Given a tuple index, is there any short, fast code to see if the index lies "in-bounds" of an array shape?
Some solutions include:
For 2D: (0 <= index[0] < shape[0] and 0 <= index[1] < shape[1])
For any dimensions: all(0 <= i < n for i, n in zip(index, shape))
Also: (np.all(0 <= np.array(index)) and np.all(np.array(index) < shape))
And: ((0 <= np.array(index)) & (np.array(index) < shape)).all()
Solution (1) is fast but hardwired to 2 dimensions.
Although (2),(3),(4) are general, they are slow as they require either generator(s) or the construction of ndarray object(s).
Are there other ideas?
Perhaps there is already a library function (imaginary "np.inshape(index, shape)") to achieve this?
Related
I'm struggling a bit finding a fast algorithm that's suitable.
I just want to minimize:
norm2(x-s)
st
G.x <= h
x >= 0
sum(x) = R
G is sparse and contains only 1s (and zeros obviously).
In the case of iterative algorithms, it would be nice to get the interim solutions to show to the user.
The context is that s is a vector of current results, and the user is saying "well the sum of these few entries (entries indicated by a few 1.0's in a row in G) should be less than this value (a row in h). So we have to remove quantities from the entries the user specified (indicated by 1.0 entries in G) in a least-squares optimal way, but since we have a global constraint on the total (R) the values removed need to be allocated in a least-squares optimal way amongst the other entries. The entries can't go negative.
All the algorithms I'm looking at are much more general, and as a result are much more complex. Also, they seem quite slow. I don't see this as a complex problem, although mixes of equality and inequality constraints always seem to make things more complex.
This has to be called from Python, so I'm looking at Python libraries like qpsolvers and scipy.optimize. But I suppose Java or C++ libraries could be used and called from Python, which might be good since multithreading is better in Java and C++.
Any thoughts on what library/package/approach to use to best solve this problem?
The size of the problem is about 150,000 rows in s, and a few dozen rows in G.
Thanks!
Your problem is a linear least squares:
minimize_x norm2(x-s)
such that G x <= h
x >= 0
1^T x = R
Thus it fits the bill of the solve_ls function in qpsolvers.
Here is an instance of how I imagine your problem matrices would look like, given what you specified. Since it is sparse we should use SciPy CSC matrices, and regular NumPy arrays for vectors:
import numpy as np
import scipy.sparse as spa
n = 150_000
# minimize 1/2 || x - s ||^2
R = spa.eye(n, format="csc")
s = np.array(range(n), dtype=float)
# such that G * x <= h
G = spa.diags(
diagonals=[
[1.0 if i % 2 == 0 else 0.0 for i in range(n)],
[1.0 if i % 3 == 0 else 0.0 for i in range(n - 1)],
[1.0 if i % 5 == 0 else 0.0 for i in range(n - 1)],
],
offsets=[0, 1, -1],
)
a_dozen_rows = np.linspace(0, n - 1, 12, dtype=int)
G = G[a_dozen_rows]
h = np.ones(12)
# such that sum(x) == 42
A = spa.csc_matrix(np.ones((1, n)))
b = np.array([42.0]).reshape((1,))
# such that x >= 0
lb = np.zeros(n)
Next, we can solve this problem with:
from qpsolvers import solve_ls
x = solve_ls(R, s, G, h, A, b, lb, solver="osqp", verbose=True)
Here I picked CVXOPT but there are other open-source solvers you can install such as ProxQP, OSQP or SCS. You can install a set of open-source solvers by: pip install qpsolvers[open_source_solvers]. After some solvers are installed, you can list those for sparse matrices by:
print(qpsolvers.sparse_solvers)
Finally, here is some code to check that the solution returned by the solver satisfies our constraints:
tol = 1e-6 # tolerance for checks
print(f"- Objective: {0.5 * (x - s).dot(x - s):.1f}")
print(f"- G * x <= h: {(G.dot(x) <= h + tol).all()}")
print(f"- x >= 0: {(x + tol >= 0.0).all()}")
print(f"- sum(x) = {x.sum():.1f}")
I just tried it with OSQP (adding the eps_rel=1e-5 keyword argument when calling solve_ls, otherwise the returned solution would be less accurate than the tol = 1e-6 tolerance) and it found a solution is 737 milliseconds on my (rather old) CPU with:
- Objective: 562494373088866.8
- G * x <= h: True
- x >= 0: True
- sum(x) = 42.0
Hoping this helps. Happy solving!
I would like to vectorize a function with a condition, meaning to calculate its values with array arithmetic. np.vectorize handles vectorization, but it does not work with array arithmetic, so it is not a complete solution
An answer was given as the solution in the question "How to vectorize a function which contains an if statement?" but did not prevent errors here; see the MWE below.
import numpy as np
def myfx(x):
return np.where(x < 1.1, 1, np.arcsin(1 / x))
y = myfx(x)
This runs but raises the following warnings:
<stdin>:2: RuntimeWarning: divide by zero encountered in true_divide
<stdin>:2: RuntimeWarning: invalid value encountered in arcsin
What is the problem, or is there a better way to do this?
I think this could be done by
Getting the indices ks of x for which x[k] > 1.1 for each k in ks.
Applying np.arcsin(1 / x[ks]) to the slice x[ks], and using 1 for the rest of the elements.
Recombining the arrays.
I am not sure about the efficiency, though.
The statement np.where(x < 1.1, 1, np.arcsin(1 / x)) is equivalent to
mask = x < 1.1
a = 1
b = np.arcsin(1 / x)
np.where(mask, a, b)
Notice that you're calling np.arcsin on all the elements of x, regardless of whether 1 / x <= 1 or not. Your basic plan is correct. You can do the operations in-place on an output array using the where keyword of np.arcsin and np.reciprocal, without having to recombine anything:
def myfx(x):
mask = (x >= 1.1)
out = np.ones(x.shape)
np.reciprocal(x, where=mask, out=out) # >= 1.1 implies != 0
return np.arcsin(out, where=mask, out=out)
Using np.ones ensures that the unmasked elements of out are initialized correctly. An equivalent method would be
out = np.empty(x.shape)
out[~mask] = 1
You can always find an arithmetic expression that prevents the "divide by zero".
Example:
def myfx(x):
return np.where( x < 1.1, 1, np.arcsin(1/np.maximum(x, 1.1)) )
The values where x<1.1 in the right wing are not used, so it's not an issue computing np.arcsin(1/1.1) where x < 1.1.
The input vector is x, which contains 5 elements and I want to optimize (minimize) a function with a constraint that all elements of x should be greater than or equal to 0.
ie., x[i] >= 0 for 1 <= i <= 5
In one of the answer I saw this and used in my code, but the answer is returning negative values also
def constraint2(x):
"""constrain all elements of a to be >= 0"""
return x
cons2 = {'type': 'ineq', 'fun': constraint2}
Where am I going wrong? How to enforce the constraint?
all(ele >= 0 for ele in x)
Try this
I looking for an elegant way to select a subset of a torch tensor which satisfies some constrains.
For example, say I have:
A = torch.rand(10,2)-1
and S is a 10x1 tensor,
sel = torch.ge(S,5) -- this is a ByteTensor
I would like to be able to do logical indexing, as follows:
A1 = A[sel]
But that doesn't work.
So there's the index function which accepts a LongTensor but I could not find a simple way to convert S to a LongTensor, except the following:
sel = torch.nonzero(sel)
which returns a K x 2 tensor (K being the number of values of S >= 5). So then I have to convert it to a 1 dimensional array, which finally allows me to index A:
A:index(1,torch.squeeze(sel:select(2,1)))
This is very cumbersome; in e.g. Matlab all I'd have to do is
A(S>=5,:)
Can anyone suggest a better way?
One possible alternative is:
sel = S:ge(5):expandAs(A) -- now you can use this mask with the [] operator
A1 = A[sel]:unfold(1, 2, 2) -- unfold to get back a 2D tensor
Example:
> A = torch.rand(3,2)-1
-0.0047 -0.7976
-0.2653 -0.4582
-0.9713 -0.9660
[torch.DoubleTensor of size 3x2]
> S = torch.Tensor{{6}, {1}, {5}}
6
1
5
[torch.DoubleTensor of size 3x1]
> sel = S:ge(5):expandAs(A)
1 1
0 0
1 1
[torch.ByteTensor of size 3x2]
> A[sel]
-0.0047
-0.7976
-0.9713
-0.9660
[torch.DoubleTensor of size 4]
> A[sel]:unfold(1, 2, 2)
-0.0047 -0.7976
-0.9713 -0.9660
[torch.DoubleTensor of size 2x2]
There are two simpler alternatives:
Use maskedSelect:
result=A:maskedSelect(your_byte_tensor)
Use a simple element-wise multiplication, for example
result=torch.cmul(A,S:gt(0))
The second one is very useful if you need to keep the shape of the original matrix (i.e A), for example to select neurons in a layer at backprop. However, since it puts zeros in the resulting matrix whenever the condition dictated by the ByteTensor doesn't apply, you can't use it to compute the product (or median, etc.). The first one only returns the elements that satisfy the condittion, so this is what I'd use to compute products or medians or any other thing where I don't want zeros.
My inner loop contains a calculation that profiling shows to be problematic.
The idea is to take a greyscale pixel x (0 <= x <= 1), and "increase its contrast". My requirements are fairly loose, just the following:
for x < .5, 0 <= f(x) < x
for x > .5, x < f(x) <= 1
f(0) = 0
f(x) = 1 - f(1 - x), i.e. it should be "symmetric"
Preferably, the function should be smooth.
So the graph must look something like this:
.
I have two implementations (their results differ but both are conformant):
float cosContrastize(float i) {
return .5 - cos(x * pi) / 2;
}
float mulContrastize(float i) {
if (i < .5) return i * i * 2;
i = 1 - i;
return 1 - i * i * 2;
}
So I request either a microoptimization for one of these implementations, or an original, faster formula of your own.
Maybe one of you can even twiddle the bits ;)
Consider the following sigmoid-shaped functions (properly translated to the desired range):
error function
normal CDF
tanh
logit
I generated the above figure using MATLAB. If interested here's the code:
x = -3:.01:3;
plot( x, 2*(x>=0)-1, ...
x, erf(x), ...
x, tanh(x), ...
x, 2*normcdf(x)-1, ...
x, 2*(1 ./ (1 + exp(-x)))-1, ...
x, 2*((x-min(x))./range(x))-1 )
legend({'hard' 'erf' 'tanh' 'normcdf' 'logit' 'linear'})
Trivially you could simply threshold, but I imagine this is too dumb:
return i < 0.5 ? 0.0 : 1.0;
Since you mention 'increasing contrast' I assume the input values are luminance values. If so, and they are discrete (perhaps it's an 8-bit value), you could use a lookup table to do this quite quickly.
Your 'mulContrastize' looks reasonably quick. One optimization would be to use integer math. Let's say, again, your input values could actually be passed as an 8-bit unsigned value in [0..255]. (Again, possibly a fine assumption?) You could do something roughly like...
int mulContrastize(int i) {
if (i < 128) return (i * i) >> 7;
// The shift is really: * 2 / 256
i = 255 - i;
return 255 - ((i * i) >> 7);
A piecewise interpolation can be fast and flexible. It requires only a few decisions followed by a multiplication and addition, and can approximate any curve. It also avoids the courseness that can be introduced by lookup tables (or the additional cost in two lookups followed by an interpolation to smooth this out), though the lut might work perfectly fine for your case.
With just a few segments, you can get a pretty good match. Here there will be courseness in the color gradients, which will be much harder to detect than courseness in the absolute colors.
As Eamon Nerbonne points out in the comments, segmentation can be optimized by "choos[ing] your segmentation points based on something like the second derivative to maximize detail", that is, where the slope is changing the most. Clearly, in my posted example, having three segments in the middle of the five segment case doesn't add much more detail.