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How can I change numbers to a letter in a string?

I'm trying to change numbers in a string to an assigned letter (0 would be 'A', 1 => 'B' and so on...).
In this case I must use a function (def) to change the numbers to the assigned letter. I tried doing the program with if's in a for cycle with indices and it worked, but apperantly there is an easier and shorter solution. I tried doing it with isdigit() instead, but my function doesn't recognize the numbers in the given word and just prints out the same word

You can use a dictionary to map digits to letters, write a generator to get those letters and join it all into a resultant string. The dictionary's get method either gets the value or returns a default. Use this to return any non-digit characters.
>>> digit_map = dict(zip('123456789', 'ABCDEFGHI'))
>>> test = 'a1b2c3'
>>> "".join(digit_map.get(c,c) for c in test)
'aAbBcC'

Use ord to get the numeric value of a character, and chr to turn the numeric value back into a character. That lets you convert a number like 1 into its offset from another character.
Then use join to put it all together:
>>> ''.join(chr(int(c)+ord('A')) if c.isdecimal() else c for c in 'a1b2c3')
'aBbCcD'

One solution would be creating an array that would store each letter as a string as such:
alphabet = ["A", "B", "C"] // and so on
Then you could loop trough the string and find all numbers, then for each numbers get the corresponding letter by accessing it from the array as such py alphabet[i] and feed it back to the string and return that

Here is a simple function that you can use to convert numbers to letters in a given string:
def convert_numbers_to_letters(s):
# Create a dictionary that maps numbers to letters
number_to_letter = {str(i): chr(i + ord('A')) for i in range(10)}
# Iterate through each character in the string
result = ""
for c in s:
# If the character is a number, convert it to a letter
if c.isdigit():
result += number_to_letter[c]
# Otherwise, just add the character to the result
else:
result += c
return result

Why does my code remove 999 in my replacement code?

I have the code below to replace all punctuation with 999 and all alphabet characters with its number position. I have included the print statement that confirms punctuation is being replaced. However I seem to override with my remaining code to replace the other characters.
import string
def encode(text):
punct = '''!()-[]{};:'"\,<>./?##$%^&*_~'''
for x in text.lower():
if x in punct:
text = text.replace(x, ".999")
print(text)
nums = [str(ord(x) - 96)
for x in text.lower()
if x >= 'a' and x <= 'z'
]
return ".".join(nums)
print(encode(str(input("Enter Text: "))))
Input: 'Morning! \n'
Output: '13.15.18.14.9.14.7 \n'
Expected Output: 13.15.18.14.9.14.7.999

No, you have two independent logical "stories" here. One replaces punctuation with 999. The other filters out all the letters and builds an independent list of their alphabetic positions.
nums = [str(ord(x) - 96)
for x in text.lower()
if x >= 'a' and x <= 'z'
]
return ".".join(nums)
Note that this does nothing to alter text, and it takes nothing but letters from text. If you want to include the numbers, do so:
nums = [str(ord(x) - 96)
if x >= 'a' and x <= 'z'
else x
for x in text.lower()
]
return ".".join(nums)
Output of print(encode("[hello]")):
..9.9.9.8.5.12.12.15...9.9.9

nums = [str(ord(x) - 96)
for x in text.lower()
if x >= 'a' and x <= 'z'
]
This means: take every character from the lowercase version of the string, and only if it is between 'a' and 'z', convert the value and put the result in nums.
In the first step, you replace a bunch of punctuation with text that includes '.' and '9' characters. But neither '9' nor '.' is between 'a' and 'z', so of course neither is preserved in the second step.
Now that I understand what you are going for: you have fundamentally the wrong approach to splitting up the problem. You want to separate the two halves of the rule for "encoding" a given part of the input. But what you want to do is separate the whole rule for encoding a single element, from the process of applying a single-element rule to the whole input. After all - that is what list comprehensions do.
This is the concept of separation of concerns. The two business rules are part of the same concern - because implementing one rule doesn't help you implement the other. Being able to encode one input character, though, does help you encode the whole string, because there is a tool for that exact job.
We can have a complicated rule for single characters - no problem. Just put it in a separate function, so that we can give it a meaningful name and keep things simple to understand. Conceptually, our individual-character encoding is a numeric value, so we will consistently encode as a number, and then let the string-encoding process do the conversion.
def encode_char(c):
if c in '''!()-[]{};:'"\,<>./?##$%^&*_~''':
return 999
if 'a' <= c.lower() <= 'z':
return ord(c) - 96
# You should think about what to do in other cases!
# In particular, you don't want digit symbols 1 through 9 to be
# confused with letters A through I.
# So I leave the rest up to you, depending on your requirements.
Now we can apply the overall encoding process: we want a string that puts '.' in between the string representations of the values. That's straightforward:
def encode(text):
return '.'.join(str(encode_char(c)) for c in text)

Find all floats or ints in a given string

Given a string, "Hello4.2this.is random 24 text42", I want to return all ints or floats, [4.2, 24, 42]. All the other questions have solutions that return just 24. I want to return a float even if non-digit characters are next to the number. Since I am new to Python, I am trying to avoid regex or other complicated imports. I have no idea how to start. Please help. Here are some research attempts: Python: Extract numbers from a string, this didn't work since it doesn't recognize 4.2 and 42. There are other questions like the one mentioned, none of which sadly recognize 4.2 and 42.

A regex from perldoc perlretut:
import re
re_float = re.compile("""(?x)
^
[+-]?\ * # first, match an optional sign *and space*
( # then match integers or f.p. mantissas:
\d+ # start out with a ...
(
\.\d* # mantissa of the form a.b or a.
)? # ? takes care of integers of the form a
|\.\d+ # mantissa of the form .b
)
([eE][+-]?\d+)? # finally, optionally match an exponent
$""")
m = re_float.match("4.5")
print m.group(0)
# -> 4.5
To get all numbers from a string:
str = "4.5 foo 123 abc .123"
print re.findall(r"[+-]? *(?:\d+(?:\.\d*)?|\.\d+)(?:[eE][+-]?\d+)?", str)
# -> ['4.5', ' 123', ' .123']

Using regular expressions is likely to give you the most concise code for this problem. It is hard to beat the conciseness of
re.findall(r"[+-]? *(?:\d+(?:\.\d*)?|\.\d+)(?:[eE][+-]?\d+)?", str)
from pythad's answer.
However, you say "I am trying to avoid regex", so here's a solution that does not use regular expressions. It is obviously a bit longer than a solution using a regular expression (and probably much slower), but it is not complicated.
The code loops through the input character by character.
As it pulls each character from the string, it appends it to current (a string that holds the number currently being parsed) if appending it still maintains a valid number. When it encounters a character that cannot be appended to current, current is saved to a list of numbers, but only if current itself isn't one of '', '.', '-' or '-.'; these are strings that could potentially begin a number but are not themselves valid numbers.
When current is saved, a trailing 'e', 'e-' or 'e+' is removed. That will happen with a string such as '1.23eA'. While parsing that string, current will eventually become '1.23e', but then 'A' is encountered, which means the string does not contain a valid exponential part, so the 'e' is discarded.
After saving current, it is reset. Usually current is reset to '', but when the character that triggered current to be saved was '.' or '-', current is set to that character, because those characters could be the beginning of a new number.
Here's the function extract_numbers(s). The line before return numbers converts the list of strings to a list of integers and floating point values. If you want just the strings, remove that line.
def extract_numbers(s):
"""
Extract numbers from a string.
Examples
--------
>>> extract_numbers("Hello4.2this.is random 24 text42")
[4.2, 24, 42]
>>> extract_numbers("2.3+45-99")
[2.3, 45, -99]
>>> extract_numbers("Avogadro's number, 6.022e23, is greater than 1 million.")
[6.022e+23, 1]
"""
numbers = []
current = ''
for c in s.lower() + '!':
if (c.isdigit() or
(c == 'e' and ('e' not in current) and (current not in ['', '.', '-', '-.'])) or
(c == '.' and ('e' not in current) and ('.' not in current)) or
(c == '+' and current.endswith('e')) or
(c == '-' and ((current == '') or current.endswith('e')))):
current += c
else:
if current not in ['', '.', '-', '-.']:
if current.endswith('e'):
current = current[:-1]
elif current.endswith('e-') or current.endswith('e+'):
current = current[:-2]
numbers.append(current)
if c == '.' or c == '-':
current = c
else:
current = ''
# Convert from strings to actual python numbers.
numbers = [float(t) if ('.' in t or 'e' in t) else int(t) for t in numbers]
return numbers

If you want to get integers or floats from a string, follow the pythad's
ways...
If you want to get both integers and floats from a single string, do this:
string = "These are floats: 10.5, 2.8, 0.5; and these are integers: 2, 1000, 1975, 308 !! :D"
for line in string:
for actualValue in line.split():
value = []
if "." in actualValue:
value = re.findall('\d+\.\d+', actualValue)
else:
value = re.findall('\d+', actualValue)
numbers += value

Manipulating string to repeat based on the length of another string

I am working on a python project, where I am required to include an input, and another value (which will be manipulated).
For example,
If I enter the string 'StackOverflow', and a value to be manipulated of 'test', the program will make the manipulatable variable equal to the number of characters, by repeating and trimming the string. This means that 'StackOverflow' and 'test' would output 'testtesttestt'.
This is the code I have so far:
originalinput = input("Please enter an input: ")
manipulateinput = input("Please enter an input to be manipulated: ")
while len(manipulateinput) < len(originalinput):
And I was thinking of including a for loop to continue the rest, but am not sure how I would use this to effectively manipulate the string. Any help would be appreciated, Thanks.

An itertools.cycle approach:
from itertools import cycle
s1 = 'Test'
s2 = 'StackOverflow'
result = ''.join(a for a, b in zip(cycle(s1), s2))
Given you mention plaintext - a is your key and b will be the character in the plaintext - so you can use this to also handily manipuate the pairing...
I'm taking a guess you're going to end up with something like:
result = ''.join(chr(ord(a) ^ ord(b)) for a, b in zip(cycle(s1), s2))
# '\x07\x11\x12\x17?*\x05\x11&\x03\x1f\x1b#'
original = ''.join(chr(ord(a) ^ ord(b)) for a,b in zip(cycle(s1), result))
# StackOverflow

There are some good, Pythonic solutions here... but if your goal is to understand while loops rather than the itertools module, they won't help. In that case, perhaps you just need to consider how to grow a string with the + operator and trim it with a slice:
originalinput = input("Please enter an input: ")
manipulateinput = input("Please enter an input to be manipulated: ")
output = ''
while len(output) < len(originalinput):
output += manipulateinput
output = output[:len(originalinput)]
(Note that this sort of string manipulation is generally frowned upon in real Python code, and you should probably use one of the others (for example, Reut Sharabani's answer).

Try something like this:
def trim_to_fit(to_trim, to_fit):
# calculate how many times the string needs
# to be self - concatenated
times_to_concatenate = len(to_fit) // len(to_trim) + 1
# slice the string to fit the target
return (to_trim * times_to_concatenate)[:len(to_fit)]
It uses slicing, and the fact that a multiplication of a X and a string in python concatenates the string X times.
Output:
>>> trim_to_fit('test', 'stackoverflow')
'testtesttestt'
You can also create an endless circular generator over the string:
# improved by Rick Teachey
def circular_gen(txt):
while True:
for c in txt:
yield c
And to use it:
>>> gen = circular_gen('test')
>>> gen_it = [next(gen) for _ in range(len('stackoverflow'))]
>>> ''.join(gen_it)
'testtesttestt'

What you need is a way to get each character out of your manipulateinput string over and over again, and so that you don't run out of characters.
You can do this by multiplying the string so it is repeated as many times as you need:
mystring = 'string'
assert 2 * mystring == 'stringstring'
But how many times to repeat it? Well, you get the length of a string using len:
assert len(mystring) == 6
So to make sure your string is at least as long as the other string, you can do this:
import math.ceil # the ceiling function
timestorepeat = ceil(len(originalinput)/len(manipulateinput))
newmanipulateinput = timestorepeat * manipulateinput
Another way to do it would be using int division, or //:
timestorepeat = len(originalinput)//len(manipulateinput) + 1
newmanipulateinput = timestorepeat * manipulateinput
Now you can use a for loop without running out of characters:
result = '' # start your result with an empty string
for character in newmanipulateinput:
# test to see if you've reached the target length yet
if len(result) == len(originalinput):
break
# update your result with the next character
result += character
# note you can concatenate strings in python with a + operator
print(result)

String manipulation weirdness when incrementing trailing digit

I got this code:
myString = 'blabla123_01_version6688_01_01Long_stringWithNumbers'
versionSplit = re.findall(r'-?\d+|[a-zA-Z!##$%^&*()_+.,<>{}]+|\W+?', myString)
for i in reversed(versionSplit):
id = versionSplit.index(i)
if i.isdigit():
digit = '%0'+str(len(i))+'d'
i = int(i) + 1
i = digit % i
versionSplit[id]=str(i)
break
final = ''
myString = final.join(versionSplit)
print myString
Which suppose to increase ONLY the last digit from the string given. But if you run that code you will see that if there is the same digit in the string as the last one it will increase it one after the other if you keep running the script. Can anyone help me find out why?
Thank you in advance for any help

Is there a reason why you aren't doing something like this instead:
prefix, version = re.match(r"(.*[^\d]+)([\d]+)$", myString).groups()
newstring = prefix + str(int(version)+1).rjust(len(version), '0')
Notes:
This will actually "carry over" the version numbers properly: ("09" -> "10") and ("99" -> "100")
This regex assumes at least one non-numeric character before the final version substring at the end. If this is not matched, it will throw an AttributeError. You could restructure it to throw a more suitable or specific exception (e.g. if re.match(...) returns None; see comments below for more info).
Adjust accordingly.

The issue is the use of the list.index() function on line 5. This returns the index of the first occurrence of a value in a list, from left to right, but the code is iterating over the reversed list (right to left). There are lots of ways to straighten this out, but here's one that makes the fewest changes to your existing code: Iterate over indices in reverse (avoids reversing the list).
for idx in range(len(versionSplit)-1, -1, -1):
i = versionSplit[idx]
if chunk.isdigit():
digit = '%0'+str(len(i))+'d'
i = int(i) + 1
i = digit % i
versionSplit[idx]=str(i)
break

myString = 'blabla123_01_version6688_01_01veryLong_stringWithNumbers01'
versionSplit = re.findall(r'-?\d+|[^\-\d]+', myString)
for i in xrange(len(versionSplit) - 1, -1, -1):
s = versionSplit[i]
if s.isdigit():
n = int(s) + 1
versionSplit[i] = "%0*d" % (len(s), n)
break
myString = ''.join(versionSplit)
print myString
Notes:
It is silly to use the .index() method to try to find the string. Just use a decrementing index to try each part of versionSplit. This was where your problem was, as commented above by #David Robinson.
Don't use id as a variable name; you are covering up the built-in function id().
This code is using the * in a format template, which will accept an integer and set the width.
I simplified the pattern: either you are matching a digit (with optional leading minus sign) or else you are matching non-digits.
I tested this and it seems to work.

First, three notes:
id is a reserved python word;
For joining, a more pythonic idiom is ''.join(), using a literal empty string
reversed() returns an iterator, not a list. That's why I use list(reversed()), in order to do rev.index(i) later.
Corrected code:
import re
myString = 'blabla123_01_version6688_01_01veryLong_stringWithNumbers01'
print myString
versionSplit = re.findall(r'-?\d+|[a-zA-Z!##$%^&*()_+.,<>{}]+|\W+?', myString)
rev = list(reversed(versionSplit)) # create a reversed list to work with from now on
for i in rev:
idd = rev.index(i)
if i.isdigit():
digit = '%0'+str(len(i))+'d'
i = int(i) + 1
i = digit % i
rev[idd]=str(i)
break
myString = ''.join(reversed(rev)) # reverse again only just before joining
print myString