H2O目标均值编码器“帧以相同的顺序发送”ERROR

Question

我遵循H2O示例在Sparking Water中运行目标均值编码（火花水2.4.2和H2O 3.22.04）。它在以下所有段落中运行良好

from h2o.targetencoder import TargetEncoder

# change label to factor
input_df_h2o['label'] = input_df_h2o['label'].asfactor()

# add fold column for Target Encoding
input_df_h2o["cv_fold_te"] = input_df_h2o.kfold_column(n_folds = 5, seed = 54321)

# find all categorical features
cat_features = [k for (k,v) in input_df_h2o.types.items() if v in ('string')]
# convert string to factor
for i in cat_features:
    input_df_h2o[i] = input_df_h2o[i].asfactor()

# target mean encode
targetEncoder = TargetEncoder(x= cat_features, y = y, fold_column = "cv_fold_te", blending_avg=True)
targetEncoder.fit(input_df_h2o)

但是当我开始使用用于拟合Target Encoder的相同数据集来运行转换代码时（请参阅下面的代码）：

ext_input_df_h2o = targetEncoder.transform(frame=input_df_h2o,
                                    holdout_type="kfold", # mean is calculating on out-of-fold data only; loo means leave one out
                                    is_train_or_valid=True,
                                    noise = 0, # determines if random noise should be added to the target average
                                    seed=54321)

我会有错误的

Traceback (most recent call last):
  File "/tmp/zeppelin_pyspark-6773422589366407956.py", line 331, in <module>
    exec(code)
  File "<stdin>", line 5, in <module>
  File "/usr/lib/envs/env-1101-ver-1619-a-4.2.9-py-3.5.3/lib/python3.5/site-packages/h2o/targetencoder.py", line 97, in transform
    assert self._encodingMap.map_keys['string'] == self._teColumns
AssertionError

我在源代码http://docs.h2o.ai/h2o/latest-stable/h2o-py/docs/_modules/h2o/targetencoder.html 中找到了代码，但是如何解决这个问题呢？它与用于运行拟合的表相同。

Answer 1

问题是因为您正在尝试编码多个分类功能。我认为这是H2O的一个错误，但你可以解决将变换器放在一个迭代所有分类名称的for循环中。

import numpy as np
import pandas as pd
import h2o
from h2o.targetencoder import TargetEncoder
h2o.init()

df = pd.DataFrame({
    'x_0': ['a'] * 5 + ['b'] * 5,
    'x_1': ['c'] * 9 + ['d'] * 1,
    'x_2': ['a'] * 3 + ['b'] * 7,
    'y_0': [1, 1, 1, 1, 0, 1, 0, 0, 0, 0]
})

hf = h2o.H2OFrame(df)
hf['cv_fold_te'] = hf.kfold_column(n_folds=2, seed=54321)
hf['y_0'] = hf['y_0'].asfactor()
cat_features = ['x_0', 'x_1', 'x_2']

for item in cat_features:
    target_encoder = TargetEncoder(x=[item], y='y_0', fold_column = 'cv_fold_te')
    target_encoder.fit(hf)
    hf = target_encoder.transform(frame=hf, holdout_type='kfold',
                                  seed=54321, noise=0.0)
hf

Answer 2

谢谢大家让我们知道。断言是一种预防措施，因为我不确定是否可能存在订单可以更改的情况。其余的代码是在考虑到这个假设的情况下编写的，因此无论如何都可以安全地使用更改的顺序，但是断言被遗忘了。添加了测试并删除了断言。现在，此问题已修复并合并。应该在即将发布的修复版本中提供。 0xdata.atlassian.net/browse/PUBDEV-6474