How to delete sub-arrays from multidimensional numpy arrays using a condition?

Question

I am trying to remove sub-arrays from a multidimensional numpy array using a condition. In this exampe I want to remove all sub-arrays which include the value 999. Here is one of my failed attempts:

a = np.array([[[1,2,3], [1,2,3]],
              [[999,5,6], [4,5,6]],
              [[999,8,9], [7,999,9]]
              ])

for i in range(0,len(a)):
    if 999 in a[i]:
        np.delete(a, i, 0)

The result I want is:

array([[1,2,3], [1,2,3]])

This is just a smaler exaple, which should help me understand a lager issue, which loos like that:

# win_list_hyper.shape -> (1449168, 233)
# win_list_multi.shape -> (1449168, 12, 5, 5)

win_list_hyper = np.where(win_list_hyper <= 0, -3.40282e+38, win_list_hyper)
win_list_multi = np.where(win_list_multi <= 0, -3.40282e+38, win_list_multi)


# fail!:
for i in range(0,len(win_list_multi)):
    
    if -3.40282e+38 in win_list_multi[i] or -3.40282e+38 in win_list_hyper[i]:
        
        np.delete(win_list_multi, i, 0)
        np.delete(win_list_hyper, i, 0)

(btw. if you know how to make this more efficient, please let me know aswell!)

Jussi Nurminen · Accepted Answer · 2020-12-15 15:30:43Z

2

Your 1st attempt fails, since np.delete does not operate inplace (i.e. it does not modify the array, it returns a new one). Also, deleting elements from an array while iterating over it is usually not a good idea (unless you know what you are doing).

You can just use np.where as follows:

inds = np.where(a == 999)  # get indices where value equals 999
np.delete(a, inds[0], axis=0)   # delete along first dimension

Result:

array([[[1, 2, 3],
        [1, 2, 3]]])

edited Dec 15, 2020 at 15:30

answered Dec 15, 2020 at 15:21

Jussi Nurminen

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1 Comment

Rebecca Over a year ago

Thank you for your help. I had to realize that my example wasn't good. I coudn't translate your solution to my data that easily. Never the less you helped me a lot. Your solution gave me the Idea to mask my arrays. I hope my solution doesn't mess with my data (e.g. shuffles it).

Rebecca · Accepted Answer · 2020-12-15 22:29:06Z

Jussi Nurminen solution works fine for my example, but I had to realize that my example wasn't good. I coudn't translate the given solution to my data that easily. Never the less Jussi Nurminen solution helped me a lot, because it gave me the Idea to mask my arrays. I hope my solution doesn't mess with my data (e.g. shuffles it). For those who are interestet...

... This is my solution for my (bad) example:

a = np.array([[[1,2,3], [1,2,3]],[[999,5,6], [4,5,6]],[[999,8,9], [7,999,9]]])

a_mask = []

for i in range(0,len(a)):
    if 999 in a[i]:
        x = 0
    else: x = 1
    
    a_mask.append(x)
 
a_mask = np.asarray(a_mask)
    
inds = np.where(a_mask == 0)
        
b = np.delete(a, inds, axis=0)

... And this is how it looks like translatet to my data:


# win_list_multi.shape -> (1449168, 12, 5, 5)
# win_list_hyper.shape -> (1449168, 233


win_list_multi = np.where(win_list_multi <= 0, -1, win_list_multi)

win_list_hyper = np.where(win_list_hyper <= 0, -1, win_list_hyper)


win_list_multi_mask = []

for i in range(0,len(win_list_multi)):
    if -1 in win_list_multi[i]:
        x = 0
    else: x = 1
    
    win_list_multi_mask.append(x)

win_list_multi_mask = np.asarray(win_list_multi_mask)



win_list_hyper_mask = []

for i in range(0,len(win_list_hyper)):
    if -1 in win_list_hyper[i]:
        x = 0
    else: x = 1
    
    win_list_hyper_mask.append(x)

win_list_hyper_mask = np.asarray(win_list_hyper_mask)



inds = np.where((win_list_multi_mask == 0) | (win_list_hyper_mask == 0))


win_list_multi_nd = np.delete(win_list_multi, inds, axis=0) 
win_list_hyper_nd = np.delete(win_list_hyper, inds, axis=0) 

# win_list_multi_nd.shape -> (9679, 12, 5, 5)
# win_list_hyper_nd.shape -> (9679, 233)

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