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LINEAR DISCRIMINANT ANALYSIS IN PYTHON

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 Linear discriminant analysis is a supervised dimensionality reduction algorithm. When dealing with large data with lot of features, it becomes difficult to compute and hence we opt for dimensionality reduction methods. The dataset I used is seed dataset from :  https://archive.ics.uci.edu/ml/datasets/seeds Here is the code: import pandas as pd import numpy as np import matplotlib.pyplot as plt #loading the dataset df = pd.read_csv('seeds_dataset.csv') df.head() X = df.iloc[:, 1:8].values y = df.iloc[:, 8].values #training the model from sklearn.model_selection import train_test_split X_train, X_test, y_train, y_test = train_test_split(X, y, test_size = 0.3, random_state = 0) #standardizing the values from sklearn.preprocessing import StandardScaler sc = StandardScaler() X_train = sc.fit_transform(X_train) X_test = sc.transform(X_test) #performing LDA from sklearn.discriminant_analysis import LinearDiscriminantAnalysis as LDA lda = LDA(n_components = 2) X_train = lda.fit_trans...
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K-MEANS CLUSTERING

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 K-Means clustering is an unsupervised centroid based algorithm. The algorithm tends to reduce the distance between the points in a cluster and the cluster centroid. The dataset I used is seeds dataset from :  https://archive.ics.uci.edu/ml/datasets/seeds import pandas as pd import numpy as np import matplotlib.pyplot as plt #loading dataset df = pd.read_csv('seeds_dataset.csv') df.head() #taking compactness and perimeter columns z= df.iloc[:,[2,3]].values #applying elbow method to find the maximum number of clusters from sklearn.cluster import KMeans   elbow_list= []  for i in range(1, 11):       kmeans = KMeans(n_clusters=i, init='k-means++', random_state= 42)       kmeans.fit(z)       elbow_list.append(kmeans.inertia_)   plt.plot(range(1, 11), elbow_list)   plt.title('The Elbow Method Graph')   plt.xlabel('Number of clusters(k)')   plt.ylabel('elbow_list')...
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SIMULATION OF AUTOREGRESSIVE PROCESS AR(2) in R

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 Here is the code in R for AR(2) process: set.seed(2017) X.ts <- arima.sim(list(ar = c(.7, .2)), n=1000) par(mfrow=c(2,1)) plot(X.ts,main="AR(2) Time Series, phi1=.7, phi2=.2") X.acf = acf(X.ts, main="Autocorrelation of AR(2) Time Series")
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SIMULATION OF AUTOREGRESSIVE PROCESS AR(1) IN R

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Here is the code in R for AR(1): set.seed(20190)  n=10000  phi = .6 Z = rnorm(n,0,1)  X=NULL  X[1] = Z[1]  for (t in 2:n) { X[t] = Z[t] + phi*X[t-1]  } X.ts = ts(X) par(mfrow=c(2,1)) plot(X.ts,main="AR(1) Time Series on White Noise, phi=.6") X.acf = acf(X.ts, main="AR(1) Time Series on White Noise, phi=.6")
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SIMULATION OF MOVING AVERAGE PROCESS IN R

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  Here is the code in R for simulation of moving average: #simulating MA(3) process noise = rnorm(10000) ma3= NULL for(i in 4:10000) { ma3[i] = noise[i] + 0.8*noise[i-1] + 0.5*noise[i-2] + 0.3*noise[i-3] } moving_average = ma3[4:10000] #changing the series into time series moving_average = ts(moving_average) par(mfrow=c(2,1)) plot(moving_average, col='blue') acf(moving_average) Conclusion: We observe the lag cuts off at 3 in the autocorrelation graph showing that the process is a MA(3) 
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SIMULATION OF A RANDOM WALK IN R

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Here is the code in R for simulation of Random walk : x=NULL x[1]=0 for( i in 2:10000) { x[i]=x[i-1] + rnorm(1) } print(x) #converting it into a time series data random_walk = ts(x) plot(random_walk, main='visualization of a random work' , xlab='days', ylab=' ') acf(random_walk) As we see there is a high correlation in the correlogram, the random walk is a non-stationary process.  # making the series stationary by differencing the values z<-diff(random_walk) plot(z)  # we get white noise acf(z) Conclusion :  We observe that there is no lag and hence no correlation. Thus we obtained stationary series by differencing the time series.
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ESTIMATION OF PI USING MONTE CARLO METHOD USING PYTHON

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  The value of pi is calculated using monte carlo method by taking a square of 1 unit and inscribing a circle in the square. The radius of circle is 0.5 units. Now the ratio of area of circle to the ratio of square multiplied by 4 gives us pi. Python code: import random n=1000000 c_points=0  #points inside circle s_points=0  #points inside square for i in range(n):     x = random.uniform(0,1)     y = random.uniform(0,1)     d = x**2 + y**2     if d<=1 :         c_points +=1     s_points +=1     pi = 4*(c_points/s_points)      print("pi value is:", pi) Conclusion: Higher the value of n, higher is the accuracy of value of pi.
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