def r2_metric(solution, prediction, task=REGRESSION):
"""
1 - Mean squared error divided by variance
:param solution:
:param prediction:
:param task:
:return:
"""
mse = mv_mean((solution - prediction) ** 2)
var = mv_mean((solution - mv_mean(solution)) ** 2)
score = 1 - mse / var
return mv_mean(score)
def a_metric(solution, prediction, task=REGRESSION):
"""
1 - Mean absolute error divided by mean absolute deviation
:param solution:
:param prediction:
:param task:
:return:
"""
mae = mv_mean(np.abs(solution - prediction)) # mean absolute error
mad = mv_mean(
np.abs(solution - mv_mean(solution))) # mean absolute deviation
score = 1 - mae / mad
return mv_mean(score)
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 = mv_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 auc_metric(solution, prediction, task=BINARY_CLASSIFICATION):
"""
Normarlized Area under ROC curve (AUC).
Return Gini index = 2*AUC-1 for binary classification problems.
Should work for a vector of binary 0/1 (or -1/1)"solution" and any discriminant values
for the predictions. If solution and prediction are not vectors, the AUC
of the columns of the matrices are computed and averaged (with no weight).
The same for all classification problems (in fact it treats well only the
binary and multilabel classification problems).
:param solution:
:param prediction:
:param task:
:return:
"""
# auc = metrics.roc_auc_score(solution, prediction, average=None)
# There is a bug in metrics.roc_auc_score: auc([1,0,0],[1e-10,0,0])
# incorrect
label_num = solution.shape[1]
auc = np.empty(label_num)
for k in range(label_num):
r_ = tied_rank(prediction[:, k])
s_ = solution[:, k]
if sum(s_) == 0:
print(
'WARNING: no positive class example in class {}'.format(k + 1))
npos = sum(s_ == 1)
nneg = sum(s_ < 1)
auc[k] = (sum(r_[s_ == 1]) - npos * (npos + 1) / 2) / (nneg * npos)
return 2 * mv_mean(auc) - 1
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.
:param solution:
:param prediction:
:param task:
:return:
"""
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 = mv_mean(np.exp(-the_log_loss))
base_pac = mv_mean(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 = -mv_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 = - mv_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 f1_metric(solution, prediction, task=BINARY_CLASSIFICATION):
"""
Compute the normalized f1 measure.
The binarization differs
for the multi-label and multi-class case.
A non-weighted average over classes is taken.
The score is normalized.
: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
true_pos_num = sp.maximum(eps, tp + fn)
found_pos_num = sp.maximum(eps, tp + fp)
tp = sp.maximum(eps, tp)
tpr = tp / true_pos_num # true positive rate (recall)
ppv = tp / found_pos_num # positive predictive value (precision)
arithmetic_mean = 0.5 * sp.maximum(eps, tpr + ppv)
# Harmonic mean:
f1 = tpr * ppv / arithmetic_mean
# Average over all classes
f1 = mv_mean(f1)
# Normalize: 0 for random, 1 for perfect
if (task != MULTICLASS_CLASSIFICATION) or (label_num == 1):
# How to choose the "base_f1"?
# For the binary/multilabel classification case, one may want to predict all 1.
# In that case tpr = 1 and ppv = frac_pos. f1 = 2 * frac_pos / (1+frac_pos)
# frac_pos = mvmean(solution.ravel())
# base_f1 = 2 * frac_pos / (1+frac_pos)
# or predict random values with probability 0.5, in which case
# base_f1 = 0.5
# the first solution is better only if frac_pos > 1/3.
# The solution in which we predict according to the class prior frac_pos gives
# f1 = tpr = ppv = frac_pos, which is worse than 0.5 if frac_pos<0.5
# So, because the f1 score is used if frac_pos is small (typically <0.1)
# the best is to assume that base_f1=0.5
base_f1 = 0.5
# For the multiclass case, this is not possible (though it does not make much sense to
# use f1 for multiclass problems), so the best would be to assign values at random to get
# tpr=ppv=frac_pos, where frac_pos=1/label_num
else:
base_f1 = 1. / label_num
score = (f1 - base_f1) / sp.maximum(eps, (1 - base_f1))
return score