[FIXED] Choosing a column which has a certain value

Issue

I have an numpy array as the following

all = [[0 0 0],[0 0 1],[0 0 2], ... , [0 0 12]]

I am trying to only show the array which has third value 12. In this case [0 0 12]. When I execute my code I get the following output

[[0 0 0],[0 0 0],[0 0 12]] 

I do not know why I get those 0 arrays. My code is below.

for i in all:
  if i[2]==12:
    print(all[i]) ```

Solution

You can use boolean mask:

When you print your array, you get:

print(arr)
array([[ 0,  0,  0],
       [ 0,  0,  1],
       [ 0,  0,  2],
       [ 0,  0, 12]])

The third row is:

array([[ 0],
       [ 1],
       [ 2],
       [12]])

Now, you can create a boolean mask as:

arr[:,2]==12

which evaluates to

array([False, False, False,  True])

this means rows 1-3 don't have 12 as a third element but 4th does.

So you use this mask on arr:

arr[arr[:,2]==12]

Output:

array([[ 0,  0, 12]])

Answered By - enke