I'm trying to cluster a group of points in a probabilistic manner. Using below, I have a single set of xy points, which are recorded in X and Y. I want to cluster into groups using a reference point, which is displayed in X2 and Y2.
With the help of an answer the current approach is to measure the distance from the reference point and group using k-means. Although, it provides a method to cluster using the reference point, the hard cutoff and adherence to k clusters makes it somewhat unsuitable when dealing with numerous datasets. For instance, the number of clusters needed for this example is probably 3. But a separate example may different. I'd have to manually go through and alter k every time.
Given the non-probabilistic nature of k-means a separate option could be GMM. Is it possible to account for the reference point when modelling? If I attach the output below the underlying model isn't clustering as I'm hoping for.
If I look at the probability each point is within a group it's not clustered as I'd hoped. With this I run into the same problem with manually altering the amount of components. Because the points are distributed randomly, using “AIC” or “BIC” to select the appropriate number of clusters doesn't work. There is no optimal number.
import pandas as pd
import matplotlib.pyplot as plt
from sklearn.cluster import KMeans
df = pd.DataFrame({
'X' : [-1.0,-1.0,0.5,0.0,0.0,2.0,3.0,5.0,0.0,-2.5,2.0,8.0,-10.5,15.0,-20.0,-32.0,-20.0,-20.0,-10.0,20.5,0.0,20.0,-30.0,-15.0,20.0,-15.0,-10.0],
'Y' : [0.0,1.0,-0.5,0.5,-0.5,0.0,1.0,4.0,5.0,-3.5,-2.0,-8.0,-0.5,-10.5,-20.5,0.0,16.0,-15.0,5.0,13.5,20.0,-20.0,2.0,-17.5,-15,19.0,20.0],
'X2' : [0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0],
'Y2' : [0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0],
})
k-means:
df['distance'] = np.sqrt(df['X']**2 + df['Y']**2)
df['distance'] = np.sqrt((df['X2'] - df['Y2'])**2 + (df['BallY'] - df['y_post'])**2)
model = KMeans(n_clusters = 2)
model_data = np.array([df['distance'].values, np.zeros(df.shape[0])])
model.fit(model_data.T)
df['group'] = model.labels_
plt.scatter(df['X'], df['Y'], c = model.labels_, cmap = 'bwr', marker = 'o', s = 5)
plt.scatter(df['X2'], df['Y2'], c ='k', marker = 'o', s = 5)
GMM:
Y_sklearn = df[['X','Y']].values
gmm = mixture.GaussianMixture(n_components=3, covariance_type='diag', random_state=42)
gmm.fit(Y_sklearn)
labels = gmm.predict(Y_sklearn)
df['group'] = labels
plt.scatter(Y_sklearn[:, 0], Y_sklearn[:, 1], c=labels, s=5, cmap='viridis');
plt.scatter(df['X2'], df['Y2'], c='red', marker = 'x', edgecolor = 'k', s = 5, zorder = 10)
proba = pd.DataFrame(gmm.predict_proba(Y_sklearn).round(2)).reset_index(drop = True)
df_pred = pd.concat([df, proba], axis = 1)
In my opinion, if you want to define clusters as "regions where points are close to each other", you should use DBSCAN.
This clustering algorithm finds clusters by looking at regions where points are close to each other (i.e. dense regions), and are separated from other clusters by regions where points are less dense.
This algorithm can categorize points as noise (outliers). Outliers are labelled -1.
They are points that do not belong to any cluster.
Here is some code to perform DBSCAN clustering, and to insert the cluster labels as a new categorical column in the original Y_sklearn DataFrame. It also prints how many clusters and how many outliers are found.
import numpy as np
import pandas as pd
from sklearn.cluster import DBSCAN
Y_sklearn = df.loc[:, ["X", "Y"]].copy()
n_points = Y_sklearn.shape[0]
dbs = DBSCAN()
labels_clusters = dbs.fit_predict(Y_sklearn)
#Number of found clusters (outliers are not considered a cluster).
n_clusters = labels_clusters.max() + 1
print(f"DBSCAN found {n_clusters} clusters in dataset with {n_points} points.")
#Number of found outliers (possibly no outliers found).
n_outliers = np.count_nonzero((labels_clusters == -1))
if n_outliers:
print(f"{n_outliers} outliers were found.\n")
else:
print(f"No outliers were found.\n")
#Add cluster labels as a new column to original DataFrame.
Y_sklearn["cluster"] = labels_clusters
#Setting `cluster` column to Categorical dtype makes seaborn function properly treat
#cluster labels as categorical, and not numerical.
Y_sklearn["cluster"] = Y_sklearn["cluster"].astype("category")
If you want to plot the results, I suggest you use Seaborn. Here is some code to plot the points of Y_sklearn DataFrame, and color them by the cluster they belong to. I also define a new color palette, which is just the default Seaborn color palette, but where outliers (with label -1) will be in black.
import matplotlib.pyplot as plt
import seaborn as sns
name_palette = "tab10"
palette = sns.color_palette(name_palette)
if n_outliers:
color_outliers = "black"
palette.insert(0, color_outliers)
else:
pass
sns.set_palette(palette)
fig, ax = plt.subplots()
sns.scatterplot(data=Y_sklearn,
x="X",
y="Y",
hue="cluster",
ax=ax,
)
Using default hyperparameters, the DBSCAN algorithm finds no cluster in the data you provided: all points are considered outliers, because there is no region where points are significantly more dense. Is that your whole dataset, or is it just a sample? If it is a sample, the whole dataset will have much more points, and DBSCAN will certainly find some high density regions.
Or you can try tweaking the hyperparameters, min_samples and eps in particular. If you want to "force" the algorithm to find more clusters, you can decrease min_samples (default is 5), or increase eps (default is 0.5). Of course, the optimal hyperparamete values depends on the specific dataset, but default values are considered quite good for DBSCAN. So, if the algorithm considers all points in your dataset to be outliers, it means that there are no "natural" clusters!
Do you mean density estimation? You can model your data as a Gaussian Mixture and then get a probability of a point to belong to the mixture. You can use sklearn.mixture.GaussianMixture for that. By changing number of components you can control how many clusters you will have. The metric to cluster on is Euclidian distance from the reference point. So the GMM model will provide you with prediction of which cluster the data point should be classified to.
Since your metric is 1d, you will get a set of Gaussian distributions, i.e. a set of means and variances. So you can easily calculate the probability of any point to be in certain cluster, just by calculating how far it is from the reference point and put the value in the normal distribution pdf formula.
To make image more clear, I'm changing the reference point to (-5, 5) and select number of clusters = 4. In order to get the best number of clusters, use some metric that minimizes total variance and penalizes growth of number of mixtures. For example argmin(model.covariances_.sum()*num_clusters)
import pandas as pd
from sklearn.mixture import GaussianMixture
import numpy as np
import matplotlib.pyplot as plt
from scipy.stats import norm
df = pd.DataFrame({
'X' : [-1.0,-1.0,0.5,0.0,0.0,2.0,3.0,5.0,0.0,-2.5,2.0,8.0,-10.5,15.0,-20.0,-32.0,-20.0,-20.0,-10.0,20.5,0.0,20.0,-30.0,-15.0,20.0,-15.0,-10.0],
'Y' : [0.0,1.0,-0.5,0.5,-0.5,0.0,1.0,4.0,5.0,-3.5,-2.0,-8.0,-0.5,-10.5,-20.5,0.0,16.0,-15.0,5.0,13.5,20.0,-20.0,2.0,-17.5,-15,19.0,20.0],
})
ref_X, ref_Y = -5, 5
dist = np.sqrt((df.X-ref_X)**2 + (df.Y-ref_Y)**2)
n_mix = 4
gmm = GaussianMixture(n_mix)
model = gmm.fit(dist.values.reshape(-1,1))
x = np.linspace(-35., 35.)
y = np.linspace(-30., 30.)
X, Y = np.meshgrid(x, y)
XX = np.sqrt((X.ravel() - ref_X)**2 + (Y.ravel() - ref_Y)**2)
Z = model.score_samples(XX.reshape(-1,1))
Z = Z.reshape(X.shape)
# plot grid points probabilities
plt.set_cmap('plasma')
plt.contourf(X, Y, Z, 40)
plt.scatter(df.X, df.Y, c=model.predict(dist.values.reshape(-1,1)), edgecolor='black')
You can read more here and here
P.S. score_samples() returns log likelihoods, use exp() to convert to probability
Taking your centre point of 0,0 we can calculate the Euclidean distance from this point to all points in your df.
df['distance'] = np.sqrt(df['X']**2 + df['Y']**2)
If you have a centre point other than zero it would be:
df['distance'] = np.sqrt((centre_point_x - df['X'])**2 + (centre_point_y - df['Y'])**2)
Using your data and chart as before, we can plot this and see the distance metric increasing as we move away from the centre.
fig, ax = plt.subplots(figsize = (6,6))
ax.scatter(df['X'], df['Y'], c = df['distance'], cmap = 'viridis', marker = 'o', s = 30)
ax.set_xlim([-35, 35])
ax.set_ylim([-35, 35])
plt.show()
K-means
We can now use this distance data and use it to calculate K-means clusters as you did before, but this time using the distance data and an array of zeros (zeros because this k-means requires a 2d-array but we only want to split the 1d aray of dimensional data. So the zeros act as 'filler'
model = KMeans(n_clusters = 2) #choose how many clusters
# create this 2d array for the KMeans model
model_data = np.array([df['distance'].values, np.zeros(df.shape[0])])
model.fit(model_data.T) # transformed array because the above code produces
# data with 27 columns and 2 rows but we want it the other way round
df['group'] = model.labels_ # put the labels into the dataframe
Then we can plot the results
fig, ax = plt.subplots(figsize = (6,6))
ax.scatter(df['X'], df['Y'], c = df['group'], cmap = 'viridis', marker = 'o', s = 30)
ax.set_xlim([-35, 35])
ax.set_ylim([-35, 35])
plt.show()
With three clusters we get the following result:
Other clustering methods
Check out SKlearn's clustering page for more options. I experimented with DBSCAN with some good results but it depends on what you are trying to achieve exactly. Check out the table underneath their example charts to see how they each compare.
So I think I might be absolutely on the wrong track here, but basically
I have a 3-d meshgrid, I find all of the distances to a testpoint at all of the points in that grid
import numpy as np
#crystal_lattice structure
x,y,z = np.linspace(-2,2,5),np.linspace(-2,2,5),np.linspace(-2,2,5)
xx,yy,zz = np.meshgrid(x,y,z)
#testpoint
point = np.array([1,1,1])
d = np.sqrt((point[0]-xx)**2 + (point[1]-yy)**2 + (point[2]-zz)**2)
#np.shape(d) = (5, 5, 5)
Then I am trying to find the coordinates of he gridpoint that is the closest to that test point.
My idea was to sort d (flatten then search), get the index of the lowest value.
low_to_hi_d = np.sort(d, axis=None) # axis=0 flattens the d, going to flatten the entire d array and then search
lowest_val = low_to_hi_d[0]
index = np.where(d == lowest_val)
#how do I get the spatial coordinates of my index, not just the position in ndarray (here the position in ndarray is (3,3,3) but the spatial position is (1,1,1), but if I do d[3,3,3] I get 0 (the value at spatial position (1,1,1))
Use that index on my 3d grid to find the point coordinates (not the d value at that point). I am trying something like this, and I am pretty sure I am overcomplicating it. How can I get the (x,y,z) of the 3-d gridpoint that is closest to my test point?
If you just want to find the coordinates of the closest point you are right, you're on the wrong track. There is no point in generating a meshgrid and calculate the distance on so many duplicates. You can do it in every dimension easily and independently:
import numpy as np
x,y,z = np.linspace(-2,2,5),np.linspace(-2,2,5),np.linspace(-2,2,5)
p=np.array([1,1,1])
closest=lambda x,p: x[np.argmin(np.abs(x-p))]
xc,yc,zc=closest(x,p[0]),closest(y,p[1]),closest(z,p[2])
I'm not completely sure that this is what you want.
You can find the index of the minimum d with:
idx = np.unravel_index(np.argmin(d), d.shape)
(3, 3, 3)
and use this to index your meshgrid:
xx[idx], yy[idx], zz[idx]
(1.0, 1.0, 1.0)
In the case of a matrix mat n x n, i can do the following
sym = 0.5 * (mat + mat.T)
the operation gives the desired result sym[i,j] = sym[j,i]
Suppose we have a 3D array ndarr[i,j,k], where i,j,k 0,1,...n,
then ndarr is n x n x n. The idea is to obtain the following "symmetric" form
nsym[i,j,k] = nsym[j,i,k] using ndarr. I tried this:
import numpy as np
# Generate some random matrix, n = 5
ndarr = np.random.beta(0.1,1,(5,5,5))
# First attempt to symmetrize
sym1 = np.array([0.5*(ndarr[:,:,k]+ndarr[:,:,k].T) for k in range(5)])
The problem here is that sym1[i,j,k] != sym1[j,i,k] as it is required. In fact I obtain sym1[i,j,k] = sym1[i,k,j], symmetric under the exchange of the last two symbols!
# Second attempt
sym2 = 0.5*(ndarr+ndarr.T)
Same problem here and sym2 is symmetric with respect the second index sym2[i,j,k]=sym2[k,j,i].
To resume, the goal is to find a symmetric form for a 3D array with respect to the third index and to preserve the values in the diagonal for the original ndarr[i,i,i].
The problem here is that you're not using the correct transpose:
sym = 0.5 * (ndarr + np.transpose(ndarr, (1, 0, 2)))
By default, np.transpose and the .T property will reverse the order of the axes. In your case, we want to only flip the first two axes: (0,1,2) -> (1,0,2).
EDIT: The reason your first attempt failed is because you were concatenating each symmetrized matrix along the first axis, not the last. It's more clear if you make ndarr with shape (5, 5, 3):
In [16]: sym = np.array([0.5*(ndarr[:,:,k]+ndarr[:,:,k].T) for k in range(3)])
In [17]: sym.shape
Out[17]: (3L, 5L, 5L)
In any case, the version above with np.transpose is cleaner and more efficient.
I don't understand why the ifft(fft(myFunction)) is not the same as my function. It seems to be the same shape but a factor of 2 out (ignoring the constant y-offset). All the documentation I can see says there is some normalisation that fft doesn't do, but that ifft should take care of that. Here's some example code below - you can see where I've bodged the factor of 2 to give me the right answer. Thanks for any help - its driving me nuts.
import numpy as np
import scipy.fftpack as fftp
import matplotlib.pyplot as plt
import matplotlib.pyplot as plt
def fourier_series(x, y, wn, n=None):
# get FFT
myfft = fftp.fft(y, n)
# kill higher freqs above wavenumber wn
myfft[wn:] = 0
# make new series
y2 = fftp.ifft(myfft).real
# find constant y offset
myfft[1:]=0
c = fftp.ifft(myfft)[0]
# remove c, apply factor of 2 and re apply c
y2 = (y2-c)*2 + c
plt.figure(num=None)
plt.plot(x, y, x, y2)
plt.show()
if __name__=='__main__':
x = np.array([float(i) for i in range(0,360)])
y = np.sin(2*np.pi/360*x) + np.sin(2*2*np.pi/360*x) + 5
fourier_series(x, y, 3, 360)
You're removing half the spectrum when you do myfft[wn:] = 0. The negative frequencies are those in the top half of the array and are required.
You have a second fudge to get your results which is taking the real part to find y2: y2 = fftp.ifft(myfft).real (fftp.ifft(myfft) has a non-negligible imaginary part due to the asymmetry in the spectrum).
Fix it with myfft[wn:-wn] = 0 instead of myfft[wn:] = 0, and remove the fudges. So the fixed code looks something like:
import numpy as np
import scipy.fftpack as fftp
import matplotlib.pyplot as plt
def fourier_series(x, y, wn, n=None):
# get FFT
myfft = fftp.fft(y, n)
# kill higher freqs above wavenumber wn
myfft[wn:-wn] = 0
# make new series
y2 = fftp.ifft(myfft)
plt.figure(num=None)
plt.plot(x, y, x, y2)
plt.show()
if __name__=='__main__':
x = np.array([float(i) for i in range(0,360)])
y = np.sin(2*np.pi/360*x) + np.sin(2*2*np.pi/360*x) + 5
fourier_series(x, y, 3, 360)
It's really worth paying attention to the interim arrays that you are creating when trying to do signal processing. Invariably, there are clues as to what is going wrong that should direct you to the problem. In this case, you taking the real part masked the problem and made your task more difficult.
Just to add another quick point: Sometimes taking the real part of the resultant array is exactly the correct thing to do. It's often the case that you end up with an imaginary part to the signal output which is just down to numerical errors in the input to the inverse FFT. Typically this manifests itself as very small imaginary values, so taking the real part is basically the same array.
You are killing the negative frequencies between 0 and -wn.
I think what you mean to do is to set myfft to 0 for all frequencies outside [-wn, wn].
Change the following line:
myfft[wn:] = 0
to:
myfft[wn:-wn] = 0