I'm running functions to create cyclical datetime features, so I have converted timestamps to sine and cosine representations for ML model training.
In one sample, x = 305.2116709309027, giving np.sin(x) = -0.459279 and np.cos(x) = -0.888292, my question is how to retrieve x from these sin and cos features later?
I assumed np.arcsin(-0.459279) == 305.2116709309027 and I could then decode the timestamp used from there but I'm not having any luck.
You should be aware that mathematically, sin(x) and cos(x) are periodic functions, meaning multiple different values as input can yield the same output.
For example, x=0, x=2pi, and x=4pi can all yield the same value. So you can't decode the x from y, except you know that the input is restricted within a period, such as between [0, 2pi].
HOWEVER, for arcsin(x), since the domain of x is limited, and each y corresponds to a unique x, you can get the x from y.
Related
I'm fairly new to python so bare with me. I have plotted a histogram using some generated data. This data has many many points. I have defined it with the variable vals. I have then plotted a histogram with these values, though I have limited it so that only values between 104 and 155 are taken into account. This has been done as follows:
bin_heights, bin_edges = np.histogram(vals, range=[104, 155], bins=30)
bin_centres = (bin_edges[:-1] + bin_edges[1:])/2.
plt.errorbar(bin_centres, bin_heights, np.sqrt(bin_heights), fmt=',', capsize=2)
plt.xlabel("$m_{\gamma\gamma} (GeV)$")
plt.ylabel("Number of entries")
plt.show()
Giving the above plot:
My next step is to take into account values from vals which are less than 120. I have done this as follows:
background_data=[j for j in vals if j <= 120] #to avoid taking the signal bump, upper limit of 120 MeV set
I need to plot a curve on the same plot as the histogram, which follows the form B(x) = Ae^(-x/λ)
I then estimated a value of λ using the maximum likelihood estimator formula:
background_data=[j for j in vals if j <= 120] #to avoid taking the signal bump, upper limit of 120 MeV set
#print(background_data)
N_background=len(background_data)
print(N_background)
sigma_background_data=sum(background_data)
print(sigma_background_data)
lamb = (sigma_background_data)/(N_background) #maximum likelihood estimator for lambda
print('lambda estimate is', lamb)
where lamb = λ. I got a value of roughly lamb = 27.75, which I know is correct. I now need to get an estimate for A.
I have been advised to do this as follows:
Given a value of λ, find A by scaling the PDF to the data such that the area beneath
the scaled PDF has equal area to the data
I'm not quite sure what this means, or how I'd go about trying to do this. PDF means probability density function. I assume an integration will have to take place, so to get the area under the data (vals), I have done this:
data_area= integrate.cumtrapz(background_data, x=None, dx=1.0)
print(data_area)
plt.plot(background_data, data_area)
However, this gives me an error
ValueError: x and y must have same first dimension, but have shapes (981555,) and (981554,)
I'm not sure how to fix it. The end result should be something like:
See the cumtrapz docs:
Returns: ... If initial is None, the shape is such that the axis of integration has one less value than y. If initial is given, the shape is equal to that of y.
So you are either to pass an initial value like
data_area = integrate.cumtrapz(background_data, x=None, dx=1.0, initial = 0.0)
or discard the first value of the background_data:
plt.plot(background_data[1:], data_area)
Suppose there are three variables that take on discrete integer values, say w1 = {1,2,3,4,5,6,7,8,9,10,11,12}, w2 = {1,2,3,4,5,6,7,8,9,10,11,12}, and w3 = {1,2,3,4,5,6,7,8,9,10,11,12}. The task is to pick one value from each set such that the resulting triplet minimizes some (black box, computationally expensive) cost function.
I've tried the surrogate optimization in Matlab but I'm not sure it is appropriate. I've also heard about simulated annealing but found no implementation applied to this instance.
Which algorithm, apart from exhaustive search, can solve this combinatorial optimization problem?
Any help would be much appreciated.
The requirement/benefit of Simulated Annealing (SA), is that the objective surface is somewhat smooth, that is, we can be close to a solution.
For a completely random spiky surface- you might as well do a random search
If it is anything smooth, or even sometimes, it makes sense to try SA.
The idea is that (sometimes) changing only 1 of the 3 values, we have little effect on out blackbox function.
Here is a basic example to do this with Simulated Annealing, using frigidum in Python
import numpy as np
w1 = np.array( [1,2,3,4,5,6,7,8,9,10,11,12] )
w2 = np.array( [1,2,3,4,5,6,7,8,9,10,11,12] )
w3 = np.array( [1,2,3,4,5,6,7,8,9,10,11,12] )
W = np.array([w1,w2,w3])
LENGTH = 12
I define a black-box using the Rastrigin function.
def rastrigin_function_n( x ):
"""
N-dimensional Rastrigin
https://en.wikipedia.org/wiki/Rastrigin_function
x_i is in [-5.12, 5.12]
"""
A = 10
n = x.shape[0]
return A*n + np.sum( x**2- A*np.cos(2*np.pi * x) )
def black_box( x ):
"""
Transform from domain [1,12] to [-5,5]
to be able to push to rastrigin
"""
x = (x - 6.5) * (5/5.5)
return rastrigin_function_n(x)
Simulated Annealing needs to modify state X. Instead of taking/modifying values directly, we keep track of indices. This simplifies creating new proposals as an index is always an integer we can simply add/subtract 1 modulo LENGTH.
def random_start():
"""
returns 3 random indices
"""
return np.random.randint(0, LENGTH, size=3)
def random_small_step(x):
"""
change only 1 index
"""
d = np.array( [1,0,0] )
if np.random.random() < .5:
d = np.array( [-1,0,0] )
np.random.shuffle(d)
return (x+d) % LENGTH
def random_big_step(x):
"""
change 2 indici
"""
d = np.array( [1,-1,0] )
np.random.shuffle(d)
return (x+d) % LENGTH
def obj(x):
"""
We have a triplet of indici,
1. Calculate corresponding values in W = [w1,w2,w3]
2. Push the values in out black-box function
"""
indices = x
values = W[np.array([0,1,2]), indices]
return black_box(values)
And throw a SA Scheme at it
import frigidum
local_opt = frigidum.sa(random_start=random_start,
neighbours=[random_small_step, random_big_step],
objective_function=obj,
T_start=10**4,
T_stop=0.000001,
repeats=10**3,
copy_state=frigidum.annealing.naked)
I am not sure what the minimum for this function should be, but it found a objective with 47.9095 with indicis np.array([9, 2, 2])
Edit:
For frigidum to change the cooling schedule, use alpha=.9. My experience is that all the work of experiment which cooling scheme works best doesn't out-weight simply let it run a little longer. The multiplication you proposed, (sometimes called geometric) is the standard one, also implemented in frigidum. So to implement Tn+1 = 0.9*Tn you need a alpha=.9. Be aware this cooling step is done after N repeats, so if repeats=100, it will first do 100 proposals before lowering the temperature with factor alpha
Simple variations on current state often works best. Since its best practice to set the initial temperature high enough to make most proposals (>90%) accepted, it doesn't matter the steps are small. But if you fear its soo small, try 2 or 3 variations. Frigidum accepts a list of proposal functions, and combinations can enforce each other.
I have no experience with MINLP. But even if, so many times experiments can surprise us. So if time/cost is small to bring another competitor to the table, yes!
Try every possible combination of the three values and see which has the lowest cost.
I have curve that initially Y increases linearly with X, then reach a plateau at point C.
In other words, the curve can be defined as:
if X < C:
Y = k * X + b
else:
Y = k * C + b
The training data is a list of X ~ Y values. I need to determine k, b and C through a machine learning approach (or similar), since the data is noisy and refection point C changes over time. I want something more robust than get C through observing the current sample data.
How can I do it using sklearn or maybe scipy?
WLOG you can say the second equation is
Y = C
looks like you have a linear regression to fit the line and then a detection point to find the constant.
You know that in the high values of X, as in X > C you are already at the constant. So just check how far back down the values of X you get the same constant.
Then do a linear regression to find the line with value of X, X <= C
Your model is nonlinear
I think the smartest way to solve this is to do these steps:
find the maximum value of Y which is equal to k*C+b
M=max(Y)
drop this maximum value from your dataset
df1 = df[df.Y != M]
and then you have simple dataset to fit your X to Y and you can use sklearn for that
I was debugging my program and I've realized that I my loss outputted NaN. These NaN values comes from the fact that I'm computing tf.log(1 + tf.exp(X))
where X is a 2d tensor. Indeed, When a value of X is large enough then tf.exp() returns +Inf and so tf.log(1 + exp(X)) will return +Inf. I was wondering if there exists a neat trick to avoid underflows and overflows in this case.
I have tried:
def log1exp(x):
maxi = tf.reduce_max(x)
return maxi + tf.log(tf.exp(x - maxi) + tf.exp(-maxi))
but it doesn't handle underflows in this case...
Also I've glanced at tf.reduce_logsumexp but it necessarily reduce the tensor along an axis... while I want to keep the same shape!
Finally I know that tf.log(1 + exp(X)) is almost equal to X for large values of X but I think that designing a function that will output X when X > threshold and log(1+exp(X)) otherwise is not very neat.
Thank you
This function is already implemented in tensorflow under the name tf.math.softplus, and takes care of overflows and underflows.
I've created a codebook using k-means of size 4000x300 (4000 centroids, each with 300 features). Using the codebook, I then want to label an input vector (for purposes of binning later on). The input vector is of size Nx300, where N is the total number of input instances I receive.
To compute the labels, I calculate the closest centroid for each of the input vectors. To do so, I compare each input vector against all centroids and pick the centroid with the minimum distance. The label is then just the index of that centroid.
My current Matlab code looks like:
function labels = assign_labels(centroids, X)
labels = zeros(size(X, 1), 1);
% for each X, calculate the distance from each centroid
for i = 1:size(X, 1)
% distance of X_i from all j centroids is: sum((X_i - centroid_j)^2)
% note: we leave off the sqrt as an optimization
distances = sum(bsxfun(#minus, centroids, X(i, :)) .^ 2, 2);
[value, label] = min(distances);
labels(i) = label;
end
However, this code is still fairly slow (for my purposes), and I was hoping there might be a way to optimize the code further.
One obvious issue is that there is a for-loop, which is the bane of good performance on Matlab. I've been trying to come up with a way to get rid of it, but with no luck (I looked into using arrayfun in conjunction with bsxfun, but haven't gotten that to work). Alternatively, if someone know of any other way to speed this up, I would be greatly appreciate it.
Update
After doing some searching, I couldn't find a great solution using Matlab, so I decided to look at what is used in Python's scikits.learn package for 'euclidean_distance' (shortened):
XX = sum(X * X, axis=1)[:, newaxis]
YY = Y.copy()
YY **= 2
YY = sum(YY, axis=1)[newaxis, :]
distances = XX + YY
distances -= 2 * dot(X, Y.T)
distances = maximum(distances, 0)
which uses the binomial form of the euclidean distance ((x-y)^2 -> x^2 + y^2 - 2xy), which from what I've read usually runs faster. My completely untested Matlab translation is:
XX = sum(data .* data, 2);
YY = sum(center .^ 2, 2);
[val, ~] = max(XX + YY - 2*data*center');
Use the following function to calculate your distances. You should see an order of magnitude speed up
The two matrices A and B have the columns as the dimenions and the rows as each point.
A is your matrix of centroids. B is your matrix of datapoints.
function D=getSim(A,B)
Qa=repmat(dot(A,A,2),1,size(B,1));
Qb=repmat(dot(B,B,2),1,size(A,1));
D=Qa+Qb'-2*A*B';
You can vectorize it by converting to cells and using cellfun:
[nRows,nCols]=size(X);
XCell=num2cell(X,2);
dist=reshape(cell2mat(cellfun(#(x)(sum(bsxfun(#minus,centroids,x).^2,2)),XCell,'UniformOutput',false)),nRows,nRows);
[~,labels]=min(dist);
Explanation:
We assign each row of X to its own cell in the second line
This piece #(x)(sum(bsxfun(#minus,centroids,x).^2,2)) is an anonymous function which is the same as your distances=... line, and using cell2mat, we apply it to each row of X.
The labels are then the indices of the minimum row along each column.
For a true matrix implementation, you may consider trying something along the lines of:
P2 = kron(centroids, ones(size(X,1),1));
Q2 = kron(ones(size(centroids,1),1), X);
distances = reshape(sum((Q2-P2).^2,2), size(X,1), size(centroids,1));
Note
This assumes the data is organized as [x1 y1 ...; x2 y2 ...;...]
You can use a more efficient algorithm for nearest neighbor search than brute force.
The most popular approach are Kd-Tree. O(log(n)) average query time instead of the O(n) brute force complexity.
Regarding a Maltab implementation of Kd-Trees, you can have a look here