I was using KNN from sklearn and predicted the labels using predict_proba. I was expecting the values in the range of 0 to 1 since it tells the probability for a particular class. But I am only getting 0 & 1.
I have put large k values also but to no gain. Though I have only 1000 samples with features around 200 and the matrix is largely sparse.
Can anybody tell me what could be the solution here?
sklearn.neighbors.KNeighborsClassifier(n_neighbors=**k**)
The reason why you're getting only 0 & 1 is because of the n_neighbors = k parameter. If k value is set to 1, then you will get 0 or 1. If it's set to 2, you will get 0, 0.5 or 1. And if it's set to 3, then the probability outputs will be 0, 0.333, 0.666, or 1.
Also note that probability values are essentially meaningless in KNN. The algorithm is based on similarity and distance.
The reason might be lack of variety of data in training and test sets.
If the features of a sample may only exist in a particular class and its features don't exist in any sample of other classes in training set, then that sample will be predicted to belong that class with probability of 100% (1) and 0% (0) for other classes.
Otherwise; let say you have 2 classes and test a sample like knn.predict_proba(sample) and expect some result like [[0.47, 0.53]] The result will yield 1 in total in either way.
If thats the case, try generating your own test sample that has features from more than one classes objects in training set.
Related
I'm trying to make a ROC curve for my model while using a Naive Bayes Classifier. To do this, I need to change the value of the threshold for my classifier. The way I interpreted it, a list must be passed with the value for the threshold of each category. So if i had two categories, and t is the threshold I want to set (0 <= t <= 1), then I would have to pass a list like this: [1-t, t].
Anyways, when i tried doing the ROC curve, I got this:
Given the result, my idea was that the idea I had for the theshold might have been wrong, so I went to check the documentation for the Naive Bayes Classifier. But when I finally found an example i dont get what the criteria is for the parameter:
nb = nb.setThresholds([0.01, 10.00])
Does anyone know what must be passed to the threshold? Supose I want the theshold to be set at 0.7 (if the probability is over 0.7 i want the prediction to be 1), what should i pass to the threshold parameter?
As it says in pyspark.ml's documentation for NaiveBayes under the thresholds parameter:
The class with largest value p/t is predicted, where p is the original
probability of that class and t is the class's threshold.
Therefore, thresholds can be thought of as handicaps on the probabilities. To keep it simple, in the case of binary classification, you can set the thresholds as a value in the range [0, 1], such that they sum to 1. This will get you the desired rule of "Classify as True if the probability is over threshold T, otherwise classify as False".
For your specific ask of a 0.7 probability threshold, this would look like:
nb = nb.setThresholds([0.3, 0.7])
assuming that the first entry is the threshold for False and the second value is the thresold for True. Using these thresholds, the model would classify a class with False and True probabilities p_false and p_true by taking the greater value out of [p_false/0.3, p_true/0.7].
You can technically set the thresholds to any value. Just remember that the probability for class X will be divided by its respective threshold and compared against the other adjusted probabilities for the other classes.
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).
I have read an article on data leakage. In a hackathon there are two sets of data, train data on which participants train their algorithm and test set on which performance is measured.
Data leakage helps in getting a perfect score in test data, with out viewing train data by exploiting the leak.
I have read the article, but I am missing the crux how the leakage is exploited.
Steps as shown in article are following:
Let's load the test data.
Note, that we don't have any training data here, just test data. Moreover, we will not even use any features of test objects. All we need to solve this task is the file with the indices for the pairs, that we need to compare.
Let's load the data with test indices.
test = pd.read_csv('../test_pairs.csv')
test.head(10)
pairId FirstId SecondId
0 0 1427 8053
1 1 17044 7681
2 2 19237 20966
3 3 8005 20765
4 4 16837 599
5 5 3657 12504
6 6 2836 7582
7 7 6136 6111
8 8 23295 9817
9 9 6621 7672
test.shape[0]
368550
For example, we can think that there is a test dataset of images, and each image is assigned a unique Id from 0 to N−1 (N -- is the number of images). In the dataframe from above FirstId and SecondId point to these Id's and define pairs, that we should compare: e.g. do both images in the pair belong to the same class or not. So, for example for the first row: if images with Id=1427 and Id=8053 belong to the same class, we should predict 1, and 0 otherwise.
But in our case we don't really care about the images, and how exactly we compare the images (as long as comparator is binary).
print(test['FirstId'].nunique())
print(test['SecondId'].nunique())
26325
26310
So the number of pairs we are given to classify is very very small compared to the total number of pairs.
To exploit the leak we need to assume (or prove), that the total number of positive pairs is small, compared to the total number of pairs. For example: think about an image dataset with 1000 classes, N images per class. Then if the task was to tell whether a pair of images belongs to the same class or not, we would have 1000*N*(N−1)/2 positive pairs, while total number of pairs was 1000*N(1000N−1)/2.
Another example: in Quora competitition the task was to classify whether a pair of qustions are duplicates of each other or not. Of course, total number of question pairs is very huge, while number of duplicates (positive pairs) is much much smaller.
Finally, let's get a fraction of pairs of class 1. We just need to submit a constant prediction "all ones" and check the returned accuracy. Create a dataframe with columns pairId and Prediction, fill it and export it to .csv file. Then submit
test['Prediction'] = np.ones(test.shape[0])
sub=pd.DataFrame(test[['pairId','Prediction']])
sub.to_csv('sub.csv',index=False)
All ones have accuracy score is 0.500000.
So, we assumed the total number of pairs is much higher than the number of positive pairs, but it is not the case for the test set. It means that the test set is constructed not by sampling random pairs, but with a specific sampling algorithm. Pairs of class 1 are oversampled.
Now think, how we can exploit this fact? What is the leak here? If you get it now, you may try to get to the final answer yourself, othewise you can follow the instructions below.
Building a magic feature
In this section we will build a magic feature, that will solve the problem almost perfectly. The instructions will lead you to the correct solution, but please, try to explain the purpose of the steps we do to yourself -- it is very important.
Incidence matrix
First, we need to build an incidence matrix. You can think of pairs (FirstId, SecondId) as of edges in an undirected graph.
The incidence matrix is a matrix of size (maxId + 1, maxId + 1), where each row (column) i corresponds i-th Id. In this matrix we put the value 1to the position [i, j], if and only if a pair (i, j) or (j, i) is present in a given set of pais (FirstId, SecondId). All the other elements in the incidence matrix are zeros.
Important! The incidence matrices are typically very very sparse (small number of non-zero values). At the same time incidence matrices are usually huge in terms of total number of elements, and it is impossible to store them in memory in dense format. But due to their sparsity incidence matrices can be easily represented as sparse matrices. If you are not familiar with sparse matrices, please see wiki and scipy.sparse reference. Please, use any of scipy.sparseconstructors to build incidence matrix.
For example, you can use this constructor: scipy.sparse.coo_matrix((data, (i, j))). We highly recommend to learn to use different scipy.sparseconstuctors, and matrices types, but if you feel you don't want to use them, you can always build this matrix with a simple for loop. You will need first to create a matrix using scipy.sparse.coo_matrix((M, N), [dtype]) with an appropriate shape (M, N) and then iterate through (FirstId, SecondId) pairs and fill corresponding elements in matrix with ones.
Note, that the matrix should be symmetric and consist only of zeros and ones. It is a way to check yourself.
import networkx as nx
import numpy as np
import pandas as pd
import scipy.sparse
import matplotlib.pyplot as plt
test = pd.read_csv('../test_pairs.csv')
x = test[['FirstId','SecondId']].rename(columns={'FirstId':'col1', 'SecondId':'col2'})
y = test[['SecondId','FirstId']].rename(columns={'SecondId':'col1', 'FirstId':'col2'})
comb = pd.concat([x,y],ignore_index=True).drop_duplicates(keep='first')
comb.head()
col1 col2
0 1427 8053
1 17044 7681
2 19237 20966
3 8005 20765
4 16837 599
data = np.ones(comb.col1.shape, dtype=int)
inc_mat = scipy.sparse.coo_matrix((data,(comb.col1,comb.col2)), shape=(comb.col1.max() + 1, comb.col1.max() + 1))
rows_FirstId = inc_mat[test.FirstId.values,:]
rows_SecondId = inc_mat[test.SecondId.values,:]
f = rows_FirstId.multiply(rows_SecondId)
f = np.asarray(f.sum(axis=1))
f.shape
(368550, 1)
f = f.sum(axis=1)
f = np.squeeze(np.asarray(f))
print (f.shape)
Now build the magic feature
Why did we build the incidence matrix? We can think of the rows in this matix as of representations for the objects. i-th row is a representation for an object with Id = i. Then, to measure similarity between two objects we can measure similarity between their representations. And we will see, that such representations are very good.
Now select the rows from the incidence matrix, that correspond to test.FirstId's, and test.SecondId's.
So do not forget to convert pd.series to np.array
These lines should normally run very quickly
rows_FirstId = inc_mat[test.FirstId.values,:]
rows_SecondId = inc_mat[test.SecondId.values,:]
Our magic feature will be the dot product between representations of a pair of objects. Dot product can be regarded as similarity measure -- for our non-negative representations the dot product is close to 0 when the representations are different, and is huge, when representations are similar.
Now compute dot product between corresponding rows in rows_FirstId and rows_SecondId matrices.
From magic feature to binary predictions
But how do we convert this feature into binary predictions? We do not have a train set to learn a model, but we have a piece of information about test set: the baseline accuracy score that you got, when submitting constant. And we also have a very strong considerations about the data generative process, so probably we will be fine even without a training set.
We may try to choose a thresold, and set the predictions to 1, if the feature value f is higer than the threshold, and 0 otherwise. What threshold would you choose?
How do we find a right threshold? Let's first examine this feature: print frequencies (or counts) of each value in the feature f.
For example use np.unique function, check for flags
Function to count frequency of each element
from scipy.stats import itemfreq
itemfreq(f)
array([[ 14, 183279],
[ 15, 852],
[ 19, 546],
[ 20, 183799],
[ 21, 6],
[ 28, 54],
[ 35, 14]])
Do you see how this feature clusters the pairs? Maybe you can guess a good threshold by looking at the values?
In fact, in other situations it can be not that obvious, but in general to pick a threshold you only need to remember the score of your baseline submission and use this information.
Choose a threshold below:
pred = f > 14 # SET THRESHOLD HERE
pred
array([ True, False, True, ..., False, False, False], dtype=bool)
submission = test.loc[:,['pairId']]
submission['Prediction'] = pred.astype(int)
submission.to_csv('submission.csv', index=False)
I want to understand the idea behind this. How we are exploiting the leak from the test data only.
There's a hint in the article. The number of positive pairs should be 1000*N*(N−1)/2, while the number of all pairs is 1000*N(1000N−1)/2. Of course, the number of all pairs is much, much larger if the test set was sampled at random.
As the author mentions, after you evaluate your constant prediction of 1s on the test set, you can tell that the sampling was not done at random. The accuracy you obtain is 50%. Had the sampling been done correctly, this value should've been much lower.
Thus, they construct the incidence matrix and calculate the dot product (the measure of similarity) between the representations of our ID features. They then reuse the information about the accuracy obtained with constant predictions (at 50%) to obtain the corresponding threshold (f > 14). It's set to be greater than 14 because that constitutes roughly half of our test set, which in turn maps back to the 50% accuracy.
The "magic" value didn't have to be greater than 14. It could have been equal to 14. You could have adjusted this value after some leader board probing (as long as you're capturing half of the test set).
It was observed that the test data was not sampled properly; same-class pairs were oversampled. Thus there is a much higher probability of each pair in the training set to have target=1 than any random pair. This led to the belief that one could construct a similarity measure based only on the pairs that are present in the test, i.e., whether a pair made it to the test is itself a strong indicator of similarity.
Using this insight one can calculate an incidence matrix and represent each id j as a binary array (the i-th element representing the presence of i-j pair in test, and thus representing the strong probability of similarity between them). This is a pretty accurate measure, allowing one to find the "similarity" between two rows just by taking their dot product.
The cutoff arrived at is purely by the knowledge of target-distribution found by leaderboard probing.
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'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.