Python - Delete duplicates in a dataframe based on two columns combinations?

By using np.sort with duplicated

df[pd.DataFrame(np.sort(df[['Name1','Name2']].values,1)).duplicated()]
Out[614]: 
  Name1 Name2  Value
1   Ale  Juan      1

Performance

df=pd.concat([df]*100000)

%timeit df[pd.DataFrame(np.sort(df[['Name1','Name2']].values,1)).duplicated()]
10 loops, best of 3: 69.3 ms per loop
%timeit df[~df[['Name1', 'Name2']].apply(frozenset, axis=1).duplicated()]
1 loop, best of 3: 3.72 s per loop

You can convert to frozenset and use pd.DataFrame.duplicated.

res = df[~df[['Name1', 'Name2']].apply(frozenset, axis=1).duplicated()]

print(res)

  Name1 Name2  Value
0  Juan   Ale      1

frozenset is necessary instead of set since duplicated uses hashing to check for duplicates.

Scales better with columns than rows. For a large number of rows, use @Wen's sort-based algorithm.

Know Im kinda late for this question but giving my contribution anyway :)

You can also use get_dummies and add for a good way of creating hashable rows

df[~(pd.get_dummies(df.a).add(pd.get_dummies(df.b), fill_value=0)).duplicated()]

Times are not as good as @Wen's answer, but it isstill way faster than apply+frozen_set

df=pd.concat([df]*1000000)
%timeit df[~(pd.get_dummies(df.a).add(pd.get_dummies(df.b), fill_value=0)).duplicated()]
1.8 s ± 85 ms per loop (mean ± std. dev. of 7 runs, 1 loop each)

%timeit df[pd.DataFrame(np.sort(df[['a','b']].values,1)).duplicated()]
1.26 s ± 19 ms per loop (mean ± std. dev. of 7 runs, 1 loop each)

%timeit df[~df[['a', 'b']].apply(frozenset, axis=1).duplicated()]
1min 9s ± 684 ms per loop (mean ± std. dev. of 7 runs, 1 loop each)