Why isn't this code working without an else block to the if statement? - Python

Question

First post in here:)

So what I'm trying to achieve is, if a character is found inside a list in 'key' list then add the 'other' character in that list to the variable 's'. If a character is not found inside a list, then add the character itself to the variable 's'.

message="hello !"
s=""
found=False
key=[['m', 'c'], ['u', 'e'], ['b', 'g'], ['a', 'k'], ['s', 'v'], ['h', 'x'],['i', 'z'], ['r', 'y'], ['p', 'w'], ['l', 'n'], ['o', 'j'], ['t', 'f'], ['q', 'd']]
for i in message:
    for j in key:
        if i==j[0]:
            found=True
            s+=j[1]
            break
        elif i==j[1]:
            found=True
            s+=j[0]
            break
    if not found:
        s+=i
return s

input:

hello !

expected output:

xunnj !

Somehow the later part(add the character itself to the 's' variable) does not work unless I put an else block to the first if statement. like this,

if i==j[0]:
        found=True
        s+=j[1]
        break
elif i==j[1]:
        found=True
        s+=j[0]
        break
else:
        found=False

output without the else block:

xunnj

I want to know why,

if not found:
        s+=i

does not work without an else block to the first if statement . Thanks.

Answer 1

You need to give found = False every time you enter in the inner loop, otherwise it sets found = False only once from the outside of all the loops, which is why if at any time in the inner loop found become True then it never become False and remains True for rest of the time, and that is why it never enter into if not found: as found remains True always.

Your code should be like:

message="hello !"
s=""
# found=False 
key=[['m', 'c'], ['u', 'e'], ['b', 'g'], ['a', 'k'], ['s', 'v'], ['h', 'x'],['i', 'z'], ['r', 'y'], ['p', 'w'], ['l', 'n'], ['o', 'j'], ['t', 'f'], ['q', 'd']]
for i in message:
    found=False  # use found here, cause you need to give `found = False` every time you enter in the inner loop, as in the inner loop you are considering found every time, so you need to set it from the outer loop
    for j in key:
        if i==j[0]:
            found=True
            s+=j[1]
            break
        elif i==j[1]:
            found=True
            s+=j[0]
            break
    if not found:
        s+=i
print(s) # you may be want to print the `s` rather return

Output:

xunnj !

Answer 2

This is much simpler using a dict , and dict s are easy to create given a list of iterables like key is.

>>> d = dict(key)
>>> d.update(map(reversed, key))
>>> ''.join(d.get(x, x) for x in message)
'xunnj !'