Alternative way to append a dataframe to itself N times and populate new column-CodePudding

Is there an alternative way to append a dataframe to itself N times where N is based on a list length, and the list contents are added as a new column to the dataframe?

For example, with this dataframe and list:

df = pd.DataFrame(
    {
        "x": [-1.089, 0, 0.3, 0.5, 0.6, 0.8],
        "y": [0, 0.3, 0.5, 0.6, 0.8, 10.089],
    }
)

z = [11, 12, 13, 14, 15, 16]

display(df)

This example does what I want, but I think it might be inefficient or nonpythonic.

df2 = pd.DataFrame()

for i in z:
    temp = df.copy()
    temp['z'] = i
    df2 = df2.append(temp)

display(df2.head(10))

CodePudding user response：

here is one way:

out = df.merge(pd.Series(z,name='z'), how='cross')

output:

>> out.head(10)

       x    y   z
0 -1.089  0.0  11
1 -1.089  0.0  12
2 -1.089  0.0  13
3 -1.089  0.0  14
4 -1.089  0.0  15
5 -1.089  0.0  16
6  0.000  0.3  11
7  0.000  0.3  12
8  0.000  0.3  13
9  0.000  0.3  14

in pandas before < 1.2 :

df_z = pd.Dataframe(z, columns='z')
df_z['key'] = 0
df['key'] = 0 
out = df.merge(df_z,on='key').drop("key", 1)

CodePudding user response：

You can try assgin the list to a new column then explode that column and at last sort the value

df = (df.assign(z=[z for _ in range(len(df))])
      .explode('z')
      .sort_values('z', kind='stable', ignore_index=True))

print(df.head(10))

       x       y   z
0 -1.089   0.000  11
1  0.000   0.300  11
2  0.300   0.500  11
3  0.500   0.600  11
4  0.600   0.800  11
5  0.800  10.089  11
6 -1.089   0.000  12
7  0.000   0.300  12
8  0.300   0.500  12
9  0.500   0.600  12