Open AI - python data wrangling - 通过单独的字典过滤 pandas dataframe 值，以元组为键答案

【问题标题】：Open AI - python data wrangling - filter pandas dataframe values by a seperate dictionary with tuples as the keysOpen AI - python data wrangling - 通过单独的字典过滤 pandas dataframe 值，以元组为键
【发布时间】：2022-12-15 15:32:32
【问题描述】：

我有一个数据整理任务，我想以某种格式保存输入。当我感到疲倦时，我决定尝试使用 OpenAI，看看代码生成能做什么。它得出了与我最初追求的不同的解决方案。如果这是一件好事，我仍然有点犹豫，特别是对于那些刚刚学习编码的人，但这是一个不同的主题。

迅速的：

create a python function to filter pandas dataframe values by a seperate dictionary with tuples as the keys 

for example:
I have a dataframe named "temp_obs"  like pd.DataFrame([[2000, January, 1,2,3],[2000, March,3,3],[2004, December,2,6,1,2] ,[1993, December,2]]
I have a dictionary named "phase" with keys as tuples corresponding to the dataframe columns and rows, like this {(2000,January):0, (2000,March):3, (2004,December):3, (1993,December):0, (1999,March):5 ,(2020,January):1}.

I want to filter the dataframe such that I only get the rows with the corresponding dictionary values greater than 2.

the output would be a dataframe with rows  [2000, March,3,3] and [2004, December,2,6,1,2]

【问题讨论】：

标签： python pandas openai

【解决方案1】：

通过 OpenAI 回答。

A：

You can use <code>pd.DataFrame.isin</code> to filter the dataframe.
<code>temp_obs = pd.DataFrame([[2000, 'January', 1,2,3],[2000, 'March',3,3],[2004, 'December',2,6,1,2] ,[1993, 'December',2]])

phase = {(2000,'January'):0, (2000,'March'):3, (2004,'December'):3, (1993,'December'):0, (1999,'March'):5 ,(2020,'January'):1}

temp_obs[temp_obs.isin(phase.keys()).sum(1) &gt; 2]
</code>
Output:
<code>  
    0      1       2  3  4  5

1  2000    March   3  3  0  0
2  2004  December  2  6  1  2
</code>

【讨论】：