I want to compute discrete difference of identity matrix.
The code below use numpy and scipy.
import numpy as np
from scipy.sparse import identity
from scipy.sparse import csc_matrix
x = identity(4).toarray()
y = csc_matrix(np.diff(x, n=2))
print(y)
I would like to improve performance or memory usage.
Since identity matrix produce many zeros, it would reduce memory usage to perform calculation in compressed sparse column(csc) format. However, np.diff() does not accept csc format, so converting between csc and normal format using csc_matrix would slow it down a bit.
Normal format
x = identity(4).toarray()
print(x)
[[1. 0. 0. 0.]
[0. 1. 0. 0.]
[0. 0. 1. 0.]
[0. 0. 0. 1.]]
csc format
x = identity(4)
print(x)
(0, 0) 1.0
(1, 1) 1.0
(2, 2) 1.0
(3, 3) 1.0
Thanks
Here is my hacky solution to get the sparse matrix as you want.
L - the length of the original identity matrix,
n - the parameter of np.diff.
In your question they are:
L = 4
n = 2
My code produces the same y as your code, but without the conversions between csc and normal formats.
Your code:
from scipy.sparse import identity, csc_matrix
x = identity(L).toarray()
y = csc_matrix(np.diff(x, n=n))
My code:
from scipy.linalg import pascal
def get_data(n, L):
nums = pascal(n + 1, kind='lower')[-1].astype(float)
minuses_from = n % 2 + 1
nums[minuses_from : : 2] *= -1
return np.tile(nums, L - n)
data = get_data(n, L)
row_ind = (np.arange(n + 1) + np.arange(L - n).reshape(-1, 1)).flatten()
col_ind = np.repeat(np.arange(L - n), n + 1)
y = csc_matrix((data, (row_ind, col_ind)), shape=(L, L - n))
I have noticed that after applying np.diff to the identity matrix n times, the values of the columns are the binomial coefficients with their signs alternating. This is my variable data.
Then I am just constructing the csc_matrix.
Unfortunately, it does not seem that SciPy provides any tools for this kind of sparse matrix manipulation. Regardless, by cleverly manipulating the indices and data of the entries one can emulate np.diff(x,n) in a straightforward fashion.
Given 2D NumPy array (matrix) of dimension MxN, np.diff() multiplies each column (of column index y) with -1 and adds the next column to it (column index y+1). Difference of order k is just the iterative application of k differences of order 1. A difference of order 0 is just the returns the input matrix.
The method below makes use of this, iterateively eliminating duplicate entries by addition through sum_duplicates(), reducing the number of columns by one, and filtering non-valid indices.
def csc_diff(x, n):
'''Emulates np.diff(x,n) for a sparse matrix by iteratively taking difference of order 1'''
assert isinstance(x, csc_matrix) or (isinstance(x, np.ndarray) & len(x.shape) == 2), "Input matrix must be a 2D np.ndarray or csc_matrix."
assert isinstance(n, int) & n >= 0, "Integer n must be larger or equal to 0."
if n >= x.shape[1]:
return csc_matrix(([], ([], [])), shape=(x.shape[0], 0))
if isinstance(x, np.ndarray):
x = csc_matrix(x)
# set-up of data/indices via column-wise difference
if(n > 0):
for k in range(1,n+1):
# extract data/indices of non-zero entries of (current) sparse matrix
M, N = x.shape
idx, idy = x.nonzero()
dat = x.data
# difference: this row (y) * (-1) + next row (y+1)
idx = np.concatenate((idx, idx))
idy = np.concatenate((idy, idy-1))
dat = np.concatenate(((-1)*dat, dat))
# filter valid indices
validInd = (0<=idy) & (idy<N-1)
# x_diff: csc_matrix emulating np.diff(x,1)'s output'
x_diff = csc_matrix((dat[validInd], (idx[validInd], idy[validInd])), shape=(M, N-1))
x_diff.sum_duplicates()
x = x_diff
return x
Moreover, the method outputs an empty csc_matrix of dimension Mx0 when the difference order is larger or equal to the number of columns of the input matrix. This is why the output is identical, see
csc_diff(x, 2).toarray()
> array([[ 1., 0.],
[-2., 1.],
[ 1., -2.],
[ 0., 1.]])
which is identical to
np.diff(x.toarray(), 2)
> array([[ 1., 0.],
[-2., 1.],
[ 1., -2.],
[ 0., 1.]])
This identity holds for other difference orders, too
(csc_diff(x, 0).toarray() == np.diff(x.toarray(), 0)).all()
>True
(csc_diff(x, 3).toarray() == np.diff(x.toarray(), 3)).all()
>True
(csc_diff(x, 13).toarray() == np.diff(x.toarray(), 13)).all()
>True
I am trying to use fancy indexing to modifying a large sparce matrix. Suppose you have the following code:
import numpy as np
import scipy.sparse as sp
a = np.array([[1,2,3,4], [5,6,7,8], [9,10,11,12]])
b = sp.lil_matrix(a)
c = sp.lil_matrix((3,4))
c[[1,2], 0] = b[[1,2], 0]
However, this code gives the following error:
ValueError: shape mismatch in assignment
I don't understand why this doesn't work. Both matrices have the same shape and this usually works if both matrices are numpy arrays. I would appreciate any help.
Yeah this is a bug with the sparse __setitem__. I've run into it before (but I just worked around it). Now I actually looked into it; first, you can fix this pretty easily:
import numpy as np
import scipy.sparse as sp
a = np.array([[1,2,3,4], [5,6,7,8], [9,10,11,12]])
b = sp.lil_matrix(a)
c = sp.lil_matrix((3,4))
c[[1,2], 0] = b[[1,2], 0]
This raises the ValueError you saw. This doesn't and works as expected:
c[[1,2], 0] = b[[1,2], [0]]
>>> c.A
array([[0., 0., 0., 0.],
[5., 0., 0., 0.],
[9., 0., 0., 0.]])
Lets just walk through the offending __setitem__ (I'm going to omit a lot of code that doesn't get called):
row, col = self._validate_indices(key)
This is fine - row = [1, 2] and col = 0
col = np.atleast_1d(col)
i, j = _broadcast_arrays(row, col)
So far so good - i = [1, 2] and j = [0, 0]
if i.ndim == 1:
# Inner indexing, so treat them like row vectors.
i = i[None]
j = j[None]
broadcast_row = x.shape[0] == 1 and i.shape[0] != 1
broadcast_col = x.shape[1] == 1 and i.shape[1] != 1
Here's our problem - i and j both got turned into row vectors with shape (1, 2). x here is what you're trying to assign (b[[1,2], 0]), which is of shape (2, 1); the next step raises a ValueError cause x and the indices don't align.
>>> c[[1,2], 0] = b[[1,2], 0].A
ValueError: cannot reshape array of size 4 into shape (2,)
Here's the same problem but __setitem__ broadcasts x into a (2,2) array, which then fails again because it's larger than the array you're assigning it to.
The workaround (b[[1,2], [0]]) has a shape of (1, 2) which is not correct, but that error ends up cancelling out the error in indexing c.
I'm not sure exactly what the logic is behind this indexing code so I'm not sure how to fix this without introducing other subtle bugs.
I have two numpy arrays of different shapes, but with the same length (leading dimension). I want to shuffle each of them, such that corresponding elements continue to correspond -- i.e. shuffle them in unison with respect to their leading indices.
This code works, and illustrates my goals:
def shuffle_in_unison(a, b):
assert len(a) == len(b)
shuffled_a = numpy.empty(a.shape, dtype=a.dtype)
shuffled_b = numpy.empty(b.shape, dtype=b.dtype)
permutation = numpy.random.permutation(len(a))
for old_index, new_index in enumerate(permutation):
shuffled_a[new_index] = a[old_index]
shuffled_b[new_index] = b[old_index]
return shuffled_a, shuffled_b
For example:
>>> a = numpy.asarray([[1, 1], [2, 2], [3, 3]])
>>> b = numpy.asarray([1, 2, 3])
>>> shuffle_in_unison(a, b)
(array([[2, 2],
[1, 1],
[3, 3]]), array([2, 1, 3]))
However, this feels clunky, inefficient, and slow, and it requires making a copy of the arrays -- I'd rather shuffle them in-place, since they'll be quite large.
Is there a better way to go about this? Faster execution and lower memory usage are my primary goals, but elegant code would be nice, too.
One other thought I had was this:
def shuffle_in_unison_scary(a, b):
rng_state = numpy.random.get_state()
numpy.random.shuffle(a)
numpy.random.set_state(rng_state)
numpy.random.shuffle(b)
This works...but it's a little scary, as I see little guarantee it'll continue to work -- it doesn't look like the sort of thing that's guaranteed to survive across numpy version, for example.
Your can use NumPy's array indexing:
def unison_shuffled_copies(a, b):
assert len(a) == len(b)
p = numpy.random.permutation(len(a))
return a[p], b[p]
This will result in creation of separate unison-shuffled arrays.
X = np.array([[1., 0.], [2., 1.], [0., 0.]])
y = np.array([0, 1, 2])
from sklearn.utils import shuffle
X, y = shuffle(X, y, random_state=0)
To learn more, see http://scikit-learn.org/stable/modules/generated/sklearn.utils.shuffle.html
Your "scary" solution does not appear scary to me. Calling shuffle() for two sequences of the same length results in the same number of calls to the random number generator, and these are the only "random" elements in the shuffle algorithm. By resetting the state, you ensure that the calls to the random number generator will give the same results in the second call to shuffle(), so the whole algorithm will generate the same permutation.
If you don't like this, a different solution would be to store your data in one array instead of two right from the beginning, and create two views into this single array simulating the two arrays you have now. You can use the single array for shuffling and the views for all other purposes.
Example: Let's assume the arrays a and b look like this:
a = numpy.array([[[ 0., 1., 2.],
[ 3., 4., 5.]],
[[ 6., 7., 8.],
[ 9., 10., 11.]],
[[ 12., 13., 14.],
[ 15., 16., 17.]]])
b = numpy.array([[ 0., 1.],
[ 2., 3.],
[ 4., 5.]])
We can now construct a single array containing all the data:
c = numpy.c_[a.reshape(len(a), -1), b.reshape(len(b), -1)]
# array([[ 0., 1., 2., 3., 4., 5., 0., 1.],
# [ 6., 7., 8., 9., 10., 11., 2., 3.],
# [ 12., 13., 14., 15., 16., 17., 4., 5.]])
Now we create views simulating the original a and b:
a2 = c[:, :a.size//len(a)].reshape(a.shape)
b2 = c[:, a.size//len(a):].reshape(b.shape)
The data of a2 and b2 is shared with c. To shuffle both arrays simultaneously, use numpy.random.shuffle(c).
In production code, you would of course try to avoid creating the original a and b at all and right away create c, a2 and b2.
This solution could be adapted to the case that a and b have different dtypes.
Very simple solution:
randomize = np.arange(len(x))
np.random.shuffle(randomize)
x = x[randomize]
y = y[randomize]
the two arrays x,y are now both randomly shuffled in the same way
James wrote in 2015 an sklearn solution which is helpful. But he added a random state variable, which is not needed. In the below code, the random state from numpy is automatically assumed.
X = np.array([[1., 0.], [2., 1.], [0., 0.]])
y = np.array([0, 1, 2])
from sklearn.utils import shuffle
X, y = shuffle(X, y)
from np.random import permutation
from sklearn.datasets import load_iris
iris = load_iris()
X = iris.data #numpy array
y = iris.target #numpy array
# Data is currently unshuffled; we should shuffle
# each X[i] with its corresponding y[i]
perm = permutation(len(X))
X = X[perm]
y = y[perm]
Shuffle any number of arrays together, in-place, using only NumPy.
import numpy as np
def shuffle_arrays(arrays, set_seed=-1):
"""Shuffles arrays in-place, in the same order, along axis=0
Parameters:
-----------
arrays : List of NumPy arrays.
set_seed : Seed value if int >= 0, else seed is random.
"""
assert all(len(arr) == len(arrays[0]) for arr in arrays)
seed = np.random.randint(0, 2**(32 - 1) - 1) if set_seed < 0 else set_seed
for arr in arrays:
rstate = np.random.RandomState(seed)
rstate.shuffle(arr)
And can be used like this
a = np.array([1, 2, 3, 4, 5])
b = np.array([10,20,30,40,50])
c = np.array([[1,10,11], [2,20,22], [3,30,33], [4,40,44], [5,50,55]])
shuffle_arrays([a, b, c])
A few things to note:
The assert ensures that all input arrays have the same length along
their first dimension.
Arrays shuffled in-place by their first dimension - nothing returned.
Random seed within positive int32 range.
If a repeatable shuffle is needed, seed value can be set.
After the shuffle, the data can be split using np.split or referenced using slices - depending on the application.
you can make an array like:
s = np.arange(0, len(a), 1)
then shuffle it:
np.random.shuffle(s)
now use this s as argument of your arrays. same shuffled arguments return same shuffled vectors.
x_data = x_data[s]
x_label = x_label[s]
There is a well-known function that can handle this:
from sklearn.model_selection import train_test_split
X, _, Y, _ = train_test_split(X,Y, test_size=0.0)
Just setting test_size to 0 will avoid splitting and give you shuffled data.
Though it is usually used to split train and test data, it does shuffle them too.
From documentation
Split arrays or matrices into random train and test subsets
Quick utility that wraps input validation and
next(ShuffleSplit().split(X, y)) and application to input data into a
single call for splitting (and optionally subsampling) data in a
oneliner.
This seems like a very simple solution:
import numpy as np
def shuffle_in_unison(a,b):
assert len(a)==len(b)
c = np.arange(len(a))
np.random.shuffle(c)
return a[c],b[c]
a = np.asarray([[1, 1], [2, 2], [3, 3]])
b = np.asarray([11, 22, 33])
shuffle_in_unison(a,b)
Out[94]:
(array([[3, 3],
[2, 2],
[1, 1]]),
array([33, 22, 11]))
One way in which in-place shuffling can be done for connected lists is using a seed (it could be random) and using numpy.random.shuffle to do the shuffling.
# Set seed to a random number if you want the shuffling to be non-deterministic.
def shuffle(a, b, seed):
np.random.seed(seed)
np.random.shuffle(a)
np.random.seed(seed)
np.random.shuffle(b)
That's it. This will shuffle both a and b in the exact same way. This is also done in-place which is always a plus.
EDIT, don't use np.random.seed() use np.random.RandomState instead
def shuffle(a, b, seed):
rand_state = np.random.RandomState(seed)
rand_state.shuffle(a)
rand_state.seed(seed)
rand_state.shuffle(b)
When calling it just pass in any seed to feed the random state:
a = [1,2,3,4]
b = [11, 22, 33, 44]
shuffle(a, b, 12345)
Output:
>>> a
[1, 4, 2, 3]
>>> b
[11, 44, 22, 33]
Edit: Fixed code to re-seed the random state
Say we have two arrays: a and b.
a = np.array([[1,2,3],[4,5,6],[7,8,9]])
b = np.array([[9,1,1],[6,6,6],[4,2,0]])
We can first obtain row indices by permutating first dimension
indices = np.random.permutation(a.shape[0])
[1 2 0]
Then use advanced indexing.
Here we are using the same indices to shuffle both arrays in unison.
a_shuffled = a[indices[:,np.newaxis], np.arange(a.shape[1])]
b_shuffled = b[indices[:,np.newaxis], np.arange(b.shape[1])]
This is equivalent to
np.take(a, indices, axis=0)
[[4 5 6]
[7 8 9]
[1 2 3]]
np.take(b, indices, axis=0)
[[6 6 6]
[4 2 0]
[9 1 1]]
If you want to avoid copying arrays, then I would suggest that instead of generating a permutation list, you go through every element in the array, and randomly swap it to another position in the array
for old_index in len(a):
new_index = numpy.random.randint(old_index+1)
a[old_index], a[new_index] = a[new_index], a[old_index]
b[old_index], b[new_index] = b[new_index], b[old_index]
This implements the Knuth-Fisher-Yates shuffle algorithm.
Shortest and easiest way in my opinion, use seed:
random.seed(seed)
random.shuffle(x_data)
# reset the same seed to get the identical random sequence and shuffle the y
random.seed(seed)
random.shuffle(y_data)
most solutions above work, however if you have column vectors you have to transpose them first. here is an example
def shuffle(self) -> None:
"""
Shuffles X and Y
"""
x = self.X.T
y = self.Y.T
p = np.random.permutation(len(x))
self.X = x[p].T
self.Y = y[p].T
With an example, this is what I'm doing:
combo = []
for i in range(60000):
combo.append((images[i], labels[i]))
shuffle(combo)
im = []
lab = []
for c in combo:
im.append(c[0])
lab.append(c[1])
images = np.asarray(im)
labels = np.asarray(lab)
I extended python's random.shuffle() to take a second arg:
def shuffle_together(x, y):
assert len(x) == len(y)
for i in reversed(xrange(1, len(x))):
# pick an element in x[:i+1] with which to exchange x[i]
j = int(random.random() * (i+1))
x[i], x[j] = x[j], x[i]
y[i], y[j] = y[j], y[i]
That way I can be sure that the shuffling happens in-place, and the function is not all too long or complicated.
Just use numpy...
First merge the two input arrays 1D array is labels(y) and 2D array is data(x) and shuffle them with NumPy shuffle method. Finally split them and return.
import numpy as np
def shuffle_2d(a, b):
rows= a.shape[0]
if b.shape != (rows,1):
b = b.reshape((rows,1))
S = np.hstack((b,a))
np.random.shuffle(S)
b, a = S[:,0], S[:,1:]
return a,b
features, samples = 2, 5
x, y = np.random.random((samples, features)), np.arange(samples)
x, y = shuffle_2d(train, test)