert it to a numpy matrix
>>> x = np.matrix(x)
>>> x
matrix([[1, 2, 3, 4, 5]])
# take the transpose of x
>>> x.T
matrix([[1],
[2],
[3],
[4],
[5]])
# use * for the matrix product
>>> x*x.T
matrix([[55]])
>>> (x*x.T)[0,0]
55
>>> x.T*x
matrix([[ 1, 2, 3, 4, 5],
[ 2, 4, 6, 8, 10],
[ 3, 6, 9, 12, 15],
[ 4, 8, 12, 16, 20],
[ 5, 10, 15, 20, 25]])
While using numpy matrices may not be the best way to represent your data from a coding perspective, it's pretty good if you are going to do a lot of matrix operations!
For starters L just means that the type is a long int. This shouldn't be an issue. You'll have to give additional information about your problem though since I cannot reproduce it with a simple test case:
In [1]: import numpy as np
In [2]: a = np.arange(12).reshape((4,3))
In [3]: a
Out[3]:
array([[ 0, 1, 2],
[ 3, 4, 5],
[ 6, 7, 8],
[ 9, 10, 11]])
In [4]: a.T #same as np.transpose(a)
Out[4]:
array([[ 0, 3, 6, 9],
[ 1, 4, 7, 10],
[ 2, 5, 8, 11]])
In [5]: a.shape
Out[5]: (4, 3)
In [6]: np.transpose(a).shape
Out[6]: (3, 4)
There is likely something subtle going on with your particular case which is causing problems. Can you post the contents of the file that you're reading into x?
This is either the inner or outer product of the two vectors, depending on the orientation you assign to them. Here is how to calculate either without changing x.
import numpy
x = numpy.array([1, 2, 3])
inner = x.dot(x)
outer = numpy.outer(x, x)
The file 'myfile.txt' contain lines such as
5.100000 3.500000 1.400000 0.200000 1
4.900000 3.000000 1.400000 0.200000 1
Here is the code I run:
import numpy as np
data = np.loadtxt('iris.txt')
x = data[1,:]
print x.shape
print np.transpose(x).shape
print x*np.transpose(x)
print np.transpose(x)*x
And I get as a result
(5L,)
(5L,)
[ 24.01 9. 1.96 0.04 1. ]
[ 24.01 9. 1.96 0.04 1. ]
I would be expecting one of the two last result to be a scalar instead of a vector, because x^T.x (or x.x^T) should give a scalar.
b = np.array([1, 2, 2])
print(b)
print(np.transpose([b]))
print("rows, cols: ", b.shape)
print("rows, cols: ", np.transpose([b]).shape)
Results in
[1 2 2]
[[1]
[2]
[2]]
rows, cols: (3,)
rows, cols: (3, 1)
Here (3,) can be thought as "(3, 0)".
However if you want the transpose of a matrix A, np.transpose(A) is the solution. Shortly, [] converts a vector to a matrix, a matrix to a higher dimension tensor.