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

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

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。

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()

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())