Get two return values ​​from Pandas.

I am trying to return two different values ​​from a method apply

, but I cannot figure out how to get the results I want.

With the as function:

def fun(row):
    s = [sum(row[i:i+2]) for i in range (len(row) -1)]
    ps = s.index(max(s))
    return max(s),ps

      

and df

how:

    6:00    6:15    6:30    
0   3       8       9       
1   60      62      116     

      

I am trying to return the maximum value of a row, but I also need to get the index of the first value that creates the maximum combination.

df["phour"] = t.apply(fun, axis=1)

      

I can get the output I want, but I don't know how I can get the index on the new column. So I get both answers intuple

    6:00    6:15    6:30    phour
0   3       8       9       (17, 1)
1   60      62      116     (178, 1)

      

How can I get the index value on my column?

+4


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6 answers


You can apply

pd.Series

df.drop('Double', 1).join(df.Double.apply(pd.Series, index=['D1', 'D2']))

   A  B  C  D1  D2
0  1  2  3   1   2
1  2  3  2   3   4
2  3  4  4   5   6
3  4  1  1   7   8

      


Equivalent to



df.drop('Double', 1).join(
    pd.DataFrame(np.array(df.Double.values.tolist()), columns=['D1', 'D2'])
)

      

Customization
using @GordonBeandf

df = pd.DataFrame({'A':[1,2,3,4], 'B':[2,3,4,1], 'C':[3,2,4,1], 'Double': [(1,2), (3,4), (5,6), (7,8)]})

      

+6


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If you are just trying to get max and argmax, I recommend using the pandas API:

DataFrame.idxmax

So:



df = pd.DataFrame({'A':[1,2,3,4], 'B':[2,3,4,1], 'C':[3,2,4,1]})
df

    A   B   C
0   1   2   3
1   2   3   2
2   3   4   4
3   4   1   1

df['Max'] = df.max(axis=1)
df['ArgMax'] = df.idxmax(axis=1)
df    

    A   B   C   Max ArgMax
0   1   2   3   3   C
1   2   3   2   3   B
2   3   4   4   4   B
3   4   1   1   4   A

      

Update :
And if you want the actual index value, you can use numpy.ndarray.argmax

:

df['ArgMaxNum'] = df[['A','B','C']].values.argmax(axis=1)


    A   B   C   Max ArgMax  ArgMaxNum
0   1   2   3   3   C   2
1   2   3   2   3   B   1
2   3   4   4   4   B   1
3   4   1   1   4   A   0

      

+2


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One way to split tuples into separate columns is with unpacking:

df = pd.DataFrame({'A':[1,2,3,4], 'B':[2,3,4,1], 'C':[3,2,4,1], 'Double': [(1,2), (3,4), (5,6), (7,8)]})
df


    A   B   C   Double
0   1   2   3   (1, 2)
1   2   3   2   (3, 4)
2   3   4   4   (5, 6)
3   4   1   1   (7, 8)

df['D1'] = [d[0] for d in df.Double]
df['D2'] = [d[1] for d in df.Double]
df


    A   B   C   Double  D1  D2
0   1   2   3   (1, 2)  1   2
1   2   3   2   (3, 4)  3   4
2   3   4   4   (5, 6)  5   6
3   4   1   1   (7, 8)  7   8

      

+2


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There should be a better way, but you can do:

df.merge(pd.DataFrame(((i,j) for 
                       i,j in df.apply(lambda x: fun(x)).values),
                      columns=['phour','index']),
         left_index=True,right_index=True)

      

+1


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You can get the index on a separate column, for example:

df[['phour','index']] = df.apply(lambda row: pd.Series(list(fun(row))), axis=1)

      

Or, if you change the fun a bit:

def fun(row):
    s = [sum(row[i:i+2]) for i in range (len(row) -1)]
    ps = s.index(max(s))
    return [max(s),ps]

      

Then the code becomes a little less confusing:

 df[['phour','index']] = df.apply(lambda row: pd.Series(fun(row)), axis=1)

      

0


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Just use result_type="expand"

df = pd.DataFrame(np.random.randint(0,10,(10,2)), columns=["random", "a"])
df[["sq_a","cube_a"]] = df.apply(lambda x: [x.a**2, x.a**3], axis=1, result_type="expand")

      

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