BEAT - robertodaza/competition-proof1/1

This algorithm is a legacy one. The API has changed since its implementation. New versions and forks will need to be updated.

This algorithm is splittable

Endpoint Groups 1

Algorithms have at least one input and one output. All algorithm endpoints are organized in groups. Groups are used by the platform to indicate which inputs and outputs are synchronized together. The first group is automatically synchronized with the channel defined by the block in which the algorithm is deployed.

Group: main

Endpoint Name	Data Format	Nature
features	tutorial/atvs_keystroke/1	Input
id	system/text/1	Input
scores	aythamimm/keystroke_model/8	Output

import numpy

def mad(lst):    
    if len(lst) < 1:
        return None
    if not(isinstance(lst[0], list)):        
        vec=[];
        m=numpy.median(lst);
        for i in range(len(lst)):
            vec.append(abs(lst[i]-m)); 
        return numpy.median(vec)
    else:
        mad_vec=[];
        for i in range(len(lst[0])):
            vec=[];            
            for j in range(len(lst)):
                vec.append(lst[j][i]);

m=numpy.mean(vec);            
            
            vec=[];            
            for j in range(len(lst)):
                vec.append(abs(lst[j][i]-m));
                
            mad_vec.append(numpy.mean(vec)+0.00001);

return mad_vec

class Algorithm:

def __init__(self):
        self._features1 = [] 
        self._features2 = [] 
        self._features3 = [] 
        self._features4 = [] 
        self._features5 = []   
    
    def process(self, inputs, outputs):

# collect all the features for the current template
        data = inputs['features'].data
        
        f1 = data.holdtime.given_name
        f2 = data.rplatency.given_name
        feature_vector=[]
        l_f=len(f1)
        for i_k in range(l_f):
            feature_vector.append(float(f1[i_k]))
        
        l_f=len(f2)
        for i_k in range(l_f):
            feature_vector.append(float(f2[i_k]))              
                
        self._features1.append(feature_vector)
                      
        f1 = data.holdtime.family_name
        f2 = data.rplatency.family_name
        feature_vector=[]
        l_f=len(f1)
        for i_k in range(l_f):
            feature_vector.append(float(f1[i_k]))
        
        l_f=len(f2)
        for i_k in range(l_f):
            feature_vector.append(float(f2[i_k]))              
                
        self._features2.append(feature_vector)
                
        f1 = data.holdtime.email
        f2 = data.rplatency.email
        feature_vector=[]
        l_f=len(f1)
        for i_k in range(l_f):
            feature_vector.append(float(f1[i_k]))
        
        l_f=len(f2)
        for i_k in range(l_f):
            feature_vector.append(float(f2[i_k]))              
                
        self._features3.append(feature_vector)
        
        f1 = data.holdtime.nationality
        f2 = data.rplatency.nationality 
        feature_vector=[]
        l_f=len(f1)
        for i_k in range(l_f):
            feature_vector.append(float(f1[i_k]))
        
        l_f=len(f2)
        for i_k in range(l_f):
            feature_vector.append(float(f2[i_k]))              
                
        self._features4.append(feature_vector)
        
        f1 = data.holdtime.id_number
        f2 = data.rplatency.id_number 
        feature_vector=[]
        l_f=len(f1)
        for i_k in range(l_f):
            feature_vector.append(float(f1[i_k]))
        
        l_f=len(f2)
        for i_k in range(l_f):
            feature_vector.append(float(f2[i_k]))              
                
        self._features5.append(feature_vector) 
                
        # Calculate mean and std
        if inputs["id"].isDataUnitDone():
            features11 = numpy.average(self._features1, axis=0)
            features12 = numpy.std(self._features1, axis=0)
            features13 = mad(self._features1)
            features21 = numpy.average(self._features2, axis=0)
            features22 = numpy.std(self._features2, axis=0)
            features23 = mad(self._features2)     
            features31 = numpy.average(self._features3, axis=0)
            features32 = numpy.std(self._features3, axis=0)
            features33 = mad(self._features3)
            features41 = numpy.average(self._features4, axis=0)
            features42 = numpy.std(self._features4, axis=0) 
            features43 = mad(self._features4) 
            features51 = numpy.average(self._features5, axis=0)
            features52 = numpy.std(self._features5, axis=0) 
            features53 = mad(self._features5) 
            # outputs data
            outputs["scores"].write({
                'given_name_average': 		features11,
                'given_name_std': 			features12,
                'given_name_mad': 			features13,
                'family_name_average': 		features21,
                'family_name_std': 			features22,
                'family_name_mad': 			features23,
                'email_average': 			features31,
                'email_std': 				features32,
                'email_mad': 				features33,
                'nationality_average': 		features41,
                'nationality_std': 			features42,
                'nationality_mad': 			features43,
                'id_number_average': 		features51,
                'id_number_std': 			features52,
                'id_number_mad': 			features53,
            }) 
            
            self._features1 = []
            self._features2 = []
            self._features3 = []
            self._features4 = []
            self._features5 = []

return True

The code for this algorithm in Python
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This algorithm is designed to be used as a simple enrollment strategy of keystroke data. It enrolls a model from several features by computing the average and standard deviation of the enrollment features.

Note

All features must have the same length.

No experiments are using this algorithm.

This algorithm was never executed.

algorithms robertodaza competition-proof1 1

Endpoint Groups 1

Group: main

algorithms

robertodaza

competition-proof1

1