Remove items from two lists based on value at index in either

Question

I need to compare two lists and conditionally remove items from both based on the value of either item.

Input:

a = np.array( [1 , 2 , 3 , -1 , 5] )
b = np.array( [1 , -1 , 3 , 4 , 5] )

Required output:

a_filter = [1 , 3 , 5]
b_filter = [1 , 3 , 5]

I can easily get an array of list indexes where a "-1" value occurs using:

np.where( ( a == -1 ) | (b == -1 ) )[0]

But what is the most efficient way to delete these indexes from both list a and b? The only way I can think of is to reverse sort the identified index's and iter through both list removing them, however this seems quite inefficient.

List order is important as need to make confusion matrix, but don't want to compare -1 results.

Answer 1

You can use numpy.delete for this:

>>> a_filter = np.delete(a, np.where( ( a == -1 ) | (b == -1 ) )[0])
>>> a_filter
array([1, 3, 5])
>>> b_filter = np.delete(b, np.where( ( a == -1 ) | (b == -1 ) )[0])
>>> b_filter
array([1, 3, 5])

Answer 2

IIUC, you could use boolean indexing:

import numpy as np

a = np.array( [1 , 2 , 3 , -1 , 5] )
b = np.array( [1 , -1 , 3 , 4 , 5] )

mask = ~((a == -1) | (b == -1))

a_filtered = a[mask]
b_filtered = b[mask]

print(a_filtered)
print(b_filtered)

Output

[1 3 5]
[1 3 5]

Answer 3

Since you tag pandas

df=pd.DataFrame([a,b]).T
s=df[~df.eq(-1).any(1)].values
a_=s[:,0]
b_=s[:,1]
a_
Out[477]: array([1, 3, 5], dtype=int64)
b_
Out[478]: array([1, 3, 5], dtype=int64)