如何在转换函数中包含两个lambda运算？ [英] How to include two lambda operations in transform function?

问题描述

我有一个如下所示的数据帧

I have a dataframe like as given below

df = pd.DataFrame({
'date' :['2173-04-03 12:35:00','2173-04-03 17:00:00','2173-04-03 20:00:00','2173-04-04 11:00:00','2173-04-04 12:00:00','2173-04-04 11:30:00','2173-04-04 16:00:00','2173-04-04 22:00:00','2173-04-05 04:00:00'],
'subject_id':[1,1,1,1,1,1,1,1,1],
'val' :[5,5,5,10,10,5,5,8,8]
 })

我想应用几个逻辑（ val 列上的 logic_1 和<在 date 列上输入了code> logic_2 ）。请在逻辑下面找到

I would like to apply couple of logic (logic_1 on val column and logic_2 on date column) to the code. Please find below the logic

logic_1 = lambda x: (x.shift(2).ge(x.shift(1))) & (x.ge(x.shift(2).add(3))) & (x.eq(x.shift(-1)))
logic_2 = lambda y: (y.shift(1).ge(1)) & (y.shift(2).ge(2)) & (y.shift(-1).ge(1)) 

向SO用户提供帮助的信用带有逻辑

credit to SO users for helping me with logic

这是我在下面尝试的

   df['label'] = ''
   df['date'] = pd.to_datetime(df['date'])
   df['tdiff'] = df['date'].shift(-1) - df['date']
   df['tdiff'] = df['tdiff'].dt.total_seconds()/3600
   df['lo_1'] = df.groupby('subject_id')['val'].transform(logic_1).map({True:'1',False:''})
   df['lo_2'] = df.groupby('subject_id')['tdiff'].transform(logic_2).map({True:'1',False:''}) 

如何使 logic_1 和 logic_2 都成为一个逻辑语句的一部分？可能吗我可能也有两种以上的逻辑。不必为每个逻辑写一行，而是可以在一个逻辑语句中将所有逻辑耦合在一起。

How can I make both the logic_1 and logic_2 be part of one logic statement? Is it even possible? I might have more than 2 logics as well. Instead of writing one line for each logic, is it possible to couple all logics together in one logic statement.

我希望我的输出是当 logic_1 和 logic_2都用 1 填充label 列满意

I expect my output to be with label column being filled with 1 when both logic_1 and logic_2 are satisfied

推荐答案

您有几件要解决的问题
首先，在 logic_2 中，您有 lambda x ，但使用 y ，因此，您必须进行如下更改

You have a few things to fix First, in logic_2, you have lambda x but use y, so, you got to change that as below

logic_2 = lambda y: (y.shift(1).ge(1)) & (y.shift(2).ge(2)) & (y.shift(-1).ge(1))

然后您可以同时使用逻辑如下所示'
无需创建空白列标签。您可以直接在下面创建标签列。

Then you can use the logic's together as below' No need to create a blank column label. You can create the '`label' column directly as below.

df['label'] = ((df.groupby('subject_id')['val'].transform(logic_1))
           & (df.groupby('subject_id')['tdiff'].transform(logic_2))).map({True:'0',False:'1'})

注意：您的逻辑生成所有错误的值。因此，如果 False 映射到1，而不是 True

Note: You logic produces all False values. So, you will get 1's if False is mapped to 1, not True

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