How to check the list in python

Question

it will display wrong wrong correct # if wrong order display wrong correct order display correct

i have tried using nested loop to compare the element but then the output display too many time which is not something i wanted

     elif first in secList : #if list1 item is in list 2 and order of item are not same

         print("wrong")
     else:

         print("nothing")

The reason you're getting three lots of print-outs is that you're comparing _every_ item in `list1` against _every_ item in `list2`, whereas you want to loop through both together at the same rate. The built-in `zip()` is perfect for exactly this - see the answer below! — Tim, Aug 31 '19 at 16:45

score 3 · Accepted Answer · edited Aug 31 '19 at 16:45

Try this :

>>> list1= ['black','red','blue']
>>> list2=['red','black','blue']
>>> print(*["correct" if i==j else "wrong" for i,j in zip(list1, list2)], sep='\n')
wrong
wrong
correct

This is equivalent to the following :

for i,j in zip(list1, list2):
  if i==j:
    print("correct")
  else:
    print("wrong")

or

for i, j in zip(list1, list2):
    print("correct" if i == j else "wrong")

Using zip function of python, you can aggregate items from multiple iterables.

How to check the list in python

1 Answers1