計算numpy ndarray中的元素數

Question

如何計算ndarray中每個數據點的元素數量？

我想做的是在ndarray中至少出現N次的所有值上運行OneHotEncoder。

我還想用數組中未出現的另一個元素替換所有出現少於N次的值（我們稱它為new_value）。

例如，我有：

import numpy as np

a = np.array([[[2], [2,3], [3,34]],
              [[3], [4,5], [3,34]],
              [[3], [2,3], [3,4] ]]])

閾值N = 2我想要類似的東西：

b = [OneHotEncoder(a[:,[i]])[0] if count(a[:,[i]])>2 
else OneHotEncoder(new_value) for i in range(a.shape(1)]

因此，僅了解我想要的替換，而不考慮onehotencoder並使用new_value = 10，我的數組應如下所示：

a = np.array([[[10], [2,3], [3,34]],
                [[3], [10], [3,34]],
                [[3], [2,3], [10] ]]])

Answer 1

這樣的事情怎么樣？

首先計算數組中的unqiue元素數：

>>> a=np.random.randint(0,5,(3,3))
>>> a
array([[0, 1, 4],
       [0, 2, 4],
       [2, 4, 0]])
>>> ua,uind=np.unique(a,return_inverse=True)
>>> count=np.bincount(uind)
>>> ua
array([0, 1, 2, 4]) 
>>> count
array([3, 1, 2, 3]) 

從ua和count數組中，它顯示0出現3次，1顯示1次，依此類推。

import numpy as np

def mask_fewest(arr,thresh,replace):
    ua,uind=np.unique(arr,return_inverse=True)
    count=np.bincount(uind)
    #Here ua has all of the unique elements, count will have the number of times 
    #each appears.


    #@Jamie's suggestion to make the rep_mask faster.
    rep_mask = np.in1d(uind, np.where(count < thresh))
    #Find which elements do not appear at least `thresh` times and create a mask

    arr.flat[rep_mask]=replace 
    #Replace elements based on above mask.

    return arr


>>> a=np.random.randint(2,8,(4,4))
[[6 7 7 3]
 [7 5 4 3]
 [3 5 2 3]
 [3 3 7 7]]


>>> mask_fewest(a,5,50)
[[10  7  7  3]
 [ 7  5 10  3]
 [ 3  5 10  3]
 [ 3  3  7  7]]

對於上面的示例：讓我知道您打算使用2D陣列還是3D陣列。

>>> a
[[[2] [2, 3] [3, 34]]
 [[3] [4, 5] [3, 34]]
 [[3] [2, 3] [3, 4]]]


>>> mask_fewest(a,2,10)
[[10 [2, 3] [3, 34]]
 [[3] 10 [3, 34]]
 [[3] [2, 3] 10]]