#! /usr/bin/env python # def laplace_cdf_values ( n_data ): #*****************************************************************************80 # ## LAPLACE_CDF_VALUES returns some values of the Laplace CDF. # # Discussion: # # In Mathematica, the function can be evaluated by: # # Needs["Statistics`ContinuousDistributions`"] # dist = LaplaceDistribution [ mu, beta ] # CDF [ dist, x ] # # Licensing: # # This code is distributed under the GNU LGPL license. # # Modified: # # 17 February 2015 # # Author: # # John Burkardt # # Reference: # # Milton Abramowitz and Irene Stegun, # Handbook of Mathematical Functions, # US Department of Commerce, 1964. # # Stephen Wolfram, # The Mathematica Book, # Fourth Edition, # Wolfram Media / Cambridge University Press, 1999. # # Parameters: # # Input/output, integer N_DATA. The user sets N_DATA to 0 before the # first call. On each call, the routine increments N_DATA by 1, and # returns the corresponding data; when there is no more data, the # output value of N_DATA will be 0 again. # # Output, real MU, the mean of the distribution. # # Output, real BETA, the shape parameter. # # Output, real X, the argument of the function. # # Output, real F, the value of the function. # import numpy as np n_max = 12 beta_vec = np.array ( ( \ 0.1000000000000000E+01, \ 0.1000000000000000E+01, \ 0.1000000000000000E+01, \ 0.1000000000000000E+01, \ 0.2000000000000000E+01, \ 0.3000000000000000E+01, \ 0.4000000000000000E+01, \ 0.5000000000000000E+01, \ 0.2000000000000000E+01, \ 0.2000000000000000E+01, \ 0.2000000000000000E+01, \ 0.2000000000000000E+01 )) f_vec = np.array ( ( \ 0.5000000000000000E+00, \ 0.8160602794142788E+00, \ 0.9323323583816937E+00, \ 0.9751064658160680E+00, \ 0.6967346701436833E+00, \ 0.6417343447131054E+00, \ 0.6105996084642976E+00, \ 0.5906346234610091E+00, \ 0.5000000000000000E+00, \ 0.3032653298563167E+00, \ 0.1839397205857212E+00, \ 0.1115650800742149E+00 )) mu_vec = np.array ( ( \ 0.0000000000000000E+01, \ 0.0000000000000000E+01, \ 0.0000000000000000E+01, \ 0.0000000000000000E+01, \ 0.0000000000000000E+01, \ 0.0000000000000000E+01, \ 0.0000000000000000E+01, \ 0.0000000000000000E+01, \ 0.1000000000000000E+01, \ 0.2000000000000000E+01, \ 0.3000000000000000E+01, \ 0.4000000000000000E+01 )) x_vec = np.array ( ( \ 0.0000000000000000E+01, \ 0.1000000000000000E+01, \ 0.2000000000000000E+01, \ 0.3000000000000000E+01, \ 0.1000000000000000E+01, \ 0.1000000000000000E+01, \ 0.1000000000000000E+01, \ 0.1000000000000000E+01, \ 0.1000000000000000E+01, \ 0.1000000000000000E+01, \ 0.1000000000000000E+01, \ 0.1000000000000000E+01 )) if ( n_data < 0 ): n_data = 0 if ( n_max <= n_data ): n_data = 0 mu = 0.0 beta = 0.0 x = 0.0 f = 0.0 else: mu = mu_vec[n_data] beta = beta_vec[n_data] x = x_vec[n_data] f = f_vec[n_data] n_data = n_data + 1 return n_data, mu, beta, x, f def laplace_cdf_values_test ( ): #*****************************************************************************80 # ## LAPLACE_CDF_VALUES_TEST demonstrates the use of LAPLACE_CDF_VALUES. # # Licensing: # # This code is distributed under the GNU LGPL license. # # Modified: # # 17 February 2015 # # Author: # # John Burkardt # import platform print ( '' ) print ( 'LAPLACE_CDF_VALUES_TEST:' ) print ( ' Python version: %s' % ( platform.python_version ( ) ) ) print ( ' LAPLACE_CDF_VALUES stores values of the LAPLACE CDF.' ) print ( '' ) print ( ' Mu Beta X F(Mu,Beta,X)' ) print ( '' ) n_data = 0 while ( True ): n_data, mu, beta, x, f = laplace_cdf_values ( n_data ) if ( n_data == 0 ): break print ( ' %12f %12f %12f %24.16g' % ( mu, beta, x, f ) ) # # Terminate. # print ( '' ) print ( 'LAPLACE_CDF_VALUES_TEST:' ) print ( ' Normal end of execution.' ) return if ( __name__ == '__main__' ): from timestamp import timestamp timestamp ( ) laplace_cdf_values_test ( ) timestamp ( )