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Extracting float or int number and substring from a string

I've just learned regex in python3 and was trying to solve a problem.
The problem is something like this:
You have given a string where the first part is a float or integer number and the next part is a substring. You must split the number and the substring and return it as a list. The substring will only contain the alphabet from a-z and A-Z. The values of numbers can be negative.
For example:
Input: 2.5ax
Output:['2.5','ax']
Input: -5bcf
Output:['-5','bcf']
Input:-69.67Gh
Output:['-69.67','Gh']
and so on.
I did several attempts with regex to solve the problem.
1st attempt:
import re
i=input()
print(re.findall(r'^(-?\d+(\.\d+)?)|[a-zA-Z]+$',i))
For the input -2.55xy, the expected output was ['-2.55','xy']
But the output came:
[('-2.55', '.55'), ('', '')]
2nd attempt:
My second attempt was similar to my first attempt just a little different:
import re
i=input()
print(re.findall(r'^(-?(\d+\.\d+)|\d+)|[a-zA-Z]+$',i))
For the same input -2.55xy, the output came as:
[('-2.55', '2.55'), ('', '')]
3rd attempt:
My next attempt was like that:
import re
i=input()
print(re.findall(r'^-?[1-9.]+|[a-z|A-Z]+$',i))
which matched the expected output for -2.55xy and also with the sample examples. But when the input is 2..5 or something like that, it considers that also as a float.
4th attempt:
import re
i=input()
value=re.findall(r"[a-zA-Z]+",i)
print([i.replace(value[0],""),value[0]])
which also matches the expected output but has the same problem as 3rd one that goes with it. Also, it doesn't look like an effective way to do it.
Conclusion:
So I don't know why my 1st and 2nd attempt isn't working. The output comes with a list of tuples which is maybe because of the groups but I don't know the exact reason and don't know how to solve them. Maybe I didn't understand the way the pattern works. Also why the substring didn't show in the output?
In the end, I want to know what's the mistake in my code and how can I write better and more efficient code to solve the problem. Thank you and sorry for my bad English.

The alternation | matches either the left part or the right part.
If the chars a-zA-Z are after the digit, you don't need the alternation | and you can use 2 capture groups to get the matches in that order.
Then using re.findall will return a list of tuples for the capture group values.
(-?\d+(?:\.\d+)?)([a-zA-Z]+)
Explanation
( Capture group 1
-?\d+ Match an optional -
(?:\.\d+)? Optionally match . and 1+ digits using a non capture group (so it is not outputted separately by re.findall)
) Close group 1
( Capture group 2
[a-zA-Z]+ Match 1+ times a char a-z or A-Z
) Close group 2
regex demo
import re
strings = [
"2.5ax",
"-5bcf",
"-69.67Gh",
]
pattern = r"(-?\d+(?:\.\d+)?)([a-zA-Z]+)"
for s in strings:
print(re.findall(pattern, s))
Output
[('2.5', 'ax')]
[('-5', 'bcf')]
[('-69.67', 'Gh')]

lookahead and lookbehind in re.sub simplify things sometimes.
(?<=\d) look behind
(?=[a-zA-Z]) look ahead
that is split between the digit and the letter.
strings = [
"2.5ax",
"-5bcf",
"-69.67Gh",
]
for s in strings:
print(re.split(r'(?<=\d)(?=[a-zA-Z])', s))
['2.5', 'ax']
['-5', 'bcf']
['-69.67', 'Gh']

Need Regex that matches all patterns with format as `{word}{.,#}{word}` with strict matching

So I have been trying to construct a regex that can detect the pattern {word}{.,#}{word} and seperate it into [word,',' (or '.','#'), word].
But i am not able to create one that does strict matching for this pattern and ignores everything else.
I used the following regex
r"[\w]+|[.]"
this one is doing well , but it doesnt do strict matching, as in if (,, # or .) characters dont occur in text, it will still give me words, which i dont want.
I would like to have a regex which strictly matches the above pattern and gives me the splits(using re.findall) and if not returns the whole word as it is.
Please Note: word on either side of the {,.#} , both words are not strictly to be present but atleast one should be present
Some example text for reference:
no.16 would give me ['no','.','16']
#400 would give me ['#,'400']
word1.word2 would give me ['word1','.','word2']
Looking forward to some help and assistance from all regex gurus out there
EDIT:
I forgot to add this. #viktor's version works as needed with only one problem, It ignores ALL other words during re.findall
eg. ONE TWO THREE #400 with the viktor's regex gives me ['','#','400']
but what was expected was ['ONE','TWO','THREE','#',400]
this can be done with NLTK or spacy, but use of those is a limitation.

I suggest using
(\w+)?([.,#])((?(1)\w*|\w+))
See the regex demo.
Details
(\w+)? - An optional group #1: one or more word chars
([.,#]) - Group #2: ., , or #
((?(1)\w*|\w+)) - Group #3: if Group 1 matched, match zero or more word chars (the word is optional on the right side then), else, match one or more word chars (there must be a word on the right side of the punctuation chars since there is no word before them).
See the Python demo:
import re
pattern = re.compile(r'(\w+)?([.,#])((?(1)\w*|\w+))')
strings = ['no.16', '#400', 'word1.word2', 'word', '123']
for s in strings:
print(s, ' -> ', pattern.findall(s))
Output:
no.16 -> [('no', '.', '16')]
#400 -> [('', '#', '400')]
word1.word2 -> [('word1', '.', 'word2')]
word -> []
123 -> []
The answer to your edit is
if re.search(r'\w[.,#]|[.,#]\w', text):
print( re.findall(r'[.,#]|[^\s.,#]+', text) )
If there is a word char, then any of the three punctuation symbols, and then a word char again in the input string, you can find and extract all occurrences of the [.,#]|[^\s.,#]+ pattern, namely a ., , or #, or one or more occurrences of any one or more chars other than whitespace, ., , and #.

I hope this code will solve your problem if you want to split the string by any of the mentioned special characters:
a='no.16'
b='#400'
c='word1.word2'
lst=[a, b, c]
for elem in lst:
result= re.split('(\.|#|,)',elem)
while('' in result):
result.remove('')
print(result)

You could do something like this:
import re
str = "no.16"
pattern = re.compile(r"(\w+)([.|#])(\w+)")
result = list(filter(None, pattern.split(str)))
The list(filter(...)) part is needed to remove the empty strings that split returns (see Python - re.split: extra empty strings that the beginning and end list).
However, this will only work if your string only contains these two words separated by one of the delimiters specified by you. If there is additional content before or after the pattern, this will also be returned by split.

Regex - Match words in pattern, except within email address

I'm looking to find words in a string that match a specific pattern.
Problem is, if the words are part of an email address, they should be ignored.
To simplify, the pattern of the "proper words" \w+\.\w+ - one or more characters, an actual period, and another series of characters.
The sentence that causes problem, for example, is a.a b.b:c.c d.d#e.e.e.
The goal is to match only [a.a, b.b, c.c] . With most Regexes I build, e.e returns as well (because I use some word boundary match).
For example:
>>> re.findall(r"(?:^|\s|\W)(?<!#)(\w+\.\w+)(?!#)\b", "a.a b.b:c.c d.d#e.e.e")
['a.a', 'b.b', 'c.c', 'e.e']
How can I match only among words that do not contain "#"?

I would definitely clean it up first and simplify the regex.
first we have
words = re.split(r':|\s', "a.a b.b:c.c d.d#e.e.e")
then filter out the words that have an # in them.
words = [re.search(r'^((?!#).)*$', word) for word in words]

Properly parsing email addresses with a regex is extremely hard, but for your simplified case, with a simple definition of word ~ \w\.\w and the email ~ any sequence that contains #, you might find this regex to do what you need:
>>> re.findall(r"(?:^|[:\s]+)(\w+\.\w+)(?=[:\s]+|$)", "a.a b.b:c.c d.d#e.e.e")
['a.a', 'b.b', 'c.c']
The trick here is not to focus on what comes in the next or previous word, but on what the word currently captured has to look like.
Another trick is in properly defining word separators. Before the word we'll allow multiple whitespaces, : and string start, consuming those characters, but not capturing them. After the word we require almost the same (except string end, instead of start), but we do not consume those characters - we use a lookahead assertion.

You may match the email-like substrings with \S+#\S+\.\S+ and match and capture your pattern with (\w+\.\w+) in all other contexts. Use re.findall to only return captured values and filter out empty items (they will be in re.findall results when there is an email match):
import re
rx = r"\S+#\S+\.\S+|(\w+\.\w+)"
s = "a.a b.b:c.c d.d#e.e.e"
res = filter(None, re.findall(rx, s))
print(res)
# => ['a.a', 'b.b', 'c.c']
See the Python demo.
See the regex demo.

regex. Find multiple occurrence of pattern

I have the following string
my_string = "this data is F56 F23 and G87"
And I would like to use regex to return the following output
['F56 F23', 'G87']
So basically, I'm interested in returning all the parts of the string that start with either F or G and are followed by two numbers. In addition, if there are multiple consecutive occurrences I would like regex to group them together.
I approached the problem with python and with this code
import re
re.findall(r'\b(F\d{2}|G\d{2})\b', my_string)
I was able to get all the occurrences
['F56', 'F23', 'G87']
But I would like to have the first two groups together since they are consecutive occurrences. Any ideas of how I can achieve that?

You can use this regex:
\b[FG]\d{2}(?:\s+[FG]\d{2})*\b
Non-capturing group (?:\s+[FG]\d{2})* will find zero or more of the following space separated F/G substrings.
Code:
>>> my_string = "this data is F56 F23 and G87"
>>> re.findall(r'\b[FG]\d{2}(?:\s+[FG]\d{2})*\b', my_string)
['F56 F23', 'G87']

So basically, I'm interested in returning all the parts of the string that start with either F or G and are followed by two numbers. In addition, if there are multiple consecutive occurrences I would like regex to group them together.
You can do this with:
\b(?:[FG]\d{2})(?:\s+[FG]\d{2})*\b
in case it is separated by at least one spacing character. If that is not a requirement, you can do this with:
\b(?:[FG]\d{2})(?:\s*[FG]\d{2})*\b
Both the first and second regex generate:
>>> re.findall(r'\b(?:[FG]\d{2})(?:\s+[FG]\d{2})*\b',my_string)
['F56 F23', 'G87']
>>> re.findall(r'\b(?:[FG]\d{2})(?:\s*[FG]\d{2})*\b',my_string)
['F56 F23', 'G87']

print map(lambda x : x[0].strip(), re.findall(r'((\b(F\d{2}|G\d{2})\b\s*)+)', my_string))
change your regex to r'((\b(F\d{2}|G\d{2})\b\s*)+)' (brackets around, /s* to find all, that are connected by whitespaces, a + after the last bracket to find more than one occurance (greedy)
now you have a list of lists, of which you need every 0th Argument. You can use map and lambda for this. To kill last blanks I used strip()

Why does this regex to find repeated characters fail?

I'm trying to build a regex to match any occurrence of two or more repeated alphanumeric characters. The following regex fails:
import re
s = '__commit__'
m = re.search(r'([a-zA-Z0-9])\1\1', s)
But when I change it to this it works:
m = re.search(r'([a-zA-A0-9])\1+', s)
I'm pretty baffled as to why this is the way it is. Can anyone provide some insight?

Look at this line.
m = re.search(r'([a-zA-Z0-9])\1\1', s)
You are using a pattern and two backreferences (A reference of already matched pattern). So, it will match only when minimum of three consecutive characters appear. You can do:
m = re.search(r'([a-zA-Z0-9])\1', s)
Which will match when minimum of two consecutive character appears.
However, the following one is much better.
m = re.search(r'([a-zA-A0-9])\1+', s)
That's because, now you are trying to match at least one or more backreferences \1+, that is minimum two consecutive characters.

The \1 is a back-reference to any of the previously matching groups. So the original regex that does not work for you essentially means :
Match alphanumeric strings that contain 3 occurences of the previously matchd group. In this case the previously matched group ([a-zA-Z0-9]) contains a single character a-z or A-Z or 0-9. You then have two '\1 in your regex which accounts for two back-references to the previously matched character.
In the second regex the back-reference \1 has a + in front of it which means match atleast one occurence of the previously captured character - which means that the string confirming to this pattern has to be atleast 2 characters in length.
Hope this helps.