Reshaping array of matrices in Python - python

I have a Numpy array X of n 2x2 matrices, arranged so that X.shape = (2,2,n), that is, to get the first matrix I call X[:,:,0]. I would like to reshape X into an array Y such that I can get the first matrix by calling Y[0] etc., but performing X.reshape(n,2,2) messes up the matrices. How can I get it to preserve the matrices while reshaping the array?
I am essentially trying to do this:
import numpy as np
Y = np.zeros([n,2,2])
for i in range(n):
Y[i] = X[:,:,i]
but without using the for loop. How can I do this with reshape or a similar function?
(To get an example array X, try X = np.concatenate([np.identity(2)[:,:,None]] * n, axis=2) for some n.)

numpy.moveaxis can be used to take a view of an array with one axis moved to a different position in the shape:
numpy.moveaxis(X, 2, 0)
numpy.moveaxis(a, source, destination) takes a view of array a where the axis originally at position source ends up at position destination, so numpy.moveaxis(X, 2, 0) makes the original axis 2 the new axis 0 in the view.
There's also numpy.transpose, which can be used to perform arbitrary rearrangements of an array's axes in one go if you pass it the optional second argument, and numpy.rollaxis, an older version of moveaxis with a more confusing calling convention.

Use swapaxis:
Y = X.swapaxes(0,2)

Related

Numpy [...,None]

I have found myself needing to add features to existing numpy arrays which has led to a question around what the last portion of the following code is actually doing:
np.ones(shape=feature_set.shape)[...,None]
Set-up
As an example, let's say I wish to solve for linear regression parameter estimates by using numpy and solving:
Assume I have a feature set shape (50,1), a target variable of shape (50,), and I wish to use the shape of my target variable to add a column for intercept values.
It would look something like this:
# Create random target & feature set
y_train = np.random.randint(0,100, size = (50,))
feature_set = np.random.randint(0,100,size=(50,1))
# Build a set of 1s after shape of target variable
int_train = np.ones(shape=y_train.shape)[...,None]
# Able to then add int_train to feature set
X = np.concatenate((int_train, feature_set),1)
What I Think I Know
I see the difference in output when I include [...,None] vs when I leave it off. Here it is:
The second version returns an error around input arrays needing the same number of dimensions, and eventually I stumbled on the solution to use [...,None].
Main Question
While I see the output of [...,None] gives me what I want, I am struggling to find any information on what it is actually supposed to do. Can anybody walk me through what this code actually means, what the None argument is doing, etc?
Thank you!
The slice of [..., None] consists of two "shortcuts":
The ellipsis literal component:
The dots (...) represent as many colons as needed to produce a complete indexing tuple. For example, if x is a rank 5 array (i.e., it has 5 axes), then
x[1,2,...] is equivalent to x[1,2,:,:,:],
x[...,3] to x[:,:,:,:,3] and
x[4,...,5,:] to x[4,:,:,5,:].
(Source)
The None component:
numpy.newaxis
The newaxis object can be used in all slicing operations to create an axis of length one. newaxis is an alias for ‘None’, and ‘None’ can be used in place of this with the same result.
(Source)
So, arr[..., None] takes an array of dimension N and "adds" a dimension "at the end" for a resulting array of dimension N+1.
Example:
import numpy as np
x = np.array([[1,2,3],[4,5,6]])
print(x.shape) # (2, 3)
y = x[...,None]
print(y.shape) # (2, 3, 1)
z = x[:,:,np.newaxis]
print(z.shape) # (2, 3, 1)
a = np.expand_dims(x, axis=-1)
print(a.shape) # (2, 3, 1)
print((y == z).all()) # True
print((y == a).all()) # True
Consider this code:
np.ones(shape=(2,3))[...,None].shape
As you see the 'None' phrase change the (2,3) matrix to a (2,3,1) tensor. As a matter of fact it put the matrix in the LAST index of the tensor.
If you use
np.ones(shape=(2,3))[None, ...].shape
it put the matrix in the FIRST‌ index of the tensor

Vector dot product along one dimension for multidimensional arrays

I want to compute the sum product along one dimension of two multidimensional arrays, using Theano.
I'll describe precisely what I want to do using numpy first. numpy.tensordot and numpy.dot seem to always do a matrix product, whereas I'm in essence looking for a batched equivalent of a vector product. Given x and y, I want to compute z like so:
x = np.random.normal(size=(200, 2, 2, 1000))
y = np.random.normal(size=(200, 2, 2))
# this is how I now approach it:
z = np.sum(y[:,:,:,np.newaxis] * x, axis=1)
# z is of shape (200, 2, 1000)
Now I know that numpy.einsum would probably be able to help me here, but again, I want to do this particular computation in Theano, which does not have an einsum equivalent. I will need to use dot, tensordot, or Theano's specialized einsum subset functions batched_dot or batched_tensordot.
The reason I'm looking to change my approach to this is performance; I suspect that using builtin (CUDA) dot products will be faster than relying on broadcasting, element-wise product, and sum.
In Theano, none of the dimensions of three and four dimensional tensors are broadcastable. You have to explicitly set them. Then the Numpy principles will work just fine. One way to do this is to use T.patternbroadcast. To read more about broadcasting, refer this.
You have three dimensions in one of the tensors. So first you need to append a singleton dimension at the end and then make that dimension broadcastable. These two things can be achieved with a single command - T.shape_padaxis. The entire code is as follows:
import theano
from theano import tensor as T
import numpy as np
X = T.ftensor4('X')
Y = T.ftensor3('Y')
Y_broadcast = T.shape_padaxis(Y, axis=-1) # appending extra dimension and making it
# broadcastable
Z = T.sum((X*Y_broadcast), axis=1) # element-wise multiplication
f = theano.function([X, Y], Z, allow_input_downcast=True)
# Making sure that it works and gives correct results
x = np.random.normal(size=(3, 2, 2, 4))
y = np.random.normal(size=(3, 2, 2))
theano_result = f(x,y)
numpy_result = np.sum(y[:,:,:,np.newaxis] * x, axis=1)
print np.amax(theano_result - numpy_result) # prints 2.7e-7 on my system, close enough!
I hope this helps.

np.bincount for 1 line, vectorized multidimensional averaging

I am trying to vectorize an operation using numpy, which I use in a python script that I have profiled, and found this operation to be the bottleneck and so needs to be optimized since I will run it many times.
The operation is on a data set of two parts. First, a large set (n) of 1D vectors of different lengths (with maximum length, Lmax) whose elements are integers from 1 to maxvalue. The set of vectors is arranged in a 2D array, data, of size (num_samples,Lmax) with trailing elements in each row zeroed. The second part is a set of scalar floats, one associated with each vector, that I have a computed and which depend on its length and the integer-value at each position. The set of scalars is made into a 1D array, Y, of size num_samples.
The desired operation is to form the average of Y over the n samples, as a function of (value,position along length,length).
This entire operation can be vectorized in matlab with use of the accumarray function: by using 3 2D arrays of the same size as data, whose elements are the corresponding value, position, and length indices of the desired final array:
sz_Y = num_samples;
sz_len = Lmax
sz_pos = Lmax
sz_val = maxvalue
ind_len = repmat( 1:sz_len ,1 ,sz_samples);
ind_pos = repmat( 1:sz_pos ,sz_samples,1 );
ind_val = data
ind_Y = repmat((1:sz_Y)',1 ,Lmax );
copiedY=Y(ind_Y);
mask = data>0;
finalarr=accumarray({ind_val(mask),ind_pos(mask),ind_len(mask)},copiedY(mask), [sz_val sz_pos sz_len])/sz_val;
I was hoping to emulate this implementation with np.bincounts. However, np.bincounts differs to accumarray in two relevant ways:
both arguments must be of same 1D size, and
there is no option to choose the shape of the output array.
In the above usage of accumarray, the list of indices, {ind_val(mask),ind_pos(mask),ind_len(mask)}, is 1D cell array of 1x3 arrays used as index tuples, while in np.bincounts it must be 1D scalars as far as I understand. I expect np.ravel may be useful but am not sure how to use it here to do what I want. I am coming to python from matlab and some things do not translate directly, e.g. the colon operator which ravels in opposite order to ravel. So my question is how might I use np.bincount or any other numpy method to achieve an efficient python implementation of this operation.
EDIT: To avoid wasting time: for these multiD index problems with complicated index manipulation, is the recommend route to just use cython to implement the loops explicity?
EDIT2: Alternative Python implementation I just came up with.
Here is a heavy ram solution:
First precalculate:
Using index units for length (i.e., length 1 =0) make a 4D bool array, size (num_samples,Lmax+1,Lmax+1,maxvalue) , holding where the conditions are satisfied for each value in Y.
ALLcond=np.zeros((num_samples,Lmax+1,Lmax+1,maxvalue+1),dtype='bool')
for l in range(Lmax+1):
for i in range(Lmax+1):
for v in range(maxvalue+!):
ALLcond[:,l,i,v]=(data[:,i]==v) & (Lvec==l)`
Where Lvec=[len(row) for row in data]. Then get the indices for these using np.where and initialize a 4D float array into which you will assign the values of Y:
[indY,ind_len,ind_pos,ind_val]=np.where(ALLcond)
Yval=np.zeros(np.shape(ALLcond),dtype='float')
Now in the loop in which I have to perform the operation, I compute it with the two lines:
Yval[ind_Y,ind_len,ind_pos,ind_val]=Y[ind_Y]
Y_avg=sum(Yval)/num_samples
This gives a factor of 4 or so speed up over the direct loop implementation. I was expecting more. Perhaps, this is a more tangible implementation for Python heads to digest. Any faster suggestions are welcome :)
One way is to convert the 3 "indices" to a linear index and then apply bincount. Numpy's ravel_multi_index is essentially the same as MATLAB's sub2ind. So the ported code could be something like:
shape = (Lmax+1, Lmax+1, maxvalue+1)
posvec = np.arange(1, Lmax+1)
ind_len = np.tile(Lvec[:,None], [1, Lmax])
ind_pos = np.tile(posvec, [n, 1])
ind_val = data
Y_copied = np.tile(Y[:,None], [1, Lmax])
mask = posvec <= Lvec[:,None] # fill-value independent
lin_idx = np.ravel_multi_index((ind_len[mask], ind_pos[mask], ind_val[mask]), shape)
Y_avg = np.bincount(lin_idx, weights=Y_copied[mask], minlength=np.prod(shape)) / n
Y_avg.shape = shape
This is assuming data has shape (n, Lmax), Lvec is Numpy array, etc. You may need to adapt the code a little to get rid of off-by-one errors.
One could argue that the tile operations are not very efficient and not very "numpythonic". Something with broadcast_arrays could be nice, but I think I prefer this way:
shape = (Lmax+1, Lmax+1, maxvalue+1)
posvec = np.arange(1, Lmax+1)
len_idx = np.repeat(Lvec, Lvec)
pos_idx = np.broadcast_to(posvec, data.shape)[mask]
val_idx = data[mask]
Y_copied = np.repeat(Y, Lvec)
mask = posvec <= Lvec[:,None] # fill-value independent
lin_idx = np.ravel_multi_index((len_idx, pos_idx, val_idx), shape)
Y_avg = np.bincount(lin_idx, weights=Y_copied, minlength=np.prod(shape)) / n
Y_avg.shape = shape
Note broadcast_to was added in Numpy 1.10.0.

numpy meshgrid of dynamic shape

I am trying to use numpy meshgrid to generate some arrays. So, I have a nd array. Let us call it data and it can have an arbitrary shape and I am trying to generate some indices array as follows:
shape = data.shape
x = np.meshgrid[1,x-1 for x in shape]
I know the syntax looks crazy but sometimes I try things like these in python and it works! Anyway, is there a way to do this dynamic meshgrid in python? This comes back with invalid syntax error:
x = np.meshgrid[1,x-1 for x in shape]
^
SyntaxError: invalid syntax
EDIT:
I would like basically to create an array of indices. For example, I can do the following when the index always begins with 0
import numpy as np
array = np.random.rand(5, 5, 5)
shape = array.shape
indices = np.indices(x-1 for x in shape)
This creates an ndarray with indices starting from 0 to (n-1) along each of the axes of my input array. Now, I wanted to have the indexing begin from 1 and could not find a good way to do this.
EDIT:
For example, a call for an array with shape (4, 5, 6) could be something like:
x = np.meshgrid(np.arange(1,4), np.arange(1,5), np.arange(1, 6))
Going off your last example, you can do something like this:
x = np.meshgrid(*[np.arange(1, x) for x in shape])
You need to explicitly create a list of the values you want to pass to meshgrid. If you want each one to start at 1, you need to put the 1 in each call to arange. You can't do something like [1, arange(x)] and have it "distribute" the 1 through all the calls.
Then the * there expands the list into separate arguments. (See here for info.)

My numpy array always ends in zero?

I think I missed something somewhere. I filled a numpy array using two for loops (x and y) and a function based on the x,y position. The only problem is that the value of the array always ends in zero irregardless of the size of the array.
thetamap = numpy.zeros(36, dtype=float)
thetamap.shape = (6, 6)
for y in range(0,5):
for x in range(0,5):
thetamap[x][y] = x+y
print thetamap
range(0, 5) produces 0, 1, 2, 3, 4. The endpoint is always omitted. You want simply range(6).
Better yet, use the awesome power of NumPy to make the array in one line:
thetamap = np.arange(6) + np.arange(6)[:,None]
This makes a row vector and a column vector, then adds them together using NumPy broadcasting to make a matrix.

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