Re: A Question Derived from Converting decimal value

[ThePorgie]

Thank you sir! Don't know how I missed that in the manual and I promise I 
did look there first.

On Wednesday, March 16, 2022 at 4:13:30 PM UTC-4 Patrick Woolsey wrote:

> On 3/16/22 at 2:06 PM, thep...@gmail.com (ThePorgie) wrote:
>
> >I was tinkering with with Patrick's question
>
> Not actually my question ;-) though close enough for the matters 
> at hand.
>
>
> >So I was looking at this and knew it would have to be several 
> >passes, but my pattern I was working with for the first pass 
> >was "(?P\d+\.)1"
> >
> >My question is since I can't use a replace pattern \1 follow by 33
>
> You can in fact do this by prefixing a zero "0" to the singular 
> backreference number, so:
>
> Replace: \0133
>
> will give you the first backreference \01 followed by the string "33".
>
>
> >I was using a named pattern, but (?P=foo) doesn't call the named pattern
> >in the replace area of a grep search? Just curious as I couldn't find
> >anything regarding this in the manual.
>
> Please see the section titled "Subpatterns Make Replacement 
> Powerful" in Chapter 8 (page 202) of the current manual:
>
> Pattern Inserts
> =
> [...]
>
> \P the text matched by the subpattern NAME
>
>
> Regards,
>
> Patrick Woolsey
> ==
> Bare Bones Software, Inc. 
>
>

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Re: Converting decimal value of some numbers using Grep

[Howard]

Patrick,

Thanks for your help,

Howard

On Wednesday, 16 March 2022 at 5:24:21 pm UTC-4 Patrick Woolsey wrote:

> On 3/16/22 at 3:10 PM, leadwi...@gmail.com (Howard) wrote:
>
> >In *Find: ^(\d+?)\.1 *what is the purpose of the question mark? 
> >I know what *\d+* does.
> >
>
> It makes matching "non-greedy", so the pattern won't try to 
> extend beyond the first possible match.
>
>
> [For details, please see "Non-Greedy Quantifiers" in Chapter 8 
> (page 201) of the ubiquitous PDF manual. :-) ]
>
>
> Regards,
>
> Patrick Woolsey
> ==
> Bare Bones Software, Inc. 
>
>

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Re: Converting decimal value of some numbers using Grep

[Patrick Woolsey]

On 3/16/22 at 3:10 PM, leadwithlo...@gmail.com (Howard) wrote:

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




It makes matching "non-greedy", so the pattern won't try to 
extend beyond the first possible match.



[For details, please see "Non-Greedy Quantifiers" in Chapter 8 
(page 201) of the ubiquitous PDF manual. :-) ]



Regards,

  Patrick Woolsey
==
Bare Bones Software, Inc. 

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Re: A Question Derived from Converting decimal value

[Patrick Woolsey]

On 3/16/22 at 2:06 PM, thepo...@gmail.com (ThePorgie) wrote:


I was tinkering with with Patrick's question


Not actually my question ;-) though close enough for the matters 
at hand.



So I was looking at this and knew it would have to be several 
passes, but my pattern I was working with for the first pass 
was "(?P\d+\.)1"


My question is since I can't use a replace pattern \1 follow by 33


You can in fact do this by prefixing a zero "0" to the singular 
backreference number, so:


Replace:\0133

will give you the first backreference \01 followed by the string "33".



I was using a named pattern, but (?P=foo) doesn't call the named pattern
in the replace area of a grep search? Just curious as I couldn't find
anything regarding this in the manual.


Please see the section titled "Subpatterns Make Replacement 
Powerful" in Chapter 8 (page 202) of the current manual:


Pattern   Inserts
=
[...]

\P  the text matched by the subpattern NAME


Regards,

 Patrick Woolsey
==
Bare Bones Software, Inc. 

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Re: Converting decimal value of some numbers using Grep

[Howard]

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. 
>
>
> > On Mar 16, 2022, at 13:02, Howard  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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A Question Derived from Converting decimal value

[ThePorgie]

I was tinkering with with Patrick's question regarding replacing the digits 
where the "n.1" is replaced with "n.33" and "n.2" is replaced with "n.67"
His sample data:
*Sample  Desired output*
0.20.67
1.11.33
1.11.33
1.21.67
1.11.33
66
3.2 3.67
88
6.2 6.67

So I was looking at this and knew it would have to be several passes, but 
my pattern I was working with for the first pass was "(?P\d+\.)1"
My question is since I can't use a replace pattern \1 follow by 33 I was 
using a named pattern, but (?P=foo) doesn't call the named pattern in the 
replace area of a grep search? Just curious as I couldn't find anything 
regarding this in the manual.

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Re: Converting decimal value of some numbers using Grep

[Patrick Woolsey]

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. 


> On Mar 16, 2022, at 13:02, Howard  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
> 0.20.67
> 1.11.33
> 1.11.33
> 1.21.67
> 1.11.33
> 66
> 3.2 3.67
> 88
> 6.2 6.67
> 

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Converting decimal value of some numbers using Grep

[Howard]

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*
0.20.67
1.11.33
1.11.33
1.21.67
1.11.33
66
3.2 3.67
88
6.2 6.67

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[ANN] BBEdit 14.1.1

[Rich Siegel]

Good { morning, afternoon, evening },

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Enjoy,

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