計算numpy ndarray中元素的數量

Question

如何計算ndarray中每個數據點的元素數量？

我想要做的是對我的ndarray中至少存在N次的所有值運行OneHotEncoder。

我也想用另一個它不出現在數組中的元素（我們稱之爲new_value）替換出現少於N次的所有值。

因此，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] ]]]) 

與閾值N = 2我想是這樣的：

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

所以才明白，我想換人，不考慮onehotencoder和使用new_value = 10我的數組應該看起來像：

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

Answer 1

這樣的事情呢？

第一計數unqiue元件的數量在一個陣列：

>>> 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陣列它表明0表示了3次，圖1示出了1次，等等。

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]] 

對於上面的例子：讓我知道你是否打算使用2D數組或3D數組。

>>> 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]]