let's say a NumPy array
a = np.array(
[[[1,2,3],
[4,5,6]],
[[7,8,9],
[10,11,12]]])
the shape will be like (2,2,3).
I'd like to make it look like this:
a = np.array(
[[1,2,3],
[7,8,9],
[4,5,6],
[10,11,12]]
)
which shape will be like (4,3).
if I use reshape, it will look like as:
a = np.array(
[[1,2,3],
[4,5,6],
[7,8,9],
[10,11,12]]
)
Which is NOT what I want. How to do this?
One way using numpy.stack and vstack:
np.vstack(np.stack(a, 1))
Output:
array([[ 1, 2, 3],
[ 7, 8, 9],
[ 4, 5, 6],
[10, 11, 12]])
By using indexing method, an idx list could be created that specifies which indices of the ex a must be placed as which indices in the new one i.e. idx is a rearranging list:
idx = [0, 2, 1, 3]
a = a.reshape(4, 3)[idx]
a is firstly reshaped to the intended shape, which is (4,3), and then rearranged by the idx. idx[1] = 2 is showing that value in index = 2 of the ex a will be replaced to index = 1 in the new a.
Here is a more pythonic version of your problem.
This uses concatenate so append the rows of your array.
a = np.array(
[[[1,2,3],
[4,5,6]],
[[7,8,9],
[10,11,12]]]
)
def transform_2d(a_arr):
nrow = len(a[:])
all = a_arr[:,0]
for i in range(1,nrow):
all = np.concatenate((all, a_arr[:,i] ))
return all
print(transform_2d(a))
First use transpose (or swapaxes) to bring the desire rows together:
In [268]: a.transpose(1,0,2)
Out[268]:
array([[[ 1, 2, 3],
[ 7, 8, 9]],
[[ 4, 5, 6],
[10, 11, 12]]])
then the reshape follows:
In [269]: a.transpose(1,0,2).reshape(-1,3)
Out[269]:
array([[ 1, 2, 3],
[ 7, 8, 9],
[ 4, 5, 6],
[10, 11, 12]])
Related
I have a 3-D array
size = (3,2,3)
[
[[1, 2, 3],[4, 5, 6]],
[[7, 8, 9],[10,11,12]],
[[13,14,15],[16,17,19]]
]
How to reshape to (3,3,2):
[
[[1,4], [2,5], [3,6]],
[[7,10], [8,11], [9,12],],
[[13,16],[14,17],[15,19]]
]
You task is not to reshape the array. You have to swap the last axis (the third dimension of your array) with the second.
import numpy as np
#input
arr = np.array([
[[1, 2, 3],[4, 5, 6]],
[[7, 8, 9],[10,11,12]],
[[13,14,15],[16,17,19]]
])
#output
np.moveaxis(arr, 2, 1)
#an alternative is
np.swapaxes(arr, 1, 2)
x = [[[1, 2, 3],[4, 5, 6]],
[[7, 8, 9],[10,11,12]],
[[13,14,15],[16,17,19]]]
res = [[[i,j] for (i,j) in zip(sub[0], sub[1])] for sub in x]
a = np.array(list(range(16).reshape((4,4))
array([[ 0, 1, 2, 3],
[ 4, 5, 6, 7],
[ 8, 9, 10, 11],
[12, 13, 14, 15]])
Say I want the middle square. It'd seem reasonable to do this:
a[[1,2],[1,2]]
but I get this:
array([5, 10])
This works, but seems inelegant:
a[[1,2],:][:,[1,2]]
array([[5, 6],
[9, 10]])
So my questions are:
Why is it this way? What premises are required to make the implemented way sensible?
Is there a canonical way to select along more than one index at once?
I think you can read more details on advanced indexing. Basically, when you slice the array by lists/arrays, the arrays will be broadcast and iterate together.
In your case, you can do:
idx = np.array([1,3])
a[idx,idx[:,None]]
Or as in the doc above:
a[np.ix_(idx, idx)]
Output:
array([[ 5, 13],
[ 7, 15]])
You can do both slicing operations at once instead of creating a view and indexing that again:
import numpy as np
a = np.arange(16).reshape((4, 4))
# preferred if possible
print(a[1:3, 1:3])
# [[ 5 6]
# [ 9 10]]
# otherwise add a second dimension to the first index to make it broadcastable
index1 = np.asarray([1, 2])
index2 = np.asarray([1, 2])
print(a[index1[:, None], index2])
# [[ 5 6]
# [ 9 10]]
You could use multiple np.take to select indices from multiple axes
a = np.arange(16).reshape((4, 4))
idx = np.array([1,2])
np.take(np.take(a, idx, axis=1), idx, axis=0)
Or (slightly more readable)
a.take(idx, axis=1).take(idx, axis=0)
Output:
array([[ 5, 6],
[ 9, 10]])
np.take also allows you to conveniently wrap around out-of-bound indices and such.
I have a multi-dimensional array in Python where there may be a repeated integer within a vector in the array. For example.
array = [[1,2,3,4],
[2,9,12,4],
[5,6,7,8],
[6,8,12,13]]
I would like to completely remove the vectors that contain any element that has appeared previously. In this case, vector [2,9,12,4] and vector [6,11,12,13] should be removed because they have an element (2 and 6 respectively) that has appeared in a previous vector within that array. Note that [6,8,12,13] contains two elements that have appeared previously, so the code should be able to work with these scenarios as well.
The resulting array should end up being:
array = [[1,2,3,4],
[5,6,7,8]]
I thought I could achieve this with np.unique(array, axis=0), but I couldnt find another function that would take care of this particular uniqueness.
Any thoughts are appreaciated.
You can work with array of sorted numbers and corresponding indices of rows that looks like so:
number_info = array([[ 0, 1],
[ 0, 2],
[ 1, 2],
[ 0, 3],
[ 0, 4],
[ 1, 4],
[ 2, 5],
[ 2, 6],
[ 3, 6],
[ 2, 7],
[ 2, 8],
[ 3, 8],
[ 1, 9],
[ 1, 12],
[ 3, 12],
[ 3, 13]])
It indicates that rows remove_idx = [2, 5, 8, 11, 14] of this array needs to be removed and it points to rows rows_idx = [1, 1, 3, 3, 3] of the original array. Now, the code:
flat_idx = np.repeat(np.arange(array.shape[0]), array.shape[1])
number_info = np.transpose([flat_idx, array.ravel()])
number_info = number_info[np.argsort(number_info[:,1])]
remove_idx = np.where((np.diff(number_info[:,1])==0) &
(np.diff(number_info[:,0])>0))[0] + 1
remove_rows = number_info[remove_idx, 0]
output = np.delete(array, remove_rows, axis=0)
Output:
array([[1, 2, 3, 4],
[5, 6, 7, 8]])
Here's a quick way to do it with a list comprehension and set intersections:
>>> array = [[1,2,3,4],
... [2,9,12,4],
... [5,6,7,8],
... [6,8,12,13]]
>>> [v for i, v in enumerate(array) if not any(set(a) & set(v) for a in array[:i])]
[[1, 2, 3, 4], [5, 6, 7, 8]]
I have two array one is 3d :
np.array([[[1,2,3],[3,2,1]],
[[2,3,2],[1,2,5]]])
and one 2d array :
np.array([[2,3],
[3,4]])
and I want to multiply these two to get
np.array([[[2,4,6],[9,6,3]],
[[6,9,6],[4,8,20]]])
How can I do this using numpy package? Thanks.
Use broadcasting:
In [129]: b[:,:,None] * a
Out[129]:
array([[[ 2, 4, 6],
[ 9, 6, 3]],
[[ 6, 9, 6],
[ 4, 8, 20]]])
With following names:
main = np.array([[[1,2,3],[3,2,1]],
[[2,3,2],[1,2,5]]])
fac = np.array([[2,3],
[3,4]])
It can managed with iteration as follows:
a1 = []
for i in [0,1]:
a2 = []
for j in [0,1]:
a2.append(main[i][j]*fac[i][j])
a1.append(a2)
print(a1)
Output:
[[array([2, 4, 6]), array([9, 6, 3])], [array([6, 9, 6]), array([ 4, 8, 20])]]
I have an array that is size (214, 144). I need it to be (214,144,1) is there a way to do this easily in Python? Basically the dimensions are supposed to be (Days, Times, Stations). Since I only have 1 station's data that dimension would be a 1. However if I could also make the code flexible enough work for say 2 stations that would be great (e.g. changing the dimension size from (428,288) to (214,144,2)) that would be great!
You could use reshape:
>>> a = numpy.array([[1,2,3,4,5,6],[7,8,9,10,11,12]])
>>> a.shape
(2, 6)
>>> a.reshape((2, 6, 1))
array([[[ 1],
[ 2],
[ 3],
[ 4],
[ 5],
[ 6]],
[[ 7],
[ 8],
[ 9],
[10],
[11],
[12]]])
>>> _.shape
(2, 6, 1)
Besides changing the shape from (x, y) to (x, y, 1), you could use (x, y/n, n) as well, but you may want to specify the column order depending on the input:
>>> a.reshape((2, 3, 2))
array([[[ 1, 2],
[ 3, 4],
[ 5, 6]],
[[ 7, 8],
[ 9, 10],
[11, 12]]])
>>> a.reshape((2, 3, 2), order='F')
array([[[ 1, 4],
[ 2, 5],
[ 3, 6]],
[[ 7, 10],
[ 8, 11],
[ 9, 12]]])
1) To add a dimension to an array a of arbitrary dimensionality:
b = numpy.reshape (a, list (numpy.shape (a)) + [1])
Explanation:
You get the shape of a, turn it into a list, concatenate 1 to that list, and use that list as the new shape in a reshape operation.
2) To specify subdivisions of the dimensions, and have the size of the last dimension calculated automatically, use -1 for the size of the last dimension. e.g.:
b = numpy.reshape(a, [numpy.size(a,0)/2, numpy.size(a,1)/2, -1])
The shape of b in this case will be [214,144,4].
(obviously you could combine the two approaches if necessary):
b = numpy.reshape (a, numpy.append (numpy.array (numpy.shape (a))/2, -1))