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。

from collections import Counter
x = array( [1,1,1,2,2,2,5,25,1,1] )
mode = counter.most_common(1)[0][0]

我对此也很感兴趣，所以我做了一点性能比较(使用perfplot，我的一个爱好项目)。结果:

y = np.bincount(a)
ii = np.nonzero(y)[0]
out = np.vstack((ii, y[ii])).T

是目前为止最快的。(请注意对数缩放。)

代码生成的情节:

import numpy as np
import pandas as pd
import perfplot
from scipy.stats import itemfreq


def bincount(a):
    y = np.bincount(a)
    ii = np.nonzero(y)[0]
    return np.vstack((ii, y[ii])).T


def unique(a):
    unique, counts = np.unique(a, return_counts=True)
    return np.asarray((unique, counts)).T


def unique_count(a):
    unique, inverse = np.unique(a, return_inverse=True)
    count = np.zeros(len(unique), dtype=int)
    np.add.at(count, inverse, 1)
    return np.vstack((unique, count)).T


def pandas_value_counts(a):
    out = pd.value_counts(pd.Series(a))
    out.sort_index(inplace=True)
    out = np.stack([out.keys().values, out.values]).T
    return out


b = perfplot.bench(
    setup=lambda n: np.random.randint(0, 1000, n),
    kernels=[bincount, unique, itemfreq, unique_count, pandas_value_counts],
    n_range=[2 ** k for k in range(26)],
    xlabel="len(a)",
)
b.save("out.png")
b.show()

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}

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]))

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。