blocks|key|1702586|text|a+=+['b','b','b','a','a','c','c']
[a[i]+for+i+in+sorted(np.unique(a,+return_index=True)[1])]|type|code-block|depth|inlineStyleRanges|entityRanges|data|syntax|javascript|1702587|unstyled|entityMap^0|0^^$0|@$1|2|3|4|5|6|7|G|8|@]|9|@]|A|$B|C]]|$1|D|3|-4|5|E|7|H|8|@]|9|@]|A|$]]]|F|$]]

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

blocks|key|1702693|text|使用np.unique的return_index功能。它返回元素在输入中首次出现的索引。然后argsort这些索引。|type|unstyled|depth|inlineStyleRanges|offset|length|style|CODE|entityRanges|data|1702694|>>>+u,+ind+=+np.unique(['b','b','b','a','a','c','c'],+return_index=True)
>>>+u[np.argsort(ind)]
array(['b',+'a',+'c'],+
++++++dtype='%7CS1')|code-block|syntax|javascript|1702695|entityMap^0|2|9|C|C|1A|7|0|0^^$0|@$1|2|3|4|5|6|7|M|8|@$9|N|A|O|B|C]|$9|P|A|Q|B|C]|$9|R|A|S|B|C]]|D|@]|E|$]]|$1|F|3|G|5|H|7|T|8|@]|D|@]|E|$I|J]]|$1|K|3|-4|5|6|7|U|8|@]|D|@]|E|$]]]|L|$]]

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

<pre><code>&gt;&gt;&gt; u, ind = np.unique(['b','b','b','a','a','c','c'], return_index=True)
&gt;&gt;&gt; u[np.argsort(ind)]
array(['b', 'a', 'c'], 
 dtype='|S1')
</code></pre>

blocks|key|1702761|text|unique()很慢，O(Nlog(N))，但您可以通过以下代码来实现：|type|unstyled|depth|inlineStyleRanges|offset|length|style|CODE|entityRanges|data|1702762|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)])|code-block|syntax|javascript|1702763|输出：|1702764|['b'+'a'+'d'+'c']|1702765|对于大数组O(N)，Pandas.unique()的速度要快得多：|1702766|import+pandas+as+pd

a+=+np.random.randint(0,+1000,+10000)
%25timeit+np.unique(a)
%25timeit+pd.unique(a)

1000+loops,+best+of+3:+644+us+per+loop
10000+loops,+best+of+3:+144+us+per+loop|1702767|entityMap^0|0|8|0|0|0|0|A|F|0|0^^$0|@$1|2|3|4|5|6|7|U|8|@$9|V|A|W|B|C]]|D|@]|E|$]]|$1|F|3|G|5|H|7|X|8|@]|D|@]|E|$I|J]]|$1|K|3|L|5|6|7|Y|8|@]|D|@]|E|$]]|$1|M|3|N|5|H|7|Z|8|@]|D|@]|E|$I|J]]|$1|O|3|P|5|6|7|10|8|@$9|11|A|12|B|C]]|D|@]|E|$]]|$1|Q|3|R|5|H|7|13|8|@]|D|@]|E|$I|J]]|$1|S|3|-4|5|6|7|14|8|@]|D|@]|E|$]]]|T|$]]

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

<pre><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)])
</code></pre>

output:

<pre><code>['b' 'a' 'd' 'c']
</code></pre>

<code>Pandas.unique()</code> is much faster for big array O(N):

<pre><code>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
</code></pre>

blocks|key|1702819|text|如果您试图删除已排序迭代器的重复项，则可以使用itertools.groupby函数：|type|unstyled|depth|inlineStyleRanges|offset|length|style|CODE|entityRanges|data|1702820|>>>+from+itertools+import+groupby
>>>+a+=+['b','b','b','a','a','c','c']
>>>+[x[0]+for+x+in+groupby(a)]
['b',+'a',+'c']|code-block|syntax|javascript|1702821|这更像unix+'uniq‘命令，因为它假定列表已经排序。当你在未排序的列表上尝试它时，你会得到类似这样的结果：|1702822|>>>+b+=+['b','b','b','a','a','c','c','a','a']
>>>+[x[0]+for+x+in+groupby(b)]
['b',+'a',+'c',+'a']|1702823|entityMap^0|N|H|0|0|0|0^^$0|@$1|2|3|4|5|6|7|Q|8|@$9|R|A|S|B|C]]|D|@]|E|$]]|$1|F|3|G|5|H|7|T|8|@]|D|@]|E|$I|J]]|$1|K|3|L|5|6|7|U|8|@]|D|@]|E|$]]|$1|M|3|N|5|H|7|V|8|@]|D|@]|E|$I|J]]|$1|O|3|-4|5|6|7|W|8|@]|D|@]|E|$]]]|P|$]]

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

<pre><code>&gt;&gt;&gt; from itertools import groupby
&gt;&gt;&gt; a = ['b','b','b','a','a','c','c']
&gt;&gt;&gt; [x[0] for x in groupby(a)]
['b', 'a', 'c']
</code></pre>

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:

<pre><code>&gt;&gt;&gt; b = ['b','b','b','a','a','c','c','a','a']
&gt;&gt;&gt; [x[0] for x in groupby(b)]
['b', 'a', 'c', 'a']
</code></pre>

blocks|key|1702888|text|如果您想要删除重复的条目，如Unix工具uniq，这是一个解决方案：|type|unstyled|depth|inlineStyleRanges|offset|length|style|CODE|entityRanges|data|1702889|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]|code-block|syntax|javascript|1702890|entityMap^0|K|4|0|0^^$0|@$1|2|3|4|5|6|7|M|8|@$9|N|A|O|B|C]]|D|@]|E|$]]|$1|F|3|G|5|H|7|P|8|@]|D|@]|E|$I|J]]|$1|K|3|-4|5|6|7|Q|8|@]|D|@]|E|$]]]|L|$]]

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

<pre><code>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]
</code></pre>

blocks|key|1916426|text|使用OrderedDict+(比列表理解更快)|type|unstyled|depth|inlineStyleRanges|entityRanges|data|1916427|from+collections+import+OrderedDict++
a+=+['b','a','b','a','a','c','c']
list(OrderedDict.fromkeys(a))|code-block|syntax|javascript|1916428|entityMap^0|0|0^^$0|@$1|2|3|4|5|6|7|I|8|@]|9|@]|A|$]]|$1|B|3|C|5|D|7|J|8|@]|9|@]|A|$E|F]]|$1|G|3|-4|5|6|7|K|8|@]|9|@]|A|$]]]|H|$]]

Use an OrderedDict (faster than a list comprehension)

<pre><code>from collections import OrderedDict 
a = ['b','a','b','a','a','c','c']
list(OrderedDict.fromkeys(a))
</code></pre>

<pre><code>['b','b','b','a','a','c','c']
</code></pre>

numpy.unique gives

<pre><code>['a','b','c']
</code></pre>

How can I get the original order preserved

<pre><code>['b','a','c']
</code></pre>

<hr>

Great answers. Bonus question. Why do none of these methods work with this dataset? <a href="http://www.uploadmb.com/dw.php?id=1364341573" rel="noreferrer">http://www.uploadmb.com/dw.php?id=1364341573</a> Here's the question <a href="https://stackoverflow.com/questions/15649097/numpy-sort-wierd-behavior?lq=1">numpy sort wierd behavior</a>

numpy.unique with order preserved

Unix

['b','b','b','a','a','c','c']numpy.unique给出['a','b','c']我怎样才能保留原来的顺序？['b','a','c']很好的答案。奖金问题。为什么这些方法都不适用于此数据集？这里是问题

问保留顺序的numpy.unique
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