python - pandas unique values multiple columns

df = pd.DataFrame({'Col1': ['Bob', 'Joe', 'Bill', 'Mary', 'Joe'],
                   'Col2': ['Joe', 'Steve', 'Bob', 'Bob', 'Steve'],
                   'Col3': np.random.random(5)})

What is the best way to return the unique values of 'Col1' and 'Col2'?

The desired output is

'Bob', 'Joe', 'Bill', 'Mary', 'Steve'

All Answers
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    pd.unique returns the unique values from an input array, or DataFrame column or index.

    The input to this function needs to be one-dimensional, so multiple columns will need to be combined. The simplest way is to select the columns you want and then view the values in a flattened NumPy array. The whole operation looks like this:

    >>> pd.unique(df[['Col1', 'Col2']].values.ravel('K'))
    array(['Bob', 'Joe', 'Bill', 'Mary', 'Steve'], dtype=object)

    Note that ravel() is an array method than returns a view (if possible) of a multidimensional array. The argument 'K' tells the method to flatten the array in the order the elements are stored in memory (pandas typically stores underlying arrays in Fortran-contiguous order; columns before rows). This can be significantly faster than using the method's default 'C' order.

    An alternative way is to select the columns and pass them to np.unique:

    >>> np.unique(df[['Col1', 'Col2']].values)
    array(['Bill', 'Bob', 'Joe', 'Mary', 'Steve'], dtype=object)

    There is no need to use ravel() here as the method handles multidimensional arrays. Even so, this is likely to be slower than pd.unique as it uses a sort-based algorithm rather than a hashtable to identify unique values.

    The difference in speed is significant for larger DataFrames (especially if there are only a handful of unique values):

    >>> df1 = pd.concat([df]*100000, ignore_index=True) # DataFrame with 500000 rows
    >>> %timeit np.unique(df1[['Col1', 'Col2']].values)
    1 loop, best of 3: 1.12 s per loop
    >>> %timeit pd.unique(df1[['Col1', 'Col2']].values.ravel('K'))
    10 loops, best of 3: 38.9 ms per loop
    >>> %timeit pd.unique(df1[['Col1', 'Col2']].values.ravel()) # ravel using C order
    10 loops, best of 3: 49.9 ms per loop

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    I have setup a DataFrame with a few simple strings in it's columns:

    >>> df
       a  b
    0  a  g
    1  b  h
    2  d  a
    3  e  e

    You can concatenate the columns you are interested in and call unique function:

    >>> pandas.concat([df['a'], df['b']]).unique()
    array(['a', 'b', 'd', 'e', 'g', 'h'], dtype=object)

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    In [5]: set(df.Col1).union(set(df.Col2))
    Out[5]: {'Bill', 'Bob', 'Joe', 'Mary', 'Steve'}


    set(df.Col1) | set(df.Col2)

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    Non-pandas solution: using set().

    import pandas as pd
    import numpy as np
    df = pd.DataFrame({'Col1' : ['Bob', 'Joe', 'Bill', 'Mary', 'Joe'],
                  'Col2' : ['Joe', 'Steve', 'Bob', 'Bob', 'Steve'],
                   'Col3' : np.random.random(5)})
    print df
    print set(df.Col1.append(df.Col2).values)


       Col1   Col2      Col3
    0   Bob    Joe  0.201079
    1   Joe  Steve  0.703279
    2  Bill    Bob  0.722724
    3  Mary    Bob  0.093912
    4   Joe  Steve  0.766027
    set(['Steve', 'Bob', 'Bill', 'Joe', 'Mary'])

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    An updated solution using numpy v1.13+ requires specifying the axis in np.unique if using multiple columns, otherwise the array is implicitly flattened.

    import numpy as np
    np.unique(df[['col1', 'col2']], axis=0)

    This change was introduced Nov 2016:

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    for those of us that love all things pandas, apply, and of course lambda functions:

    df['Col3'] = df[['Col1', 'Col2']].apply(lambda x: ''.join(x), axis=1)

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    list(set(df[['Col1', 'Col2']].as_matrix().reshape((1,-1)).tolist()[0]))

    The output will be ['Mary', 'Joe', 'Steve', 'Bob', 'Bill']

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    here's another way

    import numpy as np

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