import pandas as pd
import numpy as np

print(pd.Series(name_of_array).value_counts())

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)]

or:

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

或者你想结合计数和唯一值。

多维频率计数，即计数数组。

>>> print(color_array    )
  array([[255, 128, 128],
   [255, 128, 128],
   [255, 128, 128],
   ...,
   [255, 128, 128],
   [255, 128, 128],
   [255, 128, 128]], dtype=uint8)


>>> np.unique(color_array,return_counts=True,axis=0)
  (array([[ 60, 151, 161],
    [ 60, 155, 162],
    [ 60, 159, 163],
    [ 61, 143, 162],
    [ 61, 147, 162],
    [ 61, 162, 163],
    [ 62, 166, 164],
    [ 63, 137, 162],
    [ 63, 169, 164],
   array([     1,      2,      2,      1,      4,      1,      1,      2,
         3,      1,      1,      1,      2,      5,      2,      2,
       898,      1,      1,  

像这样的东西应该做到:

#create 100 random numbers
arr = numpy.random.random_integers(0,50,100)

#create a dictionary of the unique values
d = dict([(i,0) for i in numpy.unique(arr)])
for number in arr:
    d[j]+=1   #increment when that value is found

另外，之前的这篇关于有效计算独特元素的文章似乎与您的问题非常相似，除非我遗漏了什么。

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。

import pandas as pd
import numpy as np
x = np.array( [1,1,1,2,2,2,5,25,1,1] )
print(dict(pd.Series(x).value_counts()))

这会给你: {1,5, 2,3, 5:1, 25: 1}