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Multivariable Regression Model of FBI Property Crime Statistics

Using the FBI:UCR Crime dataset, which can be found here, build a regression model to predict property crimes.

The FBI defines property crime as including the offenses of burglary, larceny-theft, motor vehicle theft, and arson. To predict property crime, one can simply use these features.





In [1]:



    


import warnings

import pandas as pd
import numpy as np
import matplotlib.pyplot as plt
import seaborn as sns 
from sklearn import linear_model


# Suppress annoying harmless error.
warnings.filterwarnings(
    action="ignore"
)



    











In [2]:



    


data_path = "https://raw.githubusercontent.com/Thinkful-Ed/data-201-resources/master/New_York_offenses/NEW_YORK-Offenses_Known_to_Law_Enforcement_by_City_2013%20-%2013tbl8ny.csv"
data = pd.read_csv(data_path, delimiter = ',', skiprows=4, header=0, skipfooter=3, thousands=',')
data = pd.DataFrame(data)



    











In [3]:



    


data.head()



    


















    
Out[3]:










  
    

      
      City
      Population
      Violent
crime
      Murder and
nonnegligent
manslaughter
      Rape
(revised
definition)1
      Rape
(legacy
definition)2
      Robbery
      Aggravated
assault
      Property
crime
      Burglary
      Larceny-
theft
      Motor
vehicle
theft
      Arson3
    

  
  
    

      0
      Adams Village
      1861
      0
      0
      NaN
      0
      0
      0
      12
      2
      10
      0
      0.0
    

    

      1
      Addison Town and Village
      2577
      3
      0
      NaN
      0
      0
      3
      24
      3
      20
      1
      0.0
    

    

      2
      Akron Village
      2846
      3
      0
      NaN
      0
      0
      3
      16
      1
      15
      0
      0.0
    

    

      3
      Albany
      97956
      791
      8
      NaN
      30
      227
      526
      4090
      705
      3243
      142
      NaN
    

    

      4
      Albion Village
      6388
      23
      0
      NaN
      3
      4
      16
      223
      53
      165
      5
      NaN
    

  




















In [4]:



    


# Instantiate and fit our model.
regr = linear_model.LinearRegression()
Y = data['Property\ncrime'].values.reshape(-1, 1)
X = data[["Larceny-\ntheft", "Motor\nvehicle\ntheft", "Burglary"]]
regr.fit(X, Y)

# Inspect the results.
print('\nCoefficients: \n', regr.coef_)
print('\nIntercept: \n', regr.intercept_)
print('\nR-squared:')
print(regr.score(X, Y))



    


















    






Coefficients: 
 [[ 1.  1.  1.]]

Intercept: 
 [ -2.27373675e-13]

R-squared:
1.0

Perfect accuracy, as expected. However......

Predicting ALL property crimes is a more interesting question.

Building a Model to Predict Property Crimes (without using the Property Crime features)

To start, let's take a look at how each of the non-property crime features interact with property crime.





In [5]:



    


plt.figure(figsize=(15,5))
sns.pairplot(data, vars =['Property\ncrime', 'Population', 'Violent\ncrime',
       'Murder and\nnonnegligent\nmanslaughter',
       'Rape\n(legacy\ndefinition)2',
       'Robbery', 'Aggravated\nassault']) 
plt.show()



    


















    








<matplotlib.figure.Figure at 0x10c358898>

That single outlier is making the relationships difficult to view. Let's remove the outlier.





In [6]:



    


dataCleaned = data[data["Property\ncrime"] < 20000]



    











In [7]:



    


plt.figure(figsize=(15,5))
sns.pairplot(dataCleaned, vars =['Property\ncrime', 'Population', 'Violent\ncrime',
       'Murder and\nnonnegligent\nmanslaughter',
       'Rape\n(legacy\ndefinition)2',
       'Robbery', 'Aggravated\nassault']) 
plt.show()



    


















    








<matplotlib.figure.Figure at 0x10ed50320>










    
























In [8]:



    


plt.scatter(dataCleaned["Property\ncrime"], dataCleaned["Murder and\nnonnegligent\nmanslaughter"])
plt.title('Raw values')
plt.xlabel("Property Crime")
plt.ylabel("Murder")
plt.show()

There is a large number of 0's for Murder. Perhaps let's use a binary value for murder occurring vs no murder occurring.





In [9]:



    


dataCleaned["Murder"] = dataCleaned['Murder and\nnonnegligent\nmanslaughter'].apply(lambda x: 0 if x == 0 else 1)

plt.scatter(dataCleaned["Property\ncrime"], dataCleaned["Murder"])
plt.title('Raw values')
plt.xlabel("Property Crime")
plt.ylabel("Murder")
plt.show()



    


















    
























In [10]:



    


dataCleaned.head()



    


















    
Out[10]:










  
    

      
      City
      Population
      Violent
crime
      Murder and
nonnegligent
manslaughter
      Rape
(revised
definition)1
      Rape
(legacy
definition)2
      Robbery
      Aggravated
assault
      Property
crime
      Burglary
      Larceny-
theft
      Motor
vehicle
theft
      Arson3
      Murder
    

  
  
    

      0
      Adams Village
      1861
      0
      0
      NaN
      0
      0
      0
      12
      2
      10
      0
      0.0
      0
    

    

      1
      Addison Town and Village
      2577
      3
      0
      NaN
      0
      0
      3
      24
      3
      20
      1
      0.0
      0
    

    

      2
      Akron Village
      2846
      3
      0
      NaN
      0
      0
      3
      16
      1
      15
      0
      0.0
      0
    

    

      3
      Albany
      97956
      791
      8
      NaN
      30
      227
      526
      4090
      705
      3243
      142
      NaN
      1
    

    

      4
      Albion Village
      6388
      23
      0
      NaN
      3
      4
      16
      223
      53
      165
      5
      NaN
      0
    

  




















In [11]:



    


regr = linear_model.LinearRegression()
Y = dataCleaned['Property\ncrime'].values.reshape(-1, 1)
X = dataCleaned[['Population', 'Violent\ncrime',
       'Murder and\nnonnegligent\nmanslaughter',
       'Rape\n(legacy\ndefinition)2',
       'Robbery', 'Aggravated\nassault']]
regr.fit(X, Y)

# Inspect the results.
print('\nCoefficients: \n', regr.coef_)
print('\nIntercept: \n', regr.intercept_)
print('\nR-squared:')
print(regr.score(X, Y))



    


















    






Coefficients: 
 [[  1.35211120e-02   2.03018615e+01   3.63790825e+01   2.47232760e+01
   -2.12521814e+01  -1.95483156e+01]]

Intercept: 
 [ 9.48025988]

R-squared:
0.950982619239

Noice!!

What about performance when the binary Murder feature is used?





In [12]:



    


regr = linear_model.LinearRegression()
Y = dataCleaned['Property\ncrime'].values.reshape(-1, 1)
X = dataCleaned[['Population', 'Violent\ncrime',
       'Murder', 'Rape\n(legacy\ndefinition)2',
       'Robbery', 'Aggravated\nassault']]
regr.fit(X, Y)

# Inspect the results.
print('\nCoefficients: \n', regr.coef_)
print('\nIntercept: \n', regr.intercept_)
print('\nR-squared:')
print(regr.score(X, Y))



    


















    






Coefficients: 
 [[  1.39419257e-02   6.37574382e+01  -5.85043045e+01  -1.97448163e+01
   -6.52374519e+01  -6.27376441e+01]]

Intercept: 
 [ 10.74635836]

R-squared:
0.951222943644

There is a slight increase of performance when the binary indicator for murder is used.

Leave no man behind!

Reintroduce the outlier to the model.





In [13]:



    


data["Murder"] = data['Murder and\nnonnegligent\nmanslaughter'].apply(lambda x: 0 if x == 0 else 1)


regr = linear_model.LinearRegression()
Y = data['Property\ncrime'].values.reshape(-1, 1)
X = data[['Population', 'Violent\ncrime',
       'Murder', 'Rape\n(legacy\ndefinition)2',
       'Robbery', 'Aggravated\nassault']]
regr.fit(X, Y)

# Inspect the results.
print('\nCoefficients: \n', regr.coef_)
print('\nIntercept: \n', regr.intercept_)
print('\nR-squared:')
print(regr.score(X, Y))



    


















    






Coefficients: 
 [[  1.19440110e-02   7.04373894e+01   3.10866677e+01  -9.17864985e+00
   -7.07330299e+01  -7.18343944e+01]]

Intercept: 
 [ 31.38463798]

R-squared:
0.999003849335

Hmmmm....it seems that outlier result has also heavily weighted the R-squared result and coefficients. The linear model which did not incorporate the outlier is likely to be a better indicator of overall trends and accuracy.

Best Model:





In [14]:



    


regr = linear_model.LinearRegression()
Y = dataCleaned['Property\ncrime'].values.reshape(-1, 1)
X = dataCleaned[['Population', 'Violent\ncrime',
       'Murder', 'Rape\n(legacy\ndefinition)2',
       'Robbery', 'Aggravated\nassault']]
regr.fit(X, Y)

# Inspect the results.
print('\nCoefficients: \n', regr.coef_)
print('\nIntercept: \n', regr.intercept_)
print('\nR-squared:')
print(regr.score(X, Y))



    


















    






Coefficients: 
 [[  1.39419257e-02   6.37574382e+01  -5.85043045e+01  -1.97448163e+01
   -6.52374519e+01  -6.27376441e+01]]

Intercept: 
 [ 10.74635836]

R-squared:
0.951222943644

















In [ ]:



    






    











In [ ]:



    






    











In [ ]:

Validating regression models for prediction

Now let's use cross-validation to obtain a more accurate description of our accuracy:





In [15]:



    


from sklearn.cross_validation import cross_val_score

regr = linear_model.LinearRegression()

y = data['Property\ncrime'].values.reshape(-1, 1)
X = data[['Population', 'Violent\ncrime',
       'Murder', 'Rape\n(legacy\ndefinition)2',
       'Robbery', 'Aggravated\nassault']]

scores = cross_val_score(regr, X, y, cv = 10)

print("Percent accuracy within each fold:\n")
print(scores)
print("\nMean accuracy:\n")
print(scores.mean())



    


















    






Percent accuracy within each fold:

[ 0.85113397  0.97672918  0.39041084  0.76138868  0.7549484   0.72489609
  0.96157164  0.9595789   0.97762802  0.85256133]

Mean accuracy:

0.821084705201

Test the Model with Data From Another State

Now let's test our model with the 2013 Crime Rate dataset for California. Will the predictive power be similar?





In [24]:



    


data_path = "files/table_8_offenses_known_to_law_enforcement_california_by_city_2013.csv"
dataCA = pd.read_csv(data_path, delimiter = ',', skiprows=4, header=0, skipfooter=3, thousands=',')
dataCA = pd.DataFrame(dataCA)



    











In [25]:



    


dataCA.head()



    


















    
Out[25]:










  
    

      
      City
      Population
      Violent
crime
      Murder and
nonnegligent
manslaughter
      Rape
(revised
definition)1
      Rape
(legacy
definition)2
      Robbery
      Aggravated
assault
      Property
crime
      Burglary
      Larceny-
theft
      Motor
vehicle
theft
      Arson
      Unnamed: 13
    

  
  
    

      0
      Adelanto
      31165
      198
      2
      NaN
      15
      52
      129
      886
      381
      372
      133
      17
      NaN
    

    

      1
      Agoura Hills
      20762
      19
      0
      NaN
      2
      10
      7
      306
      109
      185
      12
      7
      NaN
    

    

      2
      Alameda
      76206
      158
      0
      NaN
      10
      85
      63
      1902
      287
      1285
      330
      17
      NaN
    

    

      3
      Albany
      19104
      29
      0
      NaN
      1
      24
      4
      557
      94
      388
      75
      7
      NaN
    

    

      4
      Alhambra
      84710
      163
      1
      NaN
      9
      81
      72
      1774
      344
      1196
      234
      7
      NaN
    

  




















In [26]:



    


dataCA["Murder"] = dataCA['Murder and\nnonnegligent\nmanslaughter'].apply(lambda x: 0 if x == 0 else 1)



    











In [27]:



    


y = dataCA['Property\ncrime'].values.reshape(-1, 1)
X = dataCA[['Population', 'Violent\ncrime',
       'Murder', 'Rape\n(legacy\ndefinition)2',
       'Robbery', 'Aggravated\nassault']]

scores = cross_val_score(regr, X, y, cv = 10)

print("Percent accuracy within each fold:\n")
print(scores)
print("\nMean accuracy:\n")
print(scores.mean())



    


















    






Percent accuracy within each fold:

[ 0.88256872  0.83813945  0.63594188  0.82741147  0.94607165  0.89356858
  0.59922784  0.90741413  0.86736273  0.76805549]

Mean accuracy:

0.816576193827

Results

The FBI dataset for California had similar predictive power to the New York dataset on which the Multivariable Regression Model was built. Within each cross-validation fold, there is a fair amount of variability for both California and New York.

Content source: Jackie789/JupyterNotebooks

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