從 Python 中的 H2OXGBoostEstimator 模型中提取原生 xgboost 模型

Question

是否可以從 Python 中的 H2OXGBoostEstimator 模型中提取本機 xgboost 模型泡菜文件並通過原始 XGBoost Python API 讀取？ 謝謝！

Answer 1

您可以使用以下內容：

import matplotlib.pyplot as plt
from xgboost import plot_tree

# model is your xgboost model, choose which tree in num_trees and layout direction default if top to bottom, LR is left to right
plot_tree(model, num_trees=0, rankdir='LR') 
plt.show()

這是一個完全可重現的示例，但請注意它使用的是在 xgboost 存儲庫中找到的數據集，並使用此示例創建模型

import numpy as np
import scipy.sparse
import pickle
import xgboost as xgb
import matplotlib.pyplot as plt
from xgboost import plot_tree

### simple example
# load file from text file, also binary buffer generated by xgboost
dtrain = xgb.DMatrix('../data/agaricus.txt.train')
dtest = xgb.DMatrix('../data/agaricus.txt.test')

# specify parameters via map, definition are same as c++ version
param = {'max_depth':2, 'eta':1, 'silent':1, 'objective':'binary:logistic'}

# specify validations set to watch performance
watchlist = [(dtest, 'eval'), (dtrain, 'train')]
num_round = 2
bst = xgb.train(param, dtrain, num_round, watchlist)

plot_tree(bst, num_trees=0, rankdir='LR') # bst is your xgboost model, choose which tree in num_trees and layout direction default if top to bottom, LR is left to right
plt.show()

Answer 2

您可以嘗試這兩種“h2o-to-xgboost”方法從經過訓練的 H2O 模型中提取 XGBoost 超參數和 DMatrix，這（根據文檔）將為您提供完全相同的 XGBoost 原生 python 模型。

nativeXGBoostParam = h2oModelD.convert_H2OXGBoostParams_2_XGBoostParams()
nativeXGBoostInput = data.convert_H2OFrame_2_DMatrix(myX, y, h2oModelD)
        
nativeModel = xgb.train(dtrain=nativeXGBoostInput,
                        params=nativeXGBoostParam[0],                        
                        num_boost_round=nativeXGBoostParam[1])

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