creating numerical values from parts of a string read

I have the following problem, I use the following to read in a string in a column in excel

[numerics,strings] = xlsread (filename,'spreadsheet name','Axx:Ayy')

filename is not in qoutes because it is read in from a user input using

filename = sprintf('filename',answer)

looking at the strings array I can see that the characters are read into the array

I need to extract part of the string and convert it into numerical values

the strings looks like 0(07/24/2014 10:39:10.21304) I need the last part of the string 10.21304

My approach was to first separate the string using command

strsplit(strings (2,1)) where strings (2,1)

is the location in the matrix of the strings; the error I can getting is that

First input must be a string.

So clearly I don't understand the issue - I tried the following

str1 = strings (2,1) and the output is '2 (07/24/2014 10:39:10.21304)'

Does the presence of the qoutation marks have anythings to do with the error ?

Any insight as to how to resolve the problem would be appreciated. I have google

several approaches and have not found anything that I can try. My first issue

resolving the error 'First input must be a string'

Regards, Philip

Answer 1

[Ben11]

If your strings are always of the same format, you can use regular expression to look for the colon followed by a number and followed by a dot, and then retrieve the numbers you want.

A = '0(07/24/2014 10:39:10.21304)'

ColonLocations = regexp(A,'\:\d*\.\d*','match')

which gives:

ColonLocations = 

      ':10.21304'

Get rid of the colon:

ColonLocations{1}(1) = []

Convert to number:

ColonLocations = str2num(ColonLocations{1});

You can make this shorter by combining a few lines.

ColonLocations =

     10.2130

[Evan]

Note that the str2num function makes use of eval, and should therefore be replaced by str2double where possible. This is discussed here.

[Philip]

Evan thanks for the response,

ColonLocations = regexp(A,'\:\d*\.\d*','match') works and gave ':10.21304' as

you wrote however ColonLocations{1}(1) = [] did not get rid of the colon

in checking the contents, ColonLocations showed [] only

Do you have any suggestion on what I may be doing wrong?

Thanks in advance for your feedback

Regards, Philip

[Ben11]

Mhh that's weird. You can use this as a workaround:

ColonLocations = ColonLocations{1}(2:end)

I'll play around to see what might cause the behavior you see.

[Philip]

Hi Evan so I found out what could be causing it, the first thing I have to do is to

str1 = char (strings (j,1)) where strings (j,1) contains A

then use str1 istead of A

then all the routins above works.

Thanks a lot for you help and sorry for the confusion.

Regards, Philip

[Ben11]

glad it works then!

Answer 2

[Evan]

Another option, using a lookaround operator.

A = '0(07/24/2014 10:39:10.21304)'
s = regexp(A,'(?<=:)\d+\.\d+','match');
n = str2double(s);

See more here.

[Philip]

This works as well, the key is to change A with char - see above

Agains thanks for the help.

Regards, Philip

[Evan]

No problem!