Python Pandas 使用来自同一 DF 的其他列的值创建新列

Question

It's been a long night searching for a solution, I appreciate your help.

Having the following df

proposal1_amount	proposal2_amount	proposal3_amount	accepted_proposal
1000	2000	3000	3
5000	5200	4000	2
3000	2400	1120	1

I need to build a new column with the amount coming from the accepted corresponding column, it would be like this:

proposal1_amount	proposal2_amount	proposal3_amount	accepted_proposal	accepted_amount
1000	2000	3000	3	3000
5000	5200	4000	2	5200
1450	2400	1120	1	1450

I've found some examples which work fine when the new column has a fixed value, but in this case the value comes from another column on the same df.

thanks, vv

Answer 1

Quickest solution I could think of:

df['accepted_amount'] = df.apply(lambda row: row.iloc[row['accepted_proposal']-1],axis=1)

Edit: Because I feel un-easy about the solution being contingent upon the ordering of the columns, here's a slightly wordier yet more dynamic solution:

df['accepted_amount']=df.apply(lambda row: row[['proposal1_amount','proposal2_amount','proposal3_amount']].iloc[row['accepted_proposal']-1],axis=1)

Answer 2

You can use numpy.choose to do this pretty easily.

print(df)
   proposal1_amount  proposal2_amount  proposal3_amount  accepted_proposal
0              1000              2000              3000                  3
1              5000              5200              4000                  2
2              3000              2400              1120                  1

# create 2d array of our choices (which corresponds to our amounts)
choices = df.filter(regex="proposal\d_amount").to_numpy()

# subtract 1 from "accepted_proposal" so they line up with indices in choices array
#  (we want these 0-indexed, not 1-indexed)
a = df["accepted_proposal"] - 1

# np.choose does all the heavy lifting, assign output to new column
df["accepted_amount"] = np.choose(a, choices)

print(df)
   proposal1_amount  proposal2_amount  proposal3_amount  accepted_proposal  accepted_amount
0              1000              2000              3000                  3             3000
1              5000              5200              4000                  2             5200
2              3000              2400              1120                  1             3000

np.choose will functionally iterate over each row of choices (eg iterate over each "proposalN_amount") and then take the amount that matches the index from accepted_proposal - 1 . See the docs for np.choose

Answer 3

proposal1_amount=[1000,5000,1450]
proposal2_amount=[2000,5200,2400]
proposal3_amount=[3000,4000,1120]
accepted_proposal=[3,2,1]
df=pd.DataFrame({'proposal1_amount': proposal1_amount,'proposal2_amount': proposal2_amount,'proposal3_amount':proposal3_amount,'accepted_proposal':accepted_proposal})
df['accepted_proposal']=df['accepted_proposal'].astype(int)
df=df.assign(accepted_amount=df.apply(lambda row:     row.iloc[row['accepted_proposal']-1], axis=1))    
print(df)

 output:

    proposal1_amount  proposal2_amount  proposal3_amount  accepted_proposal  
 0              1000              2000              3000                  3   
 1              5000              5200              4000                  2   
 2              1450              2400              1120                  1   

     accepted_amount  
 0             3000  
 1             5200  
 2             1450