The Logistic Regression Model

Quick Review

Basic Flow

Import LogisticRegression module from Biopython

 
from Bio import LogisticRegression

Define a list containing the distance and the score of similarity in expression profile between the 2 genes

xs =      [[-53, -200.78],
          [117, -267.14],
          [57, -163.47],
          [16, -190.30],
          [11, -220.94],
          [85, -193.94],
          [16, -182.71],
          [15, -180.41],
          [-26, -181.73],
          [58, -259.87],
          [126, -414.53],
          [191, -249.57],
          [113, -265.28],
          [145, -312.99],
          [154, -213.83],
          [147, -380.85],
          [93, -291.13]]

Define a list specifies if the gene pair belongs to the same operon (1) or different operons (0)

ys =     [1,
          1,
          1,
          1,
          1,
          1,
          1,
          1,
          1,
          1,
          0,
          0,
          0,
          0,
          0,
          0,
          0]

Run logistic regression on the training data to esitmate coefficients

model = LogisticRegression.train(xs, ys)

Output the weights estimated from logistic regression training model

model.beta

Out[1]: [8.983029015714463, -0.03596896044485087, 0.02181395662983518]

Using the logistic regression model for classification

Let's predict gene pair yxcE — yxcD

LogisticRegression.classify(model, [6, -173.143442352])

Out[2]: 1

Let's predict gene pair yxiB — yxiA

LogisticRegression.classify(model, [309, -271.005880394])

Out[3]: 0

The above prediction agrees with the biological literature

Hooray! Let's celebrate!!!

Fancier Analyses

To run the code in this section, the following keys steps must be run ahead of time

Import LogisticRegression module from Biopython
Define a list containing the distance and the score of similarity in expression profile between the 2 genes
Define a list specifies if the gene pair belongs to the same operon (1) or different operons (0)

Using arguments in train function

We can use update_fn argument in the train function to track the progress of the model calculation

First define a fundtion show_progress to print the desired outputs

def show_progress(iteration, loglikelihood):
        print("Log-likelihood function:", loglikelihood)

Then call train function with update_fn argument to output log-likelihood at each iteration

model = LogisticRegression.train(xs, ys, update_fn=show_progress)

The iteration stops once the increase in the log-likelihood function is less than 0.01. If no convergence is reached after 500 iterations, the train function returns with an AssertionError.

We can also apply typecode argument in train function to save memory for gigantic data it may be necessary to use single-precision floats rather than double

model = LogisticRegression.train(xs, ys, typecode="float")

Take a look at the estimated weights

model.beta

Out[1]: [8.983029015714463, -0.03596896044485087, 0.02181395662983518]

Print the classification output in more intuitive ways

Print the classification results in more intuitive ways

print("yxcE, yxcD:", LogisticRegression.classify(model, [6, -173.143442352]))

Output

yxcE, yxcD: 1

print("yxiB, yxiA:", LogisticRegression.classify(model, [309, -271.005880394]))

Output

yxiB, yxiA: 0

Obtain the probabilities

Obtain the probabilities to find out how confident we can be in these predictions

q, p = LogisticRegression.calculate(model, [309, -271.005880394])
print("class OP: probability =", p, "class NOP: probability =", q)

Output

class OP: probability = 0.00032121125181733316 class NOP: probability = 0.9996787887481826

Prediction Accuracy

To get some idea of the prediction accuracy of the logistic regression model, we can apply it to the training data:

for i in range(len(ys)):
        print("True:", ys[i], "Predicted:", LogisticRegression.classify(model, xs[i]))

Output

True: 1 Predicted: 1
True: 1 Predicted: 0
True: 1 Predicted: 1
True: 1 Predicted: 1
True: 1 Predicted: 1
True: 1 Predicted: 1
True: 1 Predicted: 1
True: 1 Predicted: 1
True: 1 Predicted: 1
True: 1 Predicted: 1
True: 0 Predicted: 0
True: 0 Predicted: 0
True: 0 Predicted: 0
True: 0 Predicted: 0
True: 0 Predicted: 0
True: 0 Predicted: 0
True: 0 Predicted: 0

A more reliable estimate of the prediction accuracy can be found from a leave-one-out analysis the model is recalculated from the training data after removing the gene to be predicted:

for i in range(len(ys)):
        model = LogisticRegression.train(xs[:i]+xs[i+1:], ys[:i]+ys[i+1:])
        print("True:", ys[i], "Predicted:", LogisticRegression.classify(model, xs[i]))

Output

True: 1 Predicted: 1
True: 1 Predicted: 0
True: 1 Predicted: 1
True: 1 Predicted: 1
True: 1 Predicted: 1
True: 1 Predicted: 1
True: 1 Predicted: 1
True: 1 Predicted: 1
True: 1 Predicted: 1
True: 1 Predicted: 1
True: 0 Predicted: 0
True: 0 Predicted: 0
True: 0 Predicted: 0
True: 0 Predicted: 0
True: 0 Predicted: 1
True: 0 Predicted: 0
True: 0 Predicted: 0

Using optional arguments in classify function to specify distance and weight

By default, the Euclidean distance is used.

Instead, we could use the city-block (Manhattan) distance:

def cityblock(x1, x2):
    assert len(x1)==2
    assert len(x2)==2
    distance = abs(x1[0]-x2[0]) + abs(x1[1]-x2[1])
    return distance

pair1=[6, -173.143442352]
print("yxcE, yxcD:", kNN.classify(model, pair1, distance_fn = cityblock))

Output:

yxcE, yxcD: 1

Using weight function to assign higher weight for closer pairs

def weight(x1, x2):
    assert len(x1)==2
    assert len(x2)==2
    return math.exp(-abs(x1[0]-x2[0]) - abs(x1[1]-x2[1]))

x = [6, -173.143442352]
print("yxcE, yxcD:", kNN.classify(model, x, weight_fn = weight))

Output:

yxcE, yxcD: 1

Call the calculate function to calculate the confidence in predictions

This will calculate the total weight assigned to the classes OP and NOP

pair1 = [6, -173.143442352]
weight = kNN.calculate(model, pair1)
print("class NOP: weight =", weight[0], "class OP: weight =", weight[1])

Output:

class NOP: weight = 0.0 class OP: weight = 3.0

pair2 = [117, -267.14]
weight = kNN.calculate(model, pair2)
print("class NOP: weight =", weight[0], "class OP: weight =", weight[1])

Output:

class NOP: weight = 2.0 class OP: weight = 1.0

Prediction Accuracy

for i in range(len(ys)):
        print("True:", ys[i], "Predicted:", kNN.classify(model, xs[i]))

Output:

True: 1 Predicted: 1
True: 1 Predicted: 0
True: 1 Predicted: 1
True: 1 Predicted: 1
True: 1 Predicted: 1
True: 1 Predicted: 1
True: 1 Predicted: 1
True: 1 Predicted: 1
True: 1 Predicted: 1
True: 1 Predicted: 0
True: 0 Predicted: 0
True: 0 Predicted: 0
True: 0 Predicted: 0
True: 0 Predicted: 0
True: 0 Predicted: 0
True: 0 Predicted: 0
True: 0 Predicted: 0

The prediction is correct for all but two of the gene pairs

A more reliable estimate of the prediction accuracy can be found from a leave-one-out analysis The model is recalculated from the training data after removing the gene to be predicted:

k = 3
for i in range(len(ys)):
    model = kNN.train(xs[:i]+xs[i+1:], ys[:i]+ys[i+1:], k)
    print("True:", ys[i], "Predicted:", kNN.classify(model, xs[i]))

Output:

True: 1 Predicted: 1
True: 1 Predicted: 0
True: 1 Predicted: 1
True: 1 Predicted: 1
True: 1 Predicted: 1
True: 1 Predicted: 1
True: 1 Predicted: 1
True: 1 Predicted: 1
True: 1 Predicted: 1
True: 1 Predicted: 0
True: 0 Predicted: 0
True: 0 Predicted: 0
True: 0 Predicted: 1
True: 0 Predicted: 0
True: 0 Predicted: 0
True: 0 Predicted: 0
True: 0 Predicted: 1

The prediction is correct for 13 out of 17 gene pairs, 76%

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Last updated 7 months ago