Patrick,

In *Find: ^(\d+?)\.1 *what is the purpose of the question mark? I know what 
*\d+* does.

Howard

On Wednesday, 16 March 2022 at 1:41:06 pm UTC-4 Patrick Woolsey wrote:

> Though you can do this, you'll need to perform two search & replace 
> passes: one for each value pair.
>
> Find: ^(\d+?)\.1
>
> Replace: \1.33
>
> and 
>
> Find: ^(\d+?)\.2
>
> Replace: \1.67
>
>
> Regards,
>
> Patrick Woolsey
> ==
> Bare Bones Software, Inc. <https://www.barebones.com/>
>
>
> > On Mar 16, 2022, at 13:02, Howard <[email protected]> wrote:
> > 
> > I have a column of numbers; some have a decimal value. Some numbers end 
> with `.1` and some end with `.2`. If a number's decimal part contains `.1` 
> I want to change its ending to `.33`, but keep its whole number the same, 
> so 1.1 would become 1.33. Also, if a number's decimal part contains `.2` I 
> want to change its ending to `.67`, but keep its whole number the same, so 
> 1.2 would become 1.67.
> > 
> > As the column can have hundreds of rows, using Grep is there an easy way 
> to convert the decimal portion of a number that ends with either .1 or .2 
> while retaining the whole number part?
> > 
> > Howard
> > 
> > Sample Desired output
> > 0.2 0.67
> > 1.1 1.33
> > 1.1 1.33
> > 1.2 1.67
> > 1.1 1.33
> > 6 6
> > 3.2 3.67
> > 8 8
> > 6.2 6.67
> > 
>
>

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