Select parts of number from a cell of many numbers to put in a variable?

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Ross 2014-8-12

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评论： Ross 2014-8-13

采纳的回答： Ben11

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Hi.

I have used

[num text raw] = xlsread(filename)

to get access to the cells on a bastardised xml/csv type text file. I can read values from cells using

raw(row,col)

This is fine, however my question is, one of these cells contains the date in format

20140802

I wish to select parts of the number and assign variables for month(08), day(02), year(2014).

How exactly can I achieve this? I can't seem to find an answer to what seems to be an easy problem, help appreciated!

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Ben11 2014-8-12

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编辑：Ben11 2014-8-13

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The easy way would be:

year = YourDate(1:4)
month = YourDate(5:6)
day = YourDate(7:8)

Then you can convert to digits with str2double for example. Is that what you mean?

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Ben11 2014-8-13

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When you say "it seems to work but no title is present" do you use something like this:

TitleString = sprintf('Graph Plot at %d:%d on %d/%d/%d',hours,minutes,day,month,year);
... add code for your plot
title(TitleString)

if you don't see any title it might be because matlab does not put it at the right place?

Ross 2014-8-13

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I tried this (minus hours and minutes)

titleName = sprintf('Graph Plot at on %d/%d/%d',day,month,year)
title(titleName)

No error but the title is just blank. Also, I can't seem to add my 'time' variable with this syntax without it just printing time, whereas when I use:

titleName = (['Graph Plot at', time, 'on', Day, Month, year]);
title(titleName)

It does print the value of the variables except it's as though it's carriage returning after each comma. As mentioned, it is coming out like this:

Graph Plot at
     time
      on
      06
      08
     2013

Wheras I would like it to display on the title as

Graph Plot at (time) on (day)(month)(year)

So close, it's probably a simple thing I'm missing or some title formatting option I haven't encountered.

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Azzi Abdelmalek 2014-8-12

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a=20140802
b=datevec(datenum(num2str(a),'yyyymmdd'))
b(1)  % is the year
b(2)  % the month
b(3)  % the day

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Azzi Abdelmalek 2014-8-12

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Ross, If you run my code with my data, there are no errors. You just need to use the same class of data then mine, or use this one

a={20140802  20140803} 
b=cellfun(@num2str,a,'un',0)  % convert from numeric to string
c=datevec(cellfun(@(x) datenum(x,'yyyymmdd'),b))

Ross 2014-8-13

编辑：Ross 2014-8-13

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Hi, thanks for replying.

This code sorta works, I get the date parsed but it displays in column format, so my date cell displays like so:

Original data, ie variable b looks like '20130806'. How to get it so I have variables such as:

Year = 2013
Month = 08
Day = 06

I don't quite understand this line in your code:

c=datevec(cellfun(@(x) datenum(x,'yyyymmdd'),b))

EDIT: Actually when I look at 'c' in the main window it displays as:

2013 8 6 0 0 0

This is for displaying in a title of a 2d plot. The title on the plot displays in columns and has the 3 unnecessary zero's which is why I was looking to separate the year, month and day to use easier with the title code which I have as:

titleName = (['Graph Plot at', time, 'on', day, month, year]);

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