I have need for hstacking multple arrays with with the same number of rows (although the number of rows is variable between uses) but different number of columns. However some of the arrays only have one column, eg.
array = np.array([1,2,3,4,5])
which gives
#array.shape = (5,)
but I'd like to have the shape recognized as a 2d array, eg.
#array.shape = (5,1)
So that hstack can actually combine them.
My current solution is:
array = np.atleast_2d([1,2,3,4,5]).T
#array.shape = (5,1)
So I was wondering, is there a better way to do this? Would
array = np.array([1,2,3,4,5]).reshape(len([1,2,3,4,5]), 1)
be better?
Note that my use of [1,2,3,4,5] is just a toy list to make the example concrete. In practice it will be a much larger list passed into a function as an argument. Thanks!
Check the code of hstack and vstack. One, or both of those, pass the arguments through atleast_nd. That is a perfectly acceptable way of reshaping an array.
Some other ways:
arr = np.array([1,2,3,4,5]).reshape(-1,1) # saves the use of len()
arr = np.array([1,2,3,4,5])[:,None] # adds a new dim at end
np.array([1,2,3],ndmin=2).T # used by column_stack
hstack and vstack transform their inputs with:
arrs = [atleast_1d(_m) for _m in tup]
[atleast_2d(_m) for _m in tup]
test data:
a1=np.arange(2)
a2=np.arange(10).reshape(2,5)
a3=np.arange(8).reshape(2,4)
np.hstack([a1.reshape(-1,1),a2,a3])
np.hstack([a1[:,None],a2,a3])
np.column_stack([a1,a2,a3])
result:
array([[0, 0, 1, 2, 3, 4, 0, 1, 2, 3],
[1, 5, 6, 7, 8, 9, 4, 5, 6, 7]])
If you don't know ahead of time which arrays are 1d, then column_stack is easiest to use. The others require a little function that tests for dimensionality before applying the reshaping.
Numpy: use reshape or newaxis to add dimensions
If I understand your intent correctly, you wish to convert an array of shape (N,) to an array of shape (N,1) so that you can apply np.hstack:
In [147]: np.hstack([np.atleast_2d([1,2,3,4,5]).T, np.atleast_2d([1,2,3,4,5]).T])
Out[147]:
array([[1, 1],
[2, 2],
[3, 3],
[4, 4],
[5, 5]])
In that case, you could use avoid reshaping the arrays and use np.column_stack instead:
In [151]: np.column_stack([[1,2,3,4,5], [1,2,3,4,5]])
Out[151]:
array([[1, 1],
[2, 2],
[3, 3],
[4, 4],
[5, 5]])
I followed Ludo's work and just changed the size of v from 5 to 10000. I ran the code on my PC and the result shows that atleast_2d seems to be a more efficient method in the larger scale case.
import numpy as np
import timeit
v = np.arange(10000)
print('atleast2d:',timeit.timeit(lambda:np.atleast_2d(v).T))
print('reshape:',timeit.timeit(lambda:np.array(v).reshape(-1,1))) # saves the use of len()
print('v[:,None]:', timeit.timeit(lambda:np.array(v)[:,None])) # adds a new dim at end
print('np.array(v,ndmin=2).T:', timeit.timeit(lambda:np.array(v,ndmin=2).T)) # used by column_stack
The result is:
atleast2d: 1.3809496470021259
reshape: 27.099974197000847
v[:,None]: 28.58291715100131
np.array(v,ndmin=2).T: 30.141663907001202
My suggestion is that use [:None] when dealing with a short vector and np.atleast_2d when your vector goes longer.
Just to add info on hpaulj's answer. I was curious about how fast were the four methods described. The winner is the method adding a column at the end of the 1d array.
Here is what I ran:
import numpy as np
import timeit
v = [1,2,3,4,5]
print('atleast2d:',timeit.timeit(lambda:np.atleast_2d(v).T))
print('reshape:',timeit.timeit(lambda:np.array(v).reshape(-1,1))) # saves the use of len()
print('v[:,None]:', timeit.timeit(lambda:np.array(v)[:,None])) # adds a new dim at end
print('np.array(v,ndmin=2).T:', timeit.timeit(lambda:np.array(v,ndmin=2).T)) # used by column_stack
And the results:
atleast2d: 4.455070924214851
reshape: 2.0535152913971615
v[:,None]: 1.8387219828073285
np.array(v,ndmin=2).T: 3.1735243063353664
Related
I have an error in my code and I have recreated it using the simplest example:
import numpy as np
zeros_4d = np.zeros((5,10,15,1))
zeros_3d = np.zeros((10,15,1))
ones_3d = zeros_3d + 1
bool_array = np.arange(0,15,1)>8
zeros_3d[:,9:,:] = ones_3d[:,9:,:] # This works fine
zeros_3d[:,bool_array,:] = ones_3d[:,bool_array,:] # This works fine
zeros_4d[0,:,9:,:] = ones_3d[:,9:,:] # This works fine
for i in np.arange(0,5,1):
zeros_4d[0,i,bool_array,:] = ones_3d[i,bool_array,:] # This works fine and does what I want
zeros_4d[0,:,bool_array,:] = ones_3d[:,bool_array,:] # This does not work
I get this error message: 'ValueError: shape mismatch: value array of shape (10,6,2) could not be broadcast to indexing result of shape (6,10,2)'
I do not understand why this is happening, is there anything that I can do to solve this problem or is looping the best way to do this.
That zeros_4d[0,:,bool_array,:] indexing is a mix of basic and advanced, with a slice in the middle. That's a special documented case, docs.scipy.org/doc/numpy/reference/…. All the slice dimensions are shoved to the end, after the boolean index.
So with your 4d shape, and and a mixed index:
In [213]: arr = np.ones((5,10,15,2),int)
In [214]: arr[0,:, [1,2,3],:].shape
Out[214]: (3, 10, 2)
We can get around this issue by first selecting the leading dimension:
In [216]: arr[0][:, [1,2,3],:].shape
Out[216]: (10, 3, 2)
In [217]: arr[0][:, [1,2,3],:] = 2
In [218]: arr
Out[218]:
array([[[[1, 1],
[2, 2],
[2, 2],
...,
[1, 1],
[1, 1],
[1, 1]],
...
Splitting an index during assignment can be a problem, if the first first index produces a copy rather than a view. But here arr[0] is a view, so it's ok.
I want to concatenate two numpy arrays with the shape (100,3) and (100,7) to get a (100,10) matrix.
I've tried it using hstack, concatenate but only receives a ValueError: all the int arrays must have same number of dimensions
In a dummy example like the following it works ...
x=np.arange(30).reshape(10,3)
y=np.arange(20).reshape(10,2)
np.concatenate((x,y), axis=1)
UPDATE 1:
I've created the first two metrics's with sklearn's preprocessing module (RobustScaler and OneHotEncoder).
UPDATE 2:
When using scipy.sparse.hstack it works, but why
The sparse hstack joins the coo attributes and builds a new coo sparse matrix from those. The numpy hstack knows nothing about the different sparse structure. To explain this further I'd have to explain sparse construction, and quote from the respective functions.
If you want to concatenate it vertically axis must beequal to 0. This is explained in the doc for concatenate.
In this link we have this example:
a = np.array([[1, 2], [3, 4]])
b = np.array([[5, 6]])
np.concatenate((a, b), axis=0)
array([[1, 2],
[3, 4],
[5, 6]])
np.concatenate((a, b.T), axis=1)
array([[1, 2, 5],
[3, 4, 6]])
This works perfectly fine for me:
import numpy as np
x=np.arange(100 * 3).reshape(100,3)
y=np.arange(100 * 7).reshape(100,7)
np.hstack((x,y)).shape # (100, 10)
Given an array X of shape (100,8192), I want to copy the subarrays of length 8192 for each of the 100 outer dimensions 10 times, so that the resulting array has shape (100,8192,10).
I'm kind of confused about how the tile function works, I can sort of only copy a 1d array (although probably not really elegantly), e.g. if I'm given a 1d array of shape (8192,), I can create a 2d array by copying the 1d array like this: np.tile(x,(10,1)).transpose(), but once I try to do this on a 2d array, I have no idea what the tile function is actually doing when you provide a tuple of values, the documentation is kind of unclear about that.
Can anybody tell me how to do this please?
EDIT: Example, given the 2d array:
In [229]: x
Out[229]:
array([[1, 2, 3],
[4, 5, 6]])
I want to get by copying along the columns 3 times in this case, the following array:
In [233]: y
Out[233]:
array([[[1, 1, 1],
[2, 2, 2],
[3, 3, 3]],
[[4, 4, 4],
[5, 5, 5],
[6, 6, 6]]])
One way to do this is using np.repeat, e.g.:
Let X be the array of shape (100,8192), to replicate the subarray of dimension 8192 10-times across the column dimension, do the following:
X_new = np.repeat(X,10).reshape(100,8192,10)
Are you really asking for a shape (100,8192,10)? By reading you, I would have rather thought of something like (100,10,8192)? Could you provide an example? If you're actually asking for (100,10,8192), maybe you want:
np.tile(x,10).reshape((100,10,8192))
Is it what you're asking for?
I wanted to repeat the rows of a scipy csr sparse matrix, but when I tried to call numpy's repeat method, it simply treats the sparse matrix like an object, and would only repeat it as an object in an ndarray. I looked through the documentation, but I couldn't find any utility to repeats the rows of a scipy csr sparse matrix.
I wrote the following code that operates on the internal data, which seems to work
def csr_repeat(csr, repeats):
if isinstance(repeats, int):
repeats = np.repeat(repeats, csr.shape[0])
repeats = np.asarray(repeats)
rnnz = np.diff(csr.indptr)
ndata = rnnz.dot(repeats)
if ndata == 0:
return sparse.csr_matrix((np.sum(repeats), csr.shape[1]),
dtype=csr.dtype)
indmap = np.ones(ndata, dtype=np.int)
indmap[0] = 0
rnnz_ = np.repeat(rnnz, repeats)
indptr_ = rnnz_.cumsum()
mask = indptr_ < ndata
indmap -= np.int_(np.bincount(indptr_[mask],
weights=rnnz_[mask],
minlength=ndata))
jumps = (rnnz * repeats).cumsum()
mask = jumps < ndata
indmap += np.int_(np.bincount(jumps[mask],
weights=rnnz[mask],
minlength=ndata))
indmap = indmap.cumsum()
return sparse.csr_matrix((csr.data[indmap],
csr.indices[indmap],
np.r_[0, indptr_]),
shape=(np.sum(repeats), csr.shape[1]))
and be reasonably efficient, but I'd rather not monkey patch the class. Is there a better way to do this?
Edit
As I revisit this question, I wonder why I posted it in the first place. Almost everything I could think to do with the repeated matrix would be easier to do with the original matrix, and then apply the repetition afterwards. My assumption is that post repetition will always be the better way to approach this problem than any of the potential answers.
from scipy.sparse import csr_matrix
repeated_row_matrix = csr_matrix(np.ones([repeat_number,1])) * sparse_row
It's not surprising that np.repeat does not work. It delegates the action to the hardcoded a.repeat method, and failing that, first turns a into an array (object if needed).
In the linear algebra world where sparse code was developed, most of the assembly work was done on the row, col, data arrays BEFORE creating the sparse matrix. The focus was on efficient math operations, and not so much on adding/deleting/indexing rows and elements.
I haven't worked through your code, but I'm not surprised that a csr format matrix requires that much work.
I worked out a similar function for the lil format (working from lil.copy):
def lil_repeat(S, repeat):
# row repeat for lil sparse matrix
# test for lil type and/or convert
shape=list(S.shape)
if isinstance(repeat, int):
shape[0]=shape[0]*repeat
else:
shape[0]=sum(repeat)
shape = tuple(shape)
new = sparse.lil_matrix(shape, dtype=S.dtype)
new.data = S.data.repeat(repeat) # flat repeat
new.rows = S.rows.repeat(repeat)
return new
But it is also possible to repeat using indices. Both lil and csr support indexing that is close to that of regular numpy arrays (at least in new enough versions). Thus:
S = sparse.lil_matrix([[0,1,2],[0,0,0],[1,0,0]])
print S.A.repeat([1,2,3], axis=0)
print S.A[(0,1,1,2,2,2),:]
print lil_repeat(S,[1,2,3]).A
print S[(0,1,1,2,2,2),:].A
give the same result
and best of all?
print S[np.arange(3).repeat([1,2,3]),:].A
After someone posted a really clever response for how best to do this I revisited my original question, to see if there was an even better way. I I came up with one more way that has some pros and cons. Instead of repeating all of the data (as is done with the accepted answer), we can instead instruct scipy to reuse the data of the repeated rows, creating something akin to a view of the original sparse array (as you might do with broadcast_to). This can be done by simply tiling the indptr field.
repeated = sparse.csr_matrix((orig.data, orig.indices, np.tile(orig.indptr, repeat_num)))
This technique repeats the vector repeat_num times, while only modifying the the indptr. The downside is that due to the way the csr matrices encode data, instead of creating a matrix that's repeat_num x n in dimension, it creates one that's (2 * repeat_num - 1) x n where every odd row is 0. This shouldn't be too big of a deal as any operation will be quick given that each row is 0, and they should be pretty easy to slice out afterwards (with something like [::2]), but it's not ideal.
I think the marked answer is probably still the "best" way to do this.
One of the most efficient ways to repeat the sparse matrix would be the way OP suggested. I modified indptr so that it doesn't output rows of 0s.
## original sparse matrix
indptr = np.array([0, 2, 3, 6])
indices = np.array([0, 2, 2, 0, 1, 2])
data = np.array([1, 2, 3, 4, 5, 6])
x = scipy.sparse.csr_matrix((data, indices, indptr), shape=(3, 3))
x.toarray()
array([[1, 0, 2],
[0, 0, 3],
[4, 5, 6]])
To repeat this, you need to repeat data and indices, and you need to fix-up the indptr. This is not the most elegant way, but it works.
## repeated sparse matrix
repeat = 5
new_indptr = indptr
for r in range(1,repeat):
new_indptr = np.concatenate((new_indptr, new_indptr[-1]+indptr[1:]))
x = scipy.sparse.csr_matrix((np.tile(data,repeat), np.tile(indices,repeat), new_indptr))
x.toarray()
array([[1, 0, 2],
[0, 0, 3],
[4, 5, 6],
[1, 0, 2],
[0, 0, 3],
[4, 5, 6],
[1, 0, 2],
[0, 0, 3],
[4, 5, 6],
[1, 0, 2],
[0, 0, 3],
[4, 5, 6],
[1, 0, 2],
[0, 0, 3],
[4, 5, 6]])
What exactly is the difference between numpy vstack and column_stack. Reading through the documentation, it looks as if column_stack is an implementation of vstack for 1D arrays. Is it a more efficient implementation? Otherwise, I cannot find a reason for just having vstack.
I think the following code illustrates the difference nicely:
>>> np.vstack(([1,2,3],[4,5,6]))
array([[1, 2, 3],
[4, 5, 6]])
>>> np.column_stack(([1,2,3],[4,5,6]))
array([[1, 4],
[2, 5],
[3, 6]])
>>> np.hstack(([1,2,3],[4,5,6]))
array([1, 2, 3, 4, 5, 6])
I've included hstack for comparison as well. Notice how column_stack stacks along the second dimension whereas vstack stacks along the first dimension. The equivalent to column_stack is the following hstack command:
>>> np.hstack(([[1],[2],[3]],[[4],[5],[6]]))
array([[1, 4],
[2, 5],
[3, 6]])
I hope we can agree that column_stack is more convenient.
hstack stacks horizontally, vstack stacks vertically:
The problem with hstack is that when you append a column you need convert it from 1d-array to a 2d-column first, because 1d array is normally interpreted as a vector-row in 2d context in numpy:
a = np.ones(2) # 2d, shape = (2, 2)
b = np.array([0, 0]) # 1d, shape = (2,)
hstack((a, b)) -> dimensions mismatch error
So either hstack((a, b[:, None])) or column_stack((a, b)):
where None serves as a shortcut for np.newaxis.
If you're stacking two vectors, you've got three options:
As for the (undocumented) row_stack, it is just a synonym of vstack, as 1d array is ready to serve as a matrix row without extra work.
The case of 3D and above proved to be too huge to fit in the answer, so I've included it in the article called Numpy Illustrated.
In the Notes section to column_stack, it points out this:
This function is equivalent to np.vstack(tup).T.
There are many functions in numpy that are convenient wrappers of other functions. For example, the Notes section of vstack says:
Equivalent to np.concatenate(tup, axis=0) if tup contains arrays that are at least 2-dimensional.
It looks like column_stack is just a convenience function for vstack.