Convert to date, hour, and minute

Hi,

We collect data from a data logger that collects date and time stamps in addition to the actual data. Here is the date format.

2011 95 1300

The first and third columns are self explanatory. The 2nd column is the no. of days since January the 1st.

I would like to know if there is a way to convert the information into 4/5/2011 13:00.

Thanks in advance.

Answer 1

[Andrei Bobrov]

added after the receipt of Bahram's data [edited]

d =     [ 221         2012           43         1900        13.63        2.745
          221         2012           43         2000        13.39        0.002
          221         2012           43         2100         13.2       -2.362
          221         2012           43         2200        13.04       -3.703
          221         2012           43         2300        12.87       -4.243
          221         2012           44            0        12.73       -4.555]

d1 = d(:,4)/100;
out = datestr(datenum(d(:,2), 1, d(:,3), fix(d1), 100*rem(d1,1), 0))

[Oleg Komarov]

If you're using datenum, you don't need to create a matrix of ones or zeros, simply write 1, 0, it will replicate automatically.

[Andrei Bobrov]

Thank you Oleg! Corrected.

[Bahram]

Andrei,

Thanks a lot. Your code works quite well! I get the following:

12-Feb-2012 19:00:00

12-Feb-2012 20:00:00

12-Feb-2012 21:00:00

12-Feb-2012 22:00:00

12-Feb-2012 23:00:00

13-Feb-2012 00:00:00

Now, I would like to plot the 5th column of matrix d versus out. I get an error when I try this:

plot(out,data(:,5))

??? Error using ==> plot

Invalid first data argument

Would you please help me again?

Thanks agin!

[Walter Roberson]

"out" would be an array of character. You cannot plot strings.

Skip the datestr() call (using only the datenum() call) and that will give you serial date numbers that can be used to plot. See also the datetick() routine.

[Oleg Komarov]

Then you don't need to convert your dates to strings:

out = datenum(d(:,2), 1, d(:,3), fix(d1), 100*rem(d1,1), 0);

The appearance on the plot is controlled with datetick:

plot(out, d(:,5))

datetick('x','dd-mmm-yyyy HH:MM')

Where you can customize the apperance by chosing the appropriate format. You can find the details in the documentation of datetick, datenum, datestr etc..

A note for future questions: specify what you need your task for. If you asked from the beginning that you wanted to use it in a plot, the answer could have been much more straightforward.

[Bahram]

Thanks a lot.

I am very new to this community, and I am also fairly new to MATLAB. I'll be more specific in the future!

Thanks agin.

Answer 2

[Geoff]

Yep,

First, concatenate the first and third columns to use the normal date conversion, then add the second column. Put the result into datestr:

datestr(datenum('20121300', 'yyyyHHMM')+95, 'mm/dd/yyyy HH:MM')

ans =

    04/05/2012 13:00

Obviously I've just used constants here, but you get the idea.

You can't get single-character days out of that though - they are padded with zeros.

In that case you'd want to do this:

d = datevec(datenum('20121300', 'yyyyHHMM')+95, 'mm/dd/yyyy HH:MM');
sprintf( '%d/%d/%d %d:%02d', d([2 3 1 4 5]) )

ans =

    4/5/2012 13:00

[Bahram]

Thanks for your fast response.

I tried the following code on this dataset:

221 2012 43 1900 13.63 2.745

221 2012 43 2000 13.39 0.002

221 2012 43 2100 13.2 -2.362

221 2012 43 2200 13.04 -3.703

221 2012 43 2300 12.87 -4.243

221 2012 44 0 12.73 -4.555

M=csvread('data.dat');

str=[num2str(M(1:end,2)) num2str(M(1:end,4))];

answer=datestr(datenum('str', 'yyyyHHMM')+M(1:end,3), 'mm/dd/yyyy HH:MM');

I get the following answer:

02/13/2012 00:00

02/13/2012 00:00

02/13/2012 00:00

02/13/2012 00:00

02/13/2012 00:00

02/14/2012 00:00

As you see the time stamp is all zero.

Please let me know what I did wrong.

Thanks again.

[Walter Roberson]

Do not quote str in the datenum() call: it is a variable rather than a literal.

[Bahram]

Thanks for your reply. This correction fixed the time stamp, but the time stamp gets messed up if the original value has less than four digits. I used the following code for the data shown below.

clc

clear

M=csvread('data.dat');

str=[num2str(M(1:end,2)) num2str(M(1:end,4))];

answer=datestr(datenum(str, 'yyyyHHMM')+M(1:end,3), 'mm/dd/yyyy HH:MM');

Data:

221,2012,43,1900,13.63,2.745

221,2012,43,2000,13.39,.002

221,2012,43,2100,13.2,-2.362

221,2012,43,2200,13.04,-3.703

221,2012,43,2300,12.87,-4.243

221,2012,44,0,12.73,-4.555

221,2012,44,100,12.62,-4.951

221,2012,44,200,12.53,-5.684

221,2012,44,300,12.5,-6.337

221,2012,44,400,12.49,-6.93

221,2012,44,500,12.49,-7.52

221,2012,44,600,12.5,-8.01

221,2012,44,700,12.5,-8.42

221,2012,44,800,12.51,-8.75

221,2012,44,900,12.5,-8.97

221,2012,44,1000,12.52,-8.32

Result:

02/13/2012 19:00

02/13/2012 20:00

02/13/2012 21:00

02/13/2012 22:00

02/13/2012 23:00

02/14/0002 01:02

02/14/0201 02:10

02/14/0201 02:20

02/14/0201 02:30

02/14/0201 02:40

02/14/0201 02:50

02/14/0201 03:00

02/14/0201 03:10

02/14/0201 03:20

02/14/0201 03:30

02/14/2012 10:00

Once again, I appreciate your help.

Answer 3

[Walter Roberson]

Try

datestr([M(:,1), ones(size(M,1),1), M(:,2)+1, round(M(:,3)./100), mod(M(:,3),100), zeros(size(M,1),1)], 'm/d/yyyy HH:MM')

This assumes that "days since January the 1st" will be 0 for Jan 1.

The part in [] constructs dates in the medium-length datevec format (year month day hours minutes seconds). The adjustment to convert between day number and proper month/day is handled by specifying (e.g.) Jan 95th and letting the date calculation routines do the fixup.

[Geoff]

That call to 'round' should be 'floor' or it will give incorrect results for the last 10 minutes of each hour. Unfortunately a single 'm' and 'd' in calls to 'datestr' expand to the capitalised first letter of the month and day respectively.

[Walter Roberson]

Ah yes, I was worried about floating point round-off when I used round() and didn't think it through.

Answer 4

[per isakson]

I run this. Matlab seems to more clever than I anticipated. Is this documented behavior?

    str = '2011 95 1300'
    datestr( datenum( str, 'yyyy dd HHMM' ), 31 )
    str = '2012 95 1300'
    datestr( datenum( str, 'yyyy dd HHMM' ), 31 )
    str =
    2011 95 1300
    ans =
    2011-04-05 13:00:00
    str =
    2012 95 1300
    ans =
    2012-04-04 13:00:00

[Oleg Komarov]

Expected and documented behavior.

[per isakson]

I cannot find that documentation!