numpy。

Question

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

numpy.unique给予

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

我如何保存原始订单

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

很高兴的答案。奖励问题。为什么这些方法都不适用于此数据集？

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以下是这个问题numpy sort wierd candy

unique()

Answer 1

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()

对于大阵列O（n）的速度要快得多：

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

使用

return_index

Answer 2

np.unique

这些索引。

argsort

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

如果您试图删除已经分类的重复，则可以使用

Answer 3

函数：itertools.groupby

Answer 4

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

>>> 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']

，这是一个解决方案：uniq

Answer 6

使用订单（比列表理解更快）

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]