Python: Deleting Specific value from the list

Question

Let say if

# Displaying old and reference in columns for better visual #

old=      [2,2,6,6,2,4, 6, 6, 8,2]
reference=[7,1,6,2,4,9,10,16,12,5]
ID=2

What is faster way to remove the variable ID from reference list and delete old[ reference[ID match] ] from the old list?

I happens to developed this solution while asking my question... but is there better way to do this?

def List_Value_Removal(old,reference,ID):
    counter=0
    new=[]
    for test in reference:
        if ID!=test:
           new.append(old[counter])
        checker+=1
    return new

Answer 1

Use the zip() function to pair up the lists, then a list comprehension to only keep values where the reference value does not match ID :

def List_Value_Removal(old, reference, ID):
    return [v for v, ref in zip(old, reference) if ref != ID]

 values, references = List_Value_Removal(values, references, some_id)

Demo:

>>> old = [2, 2, 6, 6, 2, 4, 6, 6, 8, 2]
>>> reference = [7, 1, 6, 2, 4, 9, 10, 16, 12, 5]
>>> ID = 2
>>> [v for v, ref in zip(old, reference) if ref != ID]
[2, 2, 6, 2, 4, 6, 6, 8, 2]

If you need to update reference as well, return two lists:

def List_Value_Removal(old, reference, ID):
    return zip(*((v, ref) for v, ref in zip(old, reference) if ref != ID))

Demo:

>>> new, newref = zip(*((v, ref) for v, ref in zip(old, reference) if ref != ID))
>>> new
(2, 2, 6, 2, 4, 6, 6, 8, 2)
>>> newref
(7, 1, 6, 4, 9, 10, 16, 12, 5)

Answer 2

You could also do:

old=      [2,2,6,6,2,4,6,6,8,2]
reference=[7,1,6,2,4,9,10,16,12,5]
ID=2

new = [old[i] for i in range(len(old)) if reference[i] != ID]

Answer 3

Just considering both the list are of same length and only one index to remove, btw we can make it more generic :

>>> old = [2,2,6,6,2,4, 6, 6, 8,2]
>>> reference=[7,1,6,2,4,9,10,16,12,5]
>>> id = 2
>>> del old[(reference.index(id))]
>>> old
[2, 2, 6, 2, 4, 6, 6, 8, 2]
>>>

Answer 4

def List_Value_Removal(old,ref, id):
    new = old[:]
    ind = ref.index(id)
    new.pop(ind)
    return new

Not sure if there is a particular reason that you are using two arrays? If you used a dict:

ref = {7:2,1:2,6:6,2:6,4:2,9:4,10:6,16:6,12:8,5:2}

You could just:

ref.pop(id)