Develop Reference

Clustering geospatial data on coordinates AND non spatial feature

Say i have the following dataframe stored as a variable called coordinates, where the first few rows look like:
business_lat business_lng business_rating
0 19.111841 72.910729 5.
1 19.111342 72.908387 5.
2 19.111342 72.908387 4.
3 19.137815 72.914085 5.
4 19.119677 72.905081 2.
5 19.119677 72.905081 2.
. . .
. . .
. . .
As you can see this data is geospatial (has a lat and a lng) AND every row has an additional value, business_rating, that corresponds to the rating of the business at the latlng in that row. I want to cluster the data, where businesses that are nearby and have similar ratings are assigned into the same cluster. Essentially I need a a geospatial cluster with an additional requirement that the clustering must consider the rating column.
I've looked online and can't really find much addressing approaches for this: only things for strict geospatial clustering (only features to cluster on are latlng) or non spatial clustering.
I have a simple DBSCAN running below, but when i plot the results of the clustering it does not seem to be doing what I want correctly.
from sklearn.cluster import DBSCAN
import numpy as np
db = DBSCAN(eps=2/6371., min_samples=5, algorithm='ball_tree', metric='haversine').fit(np.radians(coordinates))
Would I be better served trying to tweak the parameters of the DBSCAN, doing some additional processing of the data or using a different approach all together?

The tricky part about clustering two different types of information (location and rating) is determining how they should relate to each other. It's simple to ask when it is just one domain and you are comparing the same units. My approach would be to look at how to relate rows within a domain and then determine some interaction between the domains. This could be done using scaling options like MinMaxScaler mentioned, however, I think this is a bit heavy handed and we could use our knowledge of the domains to cluster better.
Handling Location
Location distance is best handled directly as this has real world meaning that we can precalculate distances for. The meaning of meters apart is direct to what we
You could use the scaling option mentioned in the previous answer but this risks distorting the location data. For example, if you have a long and thin set of locations, MinMaxScaling would give more importance to variation on the thin axis than the long axis. If you are going to use scaling, do it on the computed distance matrix, not on the lat lon themselves.
import numpy as np
from sklearn.metrics.pairwise import haversine_distances
points_in_radians = df[['business_lat','business_lng']].apply(np.radians).values
distances_in_km = haversine_distances(points_in_radians) * 6371
Adding in Rating
We can think of the problem through asking a couple of questions that relate rating to distance. We could ask, how different must ratings be to separate observations in the same place? What is the meter difference to rating difference ratio? With an idea of ratio, we can calculate another distance matrix for the rating difference for all observations and use this to scale or add on the original location distance matrix or we could increase the distance for every gap in rating. This location-plus-ratings-difference matrix can then be clustered on.
from sklearn.metrics.pairwise import euclidean_distances
added_km_per_rating_gap = 1
rating_distances = euclidean_distances(df[['business_rating']].values) * added_km_per_rating_gap
We can then simply add these together and cluster on the resulting matrix.
from sklearn.cluster import DBSCAN
distance_matrix = rating_distances + distances_in_km
clustering = DBSCAN(metric='precomputed', eps=1, min_samples=2)
clustering.fit(distance_matrix)
What we have done is cluster by location, adding a penalty for ratings difference. Making that penalty direct and controllable allows for optimisation to find the best clustering.
Testing
The problem I'm finding is that (with my test data at least) DBSCAN has a tendency to 'walk' from observation to observation forming clusters that either blend ratings together because the penalty is not high enough or separates into single rating groups. It might be that DBSCAN is not suitable for this type of clustering. If I had more time, I would look for some open data to test this on and try other clustering methods.
Here is the code I used to test. I used the square of the ratings distance to emphasise larger gaps.
import random
from sklearn.datasets import make_blobs
X, y = make_blobs(n_samples=300, centers=6, cluster_std=0.60, random_state=0)
ratings = np.array([random.randint(1,4) for _ in range(len(X)//2)] \
+[random.randint(2,5) for _ in range(len(X)//2)]).reshape(-1, 1)
distances_in_km = euclidean_distances(X)
rating_distances = euclidean_distances(ratings)
def build_clusters(multiplier, eps):
rating_addition = (rating_distances ** 2) * multiplier
distance_matrix = rating_addition + distances_in_km
clustering = DBSCAN(metric='precomputed', eps=eps, min_samples=10)
clustering.fit(distance_matrix)
return clustering.labels_

Using the DBSCAN methodology, we can calculate the distance between points (the Euclidean distance or some other distance) and look for points which are far away from others. You may want to consider using the MinMaxScaler to normalize values, so one feature doesn't overwhelm other features.
Where is your code and what are your final results? Without an actual code sample, I can only guess what you are doing.
I hacked together some sample code for you. You can see the results below.
import pandas as pd
import numpy as np
import matplotlib.pyplot as plt
from sklearn.cluster import KMeans
import seaborn as sns; sns.set()
import csv
df = pd.read_csv('C:\\your_path_here\\business.csv')
X=df.loc[:,['review_count','latitude','longitude']]
K_clusters = range(1,10)
kmeans = [KMeans(n_clusters=i) for i in K_clusters]
Y_axis = df[['latitude']]
X_axis = df[['longitude']]
score = [kmeans[i].fit(Y_axis).score(Y_axis) for i in range(len(kmeans))]# Visualize
plt.plot(K_clusters, score)
plt.xlabel('Number of Clusters')
plt.ylabel('Score')
plt.title('Elbow Curve')
plt.show()
kmeans = KMeans(n_clusters = 3, init ='k-means++')
kmeans.fit(X[X.columns[0:2]]) # Compute k-means clustering.
X['cluster_label'] = kmeans.fit_predict(X[X.columns[0:2]])
centers = kmeans.cluster_centers_ # Coordinates of cluster centers.
labels = kmeans.predict(X[X.columns[0:2]]) # Labels of each point
X.head(10)
X.plot.scatter(x = 'latitude', y = 'longitude', c=labels, s=50, cmap='viridis')
plt.scatter(centers[:, 0], centers[:, 1], c='black', s=200, alpha=0.5)
from scipy.stats import zscore
df["zscore"] = zscore(df["review_count"])
df["outlier"] = df["zscore"].apply(lambda x: x <= -2.5 or x >= 2.5)
df[df["outlier"]]
df_cord = df[["latitude", "longitude"]]
df_cord.plot.scatter(x = "latitude", y = "latitude")
from sklearn.preprocessing import MinMaxScaler
scaler = MinMaxScaler()
df_cord = scaler.fit_transform(df_cord)
df_cord = pd.DataFrame(df_cord, columns = ["latitude", "longitude"])
df_cord.plot.scatter(x = "latitude", y = "longitude")
from sklearn.cluster import DBSCAN
outlier_detection = DBSCAN(
eps = 0.5,
metric="euclidean",
min_samples = 3,
n_jobs = -1)
clusters = outlier_detection.fit_predict(df_cord)
clusters
from matplotlib import cm
cmap = cm.get_cmap('Accent')
df_cord.plot.scatter(
x = "latitude",
y = "longitude",
c = clusters,
cmap = cmap,
colorbar = False
)
The final result looks a little weird, to tell you the truth. Remember, not everything is clusterable.

Clustering data- Poor results, feature extraction

I have measured data (vibrations) from a wind turbine running under different operating conditions. My dataset consists of operating conditions as well as measurement features I have extracted from the measured data.
Dataset shape: (423, 15). Each of the 423 data points represent a measurement on a day, chronologically over 423 days.
I now want to cluster the data to see if there is any change in the measurements. Specifically, I want to examine if the vibrations change over time (which could indicate a fault in the turbine gearbox).
What I have currently done:
Scale the data between 0,1 ->
Perform PCA (reduce from 15 to 5)
Cluster using db scan since I do not know the number of clusters. I am using this code to find the optimal epsilon (eps) in dbscan:
# optimal Epsilon (distance):
X_pca = principalDf.values
neigh = NearestNeighbors(n_neighbors=2)
nbrs = neigh.fit(X_pca)
distances, indices = nbrs.kneighbors(X_pca)
distances = np.sort(distances, axis=0)
distances = distances[:,1]
plt.plot(distances,color="#0F215A")
plt.grid(True)
The result so far are not giving any clear indication that the data is changing over time:
Of course, the case could be that the data is not changing over these data points. Howver, what are some other things I could try? Kind of an open question, but I am running out of ideas.

First of all, with KMeans, if the dataset is not naturally partitioned, you may end up with some very weird results! As KMeans is unsupervised, you basically dump in all kinds of numeric variables, set the target variable, and let the machine do the lift for you. Here is a simple example using the canonical Iris dataset. You can EASILY modify this to fit your specific dataset. Just change the 'X' variables (all but the target variable) and 'y' variable (just one target variable). Try that and feedback.
import numpy as np
import pandas as pd
import matplotlib.pyplot as plt
import matplotlib as mpl
import urllib.request
import random
# seaborn is a layer on top of matplotlib which has additional visualizations -
# just importing it changes the look of the standard matplotlib plots.
# the current version also shows some warnings which we'll disable.
import seaborn as sns
sns.set(style="white", color_codes=True)
import warnings
warnings.filterwarnings("ignore")
from sklearn import datasets
iris = datasets.load_iris()
X = iris.data[:, 0:4] # we only take the first two features.
y = iris.target
from sklearn import preprocessing
scaler = preprocessing.StandardScaler()
scaler.fit(X)
X_scaled_array = scaler.transform(X)
X_scaled = pd.DataFrame(X_scaled_array)
X_scaled.sample(5)
# try clustering on the 4d data and see if can reproduce the actual clusters.
# ie imagine we don't have the species labels on this data and wanted to
# divide the flowers into species. could set an arbitrary number of clusters
# and try dividing them up into similar clusters.
# we happen to know there are 3 species, so let's find 3 species and see
# if the predictions for each point matches the label in y.
from sklearn.cluster import KMeans
nclusters = 3 # this is the k in kmeans
seed = 0
km = KMeans(n_clusters=nclusters, random_state=seed)
km.fit(X_scaled)
# predict the cluster for each data point
y_cluster_kmeans = km.predict(X_scaled)
y_cluster_kmeans
# use seaborn to make scatter plot showing species for each sample
sns.FacetGrid(data, hue="species", size=4) \
.map(plt.scatter, "sepal_length", "sepal_width") \
.add_legend();

How to choose the Chi Squared threshold in feature selection

About this:
NLP in Python: Obtain word names from SelectKBest after vectorizing
I found this code:
import pandas as pd
import numpy as np
from sklearn.feature_extraction.text import CountVectorizer
from sklearn.feature_selection import chi2
THRESHOLD_CHI = 5 # or whatever you like. You may try with
# for threshold_chi in [1,2,3,4,5,6,7,8,9,10] if you prefer
# and measure the f1 scores
X = df['text']
y = df['labels']
cv = CountVectorizer()
cv_sparse_matrix = cv.fit_transform(X)
cv_dense_matrix = cv_sparse_matrix.todense()
chi2_stat, pval = chi2(cv_dense_matrix, y)
chi2_reshaped = chi2_stat.reshape(1,-1)
which_ones_to_keep = chi2_reshaped > THRESHOLD_CHI
which_ones_to_keep = np.repeat(which_ones_to_keep ,axis=0,repeats=which_ones_to_keep.shape[1])
This code computes the chi squared test and should keep the best features within a chosen threshold.
My question is how to choose a theshold for the chi squared test scores?

Chi square does not have a specific range of outcome, so it's hard to determine a threshold beforehand. Usually what you can do is to sort the variables depending on their p values, the logic is that lower p values are better, because they imply a higher correlation between features and the target variable (we want to discard features that are independent, i.e. not predictors of the target variable). In this case you have anyway to decide how many features to keep, and that is a hyper parameter that you can tune manually or even better by using a grid search.
Be aware that you can avoid to perform the selection manually, sklearn implement already a function SelectKBest to select the best k features based on chi square, you can use it as follow:
from sklearn.feature_selection import SelectKBest, chi2
X_new = SelectKBest(chi2, k=2).fit_transform(X, y)
But if for any reason you want to rely solely on the raw chi2 value, you could calculate the minimum and maximum values among the variables, and then divide the interval in n steps to test trough a grid search.

Questions about ridge regression on python : Scaling, and interpretation

I tried to run a Ridge Regression on Boston housing data with python, but I had the following questions that I cannot find answer anywhere so I decided to post it here:
Is scaling recommended before fitting the model? Because I get the same score when I scale and when I don't scale. Also, what is the interpretation of the alpha/coeff graph in terms of choosing the best alpha?
import numpy as np
import pandas as pd
import matplotlib.pyplot as plt
import seaborn as sns
import matplotlib.pyplot as plt
%matplotlib inline
from sklearn import linear_model
df = pd.read_csv('../housing.data',delim_whitespace=True,header=None)
col_names = ['CRIM','ZN','INDUS','CHAS','NOX','RM','AGE','DIS','RAD','TAX','PTRATIO','B','LSTAT','MEDV']
df.columns = col_names
X = df.loc[:,df.columns!='MEDV']
col_X = X.columns
y = df['MEDV'].values
# Feature Scaling:
from sklearn.preprocessing import StandardScaler
scaler = StandardScaler()
X_std = scaler.fit_transform(X)
from sklearn.linear_model import Ridge
from sklearn.metrics import mean_squared_error
clf = Ridge()
coefs = []
alphas = np.logspace(-6, 6, 200)
for a in alphas:
clf.set_params(alpha=a)
clf.fit(X_std, y)
coefs.append(clf.coef_)
plt.figure(figsize=(20, 6))
plt.subplot(121)
ax = plt.gca()
ax.plot(alphas, coefs)
ax.set_xscale('log')
plt.xlabel('alpha')
plt.ylabel('weights')
plt.title('Ridge coefficients as a function of the regularization')
plt.axis('tight')
plt.show()
Alpha/coefficient graph for scaled X
Alpha/coefficient graph for unscaled X
On the scaled data, when I compute the score and choose the alpha thanks to CV, I get:
from sklearn.linear_model import RidgeCV
clf = RidgeCV(alphas=[1e-3, 1e-2, 1e-1, 1, 5, 7]).fit(X_std, y)
> clf.score(X_std, y)
> 0.74038
> clf.alpha_
> 5.0
On the non-scaled data, I even get a slightly better score with a completely different alpha:
clf = RidgeCV(alphas=[1e-3, 1e-2, 1e-1, 1, 6]).fit(X, y)
> clf.score(X, y)
> 0.74064
> clf.alpha_
> 0.01
Thanks for your insights on the matter, looking forward to reading your answers!

I think you should scale because Ridge Regularization penalizes large values, and so you don't want to lose meaningful features because of scaling issues. Perhaps you don't see a difference because the housing data is a toy dataset and is already scaled well.
A larger alpha is a stronger penalty on large values. The graph is showing you (though it has no labeling) that with a stronger alpha you send coefficients to zero more strongly. The more gradual lines are the smaller weights, so they're effected less or almost not at all until alpha becomes sufficiently large. The sharper ones are larger weights, so they drop to zero more quickly. When they do, the feature will disappear from your regression.

For the scaled data, the magnitude of design matrix is smaller, and the coefficients tend to be larger (and more L2 penalty is imposed). To minimize L2, we need more and more small coefficients. How to get more and more small coefficients? The way is to choose a very big alpha, so we can have more smaller coefficients. That is why if you scale the data, the optimal alpha is a great number.

n-dim python interpolation nonlinear

i got a function of 5 variables Fx(s,m,p,h,l)
import numpy as np
s= np.arange(0,135,15)/10
m= np.array([150,180,195,210,240,255,270,285,300])
p=np.array([-1.5,-1,-0.5,0,0.5,1,1.5])
h=np.array([0,3,6,9,12])
l=np.array([0,0.5,1,1.5,2,2.5,3,4])
and 180 values of the function in a csv file.
i would like to calculate missing value by interpolation in all points
and use radial basis function thin_plate will by great. is it possible?
for information i found here
Python 4D linear interpolation on a rectangular grid
InterpolatingFunction
but if i replace some value in data array by None, at this point f(point) give 'nan'. and i don t want to use a linear interpolation because for a set of 4 variables i got 2 points with a value.
thanks very much for helping LL

Try SVR from scikit-learn to solve your problem:
from sklearn.svm import SVR # it uses RBF as default kernel
import numpy as np
n_samples, n_features = 180, 5
y = np.random.randn(n_samples) #function values
X = np.random.randn(n_samples, n_features) #points
clf = SVR(C=1.0, epsilon=0.2)
clf.fit(X, y)
#Get value at a new point:
clf.predict([0,150,-1.5,0,0])
Because of len(s)*len(m)*len(p)*len(h)*len(l) is 22680 and function values are known only in 180 points, you have poor information about your function....