Here is an array in python:
T = np.array([[1,1,2],[2,1,1],[3,3,3]])
print np.where(T==1)
I want to find the number of appearance of each element. I tried to use np.where and then len(np.where). However the output of np.where doesn't allow to use the len() function.
You can do as following to show the number of times that the element 1 is repeated in the array T:
>>> (T == 1).sum()
4
>>> (T == 2).sum()
2
If range of your integers in the array is small you can use np.bincount
In [25]: T = np.array([[1,1,2],[2,1,1],[3,3,3]])
In [26]: np.bincount(T.reshape(-1))
Out[26]: array([0, 4, 2, 3]) # 0 showed up 0 times
# 1 showed up 4 times
# 2 showed up 2 times ...
use numpy.unique, it returns both the unique values and the number of appearances in the array. The method has a flag called return_counts which is default value is false like this:
uniqueVals, count = numpy.unique(T, return_counts = true)
numpy.unique will work without reshape, but what parameter do you pass to get the number of appearances?
return_counts = true