我如何有效地获得一个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)]

当前回答

import pandas as pd
import numpy as np

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

其他回答

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。

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

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

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。