Good way to find / replace using regex

Question

I have an value that is not being read by our OCR program correctly. It's predicable so I would like to use a find/replace in regex (because this is how we are already extracting the data).

We get the named group like this: (?<Foo>.*?)

I would like to replace 'N1123456' with 'NY123456'. We know that we expect NY when we are getting N1.

What can I try to get this done in the same regular expression?

Edit: (?<Foo>.*?)

Answer 1

Make groups of non digits and digits and add Y after non digit group.

(\D+)(\d+)

Here is demo

Enclose it inside \\b or ^ and $ for better precision.

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Sample code:

PHP:

$re = ""(\\D+)(\\d+)"";
$str = "N1123456";
$subst = '$1Y$2';

$result = preg_replace($re, $subst, $str, 1);

Python:

import re
p = re.compile(ur'(\D+)(\d+)')
test_str = u"N1123456"
subst = u"$1Y$2"

result = re.sub(p, subst, test_str)

Java:

System.out.println("N1123456".replaceAll("(\\D+)(\\d+)","$1Y$2"));

Answer 2

If you expect N1 to be always followed by 6 digits, then you can do this:

Replace this: \\bN1(\\d{6})\\b with this: NY$1 .

This will replace any N1 followed by 6 digits with NY .

Answer 3

This is what I would do:

 Dim str = Regex.Replace("N1123456", @"\bN1(\d+)", "NY$1");

The Expression to find the text is N1 followed by numbers like : \\bN1(\\d+) .

The numbers belongs to the group(1) I would like to preserve and attach to NY during replacing: NY$1