There are several ways to approach this, and the best way will depend on whether the January data is systematically different from other months. Most real-world data is likely to be somewhat seasonal, so let's use the average high temperature (Fahrenheit) of a random city in the northern hemisphere as an example.
df=pd.DataFrame({ 'month' : [10,11,12,1,2,3],
'temp' : [65,50,45,np.nan,40,43] }).set_index('month')
You could use a rolling mean as you suggest, but the issue is that you will get an average temperature over the entire year, which ignores the fact that January is the coldest month. To correct for this, you could reduce the window to 3, which results in the January temp being the average of the December and February temps. (I am also using min_periods=1
as suggested in @user394430's answer.)
df['rollmean12'] = df['temp'].rolling(12,center=True,min_periods=1).mean()
df['rollmean3'] = df['temp'].rolling( 3,center=True,min_periods=1).mean()
Those are improvements but still have the problem of overwriting existing values with rolling means. To avoid this you could combine with the update()
method (see documentation here).
df['update'] = df['rollmean3']
df['update'].update( df['temp'] ) # note: this is an inplace operation
There are even simpler approaches that leave the existing values alone while filling the missing January temps with either the previous month, next month, or the mean of the previous and next month.
df['ffill'] = df['temp'].ffill() # previous month
df['bfill'] = df['temp'].bfill() # next month
df['interp'] = df['temp'].interpolate() # mean of prev/next
In this case, interpolate()
defaults to simple linear interpretation, but you have several other intepolation options also. See documentation on pandas interpolate for more info. Or this statck overflow question:
Interpolation on DataFrame in pandas
Here is the sample data with all the results:
temp rollmean12 rollmean3 update ffill bfill interp
month
10 65.0 48.6 57.500000 65.0 65.0 65.0 65.0
11 50.0 48.6 53.333333 50.0 50.0 50.0 50.0
12 45.0 48.6 47.500000 45.0 45.0 45.0 45.0
1 NaN 48.6 42.500000 42.5 45.0 40.0 42.5
2 40.0 48.6 41.500000 40.0 40.0 40.0 40.0
3 43.0 48.6 41.500000 43.0 43.0 43.0 43.0
In particular, note that "update" and "interp" give the same results in all months. While it doesn't matter which one you use here, in other cases one way or the other might be better.