Find unique values in a Pandas dataframe, irrespective of row or column location

Solution 1:

In [1]: df = DataFrame(np.random.randint(0,10,size=100).reshape(10,10))

In [2]: df
Out[2]: 
   0  1  2  3  4  5  6  7  8  9
0  2  2  3  2  6  1  9  9  3  3
1  1  2  5  8  5  2  5  0  6  3
2  0  7  0  7  5  5  9  1  0  3
3  5  3  2  3  7  6  8  3  8  4
4  8  0  2  2  3  9  7  1  2  7
5  3  2  8  5  6  4  3  7  0  8
6  4  2  6  5  3  3  4  5  3  2
7  7  6  0  6  6  7  1  7  5  1
8  7  4  3  1  0  6  9  7  7  3
9  5  3  4  5  2  0  8  6  4  7

In [13]: Series(df.values.ravel()).unique()
Out[13]: array([9, 1, 4, 6, 0, 7, 5, 8, 3, 2])

Numpy unique sorts, so its faster to do it this way (and then sort if you need to)

In [14]: df = DataFrame(np.random.randint(0,10,size=10000).reshape(100,100))

In [15]: %timeit Series(df.values.ravel()).unique()
10000 loops, best of 3: 137 ﾵs per loop

In [16]: %timeit np.unique(df.values.ravel())
1000 loops, best of 3: 270 ﾵs per loop

Solution 2:

Or you can use:

df.stack().unique()

Then you don't need to worry if you have NaN values, as they are excluded when doing the stacking.