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Boolean indexing array through array of boolean indexes without loop

I want to index an array with a boolean mask through multiple boolean arrays without a loop.
This is what I want to achieve but without a loop and only with numpy.
import numpy as np
a = np.array([[0, 1],[2, 3]])
b = np.array([[[1, 0], [1, 0]], [[0, 0], [1, 1]]], dtype=bool)
r = []
for x in b:
print(a[x])
r.extend(a[x])
# => array([0, 2])
# => array([2, 3])
print(r)
# => [0, 2, 2, 3]
# what I would like to do is something like this
r = some_fancy_indexing_magic_with_b_and_a
print(r)
# => [0, 2, 2, 3]

Approach #1
Simply broadcast a to b's shape with np.broadcast_to and then mask it with b -
In [15]: np.broadcast_to(a,b.shape)[b]
Out[15]: array([0, 2, 2, 3])
Approach #2
Another would be getting all the indices and mod those by the size of a, which would also be the size of each 2D block in b and then indexing into flattened a -
a.ravel()[np.flatnonzero(b)%a.size]
Approach #3
On the same lines as App#2, but keeping the 2D format and using non-zero indices along the last two axes of b -
_,r,c = np.nonzero(b)
out = a[r,c]
Timings on large arrays (given sample shapes scaled up by 100x) -
In [50]: np.random.seed(0)
...: a = np.random.rand(200,200)
...: b = np.random.rand(200,200,200)>0.5
In [51]: %timeit np.broadcast_to(a,b.shape)[b]
45.5 ms ± 381 µs per loop (mean ± std. dev. of 7 runs, 10 loops each)
In [52]: %timeit a.ravel()[np.flatnonzero(b)%a.size]
94.6 ms ± 1.64 ms per loop (mean ± std. dev. of 7 runs, 10 loops each)
In [53]: %%timeit
...: _,r,c = np.nonzero(b)
...: out = a[r,c]
128 ms ± 1.46 ms per loop (mean ± std. dev. of 7 runs, 10 loops each)

How do I sort a 2D numpy array in this specific way

I realize there are quite a number of 'how to sort numpy array'-questions on here already. But I could not find how to do it in this specific way.
I have an array similar to this:
array([[1,0,1,],
[0,0,1],
[1,1,1],
[1,1,0]])
I want to sort the rows, keeping the order within the rows the same. So I expect the following output:
array([[0,0,1,],
[1,0,1],
[1,1,0],
[1,1,1]])

You can use dot and argsort:
a[a.dot(2**np.arange(a.shape[1])[::-1]).argsort()]
# array([[0, 0, 1],
# [1, 0, 1],
# [1, 1, 0],
# [1, 1, 1]])
The idea is to convert the rows into integers.
a.dot(2**np.arange(a.shape[1])[::-1])
# array([5, 1, 7, 6])
Then, find the sorted indices and use that to reorder a:
a.dot(2**np.arange(a.shape[1])[::-1]).argsort()
# array([1, 0, 3, 2])
My tests show this is slightly faster than lexsort.
a = a.repeat(1000, axis=0)
%timeit a[np.lexsort(a.T[::-1])]
%timeit a[a.dot(2**np.arange(a.shape[1])[::-1]).argsort()]
230 µs ± 18.1 µs per loop (mean ± std. dev. of 7 runs, 1000 loops each)
192 µs ± 4.1 µs per loop (mean ± std. dev. of 7 runs, 1000 loops each)
Verify correctness:
np.array_equal(a[a.dot(2**np.arange(a.shape[1])[::-1]).argsort()],
a[np.lexsort(a.T[::-1])])
# True

Python vectorization, how to get all index for every row with numpy

I'm having a hard time trying to solve this problem, the main issue is I'm running a simulation, so for lops are mainly forbidden, I have a numpy array NxN, in this case mine is about (10000x20).
stoploss = 19.9 # condition to apply
monte_carlo_simulation(20,1.08,10000,20) #which gives me that 10000x20 np array
mask_trues = np.where(np.any((simulation <= stoploss) == True, axis=1)) # boolean mask
I need some code to make a new vector of len(10000) which returns an array with all the positions for every row, lets suppose:
function([[False,True,True],[False,False,True]])
output = [[1,2],[2]]
Again, the main problem resides in not using loops.

Simply this:
list(map(np.where, my_array))
performance comparison against Kasrâmvd's solution:
def f(a):
return list(map(np.where, a))
def g(a):
x, y = np.where(a)
return np.split(y, np.where(np.diff(x) != 0)[0] + 1)
a = np.random.randint(2, size=(10000,20))
%timeit f(a)
%timeit g(a)
7.66 ms ± 38.1 µs per loop (mean ± std. dev. of 7 runs, 100 loops each)
13.3 ms ± 188 µs per loop (mean ± std. dev. of 7 runs, 100 loops each)

For completeness I'll demonstrate a sparse matrix approach:
In [57]: A = np.array([[False,True,True],[False,False,True]])
In [58]: A
Out[58]:
array([[False, True, True],
[False, False, True]])
In [59]: M = sparse.lil_matrix(A)
In [60]: M
Out[60]:
<2x3 sparse matrix of type '<class 'numpy.bool_'>'
with 3 stored elements in LInked List format>
In [61]: M.data
Out[61]: array([list([True, True]), list([True])], dtype=object)
In [62]: M.rows
Out[62]: array([list([1, 2]), list([2])], dtype=object)
And to make a large sparse one:
In [63]: BM = sparse.random(10000,20,.05, 'lil')
In [64]: BM
Out[64]:
<10000x20 sparse matrix of type '<class 'numpy.float64'>'
with 10000 stored elements in LInked List format>
In [65]: BM.rows
Out[65]:
array([list([3]), list([]), list([6, 15]), ..., list([]), list([11]),
list([])], dtype=object)
Rough time tests:
In [66]: arr = BM.A
In [67]: timeit sparse.lil_matrix(arr)
19.5 ms ± 421 µs per loop (mean ± std. dev. of 7 runs, 100 loops each)
In [68]: timeit list(map(np.where,arr))
11 ms ± 55.6 µs per loop (mean ± std. dev. of 7 runs, 100 loops each)
In [69]: %%timeit
...: x,y = np.where(arr)
...: np.split(y, np.where(np.diff(x) != 0)[0] + 1)
...:
13.8 ms ± 24.3 µs per loop (mean ± std. dev. of 7 runs, 100 loops each)
Generating a csr sparse format matrix is faster:
In [70]: timeit sparse.csr_matrix(arr)
2.68 ms ± 120 µs per loop (mean ± std. dev. of 7 runs, 100 loops each)
In [71]: Mr = sparse.csr_matrix(arr)
In [72]: Mr.indices
Out[72]: array([ 3, 6, 15, ..., 8, 16, 11], dtype=int32)
In [73]: Mr.indptr
Out[73]: array([ 0, 1, 1, ..., 9999, 10000, 10000], dtype=int32)
In [74]: np.where(arr)[1]
Out[74]: array([ 3, 6, 15, ..., 8, 16, 11])
It's indices is just like the column where, while the indptr is like the split indices.

Here is one way using np.split() and np.diff():
x, y = np.where(boolean_array)
np.split(y, np.where(np.diff(x) != 0)[0] + 1)
Demo:
In [12]: a = np.array([[False,True,True],[False,False,True]])
In [13]: x, y = np.where(a)
In [14]: np.split(y, np.where(np.diff(x) != 0)[0] + 1)
Out[14]: [array([1, 2]), array([2])]

Numpy: Can you use broadcasting to replace values by row?

I have a M x N matrix X and a 1 x N matrix Y. What I would like to do is replace any 0-entry in X with the appropriate value from Y based on its column.
So if
X = np.array([[0, 1, 2], [3, 0, 5]])
and
Y = np.array([10, 20, 30])
The desired end result would be [[10, 1, 2], [3, 20, 5]].
This can be done straightforwardly by generating a M x N matrix where every row is Y and then using filter arrays:
Y = np.ones((X.shape[0], 1)) * Y.reshape(1, -1)
X[X==0] = Y[X==0]
But could this be done using numpy's broadcasting functionality?

Sure. Instead of physically repeating Y, create a broadcasted view of Y with the shape of X, using numpy.broadcast_to:
expanded = numpy.broadcast_to(Y, X.shape)
mask = X==0
x[mask] = expanded[mask]

Expand X to make it a bit more general:
In [306]: X = np.array([[0, 1, 2], [3, 0, 5],[0,1,0]])
where identifies the 0s; the 2nd array identifies the columns
In [307]: idx = np.where(X==0)
In [308]: idx
Out[308]: (array([0, 1, 2, 2]), array([0, 1, 0, 2]))
In [309]: Z = X.copy()
In [310]: Z[idx]
Out[310]: array([0, 0, 0, 0]) # flat list of where to put the values
In [311]: Y[idx[1]]
Out[311]: array([10, 20, 10, 30]) # matching list of values by column
In [312]: Z[idx] = Y[idx[1]]
In [313]: Z
Out[313]:
array([[10, 1, 2],
[ 3, 20, 5],
[10, 1, 30]])
Not doing broadcasting, but reasonably clean numpy.
Times compared to broadcast_to approach
In [314]: %%timeit
...: idx = np.where(X==0)
...: Z[idx] = Y[idx[1]]
...:
9.28 µs ± 157 ns per loop (mean ± std. dev. of 7 runs, 100000 loops each)
In [315]: %%timeit
...: exp = np.broadcast_to(Y,X.shape)
...: mask=X==0
...: Z[mask] = exp[mask]
...:
19.5 µs ± 513 ns per loop (mean ± std. dev. of 7 runs, 10000 loops each)
Faster, though the sample size is small.
Another way to make the expanded Y, is with repeat:
In [319]: %%timeit
...: exp = np.repeat(Y[None,:],3,0)
...: mask=X==0
...: Z[mask] = exp[mask]
...:
10.8 µs ± 55.3 ns per loop (mean ± std. dev. of 7 runs, 100000 loops each)
Whose time is close to my where. It turns out that broadcast_to is relatively slow:
In [321]: %%timeit
...: exp = np.broadcast_to(Y,X.shape)
...:
10.5 µs ± 52.9 ns per loop (mean ± std. dev. of 7 runs, 100000 loops each)
In [322]: %%timeit
...: exp = np.repeat(Y[None,:],3,0)
...:
3.76 µs ± 11.6 ns per loop (mean ± std. dev. of 7 runs, 100000 loops each)
We'd have to do more tests to see whether that is just due to a setup cost, or if the relative times still apply with much larger arrays.

Index a NumPy array row-wise [duplicate]

This question already has answers here:
Indexing one array by another in numpy
(4 answers)
Closed 4 years ago.
Say I have a NumPy array:
>>> X = np.array([[1, 2, 3, 4], [5, 6, 7, 8], [9, 10, 11, 12]])
>>> X
array([[ 1, 2, 3, 4],
[ 5, 6, 7, 8],
[ 9, 10, 11, 12]])
and an array of indexes that I want to select for each row:
>>> ixs = np.array([[1, 3], [0, 1], [1, 2]])
>>> ixs
array([[1, 3],
[0, 1],
[1, 2]])
How do I index the array X so that for every row in X I select the two indices specified in ixs?
So for this case, I want to select element 1 and 3 for the first row, element 0 and 1 for the second row, and so on. The output should be:
array([[2, 4],
[5, 6],
[10, 11]])
A slow solution would be something like this:
output = np.array([row[ix] for row, ix in zip(X, ixs)])
however this can get kinda slow for extremely long arrays. Is there a faster way to do this without a loop using NumPy?
EDIT: Some very approximate speed tests on a 2.5K * 1M array with 2K wide ixs (10GB):
np.array([row[ix] for row, ix in zip(X, ixs)]) 0.16s
X[np.arange(len(ixs)), ixs.T].T 0.175s
X.take(idx+np.arange(0, X.shape[0]*X.shape[1], X.shape[1])[:,None]) 33s
np.fromiter((X[i, j] for i, row in enumerate(ixs) for j in row), dtype=X.dtype).reshape(ixs.shape) 2.4s

You can use this:
X[np.arange(len(ixs)), ixs.T].T
Here is the reference for complex indexing.

I believe you can use .take thusly:
In [185]: X
Out[185]:
array([[ 1, 2, 3, 4],
[ 5, 6, 7, 8],
[ 9, 10, 11, 12]])
In [186]: idx
Out[186]:
array([[1, 3],
[0, 1],
[1, 2]])
In [187]: X.take(idx + (np.arange(X.shape[0]) * X.shape[1]).reshape(-1, 1))
Out[187]:
array([[ 2, 4],
[ 5, 6],
[10, 11]])
If your array dimensions are massive, it might be faster, albeit uglier, to do:
idx+np.arange(0, X.shape[0]*X.shape[1], X.shape[1])[:,None]
Just for fun, see how the following performs:
np.fromiter((X[i, j] for i, row in enumerate(ixs) for j in row), dtype=X.dtype, count=ixs.size).reshape(ixs.shape)
Edit to add timings
In [15]: X = np.arange(1000*10000, dtype=np.int32).reshape(1000,-1)
In [16]: ixs = np.random.randint(0, 10000, (1000, 2))
In [17]: ixs.sort(axis=1)
In [18]: ixs
Out[18]:
array([[2738, 3511],
[3600, 7414],
[7426, 9851],
...,
[1654, 8252],
[2194, 8200],
[5497, 8900]])
In [19]: %timeit np.array([row[ix] for row, ix in zip(X, ixs)])
928 µs ± 23.8 µs per loop (mean ± std. dev. of 7 runs, 1000 loops each)
In [20]: %timeit X[np.arange(len(ixs)), ixs.T].T
23.6 µs ± 491 ns per loop (mean ± std. dev. of 7 runs, 10000 loops each)
In [21]: %timeit X.take(idx+np.arange(0, X.shape[0]*X.shape[1], X.shape[1])[:,None])
20.6 µs ± 530 ns per loop (mean ± std. dev. of 7 runs, 10000 loops each)
In [22]: %timeit np.fromiter((X[i, j] for i, row in enumerate(ixs) for j in row), dtype=X.dtype, count=ixs.size).reshape(ixs.shape)
1.42 ms ± 9.94 µs per loop (mean ± std. dev. of 7 runs, 1000 loops each)
#mxbi I've added some timings and my results aren't really consistent with yours, you should check it out
Here's a larger array:
In [33]: X = np.arange(10000*100000, dtype=np.int32).reshape(10000,-1)
In [34]: ixs = np.random.randint(0, 100000, (10000, 2))
In [35]: ixs.sort(axis=1)
In [36]: X.shape
Out[36]: (10000, 100000)
In [37]: ixs.shape
Out[37]: (10000, 2)
With some results:
In [42]: %timeit np.array([row[ix] for row, ix in zip(X, ixs)])
11.4 ms ± 177 µs per loop (mean ± std. dev. of 7 runs, 100 loops each)
In [43]: %timeit X[np.arange(len(ixs)), ixs.T].T
596 µs ± 17.8 µs per loop (mean ± std. dev. of 7 runs, 1000 loops each)
In [44]: %timeit X.take(ixs+np.arange(0, X.shape[0]*X.shape[1], X.shape[1])[:,None])
540 µs ± 16.6 µs per loop (mean ± std. dev. of 7 runs, 1000 loops each)
Now, we are using column 500 indices instead of two, and we see the list-comprehension start winning out:
In [45]: ixs = np.random.randint(0, 100000, (10000, 500))
In [46]: ixs.sort(axis=1)
In [47]: %timeit np.array([row[ix] for row, ix in zip(X, ixs)])
93 ms ± 1.2 ms per loop (mean ± std. dev. of 7 runs, 1 loop each)
In [48]: %timeit X[np.arange(len(ixs)), ixs.T].T
133 ms ± 638 µs per loop (mean ± std. dev. of 7 runs, 10 loops each)
In [49]: %timeit X.take(ixs+np.arange(0, X.shape[0]*X.shape[1], X.shape[1])[:,None])
87.5 ms ± 1.13 ms per loop (mean ± std. dev. of 7 runs, 10 loops each)

The usual suggestion for indexing items from rows is:
X[np.arange(X.shape[0])[:,None], ixs]
That is, make a row index of shape (n,1) (column vector), which will broadcast with the (n,m) shape of ixs to give a (n,m) solution.
This basically the same as:
X[np.arange(len(ixs)), ixs.T].T
which broadcasts a (n,) index against a (m,n), and transposes.
Timings are essentially the same:
In [299]: X = np.ones((1000,2000))
In [300]: ixs = np.random.randint(0,2000,(1000,200))
In [301]: timeit X[np.arange(len(ixs)), ixs.T].T
6.58 ms ± 71.6 µs per loop (mean ± std. dev. of 7 runs, 100 loops each)
In [302]: timeit X[np.arange(X.shape[0])[:,None], ixs]
6.57 ms ± 129 µs per loop (mean ± std. dev. of 7 runs, 100 loops each)
and for comparison:
In [307]: timeit np.array([row[ix] for row, ix in zip(X, ixs)])
6.63 ms ± 229 µs per loop (mean ± std. dev. of 7 runs, 100 loops each)
I'm a little surprised that this list comprehension does so well. I wonder how the relative advantages compare when the dimensions change, particularly in the relative shape of X and ixs (long, wide etc).
The first solution is the style of indexing produced by ix_:
In [303]: np.ix_(np.arange(3), np.arange(2))
Out[303]:
(array([[0],
[1],
[2]]), array([[0, 1]]))

This should work
[X[i][[y]] for i, y in enumerate(ixs)]
EDIT: I just noticed you wanted no loop solution.