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在Featuretools中计算多个训练窗口中的功能

featuretools feature-extraction

1

Tomas Greif · 技术社区 · 8 年前

我有一张有顾客和交易的桌子。有没有办法获得在过去3/6/9/12个月内过滤的功能?我想自动生成功能:

过去3个月的交易次数
....
过去12个月的交易次数
...

我试过用 training_window =["1 month", "3 months"], ,但它似乎不会为每个窗口返回多个功能。

例子:

import featuretools as ft
es = ft.demo.load_mock_customer(return_entityset=True)

window_features = ft.dfs(entityset=es,
   target_entity="customers",
   training_window=["1 hour", "1 day"],
   features_only = True)

window_features

我是否必须单独处理各个窗口,然后合并结果?

1 回复 | 直到 7 年前

1

2

Max Kanter 8 年前

如前所述,在Featuretools 0.2.1中,必须为每个训练窗口分别构建特征矩阵,然后合并结果。举个例子,你可以这样做:

import pandas as pd
import featuretools as ft
es = ft.demo.load_mock_customer(return_entityset=True)
cutoff_times = pd.DataFrame({"customer_id": [1, 2, 3, 4, 5],
                             "time": pd.date_range('2014-01-01 01:41:50', periods=5, freq='25min')})
features = ft.dfs(entityset=es,
                  target_entity="customers",
                  agg_primitives=['count'],
                  trans_primitives=[],
                  features_only = True)
fm_1 = ft.calculate_feature_matrix(features, 
                                   entityset=es, 
                                   cutoff_time=cutoff_times,
                                   training_window='1h', 
                                   verbose=True)

fm_2 = ft.calculate_feature_matrix(features, 
                                   entityset=es, 
                                   cutoff_time=cutoff_times,
                                   training_window='1d', 
                                   verbose=True)
new_df = fm_1.reset_index()
new_df = new_df.merge(fm_2.reset_index(), on="customer_id", suffixes=("_1h", "_1d"))

然后,新的数据帧将如下所示:

customer_id COUNT(sessions)_1h  COUNT(transactions)_1h  COUNT(sessions)_1d COUNT(transactions)_1d
1           1                   17                      3                 43
2           3                   36                      3                 36
3           0                   0                       1                 25
4           0                   0                       0                 0
5           1                   15                      2                 29