How to compare two lists in python using if condition-CodePudding

Using python I would need to tackle the below scenario, where we have two lists name_1 is static and name_2 is dynamic. When we compare the two lists if there is any missing or partial value in name_2 then i need to print the name_1 value corresponding to name_2.

2 scenario: If there is extra value in name_2 then i need to print it directly.

name_1 = ['mahesh','karthik','python_code','Karun']
name_2 = ['mahesh','karthik','pyth','Karun','mari']

list_match = []

i = 0
while i < len(name_2):
    if not name_2[i]:
        print("Incorrect element founded in position ", name_1[i])
        break
    elif name_2[i] not in name_1:
        print(f"'{name_2[i]}' is extra column in position ", i)
        break
    else:
        list_match.append(i)
    i =1

Expected Output1:

Incorrect element founded in position  python_code

Expected Output2:

'mari' is extra column in position  4

CodePudding user response：

You can use itertools.zip_longest:

from itertools import zip_longest

name_1 = ['mahesh','karthik','python_code','Karun']
name_2 = ['mahesh','karthik','pyth','Karun','mari']

list_match = []
for idx, (i, j) in enumerate(zip_longest(name_1, name_2)):
    if i is None:
        print("'{}' is extra column in position {}".format(j, idx))
    elif i!=j:
        print("Incorrect element '{}' found in position {}".format(i, idx))
    else:
        list_match.append(idx)

This prints:

Incorrect element 'python_code' founded in position 2
'mari' is extra column in position 4

#list_match = [0, 1, 3]

CodePudding user response：

Quick answer:

mismatches = [(i, el) for i, el in enumerate(name_1) if not el in name_2]
for idx, el in mismatches:
    print(f"Incorrect element founded in position {idx}: {el}")

Then just reverse it for the other list.

mismatches = [(i, el) for i, el in enumerate(name_2) if not el in name_1]
for idx, el in mismatches:
    print(f"Incorrect element founded in position {idx}: {el}")