XGBoost解决多分类问题_xgboost 分类数据

作者：凡人多烦事01 | 2024-04-23 09:50:00

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xgboost 分类数据

XGBoost解决多分类问题

写在前面的话

XGBoost官方给的二分类问题的例子是区别蘑菇有无毒，数据集和代码都可以在xgboost中的demo文件夹对应找到，我是用的Anaconda安装的XGBoost，实现起来比较容易。唯一的梗就是在终端中运行所给命令： ../../xgboost mushroom.conf 时会报错，是路径设置的问题，所以我干脆把xgboost文件夹下的xgboost.exe拷到了mushroom.conf配置文件所在文件夹下，这样直接定位到该文件夹下就可以运行： xgboost mushroom.conf。二分类数据预处理，也就是data wraggling部分的代码有一定的借鉴意义，值得一看。

多分类问题给的例子是根据34个特征识别6种皮肤病，由于终端中运行runexp.sh没有反应，也不报错，所以我干脆把数据集下载到对应的demo文件夹下了,主要的代码如下，原来有部分比较难懂的语句我自己加了一些注释，这样理解起来就会顺畅多了。

[python] view plain copy

#! /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)) ))

结果如下：

[python] view plain copy

[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

不管是直接返回诊断类型，还是返回各类型的概率，然后取概率最大的那个对应的类型的index，结果都是一样的。

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