我如何有效地获得一个NumPy数组中每个唯一值的频率计数?

>>> x = np.array([1,1,1,2,2,2,5,25,1,1])
>>> freq_count(x)
[(1, 5), (2, 3), (5, 1), (25, 1)]

当前回答

用这个:

>>> import numpy as np
>>> x = [1,1,1,2,2,2,5,25,1,1]
>>> np.array(np.unique(x, return_counts=True)).T
    array([[ 1,  5],
           [ 2,  3],
           [ 5,  1],
           [25,  1]])

最初的回答:

使用scipy.stats.itemfreq(警告:已弃用):

>>> from scipy.stats import itemfreq
>>> x = [1,1,1,2,2,2,5,25,1,1]
>>> itemfreq(x)
/usr/local/bin/python:1: DeprecationWarning: `itemfreq` is deprecated! `itemfreq` is deprecated and will be removed in a future version. Use instead `np.unique(..., return_counts=True)`
array([[  1.,   5.],
       [  2.,   3.],
       [  5.,   1.],
       [ 25.,   1.]])

其他回答

为了计算唯一的非整数——类似于Eelco Hoogendoorn的答案,但速度要快得多(在我的机器上是5倍),我使用了weave。内联组合numpy。只有一点c代码;

import numpy as np
from scipy import weave

def count_unique(datain):
  """
  Similar to numpy.unique function for returning unique members of
  data, but also returns their counts
  """
  data = np.sort(datain)
  uniq = np.unique(data)
  nums = np.zeros(uniq.shape, dtype='int')

  code="""
  int i,count,j;
  j=0;
  count=0;
  for(i=1; i<Ndata[0]; i++){
      count++;
      if(data(i) > data(i-1)){
          nums(j) = count;
          count = 0;
          j++;
      }
  }
  // Handle last value
  nums(j) = count+1;
  """
  weave.inline(code,
      ['data', 'nums'],
      extra_compile_args=['-O2'],
      type_converters=weave.converters.blitz)
  return uniq, nums

配置文件信息

> %timeit count_unique(data)
> 10000 loops, best of 3: 55.1 µs per loop

Eelco的纯numpy版本:

> %timeit unique_count(data)
> 1000 loops, best of 3: 284 µs per loop

Note

这里存在冗余(unique也执行排序),这意味着可以通过将唯一功能放入c-code循环中来进一步优化代码。

用这个:

>>> import numpy as np
>>> x = [1,1,1,2,2,2,5,25,1,1]
>>> np.array(np.unique(x, return_counts=True)).T
    array([[ 1,  5],
           [ 2,  3],
           [ 5,  1],
           [25,  1]])

最初的回答:

使用scipy.stats.itemfreq(警告:已弃用):

>>> from scipy.stats import itemfreq
>>> x = [1,1,1,2,2,2,5,25,1,1]
>>> itemfreq(x)
/usr/local/bin/python:1: DeprecationWarning: `itemfreq` is deprecated! `itemfreq` is deprecated and will be removed in a future version. Use instead `np.unique(..., return_counts=True)`
array([[  1.,   5.],
       [  2.,   3.],
       [  5.,   1.],
       [ 25.,   1.]])

老问题,但我想提供我自己的解决方案,这是最快的,使用普通列表而不是np。数组作为输入(或首先转移到列表),基于我的台架测试。

如果你也遇到这种情况,请检查一下。

def count(a):
    results = {}
    for x in a:
        if x not in results:
            results[x] = 1
        else:
            results[x] += 1
    return results

例如,

>>>timeit count([1,1,1,2,2,2,5,25,1,1]) would return:

100000个循环,最好的3:2.26µs每循环

>>>timeit count(np.array([1,1,1,2,2,2,5,25,1,1]))

100000个回路,最好的3:8.8µs每回路

>>>timeit count(np.array([1,1,1,2,2,2,5,25,1,1]).tolist())

100000个回路,最佳3:5.85µs每回路

而公认的答案会更慢,而scipy.stats.itemfreq解决方案更糟糕。


更深入的测试并没有证实所制定的期望。

from zmq import Stopwatch
aZmqSTOPWATCH = Stopwatch()

aDataSETasARRAY = ( 100 * abs( np.random.randn( 150000 ) ) ).astype( np.int )
aDataSETasLIST  = aDataSETasARRAY.tolist()

import numba
@numba.jit
def numba_bincount( anObject ):
    np.bincount(    anObject )
    return

aZmqSTOPWATCH.start();np.bincount(    aDataSETasARRAY );aZmqSTOPWATCH.stop()
14328L

aZmqSTOPWATCH.start();numba_bincount( aDataSETasARRAY );aZmqSTOPWATCH.stop()
592L

aZmqSTOPWATCH.start();count(          aDataSETasLIST  );aZmqSTOPWATCH.stop()
148609L

参考下面关于影响小型数据集大量重复测试结果的缓存和其他ram内副作用的评论。

numpy。Bincount可能是最好的选择。如果你的数组除了包含小的密集整数之外还包含任何东西,那么像这样包装它可能是有用的:

def count_unique(keys):
    uniq_keys = np.unique(keys)
    bins = uniq_keys.searchsorted(keys)
    return uniq_keys, np.bincount(bins)

例如:

>>> x = array([1,1,1,2,2,2,5,25,1,1])
>>> count_unique(x)
(array([ 1,  2,  5, 25]), array([5, 3, 1, 1]))

使用numpy。唯一的return_counts=True (NumPy 1.9+):

import numpy as np

x = np.array([1,1,1,2,2,2,5,25,1,1])
unique, counts = np.unique(x, return_counts=True)

>>> print(np.asarray((unique, counts)).T)
 [[ 1  5]
  [ 2  3]
  [ 5  1]
  [25  1]]

与scipy.stats.itemfreq相比:

In [4]: x = np.random.random_integers(0,100,1e6)

In [5]: %timeit unique, counts = np.unique(x, return_counts=True)
10 loops, best of 3: 31.5 ms per loop

In [6]: %timeit scipy.stats.itemfreq(x)
10 loops, best of 3: 170 ms per loop