I trained my model in keras for binary classification. I used Resnet preformer on ImageNet and I got 95% of accuracy. In my dataset, I have 9004 images for training divided into the two class and 2250 images for test divided into the two class. But the confusion matrix give me
4502 0
4502 0
can some one help me to know what is the meaning of this resul?
The interpretation for your result is the following (please note that for simplicity the indices start from 1 and not 0):
For calculating the number of correct predictions on your test dataset, one needs to sum the main diagonal of the matrix.
For the first class (class 1), all your predictions are correct.
This can be inferred from your confusion matrix, as the element on the position [1,1] (first_row,first_col) is 4502. Since you have 0 the element on the position [1,2], it means that all the predictions for class 1 are correct.
However, for the second class, which has on the position [2,2] the value 0, this means that none of your predictions for this class are correct.
Practically, we can easily verify that 4502 is on the position [2,1].
Notes:
You may have calculated the accuracy/confusion matrix on the wrong dataset. According to your description, 4502 * 2 = 9004, which means that the confusion matrix that you are giving here is for the training set not for the test set.
Whenever you see a number on the confusion matrix that does not belong to the main diagonal, that means that you have a case of FP(false positive) or FN(false negative).
Related
I have a binary classification problem for detecting AO/Non-AO images, using PyTorch for this purpose.
First, I load the data using the ImageFolder utility.
The Dataset class to label mapping in dataset.class_to_idx is {0: 'AO', 1: 'Non-AO'}.
So, my 'positive class' AO, is assigned a label 0, and my 'negative class' Non-AO is assigned a label 1.
Then I train and validate the model without any issues.
When I come to testing, I need to calculate some metrics on the test data.
Here is where I am confused.
[Method A]
fpr, tpr, thresholds = roc_curve(y_true, y_score)
roc_auc = auc(fpr, tpr)
[Method B]
# because 0 is my actual 'positive' class for this problem
fpr, tpr, thresholds = roc_curve(y_true, y_score, pos_label=0)
roc_auc = auc(fpr, tpr)
Now, this second curve is basically the mirror of the first one along the diagonal, right?
And I think, that it can't be the correct curve, because I checked the accuracy of the model by directly comparing y_true and y_pred to get the following accuracies.
Apart from this, here is what my confusion matrix looks like.
So, my first question is, am I doing something wrong? What is the significance of the curve from Method B? Can I say that Method A gives me the correct ROC curve for my classification task? If not, then how do I proceed for getting the correct curve?
What does true positive or true negative or any of the other terms signify for my confusion matrix? Does the matrix consider 0 : AO as negative and 1 : Non-AO as positive (I think so, yes) or the vice versa?
If 0 is indeed being considered as negative, when I actually want 0 to be considered as positive, how can I make changes to reflect so in the matrix (because I am using the matrix later to calculate other matrics like specificity, sensitivity, etc) ?
I am doing a project on multiclass semantic segmentation. I have formulated a model that outputs pretty descent segmented images by decreasing the loss value. However, I cannot evaluate the model performance in metrics, such as meanIoU or Dice coefficient.
In case of binary semantic segmentation it was easy just to set the threshold of 0.5, to classify the outputs as an object or background, but it does not work in the case of multiclass semantic segmentation. Could you please tell me how to obtain model performance on the aforementioned metrics? Any help will be highly appreciated!
By the way, I am using PyTorch framework and CamVid dataset.
If anyone is interested in this answer, please also look at this issue. The author of the issue points out that mIoU can be computed in a different way (and that method is more accepted in literature). So, consider that before using the implementation for any formal publication.
Basically, the other method suggested by the issue-poster is to separately accumulate the intersections and unions over the entire dataset and divide them at the final step. The method in the below original answer computes intersection and union for a batch of images, then divides them to get IoU for the current batch, and then takes a mean of the IoUs over the entire dataset.
However, this below given original method is problematic because the final mean IoU would vary with the batch-size. On the other hand, the mIoU would not vary with the batch size for the method mentioned in the issue as the separate accumulation would ensure that batch size is irrelevant (though higher batch size can definitely help speed up the evaluation).
Original answer:
Given below is an implementation of mean IoU (Intersection over Union) in PyTorch.
def mIOU(label, pred, num_classes=19):
pred = F.softmax(pred, dim=1)
pred = torch.argmax(pred, dim=1).squeeze(1)
iou_list = list()
present_iou_list = list()
pred = pred.view(-1)
label = label.view(-1)
# Note: Following for loop goes from 0 to (num_classes-1)
# and ignore_index is num_classes, thus ignore_index is
# not considered in computation of IoU.
for sem_class in range(num_classes):
pred_inds = (pred == sem_class)
target_inds = (label == sem_class)
if target_inds.long().sum().item() == 0:
iou_now = float('nan')
else:
intersection_now = (pred_inds[target_inds]).long().sum().item()
union_now = pred_inds.long().sum().item() + target_inds.long().sum().item() - intersection_now
iou_now = float(intersection_now) / float(union_now)
present_iou_list.append(iou_now)
iou_list.append(iou_now)
return np.mean(present_iou_list)
Prediction of your model will be in one-hot form, so first take softmax (if your model doesn't already) followed by argmax to get the index with the highest probability at each pixel. Then, we calculate IoU for each class (and take the mean over it at the end).
We can reshape both the prediction and the label as 1-D vectors (I read that it makes the computation faster). For each class, we first identify the indices of that class using pred_inds = (pred == sem_class) and target_inds = (label == sem_class). The resulting pred_inds and target_inds will have 1 at pixels labelled as that particular class while 0 for any other class.
Then, there is a possibility that the target does not contain that particular class at all. This will make that class's IoU calculation invalid as it is not present in the target. So, you assign such classes a NaN IoU (so you can identify them later) and not involve them in the calculation of the mean.
If the particular class is present in the target, then pred_inds[target_inds] will give a vector of 1s and 0s where indices with 1 are those where prediction and target are equal and zero otherwise. Taking the sum of all elements of this will give us the intersection.
If we add all the elements of pred_inds and target_inds, we'll get the union + intersection of pixels of that particular class. So, we subtract the already calculated intersection to get the union. Then, we can divide the intersection and union to get the IoU of that particular class and add it to a list of valid IoUs.
At the end, you take the mean of the entire list to get the mIoU. If you want the Dice Coefficient, you can calculate it in a similar fashion.
I am using NB for document classification and trying to understand threshold parameter to see how it can help to optimize algorithm.
Spark ML 2.0 thresholds doc says:
Param for Thresholds in multi-class classification to adjust the probability of predicting each class. Array must have length equal to the number of classes, with values >= 0. The class with largest value p/t is predicted, where p is the original probability of that class and t is the class' threshold.
0) Can someone explain this better? What goal it can achieve? My general idea is if you have threshold 0.7 then at least one class prediction probability should be more then 0.7 if not then prediction should return empty. Means classify it as 'uncertain' or just leave empty for prediction column. How can p/t function going to achieve that when you still pick the category with max probability?
1) What probability it adjust? default column 'probability' is actually conditional probability and 'rawPrediction' is
confidence according to document. I believe threshold will adjust 'rawPrediction' not 'probability' column. Am I right?
2) Here's how some of my probability and rawPrediction vector look like. How do I set threshold values based on this so I can remove certain uncertain classification? probability is between 0 and 1 but rawPrediction seems to be on log scale here.
Probability:
[2.233368649314982E-15,1.6429456680945863E-9,1.4377313514127723E-15,7.858651849363202E-15]
rawPrediction:
[-496.9606736723107,-483.452183395287,-497.40111830218746]
Basically I want classifier to leave Prediction column empty if it doesn't have any probability that is more then 0.7 percent.
Also, how to classify something as uncertain when more then one category has very close scores e.g. 0.812, 0.800, 0.799 . Picking max is something I may not want here but instead classify as "uncertain" or leave empty and I can do further analysis and treatment for those documents or train another model for those docs.
I haven't played with it, but the intent is to supply different threshold values for each class. I've extracted this example from the docstring:
model = nb.fit(df)
>>> result.prediction
1.0
>>> result.probability
DenseVector([0.42..., 0.57...])
>>> result.rawPrediction
DenseVector([-1.60..., -1.32...])
>>> nb = nb.setThresholds([0.01, 10.00])
>>> model3 = nb.fit(df)
>>> result = model3.transform(test0).head()
>>> result.prediction
0.0
If I understand correctly, the effect was to transform [0.42, 0.58] into [.42/.01, .58/10] = [42, 5.8], switching the prediction ("largest p/t") from column 1 (third row above) to column 0 (last row above). However, I couldn't find the logic in the source. Anyone?
Stepping back: I do not see a built-in way to do what you want: be agnostic if no class dominates. You will have to add that with something like:
def weak(probs, threshold=.7, epsilon=.01):
return np.all(probs < threshold) or np.max(np.diff(probs)) < epsilon
>>> cases = [[.5,.5],[.5,.7],[.7,.705],[.6,.1]]
>>> for case in cases:
... print '{:15s} - {}'.format(case, weak(case))
[0.5, 0.5] - True
[0.5, 0.7] - False
[0.7, 0.705] - True
[0.6, 0.1] - True
(Notice I haven't checked whether probs is a legal probability distribution.)
Alternatively, if you are not actually making a hard decision, use the predicted probabilities and a metric like Brier score, log loss, or info gain that accounts for the calibration as well as the accuracy.
I am trying to implement Expectation Maximization algorithm(Gaussian Mixture Model) on a data set data=[[x,y],...]. I am using mv_norm.pdf(data, mean,cov) function to calculate cluster responsibilities. But after calculating new values of covariance (cov matrix) after 6-7 iterations, cov matrix is becoming singular i.e determinant of cov is 0 (very small value) and hence it is giving errors
ValueError: the input matrix must be positive semidefinite
and
raise np.linalg.LinAlgError('singular matrix')
Can someone suggest any solution for this?
#E-step: Compute cluster responsibilities, given cluster parameters
def calculate_cluster_responsibility(data,centroids,cov_m):
pdfmain=[[] for i in range(0,len(data))]
for i in range(0,len(data)):
sum1=0
pdfeach=[[] for m in range(0,len(centroids))]
pdfeach[0]=1/3.*mv_norm.pdf(data[i], mean=centroids[0],cov=[[cov_m[0][0][0],cov_m[0][0][1]],[cov_m[0][1][0],cov_m[0][1][1]]])
pdfeach[1]=1/3.*mv_norm.pdf(data[i], mean=centroids[1],cov=[[cov_m[1][0][0],cov_m[1][0][1]],[cov_m[1][1][0],cov_m[0][1][1]]])
pdfeach[2]=1/3.*mv_norm.pdf(data[i], mean=centroids[2],cov=[[cov_m[2][0][0],cov_m[2][0][1]],[cov_m[2][1][0],cov_m[2][1][1]]])
sum1+=pdfeach[0]+pdfeach[1]+pdfeach[2]
pdfeach[:] = [x / sum1 for x in pdfeach]
pdfmain[i]=pdfeach
global old_pdfmain
if old_pdfmain==pdfmain:
return
old_pdfmain=copy.deepcopy(pdfmain)
softcounts=[sum(i) for i in zip(*pdfmain)]
calculate_cluster_weights(data,centroids,pdfmain,soft counts)
Initially, I've passed [[3,0],[0,3]] for each cluster covariance since expected number of clusters is 3.
Can someone suggest any solution for this?
The problem is your data lies in some manifold of dimension strictly smaller than the input data. In other words for example your data lies on a circle, while you have 3 dimensional data. As a consequence when your method tries to estimate 3 dimensional ellipsoid (covariance matrix) that fits your data - it fails since the optimal one is a 2 dimensional ellipse (third dimension is 0).
How to fix it? You will need some regularization of your covariance estimator. There are many possible solutions, all in M step, not E step, the problem is with computing covariance:
Simple solution, instead of doing something like cov = np.cov(X) add some regularizing term, like cov = np.cov(X) + eps * np.identity(X.shape[1]) with small eps
Use nicer estimator like LedoitWolf estimator from scikit-learn.
Initially, I've passed [[3,0],[0,3]] for each cluster covariance since expected number of clusters is 3.
This makes no sense, covariance matrix values has nothing to do with amount of clusters. You can initialize it with anything more or less resonable.
I've read from the relevant documentation that :
Class balancing can be done by sampling an equal number of samples from each class, or preferably by normalizing the sum of the sample weights (sample_weight) for each class to the same value.
But, it is still unclear to me how this works. If I set sample_weight with an array of only two possible values, 1's and 2's, does this mean that the samples with 2's will get sampled twice as often as the samples with 1's when doing the bagging? I cannot think of a practical example for this.
Some quick preliminaries:
Let's say we have a classification problem with K classes. In a region of feature space represented by the node of a decision tree, recall that the "impurity" of the region is measured by quantifying the inhomogeneity, using the probability of the class in that region. Normally, we estimate:
Pr(Class=k) = #(examples of class k in region) / #(total examples in region)
The impurity measure takes as input, the array of class probabilities:
[Pr(Class=1), Pr(Class=2), ..., Pr(Class=K)]
and spits out a number, which tells you how "impure" or how inhomogeneous-by-class the region of feature space is. For example, the gini measure for a two class problem is 2*p*(1-p), where p = Pr(Class=1) and 1-p=Pr(Class=2).
Now, basically the short answer to your question is:
sample_weight augments the probability estimates in the probability array ... which augments the impurity measure ... which augments how nodes are split ... which augments how the tree is built ... which augments how feature space is diced up for classification.
I believe this is best illustrated through example.
First consider the following 2-class problem where the inputs are 1 dimensional:
from sklearn.tree import DecisionTreeClassifier as DTC
X = [[0],[1],[2]] # 3 simple training examples
Y = [ 1, 2, 1 ] # class labels
dtc = DTC(max_depth=1)
So, we'll look trees with just a root node and two children. Note that the default impurity measure the gini measure.
Case 1: no sample_weight
dtc.fit(X,Y)
print dtc.tree_.threshold
# [0.5, -2, -2]
print dtc.tree_.impurity
# [0.44444444, 0, 0.5]
The first value in the threshold array tells us that the 1st training example is sent to the left child node, and the 2nd and 3rd training examples are sent to the right child node. The last two values in threshold are placeholders and are to be ignored. The impurity array tells us the computed impurity values in the parent, left, and right nodes respectively.
In the parent node, p = Pr(Class=1) = 2. / 3., so that gini = 2*(2.0/3.0)*(1.0/3.0) = 0.444..... You can confirm the child node impurities as well.
Case 2: with sample_weight
Now, let's try:
dtc.fit(X,Y,sample_weight=[1,2,3])
print dtc.tree_.threshold
# [1.5, -2, -2]
print dtc.tree_.impurity
# [0.44444444, 0.44444444, 0.]
You can see the feature threshold is different. sample_weight also affects the impurity measure in each node. Specifically, in the probability estimates, the first training example is counted the same, the second is counted double, and the third is counted triple, due to the sample weights we've provided.
The impurity in the parent node region is the same. This is just a coincidence. We can compute it directly:
p = Pr(Class=1) = (1+3) / (1+2+3) = 2.0/3.0
The gini measure of 4/9 follows.
Now, you can see from the chosen threshold that the first and second training examples are sent to the left child node, while the third is sent to the right. We see that impurity is calculated to be 4/9 also in the left child node because:
p = Pr(Class=1) = 1 / (1+2) = 1/3.
The impurity of zero in the right child is due to only one training example lying in that region.
You can extend this with non-integer sample-wights similarly. I recommend trying something like sample_weight = [1,2,2.5], and confirming the computed impurities.