Linear Regression using Graphlab

Fire up graphlab create

import graphlab

Load some house value vs. crime rate data

Dataset is from Philadelphia, PA and includes average house sales price in a number of neighborhoods. The attributes of each neighborhood we have include the crime rate (‘CrimeRate’), miles from Center City (‘MilesPhila’), town name (‘Name’), and county name (‘County’).

sales = graphlab.SFrame.read_csv('Philadelphia_Crime_Rate_noNA.csv/')
Finished parsing file /Users/feisky/machine-learning/code/basic-classifiers/Philadelphia_Crime_Rate_noNA.csv
Parsing completed. Parsed 99 lines in 0.040707 secs.
Finished parsing file /Users/feisky/machine-learning/code/basic-classifiers/Philadelphia_Crime_Rate_noNA.csv
Parsing completed. Parsed 99 lines in 0.01334 secs.
Inferred types from first 100 line(s) of file as 
If parsing fails due to incorrect types, you can correct
the inferred type list above and pass it to read_csv in
the column_type_hints argument
HousePrice HsPrc ($10,000) CrimeRate MilesPhila PopChg Name County
140463 14.0463 29.7 10.0 -1.0 Abington Montgome
113033 11.3033 24.1 18.0 4.0 Ambler Montgome
124186 12.4186 19.5 25.0 8.0 Aston Delaware
110490 11.049 49.4 25.0 2.7 Bensalem Bucks
79124 7.9124 54.1 19.0 3.9 Bristol B. Bucks
92634 9.2634 48.6 20.0 0.6 Bristol T. Bucks
89246 8.9246 30.8 15.0 -2.6 Brookhaven Delaware
195145 19.5145 10.8 20.0 -3.5 Bryn Athyn Montgome
297342 29.7342 20.2 14.0 0.6 Bryn Mawr Montgome
264298 26.4298 20.4 26.0 6.0 Buckingham Bucks
[99 rows x 7 columns]
Note: Only the head of the SFrame is printed.
You can use print_rows(num_rows=m, num_columns=n) to print more rows and columns.

Exploring the data

The house price in a town is correlated with the crime rate of that town. Low crime towns tend to be associated with higher house prices and vice versa.

graphlab.canvas.set_target('ipynb')"Scatter Plot", x="CrimeRate", y="HousePrice")

Fit the regression model using crime as the feature

crime_model = graphlab.linear_regression.create(sales, target='HousePrice', features=['CrimeRate'],validation_set=None,verbose=False)

Let’s see what our fit looks like

Matplotlib is a Python plotting library that is also useful for plotting. You can install it with:

‘pip install matplotlib’

import matplotlib.pyplot as plt
%matplotlib inline
[<matplotlib.lines.Line2D at 0x118f2d150>,
 <matplotlib.lines.Line2D at 0x118f2d390>]


Above: blue dots are original data, green line is the fit from the simple regression.

Remove Center City and redo the analysis

Center City is the one observation with an extremely high crime rate, yet house prices are not very low. This point does not follow the trend of the rest of the data very well. A question is how much including Center City is influencing our fit on the other datapoints. Let’s remove this datapoint and see what happens.

sales_noCC = sales[sales['MilesPhila'] != 0.0]"Scatter Plot", x="CrimeRate", y="HousePrice")

Refit our simple regression model on this modified dataset:

crime_model_noCC = graphlab.linear_regression.create(sales_noCC, target='HousePrice', features=['CrimeRate'],validation_set=None, verbose=False)

Look at the fit:

[<matplotlib.lines.Line2D at 0x1172d8050>,
 <matplotlib.lines.Line2D at 0x1172d8290>]


Compare coefficients for full-data fit versus no-Center-City fit

Visually, the fit seems different, but let’s quantify this by examining the estimated coefficients of our original fit and that of the modified dataset with Center City removed.

name index value stderr
(intercept) None 176626.046881 11245.5882194
CrimeRate None -576.804949058 226.90225951
[2 rows x 4 columns]
name index value stderr
(intercept) None 225204.604303 16404.0247514
CrimeRate None -2287.69717443 491.537478123
[2 rows x 4 columns]

Above: We see that for the “no Center City” version, per unit increase in crime, the predicted decrease in house prices is 2,287. In contrast, for the original dataset, the drop is only 576 per unit increase in crime. This is significantly different!

High leverage points:

Center City is said to be a “high leverage” point because it is at an extreme x value where there are not other observations. As a result, recalling the closed-form solution for simple regression, this point has the potential to dramatically change the least squares line since the center of x mass is heavily influenced by this one point and the least squares line will try to fit close to that outlying (in x) point. If a high leverage point follows the trend of the other data, this might not have much effect. On the other hand, if this point somehow differs, it can be strongly influential in the resulting fit.

Influential observations:

An influential observation is one where the removal of the point significantly changes the fit. As discussed above, high leverage points are good candidates for being influential observations, but need not be. Other observations that are not leverage points can also be influential observations (e.g., strongly outlying in y even if x is a typical value).

Remove high-value outlier neighborhoods and redo analysis

Based on the discussion above, a question is whether the outlying high-value towns are strongly influencing the fit. Let’s remove them and see what happens.

sales_nohighend = sales_noCC[sales_noCC['HousePrice'] < 350000] 
crime_model_nohighend = graphlab.linear_regression.create(sales_nohighend, target='HousePrice', features=['CrimeRate'],validation_set=None, verbose=False)

Do the coefficients change much?

name index value stderr
(intercept) None 225204.604303 16404.0247514
CrimeRate None -2287.69717443 491.537478123
[2 rows x 4 columns]
name index value stderr
(intercept) None 199073.589615 11932.5101105
CrimeRate None -1837.71280989 351.519609333
[2 rows x 4 columns]

Above: We see that removing the outlying high-value neighborhoods has some effect on the fit, but not nearly as much as our high-leverage Center City datapoint.