Prediction of letters as a function of data used

In [1]:

    

import numpy as np
import h5py
from sklearn import svm, cross_validation
from sklearn.naive_bayes import MultinomialNB


    

In [2]:

    

# First we load the file 
file_location = '../results_database/text_wall_street_big.hdf5'
f = h5py.File(file_location, 'r')

# Now we need to get the letters and align them
text_directory = '../data/wall_street_letters.npy'
letters_sequence = np.load(text_directory)
Nletters = len(letters_sequence)
symbols = set(letters_sequence)

# Load the particular example
Nspatial_clusters = 5
Ntime_clusters = 15
Nembedding = 3

run_name = '/low-resolution'
parameters_string = '/' + str(Nspatial_clusters)
parameters_string += '-' + str(Ntime_clusters)
parameters_string += '-' + str(Nembedding)

nexa = f[run_name + parameters_string]

# Now we load the time and the code vectors
time = nexa['time']
code_vectors = nexa['code-vectors']
code_vectors_distance = nexa['code-vectors-distance']
code_vectors_softmax = nexa['code-vectors-softmax']
code_vectors_winner = nexa['code-vectors-winner']


    

In [14]:

    

data_ammount = np.logspace(2, 5, 30, dtype='int')
delay = 4
accuracy = []


    

In [15]:

    

# Make prediction with scikit-learn
for N in data_ammount:
    X = code_vectors_winner[:(N - delay)]
    y = letters_sequence[delay:N]
    X_train, X_test, y_train, y_test = cross_validation.train_test_split(X, y, test_size=0.10)

    clf = svm.SVC(C=1.0, cache_size=200, kernel='linear')
    clf.fit(X_train, y_train)
    score = clf.score(X_test, y_test) * 100.0
    print('N_data', N)
    print('SVM score', score)
    accuracy.append(score)

print('Analysis donde')


    

    




N_data 100
SVM score 60.0
N_data 126
SVM score 46.1538461538
N_data 161
SVM score 43.75
N_data 204
SVM score 50.0
N_data 259
SVM score 53.8461538462
N_data 329
SVM score 60.6060606061
N_data 417
SVM score 59.5238095238
N_data 529
SVM score 60.3773584906
N_data 672
SVM score 64.1791044776
N_data 853
SVM score 67.0588235294
N_data 1082
SVM score 57.4074074074
N_data 1373
SVM score 76.6423357664
N_data 1743
SVM score 64.9425287356
N_data 2212
SVM score 67.8733031674
N_data 2807
SVM score 67.2597864769
N_data 3562
SVM score 68.5393258427
N_data 4520
SVM score 70.796460177
N_data 5736
SVM score 69.6864111498
N_data 7278
SVM score 69.0934065934
N_data 9236
SVM score 67.7489177489
N_data 11721
SVM score 66.638225256
N_data 14873
SVM score 69.3342299933
N_data 18873
SVM score 68.5214626391
N_data 23950
SVM score 68.3089770355
N_data 30391
SVM score 70.3191839421
N_data 38566
SVM score 69.7951775992
N_data 48939
SVM score 69.6771557009
N_data 62101
SVM score 68.0032206119
N_data 78804
SVM score 68.730964467
N_data 100000
SVM score 69.63
Analysis donde

In [25]:

    

# Plot this
import matplotlib.pyplot as plt
import seaborn as sns
%matplotlib inline
plt.plot(data_ammount, accuracy, 'o-', lw=2, markersize=10)
plt.xlabel('Total Data')
plt.ylabel('Accuracy %')
plt.title('Accuracy vs Amount of Data')
plt.ylim([0, 110])
plt.xscale('log')
fig = plt.gcf()
fig.set_size_inches((12, 9))


    

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