如何用条件列均值填充数据框的空/nan单元格 [英] How to fill dataframe's empty/nan cell with conditional column mean
问题描述
我正在尝试使用该特定列的平均值填充(熊猫)数据框的null/empty值.
I am trying to fill the (pandas) dataframe's null/empty value using the mean of that specific column.
数据如下:
ID Name Industry Year Revenue
1 Treslam Financial Services 2009 $5,387,469
2 Rednimdox Construction 2013
3 Lamtone IT Services 2009 $11,757,018
4 Stripfind Financial Services 2010 $12,329,371
5 Openjocon Construction 2013 $4,273,207
6 Villadox Construction 2012 $1,097,353
7 Sumzoomit Construction 2010 $7,703,652
8 Abcddd Construction 2019
.
.
我试图用行业=='建筑'的收入"列的平均值填充该空白单元格.
I am trying to fill that empty cell with the mean of Revenue column where Industry is == 'Construction'.
要得到我们的数值平均值，我做了:
To get our numerical mean value I did:
df.groupby(['Industry'], as_index = False).mean()
我正在尝试执行以下操作以就地填充空白单元格:
I am trying to do something like this to fill up that empty cell in-place:
(df[df['Industry'] == "Construction"]['Revenue']).fillna("$21212121.01", inplace = True)
..但是不起作用.谁能告诉我如何实现它！非常感谢.
..but it is not working. Can anyone tell me how to achieve it! Thanks a lot.
预期输出:
ID Name Industry Year Revenue
1 Treslam Financial Services 2009 $5,387,469
2 Rednimdox Construction 2013 $21212121.01
3 Lamtone IT Services 2009 $11,757,018
4 Stripfind Financial Services 2010 $12,329,371
5 Openjocon Construction 2013 $4,273,207
6 Villadox Construction 2012 $1,097,353
7 Sumzoomit Construction 2010 $7,703,652
8 Abcddd Construction 2019 $21212121.01
.
.
推荐答案
尽管用作平均值的数字不同，但我们提供了两种类型的平均值:正常平均值和根据包含NaN的病例数计算得出的平均值.
Although the numbers used as averages are different, we have presented two types of averages: the normal average and the average calculated on the number of cases that include NaN.
df['Revenue'] = df['Revenue'].replace({'\$':'', ',':''}, regex=True)
df['Revenue'] = df['Revenue'].astype(float)
df_mean = df.groupby(['Industry'], as_index = False)['Revenue'].mean()
df_mean
Industry Revenue
0 Construction 4.358071e+06
1 Financial Services 8.858420e+06
2 IT Services 1.175702e+07
df_mean_nan = df.groupby(['Industry'], as_index = False)['Revenue'].agg({'Sum':np.sum, 'Size':np.size})
df_mean_nan['Mean_nan'] = df_mean_nan['Sum'] / df_mean_nan['Size']
df_mean_nan
Industry Sum Size Mean_nan
0 Construction 13074212.0 5.0 2614842.4
1 Financial Services 17716840.0 2.0 8858420.0
2 IT Services 11757018.0 1.0 11757018.0
考虑到NaN的平均值
df.loc[df['Revenue'].isna(),['Revenue']] = df_mean_nan.loc[df_mean_nan['Industry'] == 'Construction',['Mean_nan']].values
df
ID Name Industry Year Revenue
0 1 Treslam Financial Services 2009 5387469.0
1 2 Rednimdox Construction 2013 2614842.4
2 3 Lamtone IT Services 2009 11757018.0
3 4 Stripfind Financial Services 2010 12329371.0
4 5 Openjocon Construction 2013 4273207.0
5 6 Villadox Construction 2012 1097353.0
6 7 Sumzoomit Construction 2010 7703652.0
7 8 Abcddd Construction 2019 2614842.4
正常平均值:(不包括NaN)
Normal average: (NaN is excluded)
df.loc[df['Revenue'].isna(),['Revenue']] = df_mean.loc[df_mean['Industry'] == 'Construction',['Revenue']].values
df
ID Name Industry Year Revenue
0 1 Treslam Financial Services 2009 5.387469e+06
1 2 Rednimdox Construction 2013 4.358071e+06
2 3 Lamtone IT Services 2009 1.175702e+07
3 4 Stripfind Financial Services 2010 1.232937e+07
4 5 Openjocon Construction 2013 4.273207e+06
5 6 Villadox Construction 2012 1.097353e+06
6 7 Sumzoomit Construction 2010 7.703652e+06
7 8 Abcddd Construction 2019 4.358071e+06
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