Pandas：估算 NaN
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Pandas: Impute NaN's
提问by Zhubarb
I have an incomplete dataframe, incomplete_df, as below. I want to impute the missing amounts with the average amountof the corresponding id. If the average for that specific idis itself NaN (see id=4), I want to use the overall average.
我有一个不完整的数据框，incomplete_df如下所示。我想amount用amount相应的id. 如果该特定的平均值id本身就是 NaN（请参阅 参考资料id=4），我想使用整体平均值。
Below are the example data and my highly inefficient solution:
以下是示例数据和我非常低效的解决方案：
import pandas as pd
import numpy as np
incomplete_df = pd.DataFrame({'id': [1,2,3,2,2,3,1,1,1,2,4],
'type': ['one', 'one', 'two', 'three', 'two', 'three', 'one', 'two', 'one', 'three','one'],
'amount': [345,928,np.NAN,645,113,942,np.NAN,539,np.NAN,814,np.NAN]
}, columns=['id','type','amount'])
# Forrest Gump Solution
for idx in incomplete_df.index[np.isnan(incomplete_df.amount)]: # loop through all rows with amount = NaN
cur_id = incomplete_df.loc[idx, 'id']
if (cur_id in means.index ):
incomplete_df.loc[idx, 'amount'] = means.loc[cur_id]['amount'] # average amount of that specific id.
else:
incomplete_df.loc[idx, 'amount'] = np.mean(means.amount) # average amount across all id's
What is the fastest and the most pythonic/pandonic way to achieve this?
实现这一目标的最快和最 Pythonic/Pandonic 的方法是什么？
回答by DSM
Disclaimer: I'm not really interested in the fastest solution but the most pandorable.
免责声明：我对最快的解决方案并不感兴趣，但对最讨人喜欢的解决方案并不感兴趣。
Here, I think that would be something like:
在这里，我认为这将是这样的：
>>> df["amount"].fillna(df.groupby("id")["amount"].transform("mean"), inplace=True)
>>> df["amount"].fillna(df["amount"].mean(), inplace=True)
which produces
产生
>>> df
id type amount
0 1 one 345.0
1 2 one 928.0
2 3 two 942.0
3 2 three 645.0
4 2 two 113.0
5 3 three 942.0
6 1 one 442.0
7 1 two 539.0
8 1 one 442.0
9 2 three 814.0
10 4 one 615.2
[11 rows x 3 columns]
There are lots of obvious tweaks depending upon exactly how you want the chained imputation process to go.
有许多明显的调整取决于您希望链式插补过程如何进行。
