regexprep does not exactly what I want

Dear all,

I have the following cell array

Charge = {'OH-1'}    {'KOH+0'}    {'K+1'}    {'I-1'}    {'HI+0'}    {'H3O+1'}    {'H2O+0'}

I want to remove all information before the + and - signs. Therefore I tried the following:

regexprep(Charge,'[^-+].','');

which produces

{'-1'}    {'0'}    {'1'}    {'1'}    {'+0'}    {'1'}    {'0'}

This works well except in case of only one character in front of the minus sign (i.e. in case of I-1). In that case, the - sign is also deleted. The - signs are crucial to be included, the + signs not.

Any suggestions?

Thanks, Tim

Answer 1

[Daniel M]

There's definitely a way to do it using regexprep, but I found this solution first, so hopefully it is sufficient.

Charge = {'OH-1','KOH+0','K+1','I-1','HI+0','H3O+1','H2O+0'};
c = regexp(Charge,'[-+]\w*','match');
cc = cat(2,c{:}); % put back into cell array

[hyble]

Great stuff, many thanks for this!

[Daniel M]

regexprep(Charge,'^\w*','')

[Guillaume]

I'm not convinced that '[-+]\w*' is the right regexp. This constrains the symbols that follow the + or - to letters, digits or _. This restriction may or many not be appropriate.

The regexp that matches exactly your specification would be

regexp(Charge, '[+-].*', 'match', 'once')

Your original regexprep did not work at all. It basically said remove pairs of characters where the first character is anything but - or + and the 2nd one is anything. So looking at 'H3O+1', the first pair is 'H3', it doesn't start with a + or -, so is removed. the 2nd pair is 'O+'. Again, it doesn't start with a + or -, so is removed. Now with 'HI+1', the first pair is 'HI', removed, the 2nd one is '+1', starts with + so not removed. If you had something like 'H3O+01', it would have removed everything since the scan would remove 'H3', then 'O+', then '01'.

A regexprep that would have worked would be:

regexprep(Charge, '[^+]+\+|[^+]+(?=\-)', '')

[Daniel M]

Thanks Guillaume, I agree that '[-+]\w*' is not robust enough (which is why I voted for Stephen's solution), but it does satisfy his test case.

As for the comment on the regexprep, I'm not sure if you're referring to me or not. I wrote

regexprep(Charge,'^\w*','')
ans = 
{'-1'}    {'+0'}    {'+1'}    {'-1'}    {'+0'}    {'+1'}    {'+0'}

which works. Your example however doesn't:

regexprep(Charge, '[^+]+\+|[^+]+(?=\-)', '')
ans = 
{'-1'}    {'0'}    {'1'}    {'-1'}    {'0'}    {'1'}    {'0'}

As you can see it drops the sign of the charge.

[Guillaume]

The comment about the regexprep referred to the original question, not your answer.

I wrote most of my comment shortly after you posted your answer but had to dash off to a meeting before posting it. When I finally posted it, it was a bit out of date. Sorry about that.

[hyble]

Many thanks to all of you for helping out here!

Answer 2

[Stephen23]

>> regexprep(Charge,'^[^-+]*','')
ans = 
    '-1'    '+0'    '+1'    '-1'    '+0'    '+1'
>> regexp(Charge,'[-+].+$','once','match')
ans = 
    '-1'    '+0'    '+1'    '-1'    '+0'    '+1'

[Daniel M]

This will handle edge cases better than my solution above. More robust.