How to recognize if there is in a character a point '.'?

Dear Matlab community,

Maybe somebody can help me with the following problem. For a project I would if there is a character a point '.' by scanning each element of the character by returning a true (1), false (0) as answer in an array. I tried the following:

xx='00-055.0'
isstrprop(xx,'punct')

ans =

  1×8 logical array

   0   0   1   0   0   0   1   0

At the moment it also shows e.g. '-', but is there a method to find explicitely the point '.'?

Thank you.

Regards

Answer 1

[madhan ravi]

contains(regexp(xx,'.','match'),'.')
%or
strcmp(regexp(xx,'.','match'),'.')
%or
cellfun(@(x)x=='.',regexp(xx,'.','match'))   

Gives:

ans =

  1×8 logical array

   0   0   0   0   0   0   1   0

[Guillaume]

??

regexp(xx,'.','match') simply splits xx into each of its character. Perhaps you meant

regexp(xx,'\.','match')  %the \ escape is critical. dot on its own matches anything

[madhan ravi]

No!, each character is split forming a cell array finally to attain logical array.

[Guillaume]

Well, then you don't need to use a regular expression engine to split a char vector into a cell array of individual characters

num2cell(xx)

will do the job faster.

[madhan ravi]

Or to use "ismember()" or "==" to directly compare.

Answer 2

[Guillaume]

Edit: I misunderstood the question. It's even simpler than what I initially wrote.

Simply:

xx == '.'

If you want to check multiple characters at once, e.g. . and +

ismember(xx, '.+')

--------

Original answer

Simply

contains(xx, '.')

or

any(xx == '.')  %probably faster than contains but only work as long as you're searching just for one character

if wanting to check if any of several characters are in xx, e.g. . and +:

any(ismember(xx, '.+'))  %are any character of xx member of the set .+ ?

[madhan ravi]

No,

contains(xx,'.') % would just give logical whether dot is present in xx it doesn't return a logical exaclty how OP wants

[Guillaume]

??

>> xx = 'abcdefgh.dd.d'
xx =
    'abcdefgh.dd.d'
>> contains(xx, '.')
ans =
  logical
   1

Just a scalar logical

[madhan ravi]

"For a project I would if there is a character a point '.' by scanning each element of the character by returning a true (1), false (0) as answer in an array."

Yes but OP wants the result to be as

ans =

  1×8 logical array

   0   0   0   0   0   0   1   0

for

xx =  '00-055.0'

[Guillaume]

Ah, well it's even simpler:

xx == '.'

[madhan ravi]

True :)

Answer 3

[Adam]

What is wrong with simply

xx == '.'

if xx is a char array rather than a string as it seems to be?