Most of simple problems get complicated because simple functionality like order() in R that gives a statistical result in both and descending order is missing in various python libraries. But if we devise our thinking that all such statistical ordering and parameters in python are easily found in pandas, we can can result sooner than looking in 100 different places. Also, development of R and pandas go hand-in-hand because they were created for same purpose. To solve this problem I use following code that gets me by anywhere:

unique, counts = np.unique(x, return_counts=True)
d = {'unique':unique, 'counts':count}  # pass the list to a dictionary
df = pd.DataFrame(d) #dictionary object can be easily passed to make a dataframe
df.sort_values(by = 'count', ascending=False, inplace = True)
df = df.reset_index(drop=True) #optional only if you want to use it further

import pandas as pd
import numpy as np

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

尽管这个问题已经得到了回答，但我建议使用一种不同的方法，即numpy.histogram。这样的函数给定一个序列，它返回其元素分组在箱子中的频率。

但是要注意:它在这个例子中是有效的，因为数字是整数。如果它们是实数，那么这个解就不适用了。

>>> from numpy import histogram
>>> y = histogram (x, bins=x.max()-1)
>>> y
(array([5, 3, 0, 0, 1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0,
       1]),
 array([  1.,   2.,   3.,   4.,   5.,   6.,   7.,   8.,   9.,  10.,  11.,
        12.,  13.,  14.,  15.,  16.,  17.,  18.,  19.,  20.,  21.,  22.,
        23.,  24.,  25.]))

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]))

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。