Why are the characters not being replaced?

Question

I'm trying to decode the encrypted message with the translations / replacements that I have assigned in the code. However, when I run the code, it soleley prints the encrypted text and not the expected outcome. How can I solve this? I am sorry if this an easy solution / problem, I am a beginner.

EDIT: I solved the original issue, however, now its printing the wrong output. Instead of BCDD, it prints AAAAAAAAAAAAA. I would prefer not using imports.

def main():
    encrypted = "**^^^****^^^^" #expected outcome is BCDD
    if "*" or "^" in encrypted:
        encrypted = encrypted.replace("*", "A")
        encrypted = encrypted.replace("^", "A")
    if "**" or "^^" in encrypted:
        encrypted = encrypted.replace("**", "B")
        encrypted = encrypted.replace("^^", "B")
    if "***" or "^^^" in encrypted:
        encrypted = encrypted.replace("***", "C")
        encrypted = encrypted.replace("^^^", "C")
    if "****" or "^^^^" in encrypted:
        encrypted = encrypted.replace("****", "D")
        encrypted = encrypted.replace("^^^^", "D")
    if "*****" or "^^^^^" in encrypted:
        encrypted = encrypted.replace("*****", "E")
        encrypted = encrypted.replace("^^^^^", "E")     
    print(encrypted)
main()

You need to save the modified string returned by `replace`, e.g. `encrypted = encrypted.replace(...)`. — sj95126, Jan 11 '22 at 02:53
In Python, strings are immutable - they can't be changed. So `replace` returns a new string. — 001, Jan 11 '22 at 02:53
You also have to start with the longest string to replace... — thebjorn, Jan 11 '22 at 02:54
Like in pretty much every recent language Replace does not work in Python - same as C#, Java, JavaScript,... — Alexei Levenkov, Jan 11 '22 at 02:54
See also: [How to test multiple variables for equality against a single value?](https://stackoverflow.com/q/15112125) Applies to `in` as well. So `if "*" or "^" in encrypted:` is always `True` but `if "*" in encrypted or "^" in encrypted:` may not be. Then again, the `if`s aren't needed at all since `replace` won't throw an error if the string isn't there. — 001, Jan 11 '22 at 02:56
Please don't update the question modifying the original code. It invalidates existing comments/answers. — 001, Jan 11 '22 at 03:02
Independently of the incorrect use of booleans, note that it is useless to check if the characters are present before replacing. This will be **less** efficient as you have to read the string twice. Also given the many replacements, you should probably use regexes. — mozway, Jan 11 '22 at 03:21

Yevgeniy Kosmak · Accepted Answer · 2022-01-11T03:42:03.657

4

This way works:

encrypted = "**^^^****^^^^"
#expected outcome is BCDD
if "*****" in encrypted or "^^^^^" in encrypted:
    encrypted = encrypted.replace("*****", "E")
    encrypted = encrypted.replace("^^^^^", "E")
if "****" in encrypted or "^^^^" in encrypted:
    encrypted = encrypted.replace("****", "D")
    encrypted = encrypted.replace("^^^^", "D")
if "***" in encrypted or "^^^" in encrypted:
    encrypted = encrypted.replace("***", "C")
    encrypted = encrypted.replace("^^^", "C")
if "**" in encrypted or "^^" in encrypted:
    encrypted = encrypted.replace("**", "B")
    encrypted = encrypted.replace("^^", "B")
if "*" in encrypted or "^" in encrypted:
    encrypted = encrypted.replace("*", "A")
    encrypted = encrypted.replace("^", "A")
print(encrypted)

What I've changed:

encrypted.replace(...) -> encrypted = encrypted.replace(...). See replace.
Rearranged replaces in reversed order. Firstly you have to replace longer strings, if some of them are substring of next ones.
Replaced conditions with correct (but still useless) ones.

This version works correctly, the versions below are basically the same but in a different form.

The same solution, but without redundant ifs:

encrypted = "**^^^****^^^^"

encrypted = encrypted.replace("*****", "E")
encrypted = encrypted.replace("^^^^^", "E")
encrypted = encrypted.replace("****", "D")
encrypted = encrypted.replace("^^^^", "D")
encrypted = encrypted.replace("***", "C")
encrypted = encrypted.replace("^^^", "C")
encrypted = encrypted.replace("**", "B")
encrypted = encrypted.replace("^^", "B")
encrypted = encrypted.replace("*", "A")
encrypted = encrypted.replace("^", "A")

print(encrypted)

The same, but without lots of encrypted = encrypted.replace(...):

encrypted = "**^^^****^^^^"

replacements = {
    "*****": "E",
    "^^^^^": "E",
    "****": "D",
    "^^^^": "D",
    "***": "C",
    "^^^": "C",
    "**": "B",
    "^^": "B",
    "*": "A",
    "^": "A",
}

for r_from, r_to in replacements.items():
    encrypted = encrypted.replace(r_from, r_to)

print(encrypted)

The same, but with more convenient replacements:

encrypted = "**^^^****^^^^"

replacements = {
    **{'*' * n: chr(n + 64) for n in reversed(range(1, 6))},
    **{'^' * n: chr(n + 64) for n in reversed(range(1, 6))}
}

for r_from, r_to in replacements.items():
    encrypted = encrypted.replace(r_from, r_to)

print(encrypted)

But the best solution, as for me, uses regex :)

edited Jan 11 '22 at 03:42

answered Jan 11 '22 at 02:57

Yevgeniy Kosmak

3,561
2
10
26

1

Please see my [comment](https://stackoverflow.com/questions/70661089/why-are-the-characters-not-being-replaced#comment124914579_70661089). The `if`s are useless and can be removed. – 001 Jan 11 '22 at 03:09
@JohnnyMopp I can solve this task in dozens of different ways. But why does it matter if my answer wouldn't be understandable and helpful for OP? – Yevgeniy Kosmak Jan 11 '22 at 03:11
1

So let the OP go on thinking the `if` statements are correct and required? They are neither. – 001 Jan 11 '22 at 03:13
@JohnnyMopp okay, right you are, it could be better to explain that in more details. I'll add some. – Yevgeniy Kosmak Jan 11 '22 at 03:15
OK. You could argue that they are required in that (when they are correct) they will avoid unnecessary calls to `replace()` but the code will work correctly without them. – 001 Jan 11 '22 at 03:17
1

@JohnnyMopp updated, looks good enough? :) – Yevgeniy Kosmak Jan 11 '22 at 03:22
1

While you have some nice content, you fail to address the only important issue (the incorrect `a or b in c`). Please read the linked duplicate. – mozway Jan 11 '22 at 03:24
@mozway didn't even look at it. Thanks a lot! I'll fix it now. – Yevgeniy Kosmak Jan 11 '22 at 03:28
All of your answers helped me; as a beginner, I didn't know if the IF statements were useful, but I appreciate you walking me through it! – Jan 11 '22 at 03:29
@JonolOakwood they were useless, see how [replace](https://docs.python.org/3/library/stdtypes.html#str.replace) works. Welcome to Stack Overflow :) – Yevgeniy Kosmak Jan 11 '22 at 03:35
@Yevgeniv please make it clear that this is the only problem with the code, All the rest is "only" optimization – mozway Jan 11 '22 at 03:35
@mozway in fact, @JonolOakwood changed their initial post, in first version they also had an issue with `encrypted.replace(...)` -> `encrypted = encrypted.replace(...)`. Anyway, I'll add it. – Yevgeniy Kosmak Jan 11 '22 at 03:39
I understand most of it now; however, I don't understand how the "but with more convenient replacements" method works. Could you explain this final thing for me? I know I'm asking for a lot so if you don't want to dont worry! Thanks again! – Jan 11 '22 at 03:51
@JonolOakwood look at these links: [dictionary comprehension](https://stackoverflow.com/questions/14507591/python-dictionary-comprehension), [multiply strings](https://www.pythoncentral.io/use-python-multiply-strings/), [chr](https://docs.python.org/3/library/functions.html#chr), [dictionary unpacking](https://python-reference.readthedocs.io/en/latest/docs/operators/dict_unpack.html). Hope that would be enough :) – Yevgeniy Kosmak Jan 11 '22 at 03:57
That's perfect, I can't thank you enough! – Jan 11 '22 at 04:07

score 1 · Answer 2 · answered Jan 11 '22 at 02:55

1

encrypted.replace does not do an in-line replace.. it returns a replaced string..

strings are immutable in python.

so do this in every line:

encrypted = encrypted.replace(..blah..)

answered Jan 11 '22 at 02:55

Bhakta Raghavan

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Awfully formatted repost from duplicate... I guess that is Python way. – Alexei Levenkov Jan 11 '22 at 02:56
I've done this. Please check my edit, the problem now is that I only get AAAAAAAAA as the output – Jan 11 '22 at 02:58

j1-lee · Answer 3 · 2022-01-11T03:08:21.113

1

If the pattern is as given, then in this specific case you can do:

import re

encrypted = "**^^^****^^^^"
output = re.sub(r"\*+|\^+", lambda m: chr(len(m.group(0)) + 64), encrypted)
print(output) # BCDD

This basically finds a pattern (r"\*+|\^+"), reads the length of the substring (len(m.group(0))), and then substitutes the pattern with a character that corresponds to the length (chr(... + 64)).

edited Jan 11 '22 at 03:08

answered Jan 11 '22 at 03:00

j1-lee

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score 1 · Answer 4 · answered Jan 11 '22 at 03:09

Everyone so far seems to have missed a repeated error. Example:

if "*" or "^" in encrypted:

is parsed as:

if ("*") or ("^" in encrypted):

and so is always True (because bool("*") is True, and the in test is never even tried).

What you wanted to write instead:

if "*" in encrypted or "^" in encrypted:

score 0 · Answer 5 · answered Jan 11 '22 at 02:56

0

You must set encrypted equal to the replace function like so:

encrypted = encrypted.replace("*", "A")

answered Jan 11 '22 at 02:56

Daniel Zamloot

128
10

I've done this. Please check my edit, the problem now is that I only get AAAAAAAAA as the output – Jan 11 '22 at 02:58

Why are the characters not being replaced?

5 Answers5