How to call a column by combining a string and another variable in a python dataframe?

Question

Imagine I have a dataframe with these variables and values:

ID	Weight	LR Weight	UR Weight	Age	LS Age	US Age	Height	LS Height	US Height
1	63	50	80	20	18	21	165	160	175
2	75	50	80	22	18	21	172	160	170
3	49	45	80	17	18	21	180	160	180

I want to create the additional following variables:

ID	Flag_Weight	Flag_Age	Flag_Height
1	1	1	1
2	1	0	0
3	1	0	1

These flags simbolize that the main variable values (eg: Weight, Age and Height) are between the correspondent Lower or Upper limits, which may start with different 2 digits (in this dataframe I gave four examples: LR, UR, LS, US, but in my real dataframe I have more), and whose limit values sometimes differ from ID to ID.

Can you help me create these flags, please?

Thank you in advance.

Answer 1

So, if I understood correctly, you want to add columns with these new variables. The simplest solution to this would be df.insert() .

You could make it something like this:

df.insert(number of column after which you want to insert the new column, name of the column, values of the new column)

You can make up the new values in pretty much everyway you can imagine. So just copying a column or simple mathematical operations like +,-,*,/, can be performed. But you can also apply a whole function, which returns the flags based on your conditions as values of the new column.

If the new columsn can just be appended, you can even just make up a new column like this:

df['new column name'] = any values you want

I hope this helped.

Answer 2

You can use reshaping using a temporary MultiIndex :

(df.set_index('ID')
   .pipe(lambda d: d.set_axis(pd.MultiIndex.from_frame(
                     d.columns.str.extract('(^[LU]?).*?\s*(\S+)$')),
                              axis=1)
        )
   .stack()
   .assign(flag=lambda d: d[''].between(d['L'], d['U']).astype(int))
   ['flag'].unstack().add_prefix('Flag_').reset_index()
)

Output:

   ID  Flag_Age  Flag_Height  Flag_Weight
0   1         1            1            1
1   2         0            0            1
2   3         0            1            1