understand how this lambda function works

Question

I thought I understood how lambda functions work, though I don't use them myself. But the lambda below from this tutorial totally stumps me:

import matplotlib.pyplot as plt
import numpy as np
import sklearn
import sklearn.datasets
import sklearn.linear_model
import matplotlib

That was easy. More:

# Generate a dataset and plot it
np.random.seed(0)
X, y = sklearn.datasets.make_moons(200, noise=0.20)
plt.scatter(X[:,0], X[:,1], s=40, c=y, cmap=plt.cm.Spectral)
clf = sklearn.linear_model.LogisticRegressionCV()
clf.fit(X, y)

# Helper function to plot a decision boundary.
# If you don't fully understand this function don't worry, it just generates the contour plot below.

def plot_decision_boundary(pred_func):

    # Set min and max values and give it some padding
    x_min, x_max = X[:, 0].min() - .5, X[:, 0].max() + .5
    y_min, y_max = X[:, 1].min() - .5, X[:, 1].max() + .5
    h = 0.01

    # Generate a grid of points with distance h between them
    xx, yy = np.meshgrid(np.arange(x_min, x_max, h), np.arange(y_min, y_max, h))

    # Predict the function value for the whole gid
    Z = pred_func(np.c_[xx.ravel(), yy.ravel()])
    Z = Z.reshape(xx.shape)

    # Plot the contour and training examples
    plt.contourf(xx, yy, Z, cmap=plt.cm.Spectral)
    plt.scatter(X[:, 0], X[:, 1], c=y, cmap=plt.cm.Spectral)

Now the line I don't understand:

plot_decision_boundary(lambda x: clf.predict(x))

I've read many times how lambdas work, but I just don't get how the x here passing the correct values from before. How is the x mapped to the relevant values?

tdelaney · Accepted Answer · 2016-01-16 17:59:49Z

11

lambdas are just anonymous functions. lambda bodies can only be an expression (as subset of what you can put into a function) because they have to fit inline with other code.

plot_decision_boundary(lambda x: clf.predict(x)) could be rewritten as

def call_clf_predict(x):
    return clf.predict(x)
plot_decision_boundary(call_clf_predict)

Here, its more clear what is going on. plot_decision_boundary gets a callable and calls it with the single parameter np.c_[xx.ravel(), yy.ravel()].

But lambda shouldn't have been used here in the first place. You could just do

plot_decision_boundary(clf.predict)

In the grand tradition of python tutorials, lambda is abused once again.

answered Jan 16, 2016 at 17:59

tdelaney

77.9k6 gold badges91 silver badges129 bronze badges

Sign up to request clarification or add additional context in comments.

4 Comments

Ada Stra Over a year ago

You are right, it works without using it. Very, very interesting.

Victor Over a year ago

I don't understand where does variable x gets its value

tdelaney Over a year ago

@VMMF - are you referring to the x parameter in the lambda function or the X in plot_decision_boundary? The lambda wasn't necessary. its x parameter is just forwarding the paramter it was called with to the function it calls. The big X is a global variable, visible in the function.

Victor Over a year ago

In the function you wrote, why isn't x passed : plot_decision_boundary(call_clf_predict). Is it because x is passed later inside plot_decision_boundary?

OneCricketeer · Accepted Answer · 2016-01-16 17:25:38Z

4

plot_decision_boundary(lambda x: clf.predict(x))

This line is passing a function that takes a single argument into the method. When the lambda is evaluated, or the method is "called" with an argument x, it'll do clf.predict(x)

Within the method, that function is named pred_func and it is called with its single argument at

Z = pred_func(np.c_[xx.ravel(), yy.ravel()])

So the code that is ran is

Z = clf.predict(np.c_[xx.ravel(), yy.ravel()])

answered Jan 16, 2016 at 17:25

OneCricketeer

193k20 gold badges146 silver badges276 bronze badges

Comments

Martijn Pieters · Accepted Answer · 2016-01-16 17:26:06Z

2

x is the concatenated Numpy object you pass in here:

Z = pred_func(np.c_[xx.ravel(), yy.ravel()])

pred_func is the argument to plot_decision_boundary(); by calling it you call the function object defined by the lambda. The above line translates to:

clf.predict(np.c_[xx.ravel(), yy.ravel()])

answered Jan 16, 2016 at 17:26

Martijn Pieters

1.1m326 gold badges4.2k silver badges3.4k bronze badges

1 Comment

Ada Stra Over a year ago

That was great, thanks. Now I'll do some homework to get a deeper insight!

user3889486 · Accepted Answer · 2019-04-20 10:01:57Z

2

plot_decision_boundary(pred_func) actually requires three input X,y,pred_func

If clf.predict(x) only takes in x as its input, then no need for lambda. However, for other prediction functions, which may require more inputs ( such as func(parameters, x) as the pred_func), and an outer function (plot_decision_boundary) that doesn't take in parameters, we need to insert them from the beginning:

plot_decision_boundary(lambda x: func(parameters, x), X, y)

edited Apr 20, 2019 at 10:01

user3889486

6562 gold badges7 silver badges22 bronze badges

answered Oct 14, 2018 at 15:00

Chelsea Chen

211 bronze badge

Comments

user11129457 · Accepted Answer · 2019-06-20 09:24:52Z

1

We want to plot the contours, so we need different coordinates for X axis, Y axis and the corresponding values of Z axis.

pred_func() is clf.predict(x), ie. the lambda function

So when you generate the meshgrid and concatenate it (heres a good link to understand concatenation) and pass it into pred_func what you're doing is actually passing the values in the x argument of the lambda function which outputs the predictions from the lambda function clf.predict()ie. Z Matrix (outputs z, corresponding to different values of feature1 and feature2) which is used to plot the contours.

answered Jun 20, 2019 at 9:24

user11129457

Collectives™ on Stack Overflow

understand how this lambda function works

5 Answers 5

4 Comments

Comments

1 Comment

Comments

Comments

Your Answer

Linked

Hot Network Questions

Collectives™ on Stack Overflow

5 Answers 5

4 Comments

Comments

1 Comment

Comments

Comments

Your Answer

Sign up or log in

Post as a guest

Linked

Related