solution-code-7.ipynb

Kernel: Python 3

from sklearn import datasets, neighbors, metrics
import pandas as pd
import seaborn as sns

%matplotlib inline

iris = datasets.load_iris()
irisdf = pd.DataFrame(iris.data, columns=iris.feature_names)
irisdf['target'] = iris.target
cmap = {'0': 'r', '1': 'g', '2': 'b' }
irisdf['ctarget'] = irisdf.target.apply(lambda x: cmap[str(x)])
irisdf.plot('petal length (cm)', 'petal width (cm)', kind='scatter', c=irisdf.ctarget)
print irisdf.plot('petal length (cm)', 'petal width (cm)', kind='scatter', c=irisdf.ctarget)
print irisdf.describe()

def my_classifier(row):
    if row['petal length (cm)'] < 2:
        return 0
    else:
        return 1

predictions = irisdf.apply(my_classifier, axis=1)

  File "<ipython-input-3-09d36c810018>", line 7
    print irisdf.plot('petal length (cm)', 'petal width (cm)', kind='scatter', c=irisdf.ctarget)
               ^
SyntaxError: invalid syntax

irisdf['predictions'] = predictions

print float(len(irisdf[irisdf.target == irisdf.predictions])) / len(irisdf)

More specific solution

For the class, this solution is as simple it really needs to be in order to get a very good prediction score. But: Why, or when, does this fail? What attributes make this a great data set for learning classification algorithms? What makes it not as great?

def my_classifier(row):
    if row['petal length (cm)'] < 2:
        return 0
    elif row['petal length (cm)'] < 5:
        return 1
    else:
        return 2

predictions = irisdf.apply(my_classifier, axis=1)

irisdf['predictions'] = predictions

print float(len(irisdf[irisdf.target == irisdf.predictions])) / len(irisdf)

Using distance: KNN implementation

from sklearn import datasets, neighbors, metrics
import pandas as pd

iris = datasets.load_iris()
# n_neighbors is our option in KNN. We'll tune this value to attempt to improve our prediction.
knn = neighbors.KNeighborsClassifier(n_neighbors=5, weights='uniform')
knn.fit(iris.data[:,2:], iris.target)
print knn.predict(iris.data[:,2:])
print iris.target

print knn.score(iris.data[:,2:], iris.target)

Do we see a change in performaance with using the distance weight?

knn = neighbors.KNeighborsClassifier(n_neighbors=5, weights='distance')
knn.fit(iris.data[:,2:], iris.target)
print knn.predict(iris.data[:,2:])
print iris.target

print knn.score(iris.data[:,2:], iris.target)

Solution to solving K

This is only one approach to the problem, but adding in the 'distance' parameter (instead of uniform) would only be additive; note that the code would need some editing to handle it properly if done in the grid search; alternatively, make the change directly in the estimator.

from sklearn import grid_search, cross_validation
import matplotlib.pyplot as plt

# some n_list! keep in mind cross validation
# recall: what's an effective way to create a numerical list in python?
k = range(2, 100)
params = {'n_neighbors': k }
kf = cross_validation.KFold(len(irisdf), n_folds = 5)
gs = grid_search.GridSearchCV(
    estimator=neighbors.KNeighborsClassifier(),
    param_grid=params,
    cv=kf,
)
gs.fit(iris.data, iris.target)
gs.grid_scores_

plt.plot(k,[s[1] for s in  gs.grid_scores_],)

Zoom in to look at fit before first dive around 25:

plt.plot(k[:25],[s[1] for s in  gs.grid_scores_][:25],)