The other really helpful method that knn provides is called findNearest. It can be used to predict the label of a new data point based on its nearest neighbors.

Thanks to our generate_data function, it is actually really easy to generate a new data point! We can think of a new data point as a dataset of size 1:

In [17]: newcomer, _ = generate_data(1)
...      newcomer
Out[17]: array([[91., 59.]], dtype=float32)

Our function also returns a random label, but we are not interested in that. Instead, we want to predict it using our trained classifier! We can tell Python to ignore an output value with an underscore (_).

Let's have a look at our town map again. We will plot the training set as we did earlier, but also add the new data point as a green circle (since we don't know yet whether it is supposed to be a blue square or a red triangle):

In [18]: plot_data(blue, red)
...      plt.plot(newcomer[0, 0], newcomer[0, 1], 'go', markersize=14);

The preceding code will produce the following diagram (minus the rings):

If you had to guess based on its neighbors, what label would you assign the new data point: blue or red to?

Well, it depends, doesn't it? If we look at the house closest to it (the one living roughly at (x, y) = (85, 75), circled with a dotted line in the preceding diagram), we would probably assign the new data point a red triangle as well. This is exactly what our classifier would predict for k=1:

In [19]: ret, results, neighbor, dist = knn.findNearest(newcomer, 1)
...      print("Predicted label:	", results)
...      print("Neighbor's label:	", neighbor)
...      print("Distance to neighbor:	", dist)
Out[19]: Predicted label:       [[ 1.]]
         Neighbor's label:      [[ 1.]]
         Distance to neighbor:  [[ 250.]]

Here, knn reports that the nearest neighbor is 250 arbitrary units away, that the neighbor has label 1 (which we said corresponds to red triangles), and that, therefore, the new data point should also have label 1. The same would be true if we looked at the k=2 nearest neighbors and the k=3 nearest neighbors. But we want to be careful not to pick arbitrary even numbers for k. Why is that? Well, you can see it in the preceding diagram (dashed circle); among the six nearest neighbors within the dashed circle, there are three blue squares and three red triangles—we have a tie!

Finally, what would happen if we dramatically widened our search window and classified the new data point based on its k=7 nearest neighbors (circled with a solid line in the diagram mentioned earlier)?

Let's find out by calling the findNearest method with k=7 neighbors:

In [20]: ret, results, neighbors, dist = knn.findNearest(newcomer, 7)
...      print("Predicted label:	", results)
...      print("Neighbors' labels:	", neighbors)
...      print("Distance to neighbors:	", dist)
Out[20]: Predicted label:        [[ 0.]]
         Neighbors' label:       [[ 1. 1. 0. 0. 0. 1. 0.]]
         Distance to neighbors:  [[ 250. 401. 784. 916. 1073. 1360. 4885.]]

Suddenly, the predicted label is 0 (blue square). The reason is that we now have four neighbors within the solid circle that are blue squares (label 0), and only three that are red triangles (label 1). So, the majority vote would suggest making the newcomer a blue square as well.

Alternatively, predictions can be made with the predict method. But first, we need to set k:

In [22]: knn.setDefaultK(1)
...      knn.predict(newcomer)
Out[22]: (1.0, array([[1.]], dtype=float32))

What if we set k = 7? Let's have a look:

In [23]: knn.setDefaultK(7)
...      knn.predict(newcomer)
Out[23]: (0.0, array([[0.]], dtype=float32))

As you can see, the outcome of k-NN changes with the number k. However, often we do not know beforehand which number k is the most suitable. A naive solution to this problem is just to try a bunch of values for k and see which one performs best. We will learn more sophisticated solutions in later chapters of this book.