CODEWITHSUNDEEP
pythonregexstring
pbreach
pbreach

Reputation: 17007

Line up decimal place when using re.sub to replace decimal numbers of varying size

It seems to me that the default behavior for replacement of matched strings with a larger string is to begin replacement at the starting position of the matched string ie.

import re

string = "Parameter Value:     0.12345"
new = re.sub(r'(\d|\.)+', '%0.5f' % 100.123, string) 
print string+'\n'+new

Gives the output:

Parameter Value:     0.12345
Parameter Value:     100.12300

Which appears to be the default behavior, but is there a way to get:

Parameter Value:     0.12345
Parameter Value:   100.12300

Instead so that the decimal place is lined up with where it was before replacement?

Upvotes: 1

Views: 590

Answers (2)

Padraic Cunningham
Padraic Cunningham

Reputation: 180461

You will have to add more logic deciding whether to ljust or rjust etc.. but this should get you started. Might pay to move '{:.5f}'.format(100.123) outside the regex and use the length of that as a deciding factor

import re

s = "Parameter Value:     0.12345"
new = re.sub(r'(\d|\.)+', '{:.5f}'.format(100.123), s)
s = s.split(" ")
new = new.split(" ")
a, b = s[-1], new[-1]
ln = len(max(a, b, key=len))
print("{}{:>{ln}}\n{}{}".format(" ".join(s[:-1]), a, " ".join(new[:-1]), b, ln=ln)))

Parameter Value:      0.12345
Parameter Value:    100.12300

Another way is getting the start and end index of the string that is going to be replaced, it is pretty easy adjust the string once you know exactly where the string is and how long the string to be replaced is in comparison to rep:

import re
f = 100.123
rep = '{:.5f}'.format(f)
s = "Parameter Value:     0.12345"
inds = next(re.finditer(r'(\d|\.)+', s))
start, end = inds.span()
print(start,end)
(21, 28)
ln = end - start

Upvotes: 2

Bhargav Rao
Bhargav Rao

Reputation: 52141

Taking a clue from Padraic's comment I tried this and it's working

>>> string = "Parameter Value:     0.12345"
>>> s = string.split(":")
>>> num = 100.123
>>> ":".join([s[0],("%0.5f"%(num)).rjust(len(s[1]))])
'Parameter Value:   100.12300'
>>> string
'Parameter Value:     0.12345'

Upvotes: 1

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