CODEWITHSUNDEEP
matlabindexing
Naga Arjun
Naga Arjun

Reputation: 1

finding a comma in string

[23567,0,0,0,0,0] and other value is [452221,0,0,0,0,0] and the value should be contineously displaying about 100 values and then i want to display only the sensor value like in first sample 23567 and in second sample 452221 , only the these values have to display . For that I have written a code
value = str2double(str(2:7));see here my attempt so I want to find the comma in the output and only display the value before first comma

Upvotes: 0

Views: 303

Answers (2)

hbaderts
hbaderts

Reputation: 14316

As proposed in a comment by excaza, MATLAB has dedicated functions, such as sscanf for such purposes.

sscanf(str,'[%d')

which matches but ignores the first [, and returns the next (i.e. the first) number as a double variable, and not as a string.

Still, I like the idea of using regular expressions to match the numbers. Instead of matching all zeros and commas, and replacing them by '' as proposed by Sardar_Usama, I would suggest directly matching the numbers using regexp.

You can return all numbers in str (still as string!) with

nums = regexp(str,'\d*','match')

and convert the first number to a double variable with

str2double(nums{1})

To match only the first number in str, we can use the regexp

nums = regexp(str,'[(\d*),','tokens')

which finds a [, then takes an arbitrary number of decimals (0-9), and stops when it finds a ,. By enclosing the \d* in brackets, only the parts in brackets are returned, i.e. only the numbers without [ and ,.

Final Note: if you continue working with strings, you could/should consider the regexp solution. If you convert it to a double anyways, using sscanf is probably faster and easier.

Upvotes: 1

Sardar Usama
Sardar Usama

Reputation: 19689

You can use regexprep as follows:

str='[23567,0,0,0,0,0]' ;
required=regexprep(str(2:end-1),',0','')
%Taking str(2:end-1) to exclude brackets, and then removing all ,0

If there can be values other than 0 after , , you can use the following more general approach instead:

required=regexprep(str(2:end-1),',[-+]?\d*\.?\d*','')

Upvotes: 0

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