How to add numbers into string for a loop

Question

I have a long list of suburbs that I want to do something to

A LOT of them have RDx (for rural Delivery) where x is a number from 1 to 30

I want to just get rid of the RDx like below

for row in WorkingData['PatientSuburb']:
    if 'RD10' in str(row):
    WorkingData['PatientSuburb'].replace(regex=True,inplace=True,to_replace=r'RD10',value=r'')

I was thinking If I could run a loop and increment the number somehow that'd be great. this wouldn't work but it's along the lines of what I'd like to do:

for rd in range(1,31,1):
    if 'RD',rd in str(row):
    WorkingData['PatientSuburb'].replace(regex=True,inplace=True,to_replace=r'RD'rd ,value=r'')

If I do this I get output with a space in between:

for rd in range(1,31,1):
    print 'RD',rd

like so:

RD 1
RD 2
RD 3
RD 4
RD 5
RD 6
RD 7
RD 8
RD 9
RD 10
RD 11
RD 12

and also I would need to figure out how this piece would work...

to_replace=r'RD'rd

I have seen someone use a % sign in labelling a plot & then it brings in a value from outside the quotes - but I don't know if that's a part of the label function (I did try it and that didn't work at all) That would look like this

to_replace=r'RD%' % rd

Any help on this would be great thanks!

This really looks like an XY problem. Looping from RD1 to RD30 to check for each of these string could be achieved by a regex substitution instead, which would be both more efficient and elegant. — Samuel Dion-Girardeau
– Samuel Dion-Girardeau, Commented May 24, 2018 at 2:40
@SamuelDion-Girardeau That was 'interesting'. You are right - I was trying to simplify the question a little. But not all that much... I'm not that familiar with regex - can you give me some insight as to how you might approach this with regex? — kiltannen
– kiltannen, Commented May 24, 2018 at 3:42
Instead of iterating through the possible values and manually matching the string, you can define a regular expression that will match them all. I'll post an answer with a code example. — Samuel Dion-Girardeau
– Samuel Dion-Girardeau, Commented May 24, 2018 at 3:51

aldnav · Accepted Answer · 2018-05-24 03:37:32Z

3

If you want to use a for loop and substitute a substring by the index then I would say you are almost there.

to_replace = 'RD%d' % i

'%' marks the start of the specifier. In the example above, "d" follows "%" which means to place here a signed integer decimal. It's the same as "printf" library function in C. If "%" is not followed by any valid conversion character, it won't change anything regardless of what's on the right-hand side.

More details and examples here: https://docs.python.org/3.6/library/stdtypes.html#printf-style-bytes-formatting

answered May 24, 2018 at 3:37

aldnav

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Samuel Dion-Girardeau · Accepted Answer · 2018-05-24 03:59:54Z

1

Even though your question is about looping over several integers to generate strings, it seems your problem would actually be more suited for a regular expression.

This would allow you to capture multiple cases in one, without looping over possible values.

>>> import re
>>> RD_PATTERN = re.compile(r'RD[1-3]?[0-9]')
>>>
>>> def strip_rd(string):
...     return re.sub(RD_PATTERN, '', string)
...
>>>
>>> strip_rd('BlablahRD5')
'Blablah'
>>> strip_rd('BlablahRD5sometext')
'Blablahsometext'
>>> strip_rd('BlablahRD10sometext')
'Blablahsometext'
>>> strip_rd('BlablahRD25sometext')
'Blablahsometext'

The regex I provided is not rock-solid by any means (e.g. it matches RD0 even though you specified [1..30]), but you can create one that fits your specific use case. For instance, it might make sense to check that the pattern is at the end of the string, if that's expected to be the case.

Also, note that re.compile-ing the pattern is not necessary (you can give the pattern string directly), but since you mentioned you have several rows, it'll be more performant.

answered May 24, 2018 at 3:59

Samuel Dion-Girardeau

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1 Comment

kiltannen Over a year ago

Thank you. This was very effective. I did try both ways and the regex was pretty much an order of magnitude faster. BUT BOTH answers helped me to learn quite a bit. Thank you for posting the link to the RE doc - and for the pointer this would be more effective!

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