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Python xarray - vectorized indexing

I'm trying to understand vectorized indexing in xarray by following this example from the docs:
import xarray as xr
import numpy as np
da = xr.DataArray(np.arange(12).reshape((3, 4)), dims=['x', 'y'],
coords={'x': [0, 1, 2], 'y': ['a', 'b', 'c', 'd']})
ind_x = xr.DataArray([0, 1], dims=['x'])
ind_y = xr.DataArray([0, 1], dims=['y'])
The output of the array da is as follows:
array([[ 0, 1, 2, 3],
[ 4, 5, 6, 7],
[ 8, 9, 10, 11]])
So far so good. Now in the example there are shown two ways of indexing. Orthogonal (not interested in this case) and vectorized (what I want). For the vectorized indexing the following is shown:
In [37]: da[ind_x, ind_x] # vectorized indexing
Out[37]:
<xarray.DataArray (x: 2)>
array([0, 5])
Coordinates:
y (x) <U1 'a' 'b'
* x (x) int64 0 1
The result seems to be what I want, but this feels very strange to me. ind_x (which in theory refers to dims=['x']) is being passed twice but somehow is capable of indexing what appears to be both in the x and y dims. As far as I understand the x dim would be the rows and y dim would be the columns, is that correct? How come the same ind_x is capable of accessing both the rows and the cols?
This seems to be the concept I need for my problem, but can't understand how it works or how to extend it to more dimensions. I was expecting this result to be given by da[ind_x, ind_y] however that seems to yield the orthogonal indexing surprisingly enough.

Having the example with ind_x being used twice is probably a little confusing: actually, the dimension of the indexer doesn't have to matter at all for the indexing behavior! Observe:
ind_a = xr.DataArray([0, 1], dims=["a"]
da[ind_a, ind_a]
Gives:
<xarray.DataArray (a: 2)>
array([0, 5])
Coordinates:
x (a) int32 0 1
y (a) <U1 'a' 'b'
Dimensions without coordinates: a
The same goes for the orthogonal example:
ind_a = xr.DataArray([0, 1], dims=["a"])
ind_b = xr.DataArray([0, 1], dims=["b"])
da[ind_a, ind_b]
Result:
<xarray.DataArray (a: 2, b: 2)>
array([[0, 2],
[4, 6]])
Coordinates:
x (a) int32 0 1
y (b) <U1 'a' 'c'
Dimensions without coordinates: a, b
The difference is purely in terms of "labeling", as in this case you end up with dimensions without coordinates.
Fancy indexing
Generally stated, I personally do not find "fancy indexing" the most intuitive concept. I did find this example in NEP 21 pretty clarifying: https://numpy.org/neps/nep-0021-advanced-indexing.html
Specifically, this:
Consider indexing a 2D array by two 1D integer arrays, e.g., x[[0, 1], [0, 1]]:
Outer indexing is equivalent to combining multiple integer indices with itertools.product(). The result in this case is another 2D
array with all combinations of indexed elements, e.g.,
np.array([[x[0, 0], x[0, 1]], [x[1, 0], x[1, 1]]])
Vectorized indexing is equivalent to combining multiple integer
indices with zip(). The result in this case is a 1D array containing
the diagonal elements, e.g., np.array([x[0, 0], x[1, 1]]).
Back to xarray
da[ind_x, ind_y]
Can also be written as:
da.isel(x=ind_x, y=ind_y)
The dimensions are implicit in the order. However, xarray still attempts to broadcast (based on dimension labels), so da[ind_y] mismatches and results in an error. da[ind_a] and da[ind_b] both work.
More dimensions
The dims you provide for the indexer are what determines the shape of the output, not the dimensions of the array you're indexing.
If you want to select single values along the dimensions (so we're zip()-ing through the indexes simultaneously), just make sure that your indexers share the dimension, here for a 3D array:
da = xr.DataArray(
data=np.arange(3 * 4 * 5).reshape(3, 4, 5),
coords={
"x": [1, 2, 3],
"y": ["a", "b", "c", "d"],
"z": [1.0, 2.0, 3.0, 4.0, 5.0],
},
dims=["x", "y", "z"],
)
ind_along_x = xr.DataArray([0, 1], dims=["new_index"])
ind_along_y = xr.DataArray([0, 2], dims=["new_index"])
ind_along_z = xr.DataArray([0, 3], dims=["new_index"])
da[ind_along_x, ind_along_y, ind_along_z]
Note that the values of the indexers do not have to the same -- that would be a pretty severe limitation, after all.
Result:
<xarray.DataArray (new_index: 2)>
array([ 0, 33])
Coordinates:
x (new_index) int32 1 2
y (new_index) <U1 'a' 'c'
z (new_index) float64 1.0 4.0
Dimensions without coordinates: new_index

numpy python - slicing rows and columns at the same time

I have a numpy matrix with 130 X 13. Say I want to select a specific set of rows meeting a condition and a subset of columns -
trainx[trainy==label,[0,6]]
The above code does not work and throws an error - IndexError: shape mismatch: indexing arrays could not be broadcast together with shapes (43,) (2,).
However if I do it in 2 steps - first subset rows and then columns, it works. Is it something weird or numpy works this way?
temp1 = trainx[trainy==label,:]
temp1 = temp1[:,[0,6]]

you can simply chain the indexing like
trainx[trainy==label][:, [0,6]]
Runable Example
arr = np.random.rand(130,13)
arr[arr[:,0]>0.5][:, [0,6]]

In [154]: x = np.arange(24).reshape(6,4)
In [155]: mask = np.array([1,0,1,0,1,0],bool)
With your two step approach:
In [156]: x[mask] # x[mask, :]
Out[156]:
array([[ 0, 1, 2, 3],
[ 8, 9, 10, 11],
[16, 17, 18, 19]])
In [157]: x[mask][:,[1,3]]
Out[157]:
array([[ 1, 3],
[ 9, 11],
[17, 19]])
Or the two indices could be combined with ix_:
In [158]: np.ix_(mask, [1,3])
Out[158]:
(array([[0],
[2],
[4]]), array([[1, 3]]))
In [159]: x[np.ix_(mask, [1,3])]
Out[159]:
array([[ 1, 3],
[ 9, 11],
[17, 19]])
Note that the first array in Out[158] is np.nonzero(mask)[0][:,None], the nonzero indices in column vector form. That (3,1) indexing array can broadcast with the (2,) column array to select a (3,2) array of elements. Or in your example a (43,2) array.
The boolean mask cannot be turned into a (6,1) array and used to mask x; that would only work if it was turned into a (6,4) mask, matching the shape of x.
So either use the 2 step indexing, or use ix_.

broadcasting arrays in numpy

I got an array and reshaped it to the following dimentions: (-1,1,1,1) and (-1,1):
Array A:
[-0.888788523827 0.11842529285 0.319928774626 0.319928774626 0.378755429421 1.225877519716 3.830653798838]
A.reshape(-1,1,1,1):
[[[[-0.888788523827]]]
[[[ 0.11842529285 ]]]
[[[ 0.319928774626]]]
[[[ 0.319928774626]]]
[[[ 0.378755429421]]]
[[[ 1.225877519716]]]
[[[ 3.830653798838]]]]
A.reshape(-1,1):
[[-0.888788523827]
[ 0.11842529285 ]
[ 0.319928774626]
[ 0.319928774626]
[ 0.378755429421]
[ 1.225877519716]
[ 3.830653798838]]
Then I have done substractig and broadcasting came in, so my resulting matrix is 7x1x7x1.
I have a hard time to visualize the intermediate step what broadcasting does. I mean I cannot imagine what elements of arrays are repeated and what they look like while broadcasting.
Could somebody shed some light on this problem,please?

In [5]: arr = np.arange(4)
In [6]: A = arr.reshape(-1,1,1,1)
In [7]: B = arr.reshape(-1,1)
In [8]: C = A + B
In [9]: C.shape
Out[9]: (4, 1, 4, 1)
In [10]: A.shape
Out[10]: (4, 1, 1, 1)
In [11]: B.shape
Out[11]: (4, 1)
There are 2 basic broadcasting rules:
expand the dimensions to match - by adding size 1 dimensions at the start
adjust all size 1 dimensions to match
So in this example:
(4,1,1,1) + (4,1)
(4,1,1,1) + (1,1,4,1) # add 2 size 1's to B
(4,1,4,1) + (4,1,4,1) # adjust 2 of the 1's to 4
(4,1,4,1)
The first step is, perhaps, the most confusing. The (4,1) is expanded to (1,1,4,1), not (4,1,1,1). The rule is intended to avoid ambiguity - by expanding in a consistent manner, not necessarily what a human might intuitively want.
Imagine the case where both arrays need expansion to match, and it could add a dimension in either direction:
(4,) and (3,)
(1,4) and (3,1) or (4,1) and (1,3)
(3,4) or (4,3)
confusion
The rule requires that the programmer choose which one expands to the right (4,1) or (3,1). numpy can then unambiguously add the other.
For a simpler example:
In [22]: A=np.arange(3).reshape(-1,1)
In [23]: B=np.arange(3)
In [24]: C = A+B (3,1)+(3,) => (3,1)+(1,3) => (3,3)
In [25]: C
Out[25]:
array([[0, 1, 2],
[1, 2, 3],
[2, 3, 4]])
In [26]: C.shape
Out[26]: (3, 3)
The [0,2,4] are present, but on the diagonal of C.
When broadcasting like this, the result is a kind of outer sum:
In [27]: np.add.outer(B,B)
Out[27]:
array([[0, 1, 2],
[1, 2, 3],
[2, 3, 4]])

Explaining the differences between dim, shape, rank, dimension and axis in numpy

I'm new to python and numpy in general. I read several tutorials and still so confused between the differences in dim, ranks, shape, aixes and dimensions. My mind seems to be stuck at the matrix representation. So if you say that A is a matrix that looks like this:
A =
1 2 3
4 5 6
then all I can think of is a 2x3 matrix (two rows and three columns). Here I understand that the shape is 2x3. But I really I am unable to go out side the thinking of a 2D matrices. I don't understand for example the dot() documentation when it says "For N dimensions it is a sum product over the last axis of a and the second-to-last of b". I'm so confused and unable to understand this. I don't understand like if V is a N:1 vector and M is N:N matrix, how dot(V,M) or dot(M,V) work and the difference between them.
Can anyone then please explain to me what is a N dimensional array, what's a shape, what's an axis and how does it relate to the documentation of the dot() function? It would be great if the explanation visualizes the ideas.

Dimensionality of NumPy arrays must be understood in the data structures sense, not the mathematical sense, i.e. it's the number of scalar indices you need to obtain a scalar value.(*)
E.g., this is a 3-d array:
>>> X = np.arange(24).reshape(2, 3, 4)
>>> X
array([[[ 0, 1, 2, 3],
[ 4, 5, 6, 7],
[ 8, 9, 10, 11]],
[[12, 13, 14, 15],
[16, 17, 18, 19],
[20, 21, 22, 23]]])
Indexing once gives a 2-d array (matrix):
>>> X[0]
array([[ 0, 1, 2, 3],
[ 4, 5, 6, 7],
[ 8, 9, 10, 11]])
Indexing twice gives a 1-d array (vector), and indexing three times gives a scalar.
The rank of X is its number of dimensions:
>>> X.ndim
3
>>> np.rank(X)
3
Axis is roughly synonymous with dimension; it's used in broadcasting operations:
>>> X.sum(axis=0)
array([[12, 14, 16, 18],
[20, 22, 24, 26],
[28, 30, 32, 34]])
>>> X.sum(axis=1)
array([[12, 15, 18, 21],
[48, 51, 54, 57]])
>>> X.sum(axis=2)
array([[ 6, 22, 38],
[54, 70, 86]])
To be honest, I find this definition of "rank" confusing since it matches neither the name of the attribute ndim nor the linear algebra definition of rank.
Now regarding np.dot, what you have to understand is that there are three ways to represent a vector in NumPy: 1-d array, a column vector of shape (n, 1) or a row vector of shape (1, n). (Actually, there are more ways, e.g. as a (1, n, 1)-shaped array, but these are quite rare.) np.dot performs vector multiplication when both arguments are 1-d, matrix-vector multiplication when one argument is 1-d and the other is 2-d, and otherwise it performs a (generalized) matrix multiplication:
>>> A = np.random.randn(2, 3)
>>> v1d = np.random.randn(2)
>>> np.dot(v1d, A)
array([-0.29269547, -0.52215117, 0.478753 ])
>>> vrow = np.atleast_2d(v1d)
>>> np.dot(vrow, A)
array([[-0.29269547, -0.52215117, 0.478753 ]])
>>> vcol = vrow.T
>>> np.dot(vcol, A)
Traceback (most recent call last):
File "<ipython-input-36-98949c6de990>", line 1, in <module>
np.dot(vcol, A)
ValueError: matrices are not aligned
The rule "sum product over the last axis of a and the second-to-last of b" matches and generalizes the common definition of matrix multiplication.
(*) Arrays of dtype=object are a bit of an exception, since they treat any Python object as a scalar.

np.dot is a generalization of matrix multiplication.
In regular matrix multiplication, an (N,M)-shape matrix multiplied with a (M,P)-shaped matrix results in a (N,P)-shaped matrix. The resultant shape can be thought of as being formed by squashing the two shapes together ((N,M,M,P)) and then removing the middle numbers, M (to produce (N,P)). This is the property that np.dot preserves while generalizing to arrays of higher dimension.
When the docs say,
"For N dimensions it is a sum product over the last axis of a and the
second-to-last of b".
it is speaking to this point. An array of shape (u,v,M) dotted with an array of shape (w,x,y,M,z) would result in an array of shape (u,v,w,x,y,z).
Let's see how this rule looks when applied to
In [25]: V = np.arange(2); V
Out[25]: array([0, 1])
In [26]: M = np.arange(4).reshape(2,2); M
Out[26]:
array([[0, 1],
[2, 3]])
First, the easy part:
In [27]: np.dot(M, V)
Out[27]: array([1, 3])
There is no surprise here; this is just matrix-vector multiplication.
Now consider
In [28]: np.dot(V, M)
Out[28]: array([2, 3])
Look at the shape of V and M:
In [29]: V.shape
Out[29]: (2,)
In [30]: M.shape
Out[30]: (2, 2)
So np.dot(V,M) is like matrix multiplication of a (2,)-shaped matrix with a (2,2)-shaped matrix, which should result in a (2,)-shaped matrix.
The last (and only) axis of V and the second-to-last axis of M (aka the first axis of M) are multiplied and summed over, leaving only the last axis of M.
If you want to visualize this: np.dot(V, M) looks as though V has 1 row and 2 columns:
[[0, 1]] * [[0, 1],
[2, 3]]
and so, when V is multiplied by M, np.dot(V, M) equals
[[0*0 + 1*2], [2,
[0*1 + 1*3]] = 3]
However, I don't really recommend trying to visualize NumPy arrays this way -- at least I never do. I focus almost exclusively on the shape.
(2,) * (2,2)
\ /
\ /
(2,)
You just think about the "middle" axes being dotted, and disappearing from the resultant shape.
np.sum(arr, axis=0) tells NumPy to sum the elements in arr eliminating the 0th axis. If arr is 2-dimensional, the 0th axis are the rows. So for example, if arr looks like this:
In [1]: arr = np.arange(6).reshape(2,3); arr
Out[1]:
array([[0, 1, 2],
[3, 4, 5]])
then np.sum(arr, axis=0) will sum along the columns, thus eliminating the 0th axis (i.e. the rows).
In [2]: np.sum(arr, axis=0)
Out[2]: array([3, 5, 7])
The 3 is the result of 0+3, the 5 equals 1+4, the 7 equals 2+5.
Notice arr had shape (2,3), and after summing, the 0th axis is removed so the result is of shape (3,). The 0th axis had length 2, and each sum is composed of adding those 2 elements. The shape (2,3) "becomes" (3,). You can know the resultant shape in advance! This can help guide your thinking.
To test your understanding, consider np.sum(arr, axis=1). Now the 1-axis is removed. So the resultant shape will be (2,), and element in the result will be the sum of 3 values.
In [3]: np.sum(arr, axis=1)
Out[3]: array([ 3, 12])
The 3 equals 0+1+2, and the 12 equals 3+4+5.
So we see that summing an axis eliminates that axis from the result. This has bearing on np.dot, since the calculation performed by np.dot is a sum of products. Since np.dot performs a summing operation along certain axes, that axis is removed from the result. That is why applying np.dot to arrays of shape (2,) and (2,2) results in an array of shape (2,). The first 2 in both arrays is summed over, eliminating both, leaving only the second 2 in the second array.

In your case,
A is a 2D array, namely a matrix, with its shape being (2, 3). From docstring of numpy.matrix:
A matrix is a specialized 2-D array that retains its 2-D nature through operations.
numpy.rank return the number of dimensions of an array, which is quite different from the concept of rank in linear algebra, e.g. A is an array of dimension/rank 2.
np.dot(V, M), or V.dot(M) multiplies matrix V with M. Note that numpy.dot do the multiplication as far as possible. If V is N:1 and M is N:N, V.dot(M) would raise an ValueError.
e.g.:
In [125]: a
Out[125]:
array([[1],
[2]])
In [126]: b
Out[126]:
array([[2, 3],
[1, 2]])
In [127]: a.dot(b)
---------------------------------------------------------------------------
ValueError Traceback (most recent call last)
<ipython-input-127-9a1f5761fa9d> in <module>()
----> 1 a.dot(b)
ValueError: objects are not aligned
EDIT:
I don't understand the difference between Shape of (N,) and (N,1) and it relates to the dot() documentation.
V of shape (N,) implies an 1D array of length N, whilst shape (N, 1) implies a 2D array with N rows, 1 column:
In [2]: V = np.arange(2)
In [3]: V.shape
Out[3]: (2,)
In [4]: Q = V[:, np.newaxis]
In [5]: Q.shape
Out[5]: (2, 1)
In [6]: Q
Out[6]:
array([[0],
[1]])
As the docstring of np.dot says:
For 2-D arrays it is equivalent to matrix multiplication, and for 1-D
arrays to inner product of vectors (without complex conjugation).
It also performs vector-matrix multiplication if one of the parameters is a vector. Say V.shape==(2,); M.shape==(2,2):
In [17]: V
Out[17]: array([0, 1])
In [18]: M
Out[18]:
array([[2, 3],
[4, 5]])
In [19]: np.dot(V, M) #treats V as a 1*N 2D array
Out[19]: array([4, 5]) #note the result is a 1D array of shape (2,), not (1, 2)
In [20]: np.dot(M, V) #treats V as a N*1 2D array
Out[20]: array([3, 5]) #result is still a 1D array of shape (2,), not (2, 1)
In [21]: Q #a 2D array of shape (2, 1)
Out[21]:
array([[0],
[1]])
In [22]: np.dot(M, Q) #matrix multiplication
Out[22]:
array([[3], #gets a result of shape (2, 1)
[5]])

numpy masked ragged array

I want to fill a masked array whose dtype is object (because I need to store masked ragged arrays) with a non scalar fill_value.
Here's an example of a 2D array whose elements are 1D numpy arrays. Of course, I would like the fill_value to be an empty array.
import numpy as np
arr = np.array([
[np.arange(10), np.arange(5), np.arange(3)],
[np.arange(1), np.arange(2), np.array([])],
])
marr = np.ma.array(arr)
marr.mask = [[True, False, False],
[True, False, True]]
marr.fill_value = np.array([])
marr.filled()
Unfortunately, it yields an error on the last line:
ValueError: could not broadcast input array from shape (0) into shape (2,3)
I could manually extract the mask, and apply it on an element-by-element algorithm; but it does not seem to be the right direction to me.
Thank you !

I would not count on MaskedArray to work well with object dtype arrays. filled is trying to copy the fill value, an array, into a subset of the slots in the data. Due to broadcasting that can be tricky, even without the masking layer.
Look at the full error:
In [39]: marr.filled()
---------------------------------------------------------------------------
ValueError Traceback (most recent call last)
<ipython-input-39-219e016a84cf> in <module>
----> 1 marr.filled()
/usr/local/lib/python3.6/dist-packages/numpy/ma/core.py in filled(self, fill_value)
3718 result = self._data.copy('K')
3719 try:
-> 3720 np.copyto(result, fill_value, where=m)
3721 except (TypeError, AttributeError):
3722 fill_value = narray(fill_value, dtype=object)
ValueError: could not broadcast input array from shape (0) into shape (2,3)
np.copyto tries to broadcast result, fill_value and m (mask) against each other, and then copy the corresponding (mask==true) elements from fill_value to result.
marr.data and marr.mask are both (2,3). But broadcasting a (0,) shape to (2,3) doesn't work, and isn't what you want anyways.
Filling with a scalar works, but not with an array (or list).
In [56]: np.broadcast_to(np.array([]),(2,3))
...
ValueError: operands could not be broadcast together with remapped shapes [original->remapped]: (0,) and requested shape (2,3)
A (1,) shape array will broadcast -
In [57]: np.broadcast_to(np.array([1]),(2,3))
Out[57]:
array([[1, 1, 1],
[1, 1, 1]])
But the filled result is not an array; it's a scalar:
In [58]: marr.filled(np.array([1]))
Out[58]:
array([[1, array([0, 1, 2, 3, 4]), array([0, 1, 2])],
[1, array([0, 1]), 1]], dtype=object)
A fill that works
I can make this fill work if I define a (1,) object dtype array, and putting the (0,) array in it (as an object).
In [97]: Ofill = np.array([None], object)
In [98]: Ofill[0] = np.array([])
In [99]: Ofill
Out[99]: array([array([], dtype=float64)], dtype=object)
In [100]: marr.filled(Ofill)
Out[100]:
array([[array([], dtype=float64), array([0, 1, 2, 3, 4]),
array([0, 1, 2])],
[array([], dtype=float64), array([0, 1]),
array([], dtype=float64)]], dtype=object)
This works because Ofill can be broadcasted to (2,3) without messing with the shape of the element
In [101]: np.broadcast_to(Ofill,(2,3))
Out[101]:
array([[array([], dtype=float64), array([], dtype=float64),
array([], dtype=float64)],
[array([], dtype=float64), array([], dtype=float64),
array([], dtype=float64)]], dtype=object)
This works, but I wouldn't say it's pretty (or recommended).
Filling with None is prettier, but even then we have to make it a list:
In [103]: marr.filled([None])
Out[103]:
array([[None, array([0, 1, 2, 3, 4]), array([0, 1, 2])],
[None, array([0, 1]), None]], dtype=object)

The function "filled" has be supplied with value to be filled for masked part.
import numpy as np
arr = np.array([
[np.arange(10), np.arange(5), np.arange(3)],
[np.arange(1), np.arange(2), np.array([])],
])
marr = np.ma.array(arr)
marr.mask = [[True, False, False],
[True, False, True]]
marr.fill_value = np.array([])
marr.filled(2)
This version of code is not giving that error.