CODEWITHSUNDEEP
pythonnumpy
siamii
siamii

Reputation: 24084

numpy.unique with order preserved

['b','b','b','a','a','c','c']

numpy.unique gives

['a','b','c']

How can I get the original order preserved

['b','a','c']

Great answers. Bonus question. Why do none of these methods work with this dataset? http://www.uploadmb.com/dw.php?id=1364341573 Here's the question numpy sort wierd behavior

Upvotes: 70

Views: 53758

Answers (7)

Vivek Ramanathan
Vivek Ramanathan

Reputation: 470

#List we need to remove duplicates from while preserving order

x = ['key1', 'key3', 'key3', 'key2'] 

thisdict = dict.fromkeys(x) #dictionary keys are unique and order is preserved

print(list(thisdict)) #convert back to list

output: ['key1', 'key3', 'key2']

Upvotes: 3

DanGoodrick
DanGoodrick

Reputation: 3198

Use an OrderedDict (faster than a list comprehension)

from collections import OrderedDict  
a = ['b','a','b','a','a','c','c']
list(OrderedDict.fromkeys(a))

Upvotes: 2

HYRY
HYRY

Reputation: 97261

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

Upvotes: 118

Albert
Albert

Reputation: 68110

If you want to delete repeated entries, like the Unix tool uniq, this is a solution:

def uniq(seq):
  """
  Like Unix tool uniq. Removes repeated entries.
  :param seq: numpy.array
  :return: seq
  """
  diffs = np.ones_like(seq)
  diffs[1:] = seq[1:] - seq[:-1]
  idx = diffs.nonzero()
  return seq[idx]

Upvotes: 2

Jan Spurny
Jan Spurny

Reputation: 5528

If you're trying to remove duplication of an already sorted iterable, you can use itertools.groupby function:

>>> from itertools import groupby
>>> a = ['b','b','b','a','a','c','c']
>>> [x[0] for x in groupby(a)]
['b', 'a', 'c']

This works more like unix 'uniq' command, because it assumes the list is already sorted. When you try it on unsorted list you will get something like this:

>>> b = ['b','b','b','a','a','c','c','a','a']
>>> [x[0] for x in groupby(b)]
['b', 'a', 'c', 'a']

Upvotes: 5

Fred Foo
Fred Foo

Reputation: 363507

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')

Upvotes: 27

YXD
YXD

Reputation: 32511

a = ['b','b','b','a','a','c','c']
[a[i] for i in sorted(np.unique(a, return_index=True)[1])]

Upvotes: 9

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