numpy.unique with order preserved

unique() is slow, O(Nlog(N)), but you can do this by following code:

import numpy as np
a = np.array(['b','a','b','b','d','a','a','c','c'])
_, idx = np.unique(a, return_index=True)
print(a[np.sort(idx)])

output:

['b' 'a' 'd' 'c']

Pandas.unique() is much faster for big array O(N):

import pandas as pd

a = np.random.randint(0, 1000, 10000)
%timeit np.unique(a)
%timeit pd.unique(a)

1000 loops, best of 3: 644 us per loop
10000 loops, best of 3: 144 us per loop

Use the return_index functionality of np.unique. That returns the indices at which the elements first occurred in the input. Then argsort those indices.

>>> u, ind = np.unique(['b','b','b','a','a','c','c'], return_index=True)
>>> u[np.argsort(ind)]
array(['b', 'a', 'c'], 
      dtype='|S1')