I have some strings like -
1. "07870 622103"
2. "(0) 07543 876545"
3. "07321 786543 - not working"
I want to get the last 10 digits of these strings. like -
1. "07870622103"
2. "07543876545"
3. "07321786543"
So far I have tried-
a = re.findall(r"\d+${10}", mobilePhone)
Please help.
It'll be easier just to filter your string for digits and picking out the last 10:
''.join([c for c in mobilePhone if c.isdigit()][-10:])
Result:
>>> mobilePhone = "07870 622103"
>>> ''.join([c for c in mobilePhone if c.isdigit()][-10:])
'7870622103'
>>> mobilePhone = "(0) 07543 876545"
>>> ''.join([c for c in mobilePhone if c.isdigit()][-10:])
'7543876545'
>>> mobilePhone = "07321 786543 - not working"
>>> ''.join([c for c in mobilePhone if c.isdigit()][-10:])
'7321786543'
The regular expression approach (filtering everything but digits), is faster though:
$ python -m timeit -s "mobilenum='07321 786543 - not working'" "''.join([c for c in mobilenum if c.isdigit()][-10:])"
100000 loops, best of 3: 6.68 usec per loop
$ python -m timeit -s "import re; notnum=re.compile(r'\D'); mobilenum='07321 786543 - not working'" "notnum.sub(mobilenum, '')[-10:]"
1000000 loops, best of 3: 0.472 usec per loop
I suggest using a regex to throw away all non-digit. Like so:
newstring = re.compile(r'\D').sub('', yourstring)
The regex is very simple - \D means non-digit. And the code above uses sub to replace any non-digit char with an empty string. So you get what you want in newstring
Oh, and for taking the last ten chars use newstring[-10:]
That was a regex answer. The answer of Martijn Pieters may be more pythonic.
