Logistic regression coefficient meaning - python

I'm trying to write my own logistic regressor (using batch/mini-batch gradient descent) for practice purposes.
I generated a random dataset (see below) with normally distributed inputs, and the output is binary (0,1). I manually used coefficients for the input and was hoping to be able to reproduce them (see below for the code snippet). However, to my surprise, neither my own code, nor sklearn LogisticRegression were able to reproduce the actual numbers (although the sign and order of magnitude are in line). Moreso, the coefficients my algorithm produced are different than the one produced by sklearn.
Am I misinterpreting what the coefficients for a logistic regression are?
I will appreciate any insight into this discrepancy.
Thank you!
edit: I tried using statsmodels Logit and got yet a third set of slightly different values for the coefficients
Some more info that might be relevant:
I wrote a linear regressor using an almost identical code and it worked perfectly, so I am fairly confident this is not a problem in the code. Also my regressor actually outperformed the sklearn one on the training set, and they have the exact same accuracy on the test set, so I have no reason to believe the regressors are wrong.
Code snippets for the generation of the dataset:
o1 = 2
o2 = -3
x[:,1]=np.random.rand(size)*2
x[:,2]=np.random.rand(size)*3
y = np.vectorize(sigmoid)(x[:,1]*o1+x[:,2]*o2 + np.random.normal(size=size))
so as can be seen, input coefficients are +2 and -3 (intercept 0);
sklearn coefficients were ~2.8 and ~-4.8;
my coefficients were ~1.7 and ~-2.6
and of the regressor (the most relevant parts of it):
for j in range(bin_size):
xs = x[i]
y_real = y[i]
z = np.dot(self.coeff,xs)
h = sigmoid(z)
dc+= (h-y_real)*xs
self.coeff-= dc * (learning_rate/n)

What was the intercept learned? It really should not be a surprise, as your y is polynomial of 3rd degree, while your model has only two coefficients, while 3 + y-intercept would be needed to model the response variable from predictors.
Furthermore, values may be different due to SGD for example.
Not really sure, but the coefficients could be different and return correct y for finite set of points. What are the metrics on each model? Do those differ?

Related

Why are probabilities hand-calculated from sklearn.linear_model.LogisticRegression coefficients different from .predict_proba()?

I am running a multinomial logistic regression in sklearn, using sklearn.linear_model.LogisticRegression(multiclass="multinomial"). The dependent categorical variable has 3 options: Agree, Disagree, Unsure. The independent variables are two categorical variables: Education and Gender (binary gender for simplicity in this example). I get different results when I hand-calculate the probabilities from the regression coefficients versus use the built-in predict_proba().
mnlr = LogisticRegression(multi_class="multinomial")
mnlr.fit(
pd.get_dummies(df[["Education","Gender"]]),
preprocessing.LabelEncoder().fit_transform(df["statement"])
)
I concatenate the outputs of mnlr.intercept_ and mnlr.coef_ into a regression coefficients table that looks like this:
Using mnlr.predict_proba(), I get results that I cast into a dataframe to which I add the independent variables like this:
These sum to 1 across the 3 potential categories for each data point.
However, I cannot seem to reproduce these results when I try to calculate the predicted probabilities by hand from the logistic regression coefficients.
First, for each Gender x Education combination, I calculate the logit (aka log-odds, if I understand correctly) by simply adding the intercept and the relevant variable terms. For example, to get the logit for a Woman with a Bachelor's degree with the Agree regression: 0.88076 + 0.21827 + 0.21687 = 1.31590. The table of logits looks like this:
From this table, as I understand it, I should be able to convert these logits (log-odds) to predicted probabilities: p = e^logit/(1+e^logit) for a given model and respondent (e.g., probability that Women with Bachelor's Agree with the statement). When I try this, however, I get much different results than I receive from .predict_proba() and the hand-calculated probabilities do not sum to 1, as indicated in the table below:
For example, Women with Bachelor's here have a 0.78850 probability to Agree with the statement, in place of the 0.7819 probability. Additionally, the hand-calculated probabilities across the 3 categories do not sum to 1, but rather to 1.47146.
I am almost certain this is a basic error on my part, but I cannot for the life of me figure it out. What am I doing incorrectly?
I figured this one out eventually. The answer is probably obvious to folks who really know multinomial logistic regression. The struggle I was having was that I needed to apply the softmax function (also known more descriptively as the normalized exponential function) to the logits. This function involves exponentiating the logit (log-odds) for each class and then dividing it by the sum of exponentiated logits for all classes. In this example, for Women with a Bachelor's degree, this would mean:
=
= 0.737007424626824
Hopefully this will be helpful to anyone else trying to understand how to do this by hand! (Which for me is really useful for trying to apply model-based inference as an alternative to design-based inference in sample surveys).
Sources that got me here:
How do I correctly manually recreate sklearn (python) logistic regression predict_proba outcome for multiple classification, https://en.wikipedia.org/wiki/Softmax_function

Coefficients for Logistic Regression scikit-learn vs statsmodels

When performed a logistic regression using the two API, they give different coefficients.
Even with this simple example it doesn't produce the same results in terms of coefficients. And I follow advice from older advice on the same topic, like setting a large value for the parameter C in sklearn since it makes the penalization almost vanish (or setting penalty="none").
import pandas as pd
import numpy as np
import sklearn as sk
from sklearn.linear_model import LogisticRegression
import statsmodels.api as sm
n = 200
x = np.random.randint(0, 2, size=n)
y = (x > (0.5 + np.random.normal(0, 0.5, n))).astype(int)
display(pd.crosstab( y, x ))
max_iter = 100
#### Statsmodels
res_sm = sm.Logit(y, x).fit(method="ncg", maxiter=max_iter)
print(res_sm.params)
#### Scikit-Learn
res_sk = LogisticRegression( solver='newton-cg', multi_class='multinomial', max_iter=max_iter, fit_intercept=True, C=1e8 )
res_sk.fit( x.reshape(n, 1), y )
print(res_sk.coef_)
For example I just run the above code and get 1.72276655 for statsmodels and 1.86324749 for sklearn. And when run multiple times it always gives different coefficients (sometimes closer than others, but anyway).
Thus, even with that toy example the two APIs give different coefficients (so odds ratios), and with real data (not shown here), it almost get "out of control"...
Am I missing something? How can I produce similar coefficients, for example at least at one or two numbers after the comma?
There are some issues with your code.
To start with, the two models you show here are not equivalent: although you fit your scikit-learn LogisticRegression with fit_intercept=True (which is the default setting), you don't do so with your statsmodels one; from the statsmodels docs:
An intercept is not included by default and should be added by the user. See statsmodels.tools.add_constant.
It seems that this is a frequent point of confusion - see for example scikit-learn & statsmodels - which R-squared is correct? (and own answer there as well).
The other issue is that, although you are in a binary classification setting, you ask for multi_class='multinomial' in your LogisticRegression, which should not be the case.
The third issue is that, as explained in the relevant Cross Validated thread Logistic Regression: Scikit Learn vs Statsmodels:
There is no way to switch off regularization in scikit-learn, but you can make it ineffective by setting the tuning parameter C to a large number.
which makes the two models again non-comparable in principle, but you have successfully addressed it here by setting C=1e8. In fact, since then (2016), scikit-learn has indeed added a way to switch regularization off, by setting penalty='none' since, according to the docs:
If ‘none’ (not supported by the liblinear solver), no regularization is applied.
which should now be considered the canonical way to switch off the regularization.
So, incorporating these changes in your code, we have:
np.random.seed(42) # for reproducibility
#### Statsmodels
# first artificially add intercept to x, as advised in the docs:
x_ = sm.add_constant(x)
res_sm = sm.Logit(y, x_).fit(method="ncg", maxiter=max_iter) # x_ here
print(res_sm.params)
Which gives the result:
Optimization terminated successfully.
Current function value: 0.403297
Iterations: 5
Function evaluations: 6
Gradient evaluations: 10
Hessian evaluations: 5
[-1.65822763 3.65065752]
with the first element of the array being the intercept and the second the coefficient of x. While for scikit learn we have:
#### Scikit-Learn
res_sk = LogisticRegression(solver='newton-cg', max_iter=max_iter, fit_intercept=True, penalty='none')
res_sk.fit( x.reshape(n, 1), y )
print(res_sk.intercept_, res_sk.coef_)
with the result being:
[-1.65822806] [[3.65065707]]
These results are practically identical, within the machine's numeric precision.
Repeating the procedure for different values of np.random.seed() does not change the essence of the results shown above.

how to use sklearn when target variable is a proportion

There are standard ways of predicting proportions such as logistic regression (without thresholding) and beta regression. There have already been discussions about this:
http://scikit-learn-general.narkive.com/4dSCktaM/using-logistic-regression-on-a-continuous-target-variable
http://scikit-learn-general.narkive.com/lLVQGzyl/beta-regression
I cannot tell if there exists a work-around within the sklearn framework.
There exists a workaround, but it is not intrinsically within the sklearn framework.
If you have a proportional target variable (value range 0-1) you run into two basic difficulties with scikit-learn:
Classifiers (such as logistic regression) deal with class labels as target variables only. As a workaround you could simply threshold your probabilities to 0/1 and interpret them as class labels, but you would lose a lot of information.
Regression models (such as linear regression) do not restrict the target variable. You can train them on proportional data, but there is no guarantee that the output on unseen data will be restricted to the 0/1 range. However, in this situation, there is a powerful work-around (below).
There are different ways to mathematically formulate logistic regression. One of them is the generalized linear model, which basically defines the logistic regression as a normal linear regression on logit-transformed probabilities. Normally, this approach requires sophisticated mathematical optimization because the probabilities are unknown and need to be estimated along with the regression coefficients.
In your case, however, the probabilities are known. This means you can simply transform them with y = log(p / (1 - p)). Now they cover the full range from -oo to oo and can serve as the target variable for a LinearRegression model [*]. Of course, the model output then needs to be transformed again to result in probabilities p = 1 / (exp(-y) + 1).
import numpy as np
from sklearn.linear_model import LinearRegression
class LogitRegression(LinearRegression):
def fit(self, x, p):
p = np.asarray(p)
y = np.log(p / (1 - p))
return super().fit(x, y)
def predict(self, x):
y = super().predict(x)
return 1 / (np.exp(-y) + 1)
if __name__ == '__main__':
# generate example data
np.random.seed(42)
n = 100
x = np.random.randn(n).reshape(-1, 1)
noise = 0.1 * np.random.randn(n).reshape(-1, 1)
p = np.tanh(x + noise) / 2 + 0.5
model = LogitRegression()
model.fit(x, p)
print(model.predict([[-10], [0.0], [1]]))
# [[ 2.06115362e-09]
# [ 5.00000000e-01]
# [ 8.80797078e-01]]
There are also numerous other alternatives. Some non-linear regression models can work naturally in the 0-1 range. For example Random Forest Regressors will never exceed the target variables' range they were trained with. Simply put probabilities in and you will get probabilities out. Neural networks with appropriate output activation functions (tanh, I guess) will also work well with probabilities, but if you want to use those there are more specialized libraries than sklearn.
[*] You could in fact plug in any linear regression model which can make the method more powerful, but then it no longer is exactly equivalent to logistic regression.

Ridge regression using stochastic gradient descent in Python

I am trying to implement a solution to Ridge regression in Python using Stochastic gradient descent as the solver. My code for SGD is as follows:
def fit(self, X, Y):
# Convert to data frame in case X is numpy matrix
X = pd.DataFrame(X)
# Define a function to calculate the error given a weight vector beta and a training example xi, yi
# Prepend a column of 1s to the data for the intercept
X.insert(0, 'intercept', np.array([1.0]*X.shape[0]))
# Find dimensions of train
m, d = X.shape
# Initialize weights to random
beta = self.initializeRandomWeights(d)
beta_prev = None
epochs = 0
prev_error = None
while (beta_prev is None or epochs < self.nb_epochs):
print("## Epoch: " + str(epochs))
indices = range(0, m)
shuffle(indices)
for i in indices: # Pick a training example from a randomly shuffled set
beta_prev = beta
xi = X.iloc[i]
errori = sum(beta*xi) - Y[i] # Error[i] = sum(beta*x) - y = error of ith training example
gradient_vector = xi*errori + self.l*beta_prev
beta = beta_prev - self.alpha*gradient_vector
epochs += 1
The data I'm testing this on is not normalized and my implementation always ends up with all the weights being Infinity, even though I initialize the weights vector to low values. Only when I set the learning rate alpha to a very small value ~1e-8, the algorithm ends up with valid values of the weights vector.
My understanding is that normalizing/scaling input features only helps reduce convergence time. But the algorithm should not fail to converge as a whole if the features are not normalized. Is my understanding correct?
You can check from scikit-learn's Stochastic Gradient Descent documentation that one of the disadvantages of the algorithm is that it is sensitive to feature scaling. In general, gradient based optimization algorithms converge faster on normalized data.
Also, normalization is advantageous for regression methods.
The updates to the coefficients during each step will depend on the ranges of each feature. Also, the regularization term will be affected heavily by large feature values.
SGD may converge without data normalization, but that is subjective to the data at hand. Therefore, your assumption is not correct.
Your assumption is not correct.
It's hard to answer this, because there are so many different methods/environments but i will try to mention some points.
Normalization
When some method is not scale-invariant (i think every linear-regression is not) you really should normalize your data
I take it that you are just ignoring this because of debugging / analyzing
Normalizing your data is not only relevant for convergence-time, the results will differ too (think about the effect within the loss-function; big values might effect in much more loss to small ones)!
Convergence
There is probably much to tell about convergence of many methods on normalized/non-normalized data, but your case is special:
SGD's convergence theory only guarantees convergence to some local-minimum (= global-minimum in your convex-opt problem) for some chosings of hyper-parameters (learning-rate and learning-schedule/decay)
Even optimizing normalized data can fail with SGD when those params are bad!
This is one of the most important downsides of SGD; dependence on hyper-parameters
As SGD is based on gradients and step-sizes, non-normalized data has a possibly huge effect on not achieving this convergence!
In order for sgd to converge in linear regression the step size should be smaller than 2/s where s is the largest singular value of the matrix (see the Convergence and stability in the mean section in https://en.m.wikipedia.org/wiki/Least_mean_squares_filter), in the case of ridge regression it should be less than 2*(1+p/s^2)/s where p is the ridge penalty.
Normalizing rows of the matrix (or gradients) changes the loss function to give each sample an equal weight and it changes the singular values of the matrix such that you can choose a step size near 1 (see the NLMS section in https://en.m.wikipedia.org/wiki/Least_mean_squares_filter). Depending on your data it might require smaller step sizes or allow for larger step sizes. It all depends on whether or not the normalization increases or deacreses the largest singular value of the matrix.
Note that when deciding whether or not to normalize the rows you shouldn't just think about the convergence rate (which is determined by the ratio between the largest and smallest singular values) or stability in the mean, but also about how it changes the loss function and whether or not it fits your needs because of that, sometimes it makes sense to normalize but sometimes (for example when you want to give different importance for different samples or when you think that a larger energy for the signal means better snr) it doesn't make sense to normalize.

Logistic regression using python

I want to implement Logisitic regression from scratch in python. Following are the functions in it:
sigmoid
cost
fminunc
Evaluating Logistic regression
I would like to know, what would be a great start to this to start from scratch in python. Any guidance on how and what would be good. I know the theory of those functions but looking for a better pythonic answer.
I used octave and I got it all right but dont know how to start in python as OCtave already has those packages setup to do the work.
You may want to try to translate your octave code to python and see what's going on. You can also use the python package to do this for you. Check out scikit-learn on logistic regression. There is also an easy example in this blog.
In order to implement Logistic Regression, You may consider the following 2 approaches:
Consider How Linear Regression Works. Apply Sigmoid Function to the Hypothesis of Linear Regression and run gradient Descent until convergence. OR Apply the Exponential based Softmax function to rule out lower possibility of occurrence.
def logistic_regression(x, y,alpha=0.05,lamda=0):
'''
Logistic regression for datasets
'''
m,n=np.shape(x)
theta=np.ones(n)
xTrans = x.transpose()
oldcost=0.0
value=True
while(value):
hypothesis = np.dot(x, theta)
logistic=hypothesis/(np.exp(-hypothesis)+1)
reg = (lamda/2*m)*np.sum(np.power(theta,2))
loss = logistic - y
cost = np.sum(loss ** 2)
#print(cost)
# avg cost per example (the 2 in 2*m doesn't really matter here.
# But to be consistent with the gradient, I include it)
# avg gradient per example
gradient = np.dot(xTrans, loss)/m
# update
if(reg):
cost=cost+reg
theta = (theta - (alpha) * (gradient+reg))
else:
theta=theta -(alpha/m) * gradient
if(oldcost==cost):
value=False
else:
oldcost=cost
print(accuracy(theta,m,y,x))
return theta,accuracy(theta,m,y,x)

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