在多类分类问题中预测值时,我想获得预测值的概率。
我试图通过使用H2O的apply函数来解决此问题:
Exception.Data
但是它不起作用:
'ValueError:unimpl字节码指令:CALL_METHOD'
也许是行不通的,因为h2o.max不像h2o.mean那样具有轴参数??? 我找不到关于apply函数支持哪些操作的文档。
我想使用类似于此熊猫代码的h2o数据操作来解决此问题:
predicted_df = modelo_assessor.predict(to_predict_h2o_frame)
predicted_df.apply((lambda x: x.max()), axis=1)
答案 0 :(得分:0)
每当使用apply时都会发生这种情况。使用H2O文档中的示例:
我能够通过降级到Python 3.6.x来解决问题
http://docs.h2o.ai/h2o/latest-stable/h2o-py/docs/frame.html#h2oframe
python_lists = [[1,2,3,4], [1,2,3,4]]
h2oframe = h2o.H2OFrame(python_obj=python_lists,
na_strings=['NA'])
colMean = h2oframe.apply(lambda x: x.mean(), axis=0)
---------------------------------------------------------------------------
ValueError Traceback (most recent call last)
<ipython-input-43-8da6b76c71bd> in <module>
2 h2oframe = h2o.H2OFrame(python_obj=python_lists,
3 na_strings=['NA'])
----> 4 colMean = h2oframe.apply(lambda x: x.mean(), axis=0)
~/anaconda3/envs/h2o1/lib/python3.7/site-packages/h2o/frame.py in apply(self, fun, axis)
4910 assert_is_type(fun, FunctionType)
4911 assert_satisfies(fun, fun.__name__ == "<lambda>")
-> 4912 res = lambda_to_expr(fun)
4913 return H2OFrame._expr(expr=ExprNode("apply", self, 1 + (axis == 0), *res))
4914
~/anaconda3/envs/h2o1/lib/python3.7/site-packages/h2o/astfun.py in lambda_to_expr(fun)
133 code = fun.__code__
134 lambda_dis = _disassemble_lambda(code)
--> 135 return _lambda_bytecode_to_ast(code, lambda_dis)
136
137 def _lambda_bytecode_to_ast(co, ops):
~/anaconda3/envs/h2o1/lib/python3.7/site-packages/h2o/astfun.py in _lambda_bytecode_to_ast(co, ops)
147 body, s = _opcode_read_arg(s, ops, keys)
148 else:
--> 149 raise ValueError("unimpl bytecode instr: " + instr)
150 if s > 0:
151 print("Dumping disassembled code: ")
ValueError: unimpl bytecode instr: CALL_METHOD