Replacing characters in a string

Hi,

I hope you can help on this little task I have. Basically I have a cell array and I want to know if this cell array contains any elements with 'ly' at the end of the word. If it does remove it from the word.

For example

strcell = {'hi' 'to' 'all' 'the' 'friendly' 'people' 'quickly' 'making' 'time' 'for' 'others'}

I understand that I can use "regexp" to find words that have these characters I want. From this I deduced *note that ive used match to just make it easy for me to see the word and not a random index value':

regexp(strcell, '\w+ly', 'match')

this should return friendly and quickly and from these words the "ly" should be removed:

* friendly -> friend
* quickly -> quick

I understand I am close with using regexprep However, from there I get confused on how to do this!

Thanks

[Sean de Wolski]

Every new poster should read this question before posting. Very well done!

Answer 1

[Sean de Wolski]

One way:

regexprep(strcell,'ly\>','')

Match ly at the end of the word \>. Replace it with nothing.

[Medhini B]

What if I want to remove ly which is in the beginning of the word? Example "lyfriend , ...."

Answer 2

[Daniel Shub]

I think regexprep does what you want. You need to modify the regexp a little bit:

s = regexprep(strcell, '(\w+)ly', '$1')

If you only want at the end

s = regexprep(strcell, '(.*)(ly\>)', '$1')

[Tanil]

I saw my problem I did try that once, but for some reason it kept outputting "$1" I had to parenthesis the \w+ part. Cheers for that

[Tanil]

O sorry this does it when 'ly' is also located in the middle of the word. I want it to do it providing its at the end only!

Answer 3

[Tanil]

after not skimming the regular expression chapter. There is another method ... I guess there is many different combinations.

[^c1c2c3] }   
Any character not contained within the brackets: anything but c1 or c2 or c3

deducing from this, the '^' is like ~ and works as a not therefore, [^a-z]

Thanks for all the suggestions.