How to take, individually, the first and last value from groupby columns? - max

Question

Inside a for loop, I am pulling the first and last value from multiple columns, so that I can use each of these values to create a scalar value that is then mathematically applied to create a new column.
My workaround was to limit the selection to the head or tail of each column, and then turn it into a numeric type (int or float) by using min() or max():
for title, group in df.groupby('Test'):
x1 = min(group["Test Reading"].head(1))
x2 = max(group["Test Reading"].tail(1))
x3 = min(group["Test Point"].head(1))
x4 = max(group["Test Point"].tail(1))
R=(x2-x1)/(x4-x3) #linearization scalar
group['Test Point Error'] =100*(group['Test Reading']- (group['Test Point']*R+x1))/(x2-x1)
There are other problems with the code, but I've attempted to address those in another question (How do I use .loc with groupby so that creating a new column based on grouped data won't be considered a copy?).

Answer 1

First we groupby the data by Test and use .agg to get the first occurrence of Test_Point and Test_reading. We save grouped df to a new variable df_agg.
df_agg=df.groupby("Test).agg({"Test_Reading":["first","last"],"Test_Point":["first","last"]})
We flatten the multindex in the columns:
df_agg.columns=["_".join(c) for c in df_agg.columns]
We make your calculation for R:
df_agg["R"]=(df_agg["Test Reading_last"]-df_agg["Test Reading_first"])/(df_agg["Test_Point_last"]-df_agg["Test_Point_first"])
We merge the aggregate df to the original one on the value of Test, so that we have R avaible for each example.
df.merge(df_agg,on="Test")
Now you have the value of R and x1 ... x4 available for all examples and you can run your formula.
df['Test Point Error'] =100*(df['Test Reading']- (df['Test Point']*df["R"]+df["Test Reading_first"]))/(df["Test Reading_last"]-df["Test Reading_first"])