How to replace a zero and nan values in dataframe based in where condition in fillna or replace

Question

I have a Data frame like this

Temp_in_C   Temp_in_F   Date          Year   Month   Day
   23          65       2011-12-12     2011    12     12
   12          72       2011-12-12    2011     12      12
   NaN         67       2011-12-12     2011    12      12
   0           0        2011-12-12     2011    12      12
   7           55       2011-12-13     2011    12       13

I am trying to get output in this format (The NaN and zero values of pertuculer day is replaced by avg temp of that day only) Output will be

Temp_in_C   Temp_in_F   Date          Year   Month   Day
   23          65       2011-12-12     2011    12     12
   12          72       2011-12-12    2011     12      12
   17.5        67       2011-12-12     2011    12      12
   17.5        68       2011-12-12     2011    12      12
   7           55       2011-12-13     2011    12       13

These vales will be replaced by mean of that perticuler day. I am trying to do this

temp_df = csv_data_df[csv_data_df["Temp_in_C"]!=0]
temp_df["Temp_in_C"] = 
temp_df["Temp_in_C"].replace('*',np.nan)
x=temp_df["Temp_in_C"].mean()   
csv_data_df["Temp_in_C"]=csv_data_df["Temp_in_C"]
.replace(0.0,x)
csv_data_df["Temp_in_C"]=csv_data_df["Temp_in_C"]
.fillna(x)

This code is taking the mean of whole columns and replacing it directly. How can i group by day and take mean and then replace values for that particular day only.

Answer 1

First, replace zeros with NaN

df = df.replace(0,np.nan)

Then fill the missing values using transform (see this post )

df.groupby('Date').transform(lambda x: x.fillna(x.mean()))

Gives:

   Temp_in_C  Temp_in_F  Year  Month  Day
0       23.0       65.0  2011     12   12
1       12.0       72.0  2011     12   12
2       17.5       67.0  2011     12   12
3       17.5       68.0  2011     12   12
4        7.0       55.0  2011     12   13