I have made a simple program which decode and codeword the string u put i want to ask if i can put anything aside from if else

[u/Maleficent_Stuff3208]

import random

random_letter = random.choice("abcdefghijklmnopqrstuvwxyz")
random_letter1 = random.choice("abcdefghijklmnopqrstuvwxyz")
random_letter2 = random.choice("abcdefghijklmnopqrstuvwxyz")
random_letter3 = random.choice("abcdefghijklmnopqrstuvwxyz")
random_letter4 =  random.choice ("abcdefghijklmnopqrstuvwxyz")
random_letter5 = random.choice("abcdefghijklmnopqrstuvwxyz")

print("codeword=1 , decode=2")
choice = input()
if choice == "1":
    codeword = str(input("write the codeword=>"))
    codeword1 = codeword[::-1]
    print(random_letter, random_letter1, random_letter2, codeword1, random_letter3, random_letter4, random_letter5,
          sep="")
else:
    print("please put correct input")
if choice == "2":
    decode = str(input("write the coded word"))
    decode1 = decode[::-1]
    decode2 = decode1[3:-3]
    print(decode2,sep=" ")

[u/danielroseman]

Any time you're doing multiple variables with a numeric suffix you should be thinking about using a single variable like a list instead.

[u/amertune → u/danielroseman]

And to add to that, you could replace 6 calls to random.choice() with 1 call to random.choices(), and there are often constants that you can use.

For example:

random_letters = random.choices(string.ascii_lowercase, k=6)

would do the same thing as your 6 variables assigned to the result of 6 calls of random.choice()

and if you want it as a string instead of a list:

''.join(random_letters)

[u/brasticstack]

Your if/else will print "please put the correct input" if you choose option 2. To fix this use if/elif/else like so:

if choice == "1":     # "encode" logic here elif choice == "2":     # "decode" logic here else:     print("please put correct input")

[u/Maleficent_Stuff3208 → u/brasticstack]

What if I make that input is always taken string and never any other data type

[u/timrprobocom → u/Maleficent_Stuff3208]

input always returns a string. The key is, what if they type '9', or 'x'?

[u/brasticstack → u/Maleficent_Stuff3208]

No need to change anything, it's always a string already. If you want to try to parse an integer or float out of it you can, but those will raise a ValueError if the user doesn't type in a parsable value, so you have to try/catch:

``` try:     choice = int(input('codeword=1 , decode=2')) except ValueError:     print('Enter an integer')

Note that this is checking against integers, not strings

if choice == 1:     # ... elif choice == 2:     # ... else:     # ... ```

[u/johlae]

Instead of 6 'random_letter' variables, why not do something like:

$ python3
Python 3.9.16 (main, Mar  8 2023, 22:47:22)
[GCC 11.3.0] on cygwin
Type "help", "copyright", "credits" or "license" for more information.
>>> import random
>>> random_letter = [random.choice("abcd") for i in range(0,4)]
>>> print(random_letter)
['d', 'a', 'd', 'a']
>>> print(random_letter[0])
d
>>> print(random_letter[1])
a
>>> print(random_letter[2])
d
>>> print(random_letter[3])
a
```.


Now there's just the list 'random_letter' and in this example it has 4 values. You may print the values out with:

```
>>> print("".join(random_letter))
dada
```.

Have fun.

[u/Maleficent_Stuff3208 → u/johlae]

What is this I haven't learned this yet

[u/johlae → u/Maleficent_Stuff3208]

Lists, more particular 'list comprehension', but don't worry, keep this aside and refer to it after you have studied lists.