I am trying to understand the predict function in Python statsmodels for a Logit model. Its documentation is here.
When I build a Logit Model and use predict, it returns values from 0 to 1 as opposed to 0 or 1. Now I read this saying these are probabilities and we need a threshold. Python statsmodel.api logistic regression (Logit)
Now, I want to produce AUC numbers and I use roc_auc_score from sklearn (docs).
Here is when I start getting confused.
When I put in the raw predicted values (probabilities) from my Logit model into the roc_auc_score as the second argument y_score, I get a reasonable AUC value of around 80%. How does the roc_auc_score function know which of my probabilities equate to 1 and which equate to 0? Nowhere was I given an opportunity to set the threshold.
When I manually convert my probabilities into 0 or 1 using a threshold of 0.5, I get an AUC of around 50%. Why would this happen?
Here's some code:
m1_result = m1.fit(disp = False)
roc_auc_score(y, m1_result.predict(X1))
AUC: 0.80
roc_auc_score(y, [1 if X >=0.5 else 0 for X in m1_result.predict(X1)])
AUC: 0.50
Why is this the case?
Your 2nd way of calculating the AUC is wrong; by definition, AUC needs probabilities, and not hard class predictions 0/1 generated after thresholding, as you do here. So, your AUC is 0.80.
You don't set a threshold yourself in AUC calculation; roughly speaking, as I have explained elsewhere, the AUC measures the performance of a binary classifier averaged across all possible decision thresholds.
It would be overkill to explain again here the rationale and details of AUC calculation; instead, these other SE threads (and the links therein) will help you get the idea:
In Classification, what is the difference between the test accuracy and the AUC score?
Advantages of AUC vs standard accuracy
Getting a low ROC AUC score but a high accuracy
Comparing AUC, log loss and accuracy scores between models
predict yields the estimated probability of event according to your fitted model. That is, each element corresponds to the predicted probability that your model calculated for each observation.
The process behind building a ROC curve consists of selecting each predicted probability as a threshold, measuring its false positive and true positive rates and plotting these results as a line graph. The area below this curve is the AUC.
To visualize this, imagine you had the following data:
observation
observed_result
predicted_prob
1
0
0.1
2
0
0.5
3
1
0.9
The function roc_auc_score will do the following:
Use 0.1 as the threshold such that all observations with predicted_prob ≤ 0.1 are classified as 0 and those with predicted_prob > 0.1 will be classified as 1
Use 0.5 as the threshold such that all observations with predicted_prob ≤ 0.5 are classified as 0 and those with predicted_prob > 0.5 will be classified as 1
Use 0.9 as the threshold such that all observations with predicted_prob ≤ 0.9 are classified as 0 and those with predicted_prob > 0.9 will be classified as 1
Each of the three different thresholds (0.1, 0.5 and 0.9) will result in its own false positive and true positive rates. The false positive rates are plotted along the x-axis, while the true positive rates are plotted in the y-axis.
As you can guess, you need to test many thresholds to plot a smooth curve. If you use 0.5 as a threshold and pass this to roc_auc_curve, you are testing out the false positive and true positive rates of a single threshold. This is incorrect and is also the reason roc_auc_curve is returning a lower AUC than before.
Instead of doing this, you may want to test the performance of a single threshold (i.e. 0.5) by calculating its corresponding accuracy, true positive rate or false positive rate.
For instance, imagine we set a threshold of 0.5 in the data above.
observation
observed_result
predicted_prob
predicted_class
1
0
0.1
0
2
0
0.5
0
3
1
0.9
1
This is a silly example, but by using 0.5 as the cutoff value, we made a perfect prediction because the observed_result matches predicted_class in all cases.
I am implementing a machine learning model in Python which predicts success or failure. I have created a dummy variable which is 1 when there is success and 0 when there is a failure. I understand the concept of confusion matrix but I have found some online where the TPs and TNs are on opposite sides of the matrix. I would like to know how to interpret the results for my variables. Is the top-left corner of the matrix predicting True Positive? If so would that translate to the amount of successes being predicted correctly or the amount of failures being predicted correctly?
Does the matrix match the diagram below and if so how?
Ideally, please describe each corner of the confusion matrix in the context where I have success as 1 and failure as 0.
Refer to the documentation: https://scikit-learn.org/stable/modules/generated/sklearn.metrics.confusion_matrix.html
Since you haven't specified the third parameter for labels in confusion_matrix, the labels in y_test_res will be used in sorted order, i.e. in this case 0 then 1. The row labels represent actual y, and column labels represent predicted y.
So the top-left corner is showing the number of failure observations, i.e. the actual y was 0 and was predicted 0, i.e. true negatives. The bottom-right corner is showing true positives, i.e. the actual y was 1 and was predicted 0.
The top-right corner would be actual y = 0 and predicted y = 1, i.e. false positive.
Using the confusion matrix plot would prettify things a little.
from sklearn.metrics import plot_confusion_matrix
plot_confusion_matrix(forest, X_test, y_test)
print(plt.show())
In the case of binary classification where classes are 0 and 1 and according to the doc :
1st row is for class 0
2nd row is for class 1
1st column is for predicted class 0
2nd column is for predicted class 1
Coefficient (0, 0) is the True Negative count (TN).
Coefficient (0, 1) is the False Positive count (FP).
Coefficient (1, 0) is the False Negative count (FN).
Coefficient (1, 1) is the True Positive count (TP).
I'm on a project for the study that I used the mean_square_error from skelarn to get my MSE and RMSE.
I can´t understand what the information means.
I'm using a dataset about House Sales and I want to predict the price of the house using Linear Regression. When I put my predict price and real price, I got the results:
MSE: 1114197668.6920328 RMSE: 33379.59958855158
What this information really means actually? That my predict will have a mean difference in price about 33379.60?
MSE = mean_squared_error(predict,testSalePrice)
RMSE = np.sqrt(MSE)
Mean Squared Error:
In statistics, the mean squared error (MSE) or mean squared deviation
(MSD) of an estimator (of a procedure for estimating an unobserved
quantity) measures the average of the squares of the errors.
So for example let's assume you have three datapoints:
Price Predicted
1900 2000
2000 2000
2100 2000
Then the MSE is: 1/3 * ((-100)*(-100)+ (0)*(0) + (100)*(100)) = 1/3 * (20000) = 6000
The perfect one would be 0, but this you will probably not reach. You have to interpret it in comparison with your actual value range.
The RMSE in this case would be: SQRT(6000) = 77,..
This is more intepretable, that means on average you are 77 away from your prediction, which makes sense if you see the three results
I have a set of independent variables X and set of values of dependent variable Y. The task at hand is binomial classification, i.e. predict whether debtor will default on his debt (1) or not (0).
After filtering out statistically insignificant variables and variables that bring about multicollinearity I am left with following summary of logistic regression model:
Accuracy ~0.87
Confusion matrix [[1038 254]
[72 1182]]
Parameters Coefficients
intercept -4.210
A 5.119
B 0.873
C -1.414
D 3.757
Now, I convert these coefficients into new continuous variable "default_probability" via log odds_ratio, i.e.
import math
e = math.e
power = (-4.210*1) + (A*5.119) + (B*0.873) + (C*-1.414) + (D*3.757)
default_probability = (e**power)/(1+(e**power))
When I divide my original dataset into quartiles according to this new continuos variable "default_probability", then:
1st quartile contains 65% of defaulted debts (577 out of 884 incidents)
2nd quartile contains 23% of defaulted debts (206 out of 884 incidents)
3rd quartile contains 9% of defaulted debts (77 out of 884 incidents)
4th quartile contains 3% of defaulted debts (24 out of 884 incidents)
At the same time:
overall quantity of debtors in 1st quartile - 1145
overall quantity of debtors in 1st quartile - 516
overall quantity of debtors in 1st quartile - 255
overall quantity of debtors in 1st quartile - 3043
I wanted to use "default probability" to surgically remove the most problematic credits by imposing the business-rule "no credit to the 1st quartile", but now I wonder whether it is "surgical" at all (by this rule I will lose (1145 - 577 = 568 "good" clients) and overall is it mathematically/logically correct to derive new continuous variables for the dataset out of the coefficients of logistic regression by the line of reasoning described above?
You have forgotten the intercept when you compute power. But supposing this is only a typo like you said in the comments, then your approach is valid. However, you might want to use scikit-learn's predict_proba function, which will save you the trouble. Example:
from sklearn.linear_model import LogisticRegression
from sklearn.datasets import load_breast_cancer
import numpy as np
data = load_breast_cancer()
X = data.data
y = data.target
lr = LogisticRegression()
lr.fit(X,y)
Suppose I then want to compute the probability of belonging to class 1 for a given observation (say observation i), I can do what you have done, essentially using the regression coefficients and the intercept like you have done:
i = 0
1/(1+np.exp(-X[i].dot(lr.coef_[0])-lr.intercept_[0]))
Or just do :
lr.predict_proba(X)[i][1]
which is faster
Closed. This question does not meet Stack Overflow guidelines. It is not currently accepting answers.
We don’t allow questions seeking recommendations for books, tools, software libraries, and more. You can edit the question so it can be answered with facts and citations.
Closed 1 year ago.
Improve this question
I am doing my graduation project in the field of computer vision, and i have only taken one course in statistics that discussed very basic concepts, and now i am facing more difficulty in rather advanced topics, so i need help (book, tutorial, course, ..etc) to grasp and review the basic ideas and concepts in statistics and then dive into the details (statistical details) used in computer vision.
You can calculate False Positives/False negatives, etc with this Confusion Matrix PyTorch example:
import torch
def confusion(prediction, truth):
""" Returns the confusion matrix for the values in the `prediction` and `truth`
tensors, i.e. the amount of positions where the values of `prediction`
and `truth` are
- 1 and 1 (True Positive)
- 1 and 0 (False Positive)
- 0 and 0 (True Negative)
- 0 and 1 (False Negative)
"""
confusion_vector = prediction / truth
# Element-wise division of the 2 tensors returns a new tensor which holds a
# unique value for each case:
# 1 where prediction and truth are 1 (True Positive)
# inf where prediction is 1 and truth is 0 (False Positive)
# nan where prediction and truth are 0 (True Negative)
# 0 where prediction is 0 and truth is 1 (False Negative)
true_positives = torch.sum(confusion_vector == 1).item()
false_positives = torch.sum(confusion_vector == float('inf')).item()
true_negatives = torch.sum(torch.isnan(confusion_vector)).item()
false_negatives = torch.sum(confusion_vector == 0).item()
return true_positives, false_positives, true_negatives, false_negatives
You could use nn.BCEWithLogitsLoss (remove the sigmoid therefore) and set the pos_weight > 1 to increase the recall. Or further optimize it with using Dice Coefficients to penalize the model for false positives, with something like:
def Dice(y_true, y_pred):
"""Returns Dice Similarity Coefficient for ground truth and predicted masks."""
#print(y_true.dtype)
#print(y_pred.dtype)
y_true = np.squeeze(y_true)/255
y_pred = np.squeeze(y_pred)/255
y_true.astype('bool')
y_pred.astype('bool')
intersection = np.logical_and(y_true, y_pred).sum()
return ((2. * intersection.sum()) + 1.) / (y_true.sum() + y_pred.sum() + 1.)
IOU Calculations Explained
Count true positives (TP)
Count false positives (FP)
Count false negatives (FN)
Intersection = TP
Union = TP + FP + FN
IOU = Intersection/Union
The left side is our ground truth, while the right side contains our predictions. The highlighted cells on the left side note which class we are looking at for statistics on the right side. The highlights on the right side note true positives in a cream color, false positives in orange, and false negatives in yellow (note that all others are true negatives — they are predicted as this individual class, and should not be based on the ground truth).
For Class 0, only the top row of the 4x4 matrix should be predicted as zeros. This is a rather simplified version of a real ground truth. In reality, the zeros could be anywhere in the matrix. On the right side, we see 1,0,0,0, meaning the first is a false negative, but the other three are true positives (aka 3 for Intersection as well). From there, we need to find anywhere else where zero was falsely predicted, and we note that happens once on the second row, and twice on the fourth row, for a total of three false positives.
To get the union, we add up TP (3), FP (3) and FN (1) to get seven. The IOU for this class, therefore, is 3/7.
If we do this for all the classes and average the IOUs, we get:
Mean IOU = [(3/7) + (2/6) + (3/4) + (1/6)] / 4 = 0.420
You will also want to learn how to pull the statistics for mAP (Mean Avg Precision):
https://www.youtube.com/watch?v=pM6DJ0ZZee0
https://towardsdatascience.com/breaking-down-mean-average-precision-map-ae462f623a52#1a59
https://medium.com/#hfdtsinghua/calculate-mean-average-precision-map-for-multi-label-classification-b082679d31be
Compute Covariance Matrixes
The variance of a variable describes how much the values are spread. The covariance is a measure that tells the amount of dependency between two variables.
A positive covariance means that the values of the first variable are large when values of the second variables are also large. A negative covariance means the opposite: large values from one variable are associated with small values of the other.
The covariance value depends on the scale of the variable so it is hard to analyze it. It is possible to use the correlation coefficient that is easier to interpret. The correlation coefficient is just the normalized covariance.
A positive covariance means that large values of one variable are associated with big values from the other (left). A negative covariance means that large values of one variable are associated with small values of the other one (right).
The covariance matrix is a matrix that summarises the variances and covariances of a set of vectors and it can tell a lot of things about your variables. The diagonal corresponds to the variance of each vector:
A matrix A and its matrix of covariance. The diagonal corresponds to the variance of each column vector. Let’s check with the formula of the variance:
With n the length of the vector, and x̄ the mean of the vector. For instance, the variance of the first column vector of A is:
This is the first cell of our covariance matrix. The second element on the diagonal corresponds of the variance of the second column vector from A and so on.
Note: the vectors extracted from the matrix A correspond to the columns of A.
The other cells correspond to the covariance between two column vectors from A. For instance, the covariance between the first and the third column is located in the covariance matrix as the column 1 and the row 3 (or the column 3 and the row 1):
The position in the covariance matrix. Column corresponds to the first variable and row to the second (or the opposite). The covariance between the first and the third column vector of A is the element in column 1 and row 3 (or the opposite = same value).
Let’s check that the covariance between the first and the third column vector of A is equal to -2.67. The formula of the covariance between two variables Xand Y is:
The variables X and Y are the first and the third column vectors in the last example. Let’s split this formula to be sure that it is crystal clear:
The sum symbol (Σ) means that we will iterate on the elements of the vectors. We will start with the first element (i=1) and calculate the first element of X minus the mean of the vector X:
Multiply the result with the first element of Y minus the mean of the vector Y:
Reiterate the process for each element of the vectors and calculate the sum of all results:
Divide by the number of elements in the vector.
EXAMPLE - Let’s start with the matrix A:
We will calculate the covariance between the first and the third column vectors:
and
Which is x̄=3, ȳ=4, and n=3 so we have:
Code example -
Using NumPy, the covariance matrix can be calculated with the function np.cov.
It is worth noting that if you want NumPy to use the columns as vectors, the parameter rowvar=False has to be used. Also, bias=True divides by n and not by n-1.
Let’s create the array first:
import numpy as np
import matplotlib.pyplot as plt
import seaborn as sns
A = np.array([[1, 3, 5], [5, 4, 1], [3, 8, 6]])
Now we will calculate the covariance with the NumPy function:
np.cov(A, rowvar=False, bias=True)
Finding the covariance matrix with the dot product
There is another way to compute the covariance matrix of A. You can center A around 0. The mean of the vector is subtracted from each element of the vector to have a vector with mean equal to 0. It is multiplied with its own transpose, and divided by the number of observations.
Let’s start with an implementation and then we’ll try to understand the link with the previous equation:
def calculateCovariance(X):
meanX = np.mean(X, axis = 0)
lenX = X.shape[0]
X = X - meanX
covariance = X.T.dot(X)/lenX
return covariance
print(calculateCovariance(A))
Output:
array([[ 2.66666667, 0.66666667, -2.66666667],
[ 0.66666667, 4.66666667, 2.33333333],
[-2.66666667, 2.33333333, 4.66666667]])
The dot product between two vectors can be expressed:
It is the sum of the products of each element of the vectors:
If we have a matrix A, the dot product between A and its transpose will give you a new matrix:
Visualize data and covariance matrices
In order to get more insights about the covariance matrix and how it can be useful, we will create a function to visualize it along with 2D data. You will be able to see the link between the covariance matrix and the data.
This function will calculate the covariance matrix as we have seen above. It will create two subplots — one for the covariance matrix and one for the data. The heatmap() function from Seaborn is used to create gradients of colour — small values will be coloured in light green and large values in dark blue. We chose one of our palette colours, but you may prefer other colours. The data is represented as a scatterplot.
def plotDataAndCov(data):
ACov = np.cov(data, rowvar=False, bias=True)
print 'Covariance matrix:\n', ACov
fig, ax = plt.subplots(nrows=1, ncols=2)
fig.set_size_inches(10, 10)
ax0 = plt.subplot(2, 2, 1)
# Choosing the colors
cmap = sns.color_palette("GnBu", 10)
sns.heatmap(ACov, cmap=cmap, vmin=0)
ax1 = plt.subplot(2, 2, 2)
# data can include the colors
if data.shape[1]==3:
c=data[:,2]
else:
c="#0A98BE"
ax1.scatter(data[:,0], data[:,1], c=c, s=40)
# Remove the top and right axes from the data plot
ax1.spines['right'].set_visible(False)
ax1.spines['top'].set_visible(False)
Uncorrelated data
Now that we have the plot function, we will generate some random data to visualize what the covariance matrix can tell us. We will start with some data drawn from a normal distribution with the NumPy function np.random.normal().
This function needs the mean, the standard deviation and the number of observations of the distribution as input. We will create two random variables of 300 observations with a standard deviation of 1. The first will have a mean of 1 and the second a mean of 2. If we randomly draw two sets of 300 observations from a normal distribution, both vectors will be uncorrelated.
np.random.seed(1234)
a1 = np.random.normal(2, 1, 300)
a2 = np.random.normal(1, 1, 300)
A = np.array([a1, a2]).T
A.shape
Note 1: We transpose the data with .T because the original shape is (2, 300) and we want the number of observations as rows (so with shape (300, 2)).
Note 2: We use np.random.seed function for reproducibility. The same random number will be used the next time we run the cell. Let’s check how the data looks like:
A[:10,:]
array([[ 2.47143516, 1.52704645],
[ 0.80902431, 1.7111124 ],
[ 3.43270697, 0.78245452],
[ 1.6873481 , 3.63779121],
[ 1.27941127, -0.74213763],
[ 2.88716294, 0.90556519],
[ 2.85958841, 2.43118375],
[ 1.3634765 , 1.59275845],
[ 2.01569637, 1.1702969 ],
[-0.24268495, -0.75170595]])
Nice, we have two column vectors; Now, we can check that the distributions are normal:
sns.distplot(A[:,0], color="#53BB04")
sns.distplot(A[:,1], color="#0A98BE")
plt.show()
plt.close()
We can see that the distributions have equivalent standard deviations but different means (1 and 2). So that’s exactly what we have asked for.
Now we can plot our dataset and its covariance matrix with our function:
plotDataAndCov(A)
plt.show()
plt.close()
Covariance matrix:
[[ 0.95171641 -0.0447816 ]
[-0.0447816 0.87959853]]
We can see on the scatterplot that the two dimensions are uncorrelated. Note that we have one dimension with a mean of 1 (y-axis) and the other with the mean of 2 (x-axis).
Also, the covariance matrix shows that the variance of each variable is very large (around 1) and the covariance of columns 1 and 2 is very small (around 0). Since we ensured that the two vectors are independent this is coherent. The opposite is not necessarily true: a covariance of 0 doesn’t guarantee independence.
Correlated data
Now, let’s construct dependent data by specifying one column from the other one.
np.random.seed(1234)
b1 = np.random.normal(3, 1, 300)
b2 = b1 + np.random.normal(7, 1, 300)/2.
B = np.array([b1, b2]).T
plotDataAndCov(B)
plt.show()
plt.close()
Covariance matrix:
[[ 0.95171641 0.92932561]
[ 0.92932561 1.12683445]]
The correlation between the two dimensions is visible on the scatter plot. We can see that a line could be drawn and used to predict y from x and vice versa. The covariance matrix is not diagonal (there are non-zero cells outside of the diagonal). That means that the covariance between dimensions is non-zero.
From this point with Covariance Matrcies, you can research further on the following:
Mean normalization
Standardization or normalization
Whitening
Zero-centering
Decorrelate
Rescaling