XGBoost解決多分類問題

XGBoost解決多分類問題

寫在前面的話

    XGBoost官方給的二分類問題的例子是區別蘑菇有無毒，數據集和代碼都可以在xgboost中的demo文件夾對應找到，我是用的Anaconda安裝的XGBoost，實現起來比較容易。唯一的梗就是在終端中運行所給命令：  ../../xgboost mushroom.conf 時會報錯，是路徑設置的問題，所以我乾脆把xgboost文件夾下的xgboost.exe拷到了mushroom.conf配置文件所在文件夾下，這樣直接定位到該文件夾下就可以運行： xgboost mushroom.conf。二分類數據預處理，也就是data wraggling部分的代碼有一定的借鑑意義，值得一看。

    多分類問題給的例子是根據34個特徵識別6種皮膚病，由於終端中運行runexp.sh沒有反應，也不報錯，所以我乾脆把數據集下載到對應的demo文件夾下了,主要的代碼如下，原來有部分比較難懂的語句我自己加了一些註釋，這樣理解起來就會順暢多了。

#! /usr/bin/python
import numpy as np
import xgboost as xgb

# label need to be 0 to num_class -1
# if col 33 is '?' let it be 1 else 0, col 34 substract 1
data = np.loadtxt('./dermatology.data', delimiter=',',converters={33: lambda x:int(x == '?'), 34: lambda x:int(x)-1 } )
sz = data.shape

train = data[:int(sz[0] * 0.7), :] # take row 1-256 as training set
test = data[int(sz[0] * 0.7):, :]  # take row 257-366 as testing set

train_X = train[:,0:33]
train_Y = train[:, 34]


test_X = test[:,0:33]
test_Y = test[:, 34]

xg_train = xgb.DMatrix( train_X, label=train_Y)
xg_test = xgb.DMatrix(test_X, label=test_Y)
# setup parameters for xgboost
param = {}
# use softmax multi-class classification
param['objective'] = 'multi:softmax'
# scale weight of positive examples
param['eta'] = 0.1
param['max_depth'] = 6
param['silent'] = 1
param['nthread'] = 4
param['num_class'] = 6

watchlist = [ (xg_train,'train'), (xg_test, 'test') ]
num_round = 5
bst = xgb.train(param, xg_train, num_round, watchlist );
# get prediction
pred = bst.predict( xg_test );

print ('predicting, classification error=%f' % (sum( int(pred[i]) != test_Y[i] for i in range(len(test_Y))) / float(len(test_Y)) ))

# do the same thing again, but output probabilities
param['objective'] = 'multi:softprob'
bst = xgb.train(param, xg_train, num_round, watchlist );
# Note: this convention has been changed since xgboost-unity
# get prediction, this is in 1D array, need reshape to (ndata, nclass)
yprob = bst.predict( xg_test ).reshape( test_Y.shape[0], 6 )
ylabel = np.argmax(yprob, axis=1)  # return the index of the biggest pro

print ('predicting, classification error=%f' % (sum( int(ylabel[i]) != test_Y[i] for i in range(len(test_Y))) / float(len(test_Y)) ))

    結果如下：

[0] train-merror:0.011719   test-merror:0.127273
[1] train-merror:0.015625   test-merror:0.127273
[2] train-merror:0.011719   test-merror:0.109091
[3] train-merror:0.007812   test-merror:0.081818
[4] train-merror:0.007812   test-merror:0.090909
predicting, classification error=0.090909
[0] train-merror:0.011719   test-merror:0.127273
[1] train-merror:0.015625   test-merror:0.127273
[2] train-merror:0.011719   test-merror:0.109091
[3] train-merror:0.007812   test-merror:0.081818
[4] train-merror:0.007812   test-merror:0.090909
predicting, classification error=0.090909

    

    不管是直接返回診斷類型，還是返回各類型的概率，然後取概率最大的那個對應的類型的index，結果都是一樣的。