Pandas merge two dataframes with different columns

Question

I'm surely missing something simple here. Trying to merge two dataframes in pandas that have mostly the same column names, but the right dataframe has some columns that the left doesn't have, and vice versa.

>df_may

  id  quantity  attr_1  attr_2
0  1        20       0       1
1  2        23       1       1
2  3        19       1       1
3  4        19       0       0

>df_jun

  id  quantity  attr_1  attr_3
0  5         8       1       0
1  6        13       0       1
2  7        20       1       1
3  8        25       1       1

I've tried joining with an outer join:

mayjundf = pd.DataFrame.merge(df_may, df_jun, how="outer")

But that yields:

Left data columns not unique: Index([....

I've also specified a single column to join on (on = "id", e.g.), but that duplicates all columns except "id" like attr_1_x, attr_1_y, which is not ideal. I've also passed the entire list of columns (there are many) to "on":

mayjundf = pd.DataFrame.merge(df_may, df_jun, how="outer", on=list(df_may.columns.values))

Which yields:

ValueError: Buffer has wrong number of dimensions (expected 1, got 2)

What am I missing? I'd like to get a df with all rows appended, and attr_1, attr_2, attr_3 populated where possible, NaN where they don't show up. This seems like a pretty typical workflow for data munging, but I'm stuck.

Thanks in advance.

Answer 1

I think in this case concat is what you want:

In [12]:

pd.concat([df,df1], axis=0, ignore_index=True)
Out[12]:
   attr_1  attr_2  attr_3  id  quantity
0       0       1     NaN   1        20
1       1       1     NaN   2        23
2       1       1     NaN   3        19
3       0       0     NaN   4        19
4       1     NaN       0   5         8
5       0     NaN       1   6        13
6       1     NaN       1   7        20
7       1     NaN       1   8        25

by passing axis=0 here you are stacking the df's on top of each other which I believe is what you want then producing NaN value where they are absent from their respective dfs.