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

ForeverLearner 0
Viewed 22

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.

list python
2 Answers

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

quamrana0

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)
Hannon qaoud0
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