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)]
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。