基于另一个（熊猫）中的值递增列中的值

Question

I'd like to find a simple way to increment values in in one column that correspond to a particular date in Pandas. This is what I have so far:

old_casual_value = tstDF['casual'].loc[tstDF['ds'] == '2012-10-08'].values[0]
old_registered_value = tstDF['registered'].loc[tstDF['ds'] == '2012-10-08'].values[0]

# Adjusting the numbers of customers for that day.
tstDF.set_value(406, 'casual', old_casual_value*1.05)
tstDF.set_value(406, 'registered', old_registered_value*1.05)

If I could find a better and simpler way to do this (a one-liner), I'd greatly appreciate it.

Thanks for your help.

Answer 1

The following one liner should work based on your limited description of your problem. If not, please provide more information.

#The code below first filters out the records based on specified date and then increase casual and regisitered column values by 1.05 times.
tstDF.loc[tstDF['ds'] == '2012-10-08',['casual','registered']]*=1.05