Develop Reference

Roll first column by 1, second column by 2, etc

I have an array in numpy. I want to roll the first column by 1, second column by 2, etc.
Here is an example.
>>> x = np.reshape(np.arange(15), (5, 3))
>>> x
array([[ 0, 1, 2],
[ 3, 4, 5],
[ 6, 7, 8],
[ 9, 10, 11],
[12, 13, 14]])
What I want to do:
>>> y = roll(x)
>>> y
array([[12, 10, 8],
[ 0, 13, 11],
[ 3, 1, 14],
[ 6, 4, 2],
[ 9, 7, 5]])
What is the best way to do it?
The real array will be very big. I'm using cupy, the GPU version of numpy. I will prefer solution fastest on GPU, but of course, any idea is welcomed.

You could use advanced indexing:
import numpy as np
x = np.reshape(np.arange(15), (5, 3))
h, w = x.shape
rows, cols = np.arange(h), np.arange(w)
offsets = cols + 1
shifted = np.subtract.outer(rows, offsets) % h
y = x[shifted, cols]
y:
array([[12, 10, 8],
[ 0, 13, 11],
[ 3, 1, 14],
[ 6, 4, 2],
[ 9, 7, 5]])

I implemented a naive solution (roll_for) and compares it to #Chrysophylaxs 's solution (roll_indexing).
Conclusion: roll_indexing is faster for small arrays, but the difference shrinks when the array goes bigger, and is eventually slower than roll_for for very large arrays.
Implementations:
import numpy as np
def roll_for(x, shifts=None, axis=-1):
if shifts is None:
shifts = np.arange(1, x.shape[axis] + 1) # OP requirement
xt = x.swapaxes(axis, 0) # https://stackoverflow.com/a/31094758/13636407
yt = np.empty_like(xt)
for idx, shift in enumerate(shifts):
yt[idx] = np.roll(xt[idx], shift=shift)
return yt.swapaxes(0, axis)
def roll_indexing(x):
h, w = x.shape
rows, cols = np.arange(h), np.arange(w)
offsets = cols + 1
shifted = np.subtract.outer(rows, offsets) % h # fix
return x[shifted, cols]
Tests:
M, N = 5, 3
x = np.arange(M * N).reshape(M, N)
expected = np.array([[12, 10, 8], [0, 13, 11], [3, 1, 14], [6, 4, 2], [9, 7, 5]])
assert np.array_equal(expected, roll_for(x))
assert np.array_equal(expected, roll_indexing(x))
M, N = 100, 200
# roll_indexing did'nt work when M < N before fix
x = np.arange(M * N).reshape(M, N)
assert np.array_equal(roll_for(x), roll_indexing(x))
Benchmark:
M, N = 100, 100
x = np.arange(M * N).reshape(M, N)
assert np.array_equal(roll_for(x), roll_indexing(x))
%timeit roll_for(x) # 859 µs ± 2.8 µs per loop (mean ± std. dev. of 7 runs, 1,000 loops each)
%timeit roll_indexing(x) # 81 µs ± 255 ns per loop (mean ± std. dev. of 7 runs, 10,000 loops each)
M, N = 1_000, 1_000
x = np.arange(M * N).reshape(M, N)
assert np.array_equal(roll_for(x), roll_indexing(x))
%timeit roll_for(x) # 12.7 ms ± 56.3 µs per loop (mean ± std. dev. of 7 runs, 100 loops each)
%timeit roll_indexing(x) # 12.4 ms ± 13.2 µs per loop (mean ± std. dev. of 7 runs, 100 loops each)
M, N = 10_000, 10_000
x = np.arange(M * N).reshape(M, N)
assert np.array_equal(roll_for(x), roll_indexing(x))
%timeit roll_for(x) # 1.3 s ± 6.46 ms per loop (mean ± std. dev. of 7 runs, 1 loop each)
%timeit roll_indexing(x) # 1.61 s ± 4.96 ms per loop (mean ± std. dev. of 7 runs, 1 loop each)

Converting multidimensional array into arrays in list for every row

I have a data set like this,
data = np.array([[ 5, 31, 61],
[ 10, 31, 67],
[ 15, 31, 69],
[ 4, 31, 72],
[ 14, 31, 73],
[ 21, 31, 77],
[ 19, 31, 78]])
I want to convert it into arrays in list for every single row. I tried,
np.split(data,len(data))
#[array([[ 5, 31, 61]]),
# array([[10, 31, 67]]),
# array([[15, 31, 69]]),
# array([[ 4, 31, 72]]),
# array([[14, 31, 73]]),
# array([[21, 31, 77]]),
# array([[19, 31, 78]])]
But as you can see it gives double [ to me. What I simply want is;
[np.array([5, 31, 61]),
np.array([10, 31, 67]),
np.array([15, 31, 69]),
np.array([4, 31, 72]),
np.array([14, 31, 73]),
np.array([21, 31, 77]),
np.array([19, 31, 78])]

What about taking advantage of unpacking?
lst = [*data]
or:
lst = list(data)
output:
[array([ 5, 31, 61]),
array([10, 31, 67]),
array([15, 31, 69]),
array([ 4, 31, 72]),
array([14, 31, 73]),
array([21, 31, 77]),
array([19, 31, 78])]

np.split could be applied too but you are required to do it one-dimensionally. So you might like to create a one-dimensional view of your data first:
%%timeit
data_ravel = data.ravel()
out = np.split(data_ravel, len(data))
>>> 14.5 µs ± 337 ns per loop (mean ± std. dev. of 7 runs, 100000 loops each)
Note that creating a view is costless (it took 0.13 µs on my computer)
Internally, it's being done like so:
out = []
div_points = range(0, data.size+1, data.shape[1])
start = div_points[:-1]
end = div_points[1:]
out = list(data_ravel[i:j] for i,j in zip(start, end))
>>> 2.31 µs ± 44.6 ns per loop (mean ± std. dev. of 7 runs, 100000 loops each)
Note that it's faster because I'd doing a couple of optimisations here:
using range instead of np.array
using lazy list comprehension instead of list.append
However, it can't compete with classical methods like in #mozway 's answer. They are optimal:
%%timeit
out = [*data]
>>> 902 ns ± 8.09 ns per loop (mean ± std. dev. of 7 runs, 1000000 loops each)
%%timeit
out = list(data)
>>> 979 ns ± 12.1 ns per loop (mean ± std. dev. of 7 runs, 1000000 loops each)
%%timeit
out = [n for n in data]
>>> 1.04 µs ± 14.3 ns per loop (mean ± std. dev. of 7 runs, 1000000 loops each)
%%timeit
out = list(n for n in data)
>>> 1.37 µs ± 80.7 ns per loop (mean ± std. dev. of 7 runs, 1000000 loops each)

Creating a tumbling windows in python

Just wondering if there is a way to construct a tumbling window in python. So for example if I have list/ndarray , listA = [3,2,5,9,4,6,3,8,7,9]. Then how could I find the maximum of the first 3 items (3,2,5) -> 5, and then the next 3 items (9,4,6) -> 9 and so on... Sort of like breaking it up to sections and finding the max. So the final result would be list [5,9,8,9]

Approach #1: One-liner for windowed-max using np.maximum.reduceat -
In [118]: np.maximum.reduceat(listA,np.arange(0,len(listA),3))
Out[118]: array([5, 9, 8, 9])
Becomes more compact with np.r_ -
np.maximum.reduceat(listA,np.r_[:len(listA):3])
Approach #2: Generic ufunc way
Here's a function for generic ufuncs and that window length as a parameter -
def windowed_ufunc(a, ufunc, W):
a = np.asarray(a)
n = len(a)
L = W*(n//W)
out = ufunc(a[:L].reshape(-1,W),axis=1)
if n>L:
out = np.hstack((out, ufunc(a[L:])))
return out
Sample run -
In [81]: a = [3,2,5,9,4,6,3,8,7,9]
In [82]: windowed_ufunc(a, ufunc=np.max, W=3)
Out[82]: array([5, 9, 8, 9])
On other ufuncs -
In [83]: windowed_ufunc(a, ufunc=np.min, W=3)
Out[83]: array([2, 4, 3, 9])
In [84]: windowed_ufunc(a, ufunc=np.sum, W=3)
Out[84]: array([10, 19, 18, 9])
In [85]: windowed_ufunc(a, ufunc=np.mean, W=3)
Out[85]: array([3.33333333, 6.33333333, 6. , 9. ])
Benchmarking
Timings on NumPy solutions on array data with sample data scaled up by 10000x -
In [159]: a = [3,2,5,9,4,6,3,8,7,9]
In [160]: a = np.tile(a, 10000)
# #yatu's soln
In [162]: %timeit moving_maxima(a, w=3)
435 µs ± 8.54 µs per loop (mean ± std. dev. of 7 runs, 1000 loops each)
# From this post - app#1
In [167]: %timeit np.maximum.reduceat(a,np.arange(0,len(a),3))
353 µs ± 2.55 µs per loop (mean ± std. dev. of 7 runs, 1000 loops each)
# From this post - app#2
In [165]: %timeit windowed_ufunc(a, ufunc=np.max, W=3)
379 µs ± 6.44 µs per loop (mean ± std. dev. of 7 runs, 1000 loops each)

If you want a one-liner, you can use list comprehension:
listA = [3,2,5,9,4,6,3,8,7,9]
listB=[max(listA[i:i+3]) for i in range(0,len(listA),3)]
print (listB)
it returns:
[5, 9, 8, 9]
Of course the codes can be written more dynamically: if you want a different window size, just change 3 to any integer.

Using numpy, you can extend the list with zeroes so its length is divisible by the window size, and reshape and compute the maxalong the second axis:
def moving_maxima(a, w):
mod = len(a)%w
d = w if mod else mod
x = np.r_[a, [0]*(d-mod)]
return x.reshape(-1,w).max(1)
Some examples:
moving_maxima(listA,2)
# array([3., 9., 6., 8., 9.])
moving_maxima(listA,3)
#array([5, 9, 8, 9])
moving_maxima(listA,4)
#array([9, 8, 9])

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.

Summarize 2darray by index

I want to sum columns of a 2d array dat by row index idx. The following example works but is slow for large arrays. Any idea to speed it up?
import numpy as np
dat = np.arange(18).reshape(6, 3, order = 'F')
idx = np.array([0, 1, 1, 1, 2, 2])
for i in np.unique(idx):
print(np.sum(dat[idx==i], axis = 0))
Output
[ 0 6 12]
[ 6 24 42]
[ 9 21 33]

Approach #1
We can leverage matrix-multiplication with np.dot -
In [56]: mask = idx[:,None] == np.unique(idx)
In [57]: mask.T.dot(dat)
Out[57]:
array([[ 0, 6, 12],
[ 6, 24, 42],
[ 9, 21, 33]])
Approach #2
For the case with idx already sorted, we can use np.add.reduceat -
In [52]: p = np.flatnonzero(np.r_[True,idx[:-1] != idx[1:]])
In [53]: np.add.reduceat(dat, p, axis=0)
Out[53]:
array([[ 0, 6, 12],
[ 6, 24, 42],
[ 9, 21, 33]])

A bit faster approach with set object and ndarray.sum() method:
In [216]: for i in set(idx):
...: print(dat[idx == i].sum(axis=0))
...:
[ 0 6 12]
[ 6 24 42]
[ 9 21 33]
Time execution comparison:
In [217]: %timeit for i in np.unique(idx): r = np.sum(dat[idx==i], axis = 0)
109 µs ± 1.1 µs per loop (mean ± std. dev. of 7 runs, 10000 loops each)
In [218]: %timeit for i in set(idx): r = dat[idx == i].sum(axis=0)
71.1 µs ± 1.98 µs per loop (mean ± std. dev. of 7 runs, 10000 loops each)