I have two lists, let's say:

l1 = ['c', 'o', 'k', 'e']
l2 = ['a', 'b', 'c', 'd']

I would like to create a loop, that would check if each letter in l1 is in l2 (if that's the case I would get " ") or if it's there and on the same position (in that case I would get "Y").

I started with this code (unfortunately I failed). Could you please advise what's missing?

     for i in l1:
        for j in range(0,4):
            if l1[j] == l2[j]:
                v = "Y"
            elif i in l2:
                v = "."
            else:
                v = "N"
            text = "".join(v)
        

Which those lists in the example, I would assume to get:

text = .NNN

I understand that this might be an easy question, but I'm a beginner and it's driving me crazy :)

CodePudding user response：

Looking at your code, you can use zip() to iterate over the two lists simultaneously. Also use str.join after the loop:

l1 = ["c", "o", "k", "e"]
l2 = ["a", "b", "c", "d"]

out = []
for a, b in zip(l1, l2):
    if a == b:
        out.append("Y")
    elif a in l2:
        out.append(".")
    else:
        out.append("N")

print("".join(out))

Prints:

.NNN

CodePudding user response：

l1 = ['c', 'o', 'k', 'e']
l2 = ['a', 'b', 'c', 'd']

text = ""
for i in l1:
    if i in l2:
        v = "."
    else:
        v = "N"
    text  = v
print(text)

First we define text="" so that we can just append. Then we loop all the letters in l1. We then check if that letter is in l2. if it is we add '.' to text if its not we add N. And finally we print the text

CodePudding user response：

You can do something like:

ans = ""
for i, val in enumerate(l1):
    if l2[i] == val:
        ans  = "Y"
    elif val in l2:
        ans  = "."
    else
        ans  = "N"

CodePudding user response：

You don't have to loop trough the second array with the operations you are performing. You can simplify your code like this.

l1 = ['c', 'o', 'k', 'e']
l2 = ['a', 'b', 'c', 'd']

text = ""
for i in range(len(l1)):
    if l1[i] == l2[i]:
        text  = "Y"
    elif l1[i] in l2:
        text  = "."
    else:
        text  = "N"
        
print(text)

Looping twice is not needed and therefore not the most efficient solution to your problem.

CodePudding user response：

Use python sets.

Simple ./N check

l1 = ['c', 'o', 'k', 'e']
l2 = ['a', 'b', 'c', 'd']

S = set(l2)

out = ''.join('.' if x in S else 'N' for x in l1)

output: .NNN

more complex ./Y/N check:

l1 = ['c', 'o', 'k', 'e', 'z']
l2 = ['a', 'b', 'c', 'd', 'z']

S = set(l2)

out = ''.join(('Y' if l2[i]==x else '.') if x in S else 'N'
              for i, x in enumerate(l1))

output: .NNNY

CodePudding user response：

l1 = ['c', 'o', 'k', 'e']
l2 = ['a', 'b', 'c', 'd']

text = ""
for char in l1:
  if char in l2:
    index = l1.index(char)
    if char == l2[index]:
      v = "Y"
    else:
      v = "."
  else:
    v = "N"
  text  = v

print(text)