def bac_metric (solution, prediction, task='binary.classification'):
''' Compute the normalized balanced accuracy. The binarization and
the normalization differ for the multi-label and multi-class case. '''
label_num = solution.shape[1]
score = np.zeros(label_num)
bin_prediction = binarize_predictions(prediction, task)
[tn,fp,tp,fn] = acc_stat(solution, bin_prediction)
# Bounding to avoid division by 0
eps = 1e-15
tp = sp.maximum (eps, tp)
pos_num = sp.maximum (eps, tp+fn)
tpr = tp / pos_num # true positive rate (sensitivity)
if (task != 'multiclass.classification') or (label_num==1):
tn = sp.maximum (eps, tn)
neg_num = sp.maximum (eps, tn+fp)
tnr = tn / neg_num # true negative rate (specificity)
bac = 0.5*(tpr + tnr)
base_bac = 0.5 # random predictions for binary case
else:
bac = tpr
base_bac = 1./label_num # random predictions for multiclass case
bac = mvmean(bac) # average over all classes
# Normalize: 0 for random, 1 for perfect
score = (bac - base_bac) / sp.maximum(eps, (1 - base_bac))
return score
python类maximum()的实例源码
def bac_metric (solution, prediction, task='binary.classification'):
''' Compute the normalized balanced accuracy. The binarization and
the normalization differ for the multi-label and multi-class case. '''
label_num = solution.shape[1]
score = np.zeros(label_num)
bin_prediction = binarize_predictions(prediction, task)
[tn,fp,tp,fn] = acc_stat(solution, bin_prediction)
# Bounding to avoid division by 0
eps = 1e-15
tp = sp.maximum (eps, tp)
pos_num = sp.maximum (eps, tp+fn)
tpr = tp / pos_num # true positive rate (sensitivity)
if (task != 'multiclass.classification') or (label_num==1):
tn = sp.maximum (eps, tn)
neg_num = sp.maximum (eps, tn+fp)
tnr = tn / neg_num # true negative rate (specificity)
bac = 0.5*(tpr + tnr)
base_bac = 0.5 # random predictions for binary case
else:
bac = tpr
base_bac = 1./label_num # random predictions for multiclass case
bac = mvmean(bac) # average over all classes
# Normalize: 0 for random, 1 for perfect
score = (bac - base_bac) / sp.maximum(eps, (1 - base_bac))
return score
def prior_log_loss(frac_pos, task = 'binary.classification'):
''' Baseline log loss. For multiplr classes ot labels return the volues for each column'''
eps = 1e-15
frac_pos_ = sp.maximum (eps, frac_pos)
if (task != 'multiclass.classification'): # binary case
frac_neg = 1-frac_pos
frac_neg_ = sp.maximum (eps, frac_neg)
pos_class_log_loss_ = - frac_pos * np.log(frac_pos_)
neg_class_log_loss_ = - frac_neg * np.log(frac_neg_)
base_log_loss = pos_class_log_loss_ + neg_class_log_loss_
# base_log_loss = mvmean(base_log_loss)
# print('binary {}'.format(base_log_loss))
# In the multilabel case, the right thing i to AVERAGE not sum
# We return all the scores so we can normalize correctly later on
else: # multiclass case
fp = frac_pos_ / sum(frac_pos_) # Need to renormalize the lines in multiclass case
# Only ONE label is 1 in the multiclass case active for each line
pos_class_log_loss_ = - frac_pos * np.log(fp)
base_log_loss = np.sum(pos_class_log_loss_)
return base_log_loss
# sklearn implementations for comparison
def bac_metric (solution, prediction, task='binary.classification'):
''' Compute the normalized balanced accuracy. The binarization and
the normalization differ for the multi-label and multi-class case. '''
label_num = solution.shape[1]
score = np.zeros(label_num)
bin_prediction = binarize_predictions(prediction, task)
[tn,fp,tp,fn] = acc_stat(solution, bin_prediction)
# Bounding to avoid division by 0
eps = 1e-15
tp = sp.maximum (eps, tp)
pos_num = sp.maximum (eps, tp+fn)
tpr = tp / pos_num # true positive rate (sensitivity)
if (task != 'multiclass.classification') or (label_num==1):
tn = sp.maximum (eps, tn)
neg_num = sp.maximum (eps, tn+fp)
tnr = tn / neg_num # true negative rate (specificity)
bac = 0.5*(tpr + tnr)
base_bac = 0.5 # random predictions for binary case
else:
bac = tpr
base_bac = 1./label_num # random predictions for multiclass case
bac = mvmean(bac) # average over all classes
# Normalize: 0 for random, 1 for perfect
score = (bac - base_bac) / sp.maximum(eps, (1 - base_bac))
return score
def prior_log_loss(frac_pos, task = 'binary.classification'):
''' Baseline log loss. For multiplr classes ot labels return the volues for each column'''
eps = 1e-15
frac_pos_ = sp.maximum (eps, frac_pos)
if (task != 'multiclass.classification'): # binary case
frac_neg = 1-frac_pos
frac_neg_ = sp.maximum (eps, frac_neg)
pos_class_log_loss_ = - frac_pos * np.log(frac_pos_)
neg_class_log_loss_ = - frac_neg * np.log(frac_neg_)
base_log_loss = pos_class_log_loss_ + neg_class_log_loss_
# base_log_loss = mvmean(base_log_loss)
# print('binary {}'.format(base_log_loss))
# In the multilabel case, the right thing i to AVERAGE not sum
# We return all the scores so we can normalize correctly later on
else: # multiclass case
fp = frac_pos_ / sum(frac_pos_) # Need to renormalize the lines in multiclass case
# Only ONE label is 1 in the multiclass case active for each line
pos_class_log_loss_ = - frac_pos * np.log(fp)
base_log_loss = np.sum(pos_class_log_loss_)
return base_log_loss
# sklearn implementations for comparison
def pac_metric (solution, prediction, task='binary.classification'):
''' Probabilistic Accuracy based on log_loss metric.
We assume the solution is in {0, 1} and prediction in [0, 1].
Otherwise, run normalize_array.'''
debug_flag=False
[sample_num, label_num] = solution.shape
if label_num==1: task='binary.classification'
eps = 1e-15
the_log_loss = log_loss(solution, prediction, task)
# Compute the base log loss (using the prior probabilities)
pos_num = 1.* sum(solution) # float conversion!
frac_pos = pos_num / sample_num # prior proba of positive class
the_base_log_loss = prior_log_loss(frac_pos, task)
# Alternative computation of the same thing (slower)
# Should always return the same thing except in the multi-label case
# For which the analytic solution makes more sense
if debug_flag:
base_prediction = np.empty(prediction.shape)
for k in range(sample_num): base_prediction[k,:] = frac_pos
base_log_loss = log_loss(solution, base_prediction, task)
diff = np.array(abs(the_base_log_loss-base_log_loss))
if len(diff.shape)>0: diff=max(diff)
if(diff)>1e-10:
print('Arrggh {} != {}'.format(the_base_log_loss,base_log_loss))
# Exponentiate to turn into an accuracy-like score.
# In the multi-label case, we need to average AFTER taking the exp
# because it is an NL operation
pac = mvmean(np.exp(-the_log_loss))
base_pac = mvmean(np.exp(-the_base_log_loss))
# Normalize: 0 for random, 1 for perfect
score = (pac - base_pac) / sp.maximum(eps, (1 - base_pac))
return score
def log_loss(solution, prediction, task = 'binary.classification'):
''' Log loss for binary and multiclass. '''
[sample_num, label_num] = solution.shape
eps = 1e-15
pred = np.copy(prediction) # beware: changes in prediction occur through this
sol = np.copy(solution)
if (task == 'multiclass.classification') and (label_num>1):
# Make sure the lines add up to one for multi-class classification
norma = np.sum(prediction, axis=1)
for k in range(sample_num):
pred[k,:] /= sp.maximum (norma[k], eps)
# Make sure there is a single label active per line for multi-class classification
sol = binarize_predictions(solution, task='multiclass.classification')
# For the base prediction, this solution is ridiculous in the multi-label case
# Bounding of predictions to avoid log(0),1/0,...
pred = sp.minimum (1-eps, sp.maximum (eps, pred))
# Compute the log loss
pos_class_log_loss = - mvmean(sol*np.log(pred), axis=0)
if (task != 'multiclass.classification') or (label_num==1):
# The multi-label case is a bunch of binary problems.
# The second class is the negative class for each column.
neg_class_log_loss = - mvmean((1-sol)*np.log(1-pred), axis=0)
log_loss = pos_class_log_loss + neg_class_log_loss
# Each column is an independent problem, so we average.
# The probabilities in one line do not add up to one.
# log_loss = mvmean(log_loss)
# print('binary {}'.format(log_loss))
# In the multilabel case, the right thing i to AVERAGE not sum
# We return all the scores so we can normalize correctly later on
else:
# For the multiclass case the probabilities in one line add up one.
log_loss = pos_class_log_loss
# We sum the contributions of the columns.
log_loss = np.sum(log_loss)
#print('multiclass {}'.format(log_loss))
return log_loss
def prior_log_loss(frac_pos, task = 'binary.classification'):
''' Baseline log loss. For multiplr classes ot labels return the volues for each column'''
eps = 1e-15
frac_pos_ = sp.maximum (eps, frac_pos)
if (task != 'multiclass.classification'): # binary case
frac_neg = 1-frac_pos
frac_neg_ = sp.maximum (eps, frac_neg)
pos_class_log_loss_ = - frac_pos * np.log(frac_pos_)
neg_class_log_loss_ = - frac_neg * np.log(frac_neg_)
base_log_loss = pos_class_log_loss_ + neg_class_log_loss_
# base_log_loss = mvmean(base_log_loss)
# print('binary {}'.format(base_log_loss))
# In the multilabel case, the right thing i to AVERAGE not sum
# We return all the scores so we can normalize correctly later on
else: # multiclass case
fp = frac_pos_ / sum(frac_pos_) # Need to renormalize the lines in multiclass case
# Only ONE label is 1 in the multiclass case active for each line
pos_class_log_loss_ = - frac_pos * np.log(fp)
base_log_loss = np.sum(pos_class_log_loss_)
return base_log_loss
# sklearn implementations for comparison
def binary_logloss(p, y):
epsilon = 1e-15
p = sp.maximum(epsilon, p)
p = sp.minimum(1-epsilon, p)
res = sum(y * sp.log(p) + sp.subtract(1, y) * sp.log(sp.subtract(1, p)))
res *= -1.0/len(y)
return res
def pac_metric (solution, prediction, task='binary.classification'):
''' Probabilistic Accuracy based on log_loss metric.
We assume the solution is in {0, 1} and prediction in [0, 1].
Otherwise, run normalize_array.'''
debug_flag=False
[sample_num, label_num] = solution.shape
if label_num==1: task='binary.classification'
eps = 1e-15
the_log_loss = log_loss(solution, prediction, task)
# Compute the base log loss (using the prior probabilities)
pos_num = 1.* sum(solution) # float conversion!
frac_pos = pos_num / sample_num # prior proba of positive class
the_base_log_loss = prior_log_loss(frac_pos, task)
# Alternative computation of the same thing (slower)
# Should always return the same thing except in the multi-label case
# For which the analytic solution makes more sense
if debug_flag:
base_prediction = np.empty(prediction.shape)
for k in range(sample_num): base_prediction[k,:] = frac_pos
base_log_loss = log_loss(solution, base_prediction, task)
diff = np.array(abs(the_base_log_loss-base_log_loss))
if len(diff.shape)>0: diff=max(diff)
if(diff)>1e-10:
print('Arrggh {} != {}'.format(the_base_log_loss,base_log_loss))
# Exponentiate to turn into an accuracy-like score.
# In the multi-label case, we need to average AFTER taking the exp
# because it is an NL operation
pac = mvmean(np.exp(-the_log_loss))
base_pac = mvmean(np.exp(-the_base_log_loss))
# Normalize: 0 for random, 1 for perfect
score = (pac - base_pac) / sp.maximum(eps, (1 - base_pac))
return score
def log_loss(solution, prediction, task = 'binary.classification'):
''' Log loss for binary and multiclass. '''
[sample_num, label_num] = solution.shape
eps = 1e-15
pred = np.copy(prediction) # beware: changes in prediction occur through this
sol = np.copy(solution)
if (task == 'multiclass.classification') and (label_num>1):
# Make sure the lines add up to one for multi-class classification
norma = np.sum(prediction, axis=1)
for k in range(sample_num):
pred[k,:] /= sp.maximum (norma[k], eps)
# Make sure there is a single label active per line for multi-class classification
sol = binarize_predictions(solution, task='multiclass.classification')
# For the base prediction, this solution is ridiculous in the multi-label case
# Bounding of predictions to avoid log(0),1/0,...
pred = sp.minimum (1-eps, sp.maximum (eps, pred))
# Compute the log loss
pos_class_log_loss = - mvmean(sol*np.log(pred), axis=0)
if (task != 'multiclass.classification') or (label_num==1):
# The multi-label case is a bunch of binary problems.
# The second class is the negative class for each column.
neg_class_log_loss = - mvmean((1-sol)*np.log(1-pred), axis=0)
log_loss = pos_class_log_loss + neg_class_log_loss
# Each column is an independent problem, so we average.
# The probabilities in one line do not add up to one.
# log_loss = mvmean(log_loss)
# print('binary {}'.format(log_loss))
# In the multilabel case, the right thing i to AVERAGE not sum
# We return all the scores so we can normalize correctly later on
else:
# For the multiclass case the probabilities in one line add up one.
log_loss = pos_class_log_loss
# We sum the contributions of the columns.
log_loss = np.sum(log_loss)
#print('multiclass {}'.format(log_loss))
return log_loss
def open_file(maindir):
"""
Creates the digital RF reading object.
Args:
maindir (:obj:'str'): The directory where the data is located.
Returns:
drfObj (obj:"DigitalRFReader"): Digital RF Reader object.
chandict (obj:"dict"): Dictionary that holds info for the channels.
start_indx (obj:'long'): Start index in samples.
end_indx (obj:'long'): End index in samples.
"""
mainpath = os.path.expanduser(maindir)
drfObj = drf.DigitalRFReader(mainpath)
chans = drfObj.get_channels()
chandict={}
start_indx, end_indx=[0, sp.inf]
# Get channel info
for ichan in chans:
curdict = {}
curdict['sind'], curdict['eind'] = drfObj.get_bounds(ichan)
# determine the read boundrys assuming the sampling is the same.
start_indx = sp.maximum(curdict['sind'], start_indx)
end_indx = sp.minimum(curdict['eind'], end_indx)
dmetadict = drfObj.read_metadata(start_indx, end_indx, ichan)
dmetakeys = dmetadict.keys()
curdict['sps'] = dmetadict[dmetakeys[0]]['samples_per_second']
curdict['fo'] = dmetadict[dmetakeys[0]]['center_frequencies'].ravel()[0]
chandict[ichan] = curdict
return (drfObj, chandict, start_indx, end_indx)
def binary_logloss(p, y):
epsilon = 1e-15
p = sp.maximum(epsilon, p)
p = sp.minimum(1-epsilon, p)
res = sum(y * sp.log(p) + sp.subtract(1, y) * sp.log(sp.subtract(1, p)))
res *= -1.0/len(y)
return res
def pac_metric (solution, prediction, task='binary.classification'):
''' Probabilistic Accuracy based on log_loss metric.
We assume the solution is in {0, 1} and prediction in [0, 1].
Otherwise, run normalize_array.'''
debug_flag=False
[sample_num, label_num] = solution.shape
if label_num==1: task='binary.classification'
eps = 1e-15
the_log_loss = log_loss(solution, prediction, task)
# Compute the base log loss (using the prior probabilities)
pos_num = 1.* sum(solution) # float conversion!
frac_pos = pos_num / sample_num # prior proba of positive class
the_base_log_loss = prior_log_loss(frac_pos, task)
# Alternative computation of the same thing (slower)
# Should always return the same thing except in the multi-label case
# For which the analytic solution makes more sense
if debug_flag:
base_prediction = np.empty(prediction.shape)
for k in range(sample_num): base_prediction[k,:] = frac_pos
base_log_loss = log_loss(solution, base_prediction, task)
diff = np.array(abs(the_base_log_loss-base_log_loss))
if len(diff.shape)>0: diff=max(diff)
if(diff)>1e-10:
print('Arrggh {} != {}'.format(the_base_log_loss,base_log_loss))
# Exponentiate to turn into an accuracy-like score.
# In the multi-label case, we need to average AFTER taking the exp
# because it is an NL operation
pac = mvmean(np.exp(-the_log_loss))
base_pac = mvmean(np.exp(-the_base_log_loss))
# Normalize: 0 for random, 1 for perfect
score = (pac - base_pac) / sp.maximum(eps, (1 - base_pac))
return score
def log_loss(solution, prediction, task = 'binary.classification'):
''' Log loss for binary and multiclass. '''
[sample_num, label_num] = solution.shape
eps = 1e-15
pred = np.copy(prediction) # beware: changes in prediction occur through this
sol = np.copy(solution)
if (task == 'multiclass.classification') and (label_num>1):
# Make sure the lines add up to one for multi-class classification
norma = np.sum(prediction, axis=1)
for k in range(sample_num):
pred[k,:] /= sp.maximum (norma[k], eps)
# Make sure there is a single label active per line for multi-class classification
sol = binarize_predictions(solution, task='multiclass.classification')
# For the base prediction, this solution is ridiculous in the multi-label case
# Bounding of predictions to avoid log(0),1/0,...
pred = sp.minimum (1-eps, sp.maximum (eps, pred))
# Compute the log loss
pos_class_log_loss = - mvmean(sol*np.log(pred), axis=0)
if (task != 'multiclass.classification') or (label_num==1):
# The multi-label case is a bunch of binary problems.
# The second class is the negative class for each column.
neg_class_log_loss = - mvmean((1-sol)*np.log(1-pred), axis=0)
log_loss = pos_class_log_loss + neg_class_log_loss
# Each column is an independent problem, so we average.
# The probabilities in one line do not add up to one.
# log_loss = mvmean(log_loss)
# print('binary {}'.format(log_loss))
# In the multilabel case, the right thing i to AVERAGE not sum
# We return all the scores so we can normalize correctly later on
else:
# For the multiclass case the probabilities in one line add up one.
log_loss = pos_class_log_loss
# We sum the contributions of the columns.
log_loss = np.sum(log_loss)
#print('multiclass {}'.format(log_loss))
return log_loss
def loglossl(act, pred):
epsilon = 1e-15
pred = sp.maximum(epsilon, pred)
pred = sp.minimum(1-epsilon, pred)
ll = sum(act*sp.log(pred) + sp.subtract(1,act)*sp.log(sp.subtract(1,pred)))
ll = ll * -1.0/len(act)
return ll
def my_logloss(act, pred):
epsilon = 1e-15
pred = K.maximum(epsilon, pred)
pred = K.minimum(1 - epsilon, pred)
ll = K.sum(act * K.log(pred) + (1 - act) * K.log(1 - pred))
ll = ll * -1.0 / K.shape(act)[0]
return ll
def logloss(act, pred):
'''
????????
:param act:
:param pred:
:return:
'''
epsilon = 1e-15
pred = sp.maximum(epsilon, pred)
pred = sp.minimum(1 - epsilon, pred)
ll = sum(act * sp.log(pred) + sp.subtract(1, act) * sp.log(sp.subtract(1, pred)))
ll = ll * -1.0 / len(act)
return ll
def binary_logloss(p, y):
epsilon = 1e-15
p = sp.maximum(epsilon, p)
p = sp.minimum(1-epsilon, p)
res = sum(y * sp.log(p) + sp.subtract(1, y) * sp.log(sp.subtract(1, p)))
res *= -1.0/len(y)
return res
classification_metrics.py 文件源码
项目：AutoML-Challenge
作者: postech-mlg-exbrain
项目源码
文件源码
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def acc_metric(solution, prediction, task=BINARY_CLASSIFICATION):
"""
Compute the accuracy.
Get the accuracy stats
acc = (tpr + fpr) / (tn + fp + tp + fn)
Normalize, so 1 is the best and zero mean random...
:param solution:
:param prediction:
:param task:
:return:
"""
label_num = solution.shape[1]
bin_predictions = binarize_predictions(prediction, task)
tn, fp, tp, fn = acc_stat(solution, bin_predictions)
# Bounding to avoid division by 0
eps = np.float(1e-15)
tp = np.sum(tp)
fp = np.sum(fp)
tn = np.sum(tn)
fn = np.sum(fn)
if (task != MULTICLASS_CLASSIFICATION) or (label_num == 1):
accuracy = (np.sum(tp) + np.sum(tn)) / (
np.sum(tp) + np.sum(fp) + np.sum(tn) + np.sum(fn)
)
else:
accuracy = np.sum(tp) / (np.sum(tp) + np.sum(fp))
if (task != MULTICLASS_CLASSIFICATION) or (label_num == 1):
base_accuracy = 0.5 # random predictions for binary case
else:
base_accuracy = 1. / label_num
# Normalize: 0 for random, 1 for perfect
score = (accuracy - base_accuracy) / sp.maximum(eps, (1 - base_accuracy))
return score
classification_metrics.py 文件源码
项目：AutoML-Challenge
作者: postech-mlg-exbrain
项目源码
文件源码
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def bac_metric(solution, prediction, task=BINARY_CLASSIFICATION):
"""
Compute the normalized balanced accuracy.
The binarization and
the normalization differ for the multi-label and multi-class case.
:param solution:
:param prediction:
:param task:
:return:
"""
label_num = solution.shape[1]
score = np.zeros(label_num)
bin_prediction = binarize_predictions(prediction, task)
[tn, fp, tp, fn] = acc_stat(solution, bin_prediction)
# Bounding to avoid division by 0
eps = 1e-15
tp = sp.maximum(eps, tp)
pos_num = sp.maximum(eps, tp + fn)
tpr = tp / pos_num # true positive rate (sensitivity)
if (task != MULTICLASS_CLASSIFICATION) or (label_num == 1):
tn = sp.maximum(eps, tn)
neg_num = sp.maximum(eps, tn + fp)
tnr = tn / neg_num # true negative rate (specificity)
bac = 0.5 * (tpr + tnr)
base_bac = 0.5 # random predictions for binary case
else:
bac = tpr
base_bac = 1. / label_num # random predictions for multiclass case
bac = np.mean(bac) # average over all classes
# Normalize: 0 for random, 1 for perfect
score = (bac - base_bac) / sp.maximum(eps, (1 - base_bac))
return score
def log_loss(solution, prediction, task=BINARY_CLASSIFICATION):
"""Log loss for binary and multiclass."""
[sample_num, label_num] = solution.shape
eps = 1e-15
pred = np.copy(prediction) # beware: changes in prediction occur through this
sol = np.copy(solution)
if (task == MULTICLASS_CLASSIFICATION) and (label_num > 1):
# Make sure the lines add up to one for multi-class classification
norma = np.sum(prediction, axis=1)
for k in range(sample_num):
pred[k, :] /= sp.maximum(norma[k], eps)
# Make sure there is a single label active per line for multi-class
# classification
sol = binarize_predictions(solution, task=MULTICLASS_CLASSIFICATION)
# For the base prediction, this solution is ridiculous in the
# multi-label case
# Bounding of predictions to avoid log(0),1/0,...
pred = sp.minimum(1 - eps, sp.maximum(eps, pred))
# Compute the log loss
pos_class_log_loss = -np.mean(sol * np.log(pred), axis=0)
if (task != MULTICLASS_CLASSIFICATION) or (label_num == 1):
# The multi-label case is a bunch of binary problems.
# The second class is the negative class for each column.
neg_class_log_loss = -np.mean((1 - sol) * np.log(1 - pred), axis=0)
log_loss = pos_class_log_loss + neg_class_log_loss
# Each column is an independent problem, so we average.
# The probabilities in one line do not add up to one.
# log_loss = mvmean(log_loss)
# print('binary {}'.format(log_loss))
# In the multilabel case, the right thing i to AVERAGE not sum
# We return all the scores so we can normalize correctly later on
else:
# For the multiclass case the probabilities in one line add up one.
log_loss = pos_class_log_loss
# We sum the contributions of the columns.
log_loss = np.sum(log_loss)
# print('multiclass {}'.format(log_loss))
return log_loss
def binary_logloss(p, y):
epsilon = 1e-15
p = sp.maximum(epsilon, p)
p = sp.minimum(1-epsilon, p)
res = sum(y * sp.log(p) + sp.subtract(1, y) * sp.log(sp.subtract(1, p)))
res *= -1.0/len(y)
return res
def logloss(act, pred):
epsilon = 1e-15
pred = sp.maximum(epsilon, pred)
pred = sp.minimum(1-epsilon, pred)
ll = sum(act*sp.log(pred) + sp.subtract(1,act)*sp.log(sp.subtract(1,pred)))
ll = ll * -1.0/len(act)
return ll
def logloss(act, pred):
epsilon = 1e-15
pred = sp.maximum(epsilon, pred)
pred = sp.minimum(1-epsilon, pred)
ll = sum(act*sp.log(pred) + sp.subtract(1,act)*sp.log(sp.subtract(1,pred)))
ll = ll * -1.0/len(act)
return ll
def binary_logloss(p, y):
epsilon = 1e-15
p = sp.maximum(epsilon, p)
p = sp.minimum(1-epsilon, p)
res = sum(y*sp.log(p) + sp.subtract(1,y)*sp.log(sp.subtract(1,p)))
res *= -1.0/len(y)
return res
def binary_logloss(p, y):
epsilon = 1e-15
p = sp.maximum(epsilon, p)
p = sp.minimum(1-epsilon, p)
res = sum(y * sp.log(p) + sp.subtract(1, y) * sp.log(sp.subtract(1, p)))
res *= -1.0/len(y)
return res
def logloss(act, preds):
epsilon = 1e-15
preds = sp.maximum(epsilon, preds)
preds = sp.minimum(1 - epsilon, preds)
ll = sum(act * sp.log(preds) + sp.subtract(1, act) * sp.log(sp.subtract(1, preds)))
ll = ll * -1.0 / len(act)
return ll
def logloss(act, preds):
epsilon = 1e-15
preds = sp.maximum(epsilon, preds)
preds = sp.minimum(1 - epsilon, preds)
ll = sum(act * sp.log(preds) + sp.subtract(1, act) * sp.log(sp.subtract(1, preds)))
ll = ll * -1.0 / len(act)
return ll
def logloss(act, preds):
epsilon = 1e-15
preds = sp.maximum(epsilon, preds)
preds = sp.minimum(1 - epsilon, preds)
ll = sum(act * sp.log(preds) + sp.subtract(1, act) * sp.log(sp.subtract(1, preds)))
ll = ll * -1.0 / len(act)
return ll
