Split a string based on a delimiter but shift across 1

Question

I trying to split a string (although numbers currently a string in df column) but am struggling to find an answer anywhere. I think using expressions might be the way forward but haven't quite got my head around them.

example 1) 12.540%
example 2) 4.555.6%

I would like to take everything to the left of the first '.' and only one number going to the right of the same first '.'
I need to apply it to all different number lengths and the above statement is the only constant.

example 1 ) 12.5 and 40%
example 2) 4.5 and 55.6%

Thank you

archer · Accepted Answer

The following function should do what you want:
def split_string(num):
  s=num.split('.', 1)
  s1=s[0]+'.'+s[1][0]
  s2=s[1][1:]
  return (s1, s2)

Anurag Reddy · Answer

If you want regular expressions, catch the groups using this.
^(d+..)(.*)$

Use this with either re.search if you want.
b = re.search(r'^(d+..)(.*)$', string)
b.group(1)
b.group(2)

Ex-
val = '12.445.6'
b = re.search(r'^(d+..)(.*)$', val)
b.group(1)
Out[24]: '12.4'
b.group(2)
Out[25]: '45.6'

Prune · Answer

This is a straightforward problem in string manipulation.  Any string tutorial will teach you the basic operations.

Find the location of the period.
Add 1.
Split the string at that point: grab one slice through that index; a second slice from there to the end.

For instance, one you find the location loc and adjust 1 or 2 spots to the right:
num, pct = str[:loc], str[loc:]

Split a string based on a delimiter but shift across 1

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