reorder a python list to match the order of another python list

Question

I have two lists:

listA = ['market', 'fraud', 'crime', 'security', 'public', 
'state']

listB = ['security','fraud', 'state']

the elements present in listB needs to be in the same position of the elements of listA. If listA contains any element that listB does not contain, then I need to fill this empty position with the number 0. So far, I only added the zeros to listB (see below). How can I achieve this goal?

difference = len(listA) - len(listB)
c= 0 
while c < b:
  c+=1
  listB.append(0)

my final output should look like:

listC = [0, 'fraud', 0, 'security', 0, 'state']

listC should contain 0 replacing the element that is not in listB. Strings that are present in listB should be in the listC in the same order (index position) they are in listA.

Asked By: ForeverLearner

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Source

Answer 1

Just create the new list from listA, but substitute 0 if items are not found in listB:

listA = ['market', 'fraud', 'crime', 'security', 'public', 'state']

listB = ['security','fraud', 'state']

listC = [item if item in listB else 0 for item in listA]
print(listC)

Output as requested

Update:

As Mark Ransom points out, if listB is very large, it would be more efficient to search through a set of those elements:

listA = ['market', 'fraud', 'crime', 'security', 'public', 'state']

listB = ['security','fraud', 'state']
slistB = set(listB)

listC = [item if item in slistB else 0 for item in listA]
print(listC)

Same output

Answered By: quamrana

Answer 2

the other solution is correct here’s another simpler format:

listA = ['market', 'fraud', 'crime', 'security', 'public', 
'state']

listB = ['security','fraud', 'state']
listC=[]
for i in listA:
    if i in listB:
        listC.append(i)
    else:
        listC.append('0')
        
print(listC)

Answered By: Hannon qaoud

reorder a python list to match the order of another python list

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