Reputation: 1591
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?
Upvotes: 4
Views: 13669
Reputation:
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.
Upvotes: 1
Reputation: 21
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)
Upvotes: 2
Reputation: 77407
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.
Upvotes: 11
Reputation: 1125408
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()])
Upvotes: 2
Reputation: 192043
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()])
Upvotes: 4