kaggle之旧金山犯罪

时间：2021-05-22 12:27:17

kaggle地址

github地址

特点：
1. 离散特征
2. 离散特征二值化处理

## 数据概览 
import pandas as pd
import numpy as np

# 载入数据
train = pd.read_csv('~/kaggle/dataset/San_Francisco_Crime_Classification/train.csv', parse_dates = ['Dates'])
test = pd.read_csv('~/kaggle/dataset/San_Francisco_Crime_Classification/test.csv', parse_dates = ['Dates'])
预览训练集 
print train.head(10)
Dates Category Descript \ 0 2015-05-13 23:53:00 WARRANTS WARRANT ARREST 1 2015-05-13 23:53:00 OTHER OFFENSES TRAFFIC VIOLATION ARREST 2 2015-05-13 23:33:00 OTHER OFFENSES TRAFFIC VIOLATION ARREST 3 2015-05-13 23:30:00 LARCENY/THEFT GRAND THEFT FROM LOCKED AUTO 4 2015-05-13 23:30:00 LARCENY/THEFT GRAND THEFT FROM LOCKED AUTO 5 2015-05-13 23:30:00 LARCENY/THEFT GRAND THEFT FROM UNLOCKED AUTO 6 2015-05-13 23:30:00 VEHICLE THEFT STOLEN AUTOMOBILE 7 2015-05-13 23:30:00 VEHICLE THEFT STOLEN AUTOMOBILE 8 2015-05-13 23:00:00 LARCENY/THEFT GRAND THEFT FROM LOCKED AUTO 9 2015-05-13 23:00:00 LARCENY/THEFT GRAND THEFT FROM LOCKED AUTO DayOfWeek PdDistrict Resolution Address \ 0 Wednesday NORTHERN ARREST, BOOKED OAK ST / LAGUNA ST 1 Wednesday NORTHERN ARREST, BOOKED OAK ST / LAGUNA ST 2 Wednesday NORTHERN ARREST, BOOKED VANNESS AV / GREENWICH ST 3 Wednesday NORTHERN NONE 1500 Block of LOMBARD ST 4 Wednesday PARK NONE 100 Block of BRODERICK ST 5 Wednesday INGLESIDE NONE 0 Block of TEDDY AV 6 Wednesday INGLESIDE NONE AVALON AV / PERU AV 7 Wednesday BAYVIEW NONE KIRKWOOD AV / DONAHUE ST 8 Wednesday RICHMOND NONE 600 Block of 47TH AV 9 Wednesday CENTRAL NONE JEFFERSON ST / LEAVENWORTH ST X Y 0 -122.425892 37.774599 1 -122.425892 37.774599 2 -122.424363 37.800414 3 -122.426995 37.800873 4 -122.438738 37.771541 5 -122.403252 37.713431 6 -122.423327 37.725138 7 -122.371274 37.727564 8 -122.508194 37.776601 9 -122.419088 37.807802 预览测试集合 
print test.head(10)
Id Dates DayOfWeek PdDistrict Address \ 0 0 2015-05-10 23:59:00 Sunday BAYVIEW 2000 Block of THOMAS AV 1 1 2015-05-10 23:51:00 Sunday BAYVIEW 3RD ST / REVERE AV 2 2 2015-05-10 23:50:00 Sunday NORTHERN 2000 Block of GOUGH ST 3 3 2015-05-10 23:45:00 Sunday INGLESIDE 4700 Block of MISSION ST 4 4 2015-05-10 23:45:00 Sunday INGLESIDE 4700 Block of MISSION ST 5 5 2015-05-10 23:40:00 Sunday TARAVAL BROAD ST / CAPITOL AV 6 6 2015-05-10 23:30:00 Sunday INGLESIDE 100 Block of CHENERY ST 7 7 2015-05-10 23:30:00 Sunday INGLESIDE 200 Block of BANKS ST 8 8 2015-05-10 23:10:00 Sunday MISSION 2900 Block of 16TH ST 9 9 2015-05-10 23:10:00 Sunday CENTRAL TAYLOR ST / GREEN ST X Y 0 -122.399588 37.735051 1 -122.391523 37.732432 2 -122.426002 37.792212 3 -122.437394 37.721412 4 -122.437394 37.721412 5 -122.459024 37.713172 6 -122.425616 37.739351 7 -122.412652 37.739750 8 -122.418700 37.765165 9 -122.413935 37.798886 我们看到训练集和测试集都有Dates、DayOfWeek、PdDistrict三个特征，我们先从这三个特征入手。训练集中的Category是我们的预测目标，我们先对其进行编码，这里用到sklearn的LabelEncoder()，示例如下： 
from sklearn import preprocessing
label = preprocessing.LabelEncoder()
label.fit([1, 2, 2, 6])
print label.transform([1, 1, 2, 6])
[0 0 1 2]接下来我们对类别进行编码： 
crime = label.fit_transform(train.Category)
对于离散化的特征，有一种常用的特征处理方式是二值化处理，pandas中有get_dummies()函数，函数示例如下： 
pd.get_dummies(pd.Series(list('abca')))
a b c
0 1.0 0.0 0.0
1 0.0 1.0 0.0
2 0.0 0.0 1.0
3 1.0 0.0 0.0

接下来对Dates、DayOfWeek、PdDistrict三个特征进行二值化处理：

days = pd.get_dummies(train.DayOfWeek)
district = pd.get_dummies(train.PdDistrict)
hour = pd.get_dummies(train.Dates.dt.hour)

接下来重新组合训练集，并把类别附加上：

train_data = pd.concat([days, district, hour], axis=1)
train_data['crime'] = crime

针对测试集做同样的处理：

days = pd.get_dummies(test.DayOfWeek)
district = pd.get_dummies(test.PdDistrict)
hour = pd.get_dummies(test.Dates.dt.hour)
test_data = pd.concat([days, district, hour], axis=1)

预览新的训练集和测试集：

print train_data.head(10)
print test_data.head(10)
   Friday  Monday  Saturday  Sunday  Thursday  Tuesday  Wednesday  BAYVIEW  \
0     0.0     0.0       0.0     0.0       0.0      0.0        1.0      0.0   
1     0.0     0.0       0.0     0.0       0.0      0.0        1.0      0.0   
2     0.0     0.0       0.0     0.0       0.0      0.0        1.0      0.0   
3     0.0     0.0       0.0     0.0       0.0      0.0        1.0      0.0   
4     0.0     0.0       0.0     0.0       0.0      0.0        1.0      0.0   
5     0.0     0.0       0.0     0.0       0.0      0.0        1.0      0.0   
6     0.0     0.0       0.0     0.0       0.0      0.0        1.0      0.0   
7     0.0     0.0       0.0     0.0       0.0      0.0        1.0      1.0   
8     0.0     0.0       0.0     0.0       0.0      0.0        1.0      0.0   
9     0.0     0.0       0.0     0.0       0.0      0.0        1.0      0.0   

   CENTRAL  INGLESIDE  ...     15   16   17   18   19   20   21   22   23  \
0      0.0        0.0  ...    0.0  0.0  0.0  0.0  0.0  0.0  0.0  0.0  1.0   
1      0.0        0.0  ...    0.0  0.0  0.0  0.0  0.0  0.0  0.0  0.0  1.0   
2      0.0        0.0  ...    0.0  0.0  0.0  0.0  0.0  0.0  0.0  0.0  1.0   
3      0.0        0.0  ...    0.0  0.0  0.0  0.0  0.0  0.0  0.0  0.0  1.0   
4      0.0        0.0  ...    0.0  0.0  0.0  0.0  0.0  0.0  0.0  0.0  1.0   
5      0.0        1.0  ...    0.0  0.0  0.0  0.0  0.0  0.0  0.0  0.0  1.0   
6      0.0        1.0  ...    0.0  0.0  0.0  0.0  0.0  0.0  0.0  0.0  1.0   
7      0.0        0.0  ...    0.0  0.0  0.0  0.0  0.0  0.0  0.0  0.0  1.0   
8      0.0        0.0  ...    0.0  0.0  0.0  0.0  0.0  0.0  0.0  0.0  1.0   
9      1.0        0.0  ...    0.0  0.0  0.0  0.0  0.0  0.0  0.0  0.0  1.0   

   crime  
0     37  
1     21  
2     21  
3     16  
4     16  
5     16  
6     36  
7     36  
8     16  
9     16  

[10 rows x 42 columns]
   Friday  Monday  Saturday  Sunday  Thursday  Tuesday  Wednesday  BAYVIEW  \
0     0.0     0.0       0.0     1.0       0.0      0.0        0.0      1.0   
1     0.0     0.0       0.0     1.0       0.0      0.0        0.0      1.0   
2     0.0     0.0       0.0     1.0       0.0      0.0        0.0      0.0   
3     0.0     0.0       0.0     1.0       0.0      0.0        0.0      0.0   
4     0.0     0.0       0.0     1.0       0.0      0.0        0.0      0.0   
5     0.0     0.0       0.0     1.0       0.0      0.0        0.0      0.0   
6     0.0     0.0       0.0     1.0       0.0      0.0        0.0      0.0   
7     0.0     0.0       0.0     1.0       0.0      0.0        0.0      0.0   
8     0.0     0.0       0.0     1.0       0.0      0.0        0.0      0.0   
9     0.0     0.0       0.0     1.0       0.0      0.0        0.0      0.0   

   CENTRAL  INGLESIDE ...    14   15   16   17   18   19   20   21   22   23  
0      0.0        0.0 ...   0.0  0.0  0.0  0.0  0.0  0.0  0.0  0.0  0.0  1.0  
1      0.0        0.0 ...   0.0  0.0  0.0  0.0  0.0  0.0  0.0  0.0  0.0  1.0  
2      0.0        0.0 ...   0.0  0.0  0.0  0.0  0.0  0.0  0.0  0.0  0.0  1.0  
3      0.0        1.0 ...   0.0  0.0  0.0  0.0  0.0  0.0  0.0  0.0  0.0  1.0  
4      0.0        1.0 ...   0.0  0.0  0.0  0.0  0.0  0.0  0.0  0.0  0.0  1.0  
5      0.0        0.0 ...   0.0  0.0  0.0  0.0  0.0  0.0  0.0  0.0  0.0  1.0  
6      0.0        1.0 ...   0.0  0.0  0.0  0.0  0.0  0.0  0.0  0.0  0.0  1.0  
7      0.0        1.0 ...   0.0  0.0  0.0  0.0  0.0  0.0  0.0  0.0  0.0  1.0  
8      0.0        0.0 ...   0.0  0.0  0.0  0.0  0.0  0.0  0.0  0.0  0.0  1.0  
9      1.0        0.0 ...   0.0  0.0  0.0  0.0  0.0  0.0  0.0  0.0  0.0  1.0  

[10 rows x 41 columns]

分割训练集和验证集(70%训练,30%验证)准备建模：

from sklearn.cross_validation import train_test_split
training, validation = train_test_split(train_data, train_size=0.6)

贝叶斯训练

from sklearn.metrics import log_loss
from sklearn.naive_bayes import BernoulliNB

model = BernoulliNB()
feature_list = training.columns.tolist()
feature_list = feature_list[:len(feature_list) - 1]
print '选取的特征列：', feature_list
model.fit(training[feature_list], training['crime'])

predicted = np.array(model.predict_proba(validation[feature_list]))
print "朴素贝叶斯log损失为 %f" % (log_loss(validation['crime'], predicted))
选取的特征列： ['Friday', 'Monday', 'Saturday', 'Sunday', 'Thursday', 'Tuesday', 'Wednesday', 'BAYVIEW', 'CENTRAL', 'INGLESIDE', 'MISSION', 'NORTHERN', 'PARK', 'RICHMOND', 'SOUTHERN', 'TARAVAL', 'TENDERLOIN', 0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15, 16, 17, 18, 19, 20, 21, 22, 23]
朴素贝叶斯log损失为 2.581561

逻辑回归

from sklearn.linear_model import LogisticRegression
model = LogisticRegression(C=0.1)
model.fit(training[feature_list], training['crime'])

predicted = np.array(model.predict_proba(validation[feature_list]))
print "逻辑回归log损失为 %f" %(log_loss(validation['crime'], predicted))
逻辑回归log损失为 2.580102

在测试集上运行：

test_predicted = np.array(model.predict_proba(test_data[feature_list]))

保存结果：

col_names = np.sort(train['Category'].unique())
print col_names
result = pd.DataFrame(data=test_predicted, columns=col_names)
result['Id'] = test['Id'].astype(int)
result.to_csv('output.csv', index=False)
['ARSON' 'ASSAULT' 'BAD CHECKS' 'BRIBERY' 'BURGLARY' 'DISORDERLY CONDUCT'
 'DRIVING UNDER THE INFLUENCE' 'DRUG/NARCOTIC' 'DRUNKENNESS' 'EMBEZZLEMENT'
 'EXTORTION' 'FAMILY OFFENSES' 'FORGERY/COUNTERFEITING' 'FRAUD' 'GAMBLING'
 'KIDNAPPING' 'LARCENY/THEFT' 'LIQUOR LAWS' 'LOITERING' 'MISSING PERSON'
 'NON-CRIMINAL' 'OTHER OFFENSES' 'PORNOGRAPHY/OBSCENE MAT' 'PROSTITUTION'
 'RECOVERED VEHICLE' 'ROBBERY' 'RUNAWAY' 'SECONDARY CODES'
 'SEX OFFENSES FORCIBLE' 'SEX OFFENSES NON FORCIBLE' 'STOLEN PROPERTY'
 'SUICIDE' 'SUSPICIOUS OCC' 'TREA' 'TRESPASS' 'VANDALISM' 'VEHICLE THEFT'
 'WARRANTS' 'WEAPON LAWS']
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