Python split before a certain character

Question

I have following string:

BUCKET1:/dir1/dir2/BUCKET1:/dir3/dir4/BUCKET2:/dir5/dir6

I am trying to split it in a way I would get back the following dict / other data structure:

BUCKET1 -> /dir1/dir2/, BUCKET1 -> /dir3/dir4/, BUCKET2 -> /dir5/dir6/

I can somehow split it if I only have one BUCKET, not multiple, like this:

res.split(res.split(':', 1)[0].replace('.', '').upper()) -> it's not perfect 

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Input: ADRIAN:/dir1/dir11/DANIEL:/dir2/ADI_BUCKET:/dir3/CULEA:/dir4/ADRIAN:/dir5/ADRIAN:/dir6/

Output: [(ADRIAN, /dir1/dir11), (DANIEL, /dir2/), (CULEA, /dir3/), (ADRIAN, /dir5/), (ADRIAN, /dir6/)

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As per Wiktor Stribiżew comments, the following regex does the job:

 r"(BUCKET1|BUCKET2):(.*?)(?=(?:BUCKET1|BUCKET2)|$)"

Answer 1

If you're experienced, I'd recommend learning Regex just as the others have suggested. However, if you're looking for an alternative, here's a way of doing such without Regex. It also produces the output you're looking for.

string = input("Enter:") #Put your own input here.

tempList = string.replace("BUCKET",':').split(":")
outputList = []
for i in range(1,len(tempList)-1,2):
    someTuple = ("BUCKET"+tempList[i],tempList[i+1])
    outputList.append(someTuple)

print(outputList) #Put your own output here.

This will produce:

[('BUCKET1', '/dir1/dir2/'), ('BUCKET1', '/dir3/dir4/'), ('BUCKET2', '/dir5/dir6')]

This code is hopefully easier to understand and manipulate if you're unfamiliar with Regex, although I'd still personally recommend Regex to solve this if you're familiar with how to use it.

Answer 2

Use re.findall() function:

s = "ADRIAN:/dir1/dir11/DANIEL:/dir2/ADI_BUCKET:/dir3/CULEA:/dir4/ADRIAN:/dir5/ADRIAN:/dir6/"
result = re.findall(r'(\w+):([^:]+\/)', s)

print(result)

The output:

[('ADRIAN', '/dir1/dir11/'), ('DANIEL', '/dir2/'), ('ADI_BUCKET', '/dir3/'), ('CULEA', '/dir4/'), ('ADRIAN', '/dir5/'), ('ADRIAN', '/dir6/')]