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I'm getting a string from a form format 07123456789 (it's a phone number) and I would like to remove the first number/character only if it's a zero. And after that add 44 in front. So the output will be 447123456789.

Which is the best way to do it?

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6
  • You should really add +44 to the front. Otherwise you cannot distinguish betewen phone numbers which started to have no 0, but a 44 at the start. (I suppose adding the 44 should only happen if there was a 0, otherwise not.) Commented Jan 3, 2015 at 10:24
  • Im using an API to send text messages. The accepted format of number is 447123456789. Commented Jan 3, 2015 at 10:42
  • Ok. But what happens (or: what should happen) if the number didn't start with 0 at the first place? Is it supposed to contain the prefix already or is it an error case? Commented Jan 3, 2015 at 10:45
  • the form that Im requesting the number has validation rules and will always starts with zero. Actually here I converting the local number to an international accepted number for the API. I would like the user to type his number as local to be much easier for him. Commented Jan 3, 2015 at 10:47
  • Ok, then all is fine. I just stumbled over your "I would like to remove the first number/character only if its a zero" part and wondered what would happen when it is not a zero. BTW, with these restrictions you disallow entering numbers from abroad. That may or may not be wanted, just wanted to point out... Commented Jan 3, 2015 at 10:52

5 Answers 5

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13

Use str.startswith

In [11]: s = '07123456789'

In [12]: '44{}'.format(s[1:] if s.startswith('0') else s)
Out[12]: '447123456789'

Also, instead of formatting, you can join the strings together with + operator:

'44' + (s[1:] if s.startswith('0') else s)

If you're sure there's at most 1 zero at the beginning of the number, you can safely use str.lstrip or int conversion (see other answers).

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Comments

5

We can convert string into integer e.g.

>>> s = "07123456789"
>>> "44%s"%(int(s))
'447123456789'

We can use lstrip method of string which will remove all "0" from left of string. e.g.

>>> "44%s"%(s.lstrip("0"))
'447123456789'

If we want to consider only the first character from string then we can try following: (above two solution will not when more then one "0" at starting of string.)

>>> if s[0]=="0":
...   s = s[1:]
... 
>>> "44%s"%s
'447123456789'

Or go with solution from jamylak 

>>> s = '07123456789'
>>> "44%s%s"%(s[0].strip("0"), s[1:])
'447123456789'
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4 Comments

I like lstrip in this case. seems like the exact tool for the job
@jamylak not when the phone number has more than one 0 at the beginning... I removed the lstrip from my answer for this very reason.
I thought about that, then i thought the phone numbers probably don't have more than 1 0 but that's being naiive
yes, lstrip and int are not work when string has more than one "0" at the beginning, it will remove all "0" from the beginning.
4

Just another possible option:

>>> s = '07123456789'
>>> '44' + s[0].strip('0') + s[1:]
'447123456789'
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Comments

3

str.lstrip is best for this case:

>>> s = '07123456789'
>>> '44'+s.lstrip('0')
'447123456789'
>>> s = '7123456789'
>>> '44'+s.lstrip('0')
'447123456789'
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Comments

1

One way to go about this is to cast to int to remove the zero at the beginning and then cast back to a str:

s = '07123456789'
s = '44'+str(int(s))
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2 Comments

Plus 1. More general approach than using startswith, probably.
This will remove all the leading zeroes, while he asked for the first one.

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