python - numpy.unique 保留顺序

Question

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

numpy.unique 给出

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

我怎样才能保留原始订单

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

---

很好的答案。奖金问题。为什么这些方法都不适用于这个数据集？http://www.uploadmb.com/dw.php?id=1364341573这是问题numpy sort wierd behavior

Answer 1

unique()很慢，O(Nlog(N))，但你可以通过以下代码来做到这一点：

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)])

输出：

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

Pandas.unique()对于大数组 O(N) 来说要快得多：

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

Answer 2

使用 的return_index功能np.unique。这将返回元素首次出现在输入中的索引。然后argsort是那些指数。

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

Answer 3

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

Answer 4

如果您尝试删除已排序的可迭代对象的重复项，则可以使用itertools.groupby函数：

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

这更像是 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，你会得到这样的东西：

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

Answer 5

#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']

Answer 6

如果你想删除重复的条目，比如 Unix tool uniq，这是一个解决方案：

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]

Answer 7

使用 OrderedDict（比列表理解更快）

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