NumPy数组中唯一值的频率计数

Question

如何有效地获取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)]

Answer 1

使用这个：

>>> 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]])

原答案：

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.]])

Answer 2

看看 np.bincount：

http://docs.scipy.org/doc/numpy/reference/generated/numpy.bincount.html

import numpy as np
x = np.array([1,1,1,2,2,2,5,25,1,1])
y = np.bincount(x)
ii = np.nonzero(y)[0]

然后：

zip(ii,y[ii]) 
# [(1, 5), (2, 3), (5, 1), (25, 1)]

或者:

np.vstack((ii,y[ii])).T
# array([[ 1,  5],
         [ 2,  3],
         [ 5,  1],
         [25,  1]])

或者您想要组合计数和唯一值。

Answer 3

使用numpy。唯一，return_counts=True（对于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