Apply pandas function to column to create multiple new columns error-CodePudding

For this question i have found this example:

df = pd.DataFrame([[i] for i in range(5)], columns=['num'])
def powers(x):
    return x, x**2, x**3, x**4, x**5, x**6
df['p1'], df['p2'], df['p3'], df['p4'], df['p5'], df['p6'] = zip(*df['num'].apply(powers))
df

i changed the map() function to apply() function, it worked the same.

as you see we have passed a series for the apply() function: zip(*df['num'].apply(powers)).

This question's answer is good, But in my case of study i want to pass a dataFrame to the apply() function as: zip(*df[['num']].apply(powers)) by adding double*double brackets df[['num']] , but i got the following error: ValueError: not enough values to unpack (expected 6, got 3).

i didn't understand where the mistake is, can you help me please?

CodePudding user response：

In my opinion zip with apply is not recommended combine, for add multiple new columns is possible use:

df = pd.DataFrame([[i] for i in range(5)], columns=['num'])
def powers(x):
    
    return pd.Series([x, x**2, x**3, x**4, x**5, x**6])
df[['p1','p2','p3','p4','p5','p6']] = df['num'].apply(powers)

print (df)
   num  p1  p2  p3   p4    p5    p6
0    0   0   0   0    0     0     0
1    1   1   1   1    1     1     1
2    2   2   4   8   16    32    64
3    3   3   9  27   81   243   729
4    4   4  16  64  256  1024  4096

For pass one column DataFrame is possible use:

df = pd.DataFrame([[i] for i in range(5)], columns=['num'])
def powers(x):
    
    return [x, x**2, x**3, x**4, x**5, x**6]
df[['p1','p2','p3','p4','p5','p6']] = df[['num']].pipe(powers)

print (df)
   num  p1  p2  p3   p4    p5    p6
0    0   0   0   0    0     0     0
1    1   1   1   1    1     1     1
2    2   2   4   8   16    32    64
3    3   3   9  27   81   243   729
4    4   4  16  64  256  1024  4096

For multiple columns:

df = pd.DataFrame([[i] for i in range(5)], columns=['num'])
df['new'] = df['num'] * 2
def powers(x):
    
    return [x, x**2, x**3, x**4, x**5, x**6]


df = pd.concat(df[['num','new']].pipe(powers), axis=1, keys=['p1','p2','p3','p4','p5','p6'])
df.columns = df.columns.map(lambda x: f'{x[0]}_{x[1]}')
print (df)
   p1_num  p1_new  p2_num  p2_new  p3_num  p3_new  p4_num  p4_new  p5_num  \
0       0       0       0       0       0       0       0       0       0   
1       1       2       1       4       1       8       1      16       1   
2       2       4       4      16       8      64      16     256      32   
3       3       6       9      36      27     216      81    1296     243   
4       4       8      16      64      64     512     256    4096    1024   

   p5_new  p6_num  p6_new  
0       0       0       0  
1      32       1      64  
2    1024      64    4096  
3    7776     729   46656  
4   32768    4096  262144

CodePudding user response：

Alternative would be the following.

def powers(n):
    n=6
    cols=[x for x in np.arange(0,n 1)]
    for col in cols:
        df[f'num_{str(col)}'] = df['num'].apply(lambda x:x**col)
    return df
powers(df)

outcome1

    num  num_0  num_1  num_2  num_3  num_4  num_5  num_6
0    0      1      0      0      0      0      0      0
1    1      1      1      1      1      1      1      1
2    2      1      2      4      8     16     32     64
3    3      1      3      9     27     81    243    729
4    4      1      4     16     64    256   1024   4096

And if you needed p, this could do

def powers(n):
    n=6
    cols=[x for x in np.arange(1,n 1)]
    for col in cols:
        df[f'p{str(col)}'] = df['num'].apply(lambda x:x**col)
    return df
print(powers(df))

outcome 2

    num  p1  p2  p3   p4    p5    p6
0    0   0   0   0    0     0     0
1    1   1   1   1    1     1     1
2    2   2   4   8   16    32    64
3    3   3   9  27   81   243   729
4    4   4  16  64  256  1024  4096