I'd like to print unique values in each column of a grouped dataframe and the following code snippet doesn't work as expected:
df = pd.DataFrame({'a' : [1, 2, 1, 2], 'b' : [5, 5, 5, 5], 'c' : [11, 12, 13, 14]})
print(
df.groupby(['a']).apply(
lambda df: df.apply(
lambda col: col.unique(), axis=0))
)
I'd expect it to print
1 [5] [11, 13]
2 [5] [12, 14]
While there are other ways of doing so, I'd like to understand what's wrong with this approach. Any ideas?
This should do the trick:
print(df.groupby(['a', 'b'])['c'].unique())
a | b |
--+---+---------
1 | 5 | [11, 13]
2 | 5 | [12, 14]
As to what's wrong with your approach - when you groupby on df and then apply some function f, the input for f will be a DataFrame with all of df's columns, unless otherwise specified (as is in my code snippet with ['c']). So your first apply is passing a DataFrame with 3 columns, and so is your second apply. Then your function also_print iterates over each of those 3 columns and prints them out, so you get 3 prints for every group.
df.groupby('a').apply(lambda df: pd.Series([df[col].unique() for col in df.columns[1:]], index=df.columns[1:]))