Grouping by and applying lambda with condition for the first row - Pandas - pandas

Question

I have a data frame with IDs, and choices that have made by those IDs.
The alternatives (choices) set is a list of integers: [10, 20, 30, 40].
Note: That's important to use this list. Let's call it 'choice_list'.
This is the data frame:
ID Choice
1 10
1 30
1 10
2 40
2 40
2 40
3 20
3 40
3 10
I want to create a variable for each alternative: '10_Var', '20_Var', '30_Var', '40_Var'.
At the first row of each ID, if the first choice was '10' for example, so the variable '10_Var' will get the value 0.6 (some parameter), and each of the other variables ('20_Var', '30_Var', '40_Var') will get the value (1 - 0.6) / 4.
The number 4 stands for the number of alternatives.
Expected result:
ID Choice 10_Var 20_Var 30_Var 40_Var
1 10 0.6 0.1 0.1 0.1
1 30
1 10
2 40 0.1 0.1 0.1 0.6
2 40
2 40
3 20 0.1 0.6 0.1 0.1
3 40
3 10

Answer 1

you can use np.where to do this. It is efficient that df.where
df = pd.DataFrame([['1', 10], ['1', 30], ['1', 10], ['2', 40], ['2', 40], ['2', 40], ['3', 20], ['3', 40], ['3', 10]], columns=('ID', 'Choice'))
choices = np.unique(df.Choice)
for choice in choices:
df[f"var_{choice}"] = np.where(df.Choice==choice, 0.6, (1 - 0.6) / 4)
df
Result
ID Choice var_10 var_20 var_30 var_40
0 1 10 0.6 0.1 0.1 0.1
1 1 30 0.1 0.1 0.6 0.1
2 1 10 0.6 0.1 0.1 0.1
3 2 40 0.1 0.1 0.1 0.6
4 2 40 0.1 0.1 0.1 0.6
5 2 40 0.1 0.1 0.1 0.6
6 3 20 0.1 0.6 0.1 0.1
7 3 40 0.1 0.1 0.1 0.6
8 3 10 0.6 0.1 0.1 0.1
Edit
To set values to 1st row of group only
df = pd.DataFrame([['1', 10], ['1', 30], ['1', 10], ['2', 40], ['2', 40], ['2', 40], ['3', 20], ['3', 40], ['3', 10]], columns=('ID', 'Choice'))
df=df.set_index("ID")
## create unique index for each row if not already
df = df.reset_index()
choices = np.unique(df.Choice)
## get unique id of 1st row of each group
grouped = df.loc[df.reset_index().groupby("ID")["index"].first()]
## set value for each new variable
for choice in choices:
grouped[f"var_{choice}"] = np.where(grouped.Choice==choice, 0.6, (1 - 0.6) / 4)
pd.concat([df, grouped.iloc[:, -len(choices):]], axis=1)

Answer 2

We can use insert o create the rows based on the unique ID values ​​obtained through Series.unique.We can also create a mask to fill only the first row using np.where.
At the beginning sort_values ​​is used to sort the values ​​based on the ID. You can skip this step if your data frame is already well sorted (like the one shown in the example):
df=df.sort_values('ID')
n=df['Choice'].nunique()
mask=df['ID'].ne(df['ID'].shift())
for choice in df['Choice'].sort_values(ascending=False).unique():
df.insert(2,column=f'{choice}_Var',value=np.nan)
df.loc[mask,f'{choice}_Var']=np.where(df.loc[mask,'Choice'].eq(choice),0.6,0.4/n)
print(df)
ID Choice 10_Var 20_Var 30_Var 40_Var
0 1 10 0.6 0.1 0.1 0.1
1 1 30 NaN NaN NaN NaN
2 1 10 NaN NaN NaN NaN
3 2 40 0.1 0.1 0.1 0.6
4 2 40 NaN NaN NaN NaN
5 2 40 NaN NaN NaN NaN
6 3 20 0.1 0.6 0.1 0.1
7 3 40 NaN NaN NaN NaN
8 3 10 NaN NaN NaN NaN

Answer 3

A mix of numpy and pandas solution:
rows = np.unique(df.ID.values, return_index=1)[1]
df1 = df.loc[rows].assign(val=0.6)
df2 = (pd.crosstab([df1.index, df1.ID, df1.Choice], df1.Choice, df1.val, aggfunc='first')
.reindex(choice_list, axis=1)
.fillna((1-0.6)/len(choice_list)).reset_index(level=[1,2], drop=True))
pd.concat([df, df2], axis=1)
Out[217]:
ID Choice 10 20 30 40
0 1 10 0.6 0.1 0.1 0.1
1 1 30 NaN NaN NaN NaN
2 1 10 NaN NaN NaN NaN
3 2 40 0.1 0.1 0.1 0.6
4 2 40 NaN NaN NaN NaN
5 2 40 NaN NaN NaN NaN
6 3 20 0.1 0.6 0.1 0.1
7 3 40 NaN NaN NaN NaN
8 3 10 NaN NaN NaN NaN