Say I have an array
np.zeros((4,2))
I have a list of values [4,3,2,1], which I want to assign to the following positions:
[(0,0),(1,1),(2,1),(3,0)]
How can I do that without using the for loop or flattening the array?
I can use fancy index to retrieve the value, but not to assign them.
======Update=========
Thanks to #hpaulj, I realize the bug in my original code is.
When I use zeros_like to initiate the array, it defaults to int and truncates values. Therefore, it looks like I did not assign anything!
You can use tuple indexing:
>>> import numpy as np
>>> a = np.zeros((4,2))
>>> vals = [4,3,2,1]
>>> pos = [(0,0),(1,1),(2,0),(3,1)]
>>> rows, cols = zip(*pos)
>>> a[rows, cols] = vals
>>> a
array([[ 4., 0.],
[ 0., 3.],
[ 2., 0.],
[ 0., 1.]])
Here is a streamlined version of #wim's answer based on #hpaulj's comment. np.transpose automatically converts the Python list of tuples into a NumPy array and transposes it. tuple casts the index coordinates to tuples which works because a[rows, cols] is equivalent to a[(rows, cols)] in NumPy.
import numpy as np
a = np.zeros((4, 2))
vals = range(4)
indices = [(0, 0), (1, 1), (2, 0), (3, 1)]
a[tuple(np.transpose(indices))] = vals
print(a)
I am constructing a sparse vector using a scipy.sparse.csr_matrix like so:
csr_matrix((values, (np.zeros(len(indices)), indices)), shape = (1, max_index))
This works fine for most of my data, but occasionally I get a ValueError: could not convert integer scalar.
This reproduces the problem:
In [145]: inds
Out[145]:
array([ 827969148, 996833913, 1968345558, 898183169, 1811744124,
2101454109, 133039182, 898183170, 919293479, 133039089])
In [146]: vals
Out[146]:
array([ 1., 1., 1., 1., 1., 2., 1., 1., 1., 1.])
In [147]: max_index
Out[147]:
2337713000
In [143]: csr_matrix((vals, (np.zeros(10), inds)), shape = (1, max_index+1))
...
996 fn = _sparsetools.csr_sum_duplicates
997 M,N = self._swap(self.shape)
--> 998 fn(M, N, self.indptr, self.indices, self.data)
999
1000 self.prune() # nnz may have changed
ValueError: could not convert integer scalar
inds is a np.int64 array and vals is a np.float64 array.
The relevant part of the scipy sum_duplicates code is here.
Note that this works:
In [235]: csr_matrix(([1,1], ([0,0], [1,2])), shape = (1, 2**34))
Out[235]:
<1x17179869184 sparse matrix of type '<type 'numpy.int64'>'
with 2 stored elements in Compressed Sparse Row format>
So the problem is not that one of the dimensions is > 2^31
Any thoughts why these values should be causing a problem?
Might it be that max_index > 2**31 ?
Try this, just to make sure:
csr_matrix((vals, (np.zeros(10), inds/2)), shape = (1, max_index/2))
The max index you are giving is less than the maximum index of the rows you are supplying.
This
sparse.csr_matrix((vals, (np.zeros(10), inds)), shape = (1, np.max(inds)+1))
works fine with me.
Although making a .todense() results in memory error for the large size of the matrix
Uncommenting the sum_duplicates - function will lead to other errors. But this fix: strange error when creating csr_matrix also solves your problem. You can extend the version_check to newer versions of scipy.
import scipy
import scipy.sparse
if scipy.__version__ in ("0.14.0", "0.14.1", "0.15.1"):
_get_index_dtype = scipy.sparse.sputils.get_index_dtype
def _my_get_index_dtype(*a, **kw):
kw.pop('check_contents', None)
return _get_index_dtype(*a, **kw)
scipy.sparse.compressed.get_index_dtype = _my_get_index_dtype
scipy.sparse.csr.get_index_dtype = _my_get_index_dtype
scipy.sparse.bsr.get_index_dtype = _my_get_index_dtype
Is there a filter similar to ndimage's generic_filter that supports vector output? I did not manage to make scipy.ndimage.filters.generic_filter return more than a scalar. Uncomment the line in the code below to get the error: TypeError: only length-1 arrays can be converted to Python scalars.
I'm looking for a generic filter that process 2D or 3D arrays and returns a vector at each point. Thus the output would have one added dimension. For the example below I'd expect something like this:
m.shape # (10,10)
res.shape # (10,10,2)
Example Code
import numpy as np
from scipy import ndimage
a = np.ones((10, 10)) * np.arange(10)
footprint = np.array([[1,1,1],
[1,0,1],
[1,1,1]])
def myfunc(x):
r = sum(x)
#r = np.array([1,1]) # uncomment this
return r
res = ndimage.generic_filter(a, myfunc, footprint=footprint)
The generic_filter expects myfunc to return a scalar, never a vector.
However, there is nothing that precludes myfunc from also adding information
to, say, a list which is passed to myfunc as an extra argument.
Instead of using the array returned by generic_filter, we can generate our vector-valued array by reshaping this list.
For example,
import numpy as np
from scipy import ndimage
a = np.ones((10, 10)) * np.arange(10)
footprint = np.array([[1,1,1],
[1,0,1],
[1,1,1]])
ndim = 2
def myfunc(x, out):
r = np.arange(ndim, dtype='float64')
out.extend(r)
return 0
result = []
ndimage.generic_filter(
a, myfunc, footprint=footprint, extra_arguments=(result,))
result = np.array(result).reshape(a.shape+(ndim,))
I think I get what you're asking, but I'm not completely sure how does the ndimage.generic_filter work (how abstruse is the source!).
Here's just a simple wrapper function. This function will take in an array, all the parameters ndimage.generic_filter needs. Function returns an array where each element of the former array is now represented by an array with shape (2,), result of the function is stored as the second element of that array.
def generic_expand_filter(inarr, func, **kwargs):
shape = inarr.shape
res = np.empty(( shape+(2,) ))
temp = ndimage.generic_filter(inarr, func, **kwargs)
for row in range(shape[0]):
for val in range(shape[1]):
res[row][val][0] = inarr[row][val]
res[row][val][1] = temp[row][val]
return res
Output, where res denotes just the generic_filter and res2 denotes generic_expand_filter, of this function is:
>>> a.shape #same as res.shape
(10, 10)
>>> res2.shape
(10, 10, 2)
>>> a[0]
array([ 0., 1., 2., 3., 4., 5., 6., 7., 8., 9.])
>>> res[0]
array([ 3., 8., 16., 24., 32., 40., 48., 56., 64., 69.])
>>> print(*res2[0], sep=", ") #this is just to avoid the vertical default output
[ 0. 3.], [ 1. 8.], [ 2. 16.], [ 3. 24.], [ 4. 32.], [ 5. 40.], [ 6. 48.], [ 7. 56.], [ 8. 64.], [ 9. 69.]
>>> a[0][0]
0.0
>>> res[0][0]
3.0
>>> res2[0][0]
array([ 0., 3.])
Of course you probably don't want to save the old array, but instead have both fields as new results. Except I don't know what exactly you had in mind, if the two values you want stored are unrelated, just add a temp2 and func2 and call another generic_filter with the same **kwargs and store that as the first value.
However if you want an actual vector quantity that is calculated using multiple inarr elements, meaning that the two new created fields aren't independent, you are just going to have to write that kind of a function, one that takes in an array, idx, idy indices and returns a tuple\list\array value which you can then unpack and assign to the result.
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)