I have a matrix
M =
hh:mm:ss ms
'12:00:01' 1
'12:00:02' 2
'12:00:03' 3
'12:00:04' 4
'12:00:05' 5
Now I want to add the array of milliseconds ms to my time vector. Like
N =
hh:mm:ss
'12:00:01.001'
'12:00:02.002'
'12:00:03.003'
'12:00:04.004'
'12:00:05.005'
How can I do this? What I tried was:
for k=1:length(M)
t1 = datenum(M{k,1},'HH:MM:SS');
c = num2str(M{k,2});
t2 = datenum(c,'FFF');
time = t1+t2;
N{k,1} = datestr(time,'HH:MM:SS.FFF');
end
But this did not do the job right. What I get is:
N =
hh:mm:ss
'12:00:01.100'
'12:00:02.200'
'12:00:03.300'
...
'12:00:04.100'
'12:00:05.110'
'12:00:05.120'
I think this problem is simple to solve. But at the moment I just do not know how solve it.
It's a string formatting problem.
In your code, instead of
c = num2str(M{k,2});
use
c = sprintf('%03d',M{k,2});
In the above usage, sprintf pads zeros in front of a string if it's less than 3 characters long.
Related
I have some text files with dates of the form
2015-09-08 14:38:03
2015-09-08 14:38:03.1
2015-09-08 14:38:03.2
that I want to convert into Matlab date/time format. As you can see, the text strings have a different time format regarding the milliseconds. In the first case, no milliseconds are given; in the seconds case, milliseconds are given with one digit only. This gives a sampling rate of 20Hz for measuring data.
So far, only
x = datenum(file.dateColumn, 'yyyy-mm-dd HH:MM:SS');
is working, but of course misses the milliseconds. A conversion like
x = datenum(file.dateColumn, 'yyyy-mm-dd HH:MM:SS.FFF');
does not work as the milliseconds are either zero (full seconds) or have one digit after the '.' delimiter. A workaround like
x = datestr(file.dateColumn, 'yyyy-mm-dd HH:MM:SS.FFF');
x = datenum(file.dateColumn, 'yyyy-mm-dd HH:MM:SS.FFF');
i.e. converting the text string into a Matlab string (and giving it additional FFF/FF digits) and then converting that one into a date/time number works - but this is such time-cosuming that I cannot use it for my data. I have millions of time rows in different files.
Do you have any ideas?
Greetings, Arne
Thanks to Nick I found a way to solve it:
dataVelPres = readtable(fileName, ...
'ReadVariableNames', false, ...
'HeaderLines', 4 );
dataVelPres.Properties.VariableNames = {'date' 'velU' 'velV' 'velW' 'pres'};
dateMill = datetime(dataVelPres.date, 'inputformat', 'yyyy-MM-dd HH:mm:ss.S');
dateFull = datetime(dataVelPres.date, 'inputformat', 'yyyy-MM-dd HH:mm:ss');
dateNaT = isnat(dateMill);
dateMill(dateNaT) = dateFull(dateNaT);
dataVelPres.dateTime = dateMill;
dataVelPres.date = datenum(dataVelPres.dateTime); % Convert to number format if needed
This works with two tables - one for millisec. and one without - and puts both together, as both give NaT entries in case the inputformat does not match.
Is there a more elegant way?
You may try something like:
a='2015-09-08 14:38:03';
s=strsplit(a,{'-',':',' '})
x=datenum(cellfun(#str2num,s(1:end)))
I highly recommend using the new datetime object:
strings = {'2015-09-08 14:38:03', '2015-09-08 14:38:03.1', '2015-09-08 14:38:03.2'};
dates = {};
for d = strings
d = d{1};
try
dt = datetime(d, 'inputformat', 'yyyy-MM-dd HH:mm:ss.S');
catch
dt = datetime(d, 'inputformat', 'yyyy-MM-dd HH:mm:ss');
end
dates{end + 1} = dt;
end
>> dates
dates =
[08-Sep-2015 14:38:03] [08-Sep-2015 14:38:03] [08-Sep-2015 14:38:03]
>> dates{end}.Second
ans =
3.2000
It's also easy convert from a datetime object to a datenum:
>> x = [datetime('now'), datetime('yesterday')]
x =
10-Dec-2015 12:53:40 09-Dec-2015 00:00:00
>> datenum(x)
ans =
1.0e+05 *
7.3631 7.3631
>>
I have a date vector in form
20140117
20130325
20130530
etc.
There are 5,000,000 lines in the double vector.
How can I transfer it to a datevector recognized by matlab?
I don't like changing it to string and putting the parts in separately. It takes too long!
Please Help!
a combination of fix and mod let's you extract the digits you want:
%Matrix Columns YY,MM,DD,hh,mm,ss
[mod(fix(x/10000),100000),mod(fix(x/100),100),mod(x,100),zeros(size(x,1),3)]
%datenum
datenum(mod(fix(x/10000),10000),mod(fix(x/100),100),mod(x,100))
If you really want to avoid casting to string then back to number then you can use this method:
D = [20140117; 20130325; 20130530];
YY = fix( D./10000 ) ;
MM = fix( (D-YY.*10000) /100 ) ;
DD = fix( (D-YY.*10000-MM.*100 ) );
DateInMatlabformat = datenum( YY , MM , DD ) ;
You can package that in a one liner if you want, but basically what it does is:
Divide by 10000 to get the year in the variable YY
Remove this part from your original date ((D-YY.*10000)), then divide by 100 to get the month.
remove all of that, you obtain the day.
The last line merge all of that in a Matlab standard time serial format. Read the doc on datenum and datestr for more information.
I have some time as string format in my data. Can anyone help me to convert this date to milliseconds in Matlab.
This is an example how date looks like '00:26:16:926', So, that is 0 hours 26 minutes 16 seconds and 926 milliseconds. After converting this time, I need to get only milliseconds such as 1576926 milliseconds for the time that I gave above. Thank you in advance.
Why don't you try using datevec instead? datevec is designed to take in various time and date strings and it parses the string and spits out useful information for you. There's no need to use regexp or split up your string in any way. Here's a quick example:
[~,~,~,hours,minutes,seconds] = datevec('00:26:16:926', 'HH:MM:SS:FFF');
out = 1000*(3600*hours + 60*minutes + seconds);
out =
1576926
In this format, the output of datevec will be a 6 element vector which outputs the year, month, day, hours, minutes and seconds respectively. The millisecond resolution will be added on to the sixth element of datevec's output, so all you have to do is convert the fourth to sixth elements into milliseconds and add them all up, which is what is done above. If you don't specify the actual day, it just defaults to January 1st of the current year... but we're not using the date anyway... we just want the time!
The beauty with datevec is that it can accept multiple strings so you're not just limited to a single input. Simply put all of your strings into a single cell array, then use datevec in the following way:
times = {'00:26:16:926','00:27:16:926', '00:28:16:926'};
[~,~,~,hours,minutes,seconds] = datevec(times, 'HH:MM:SS:FFF');
out = 1000*(3600*hours + 60*minutes + seconds);
out =
1576926
1636926
1696926
One solution could be:
timeString = '00:26:16:926';
cellfun(#(x)str2num(x),regexp(timeString,':','split'))*[3600000;60000;1000;1]
Result:
1576926
Assuming that your date string comes in that format consistently, you could use something as simple as this:
test = '00:26:16:926';
H = str2num(test(1:2)); % hours
M = str2num(test(4:5)); % minutes
S = str2num(test(7:8)); % seconds
MS = str2num(test(10:12)); % milliseconds
totalMS = MS + 1000*S + 1000*60*M + 1000*60*60*H;
Output:
1576926.00
you can convert a single string with a date or even a vector by using datevec for conversion and the dot product
a = ['00:26:16:926' ; '08:42:12:936']
datevec(a,'HH:MM:SS:FFF') * [0 0 0 3600e3 60e3 1e3]'
ans =
1576926
31332936
I simply want to generate a series of dates 1 year apart from today.
I tried this
CurveLength=30;
t=zeros(CurveLength);
t(1)=datestr(today);
x=2:CurveLength-1;
t=addtodate(t(1),x,'year');
I am getting two errors so far?
??? In an assignment A(I) = B, the number of elements in B and
Which I am guessing is related to the fact that the date is a string, but when I modified the string to be the same length as the date dd-mmm-yyyy i.e. 11 letters I still get the same error.
Lsstly I get the error
??? Error using ==> addtodate at 45
Quantity must be a numeric scalar.
Which seems to suggest that the function can't be vectorised? If this is true is there anyway to tell in advance which functions can be vectorised and which can not?
To add n years to a date x, you do this:
y = addtodate(x, n, 'year');
However, addtodate requires the following:
x must be a scalar number, not a string.
n must be a scalar number, not a vector.
Hence the errors you get.
I suggest you use a loop to do this:
CurveLength = 30;
t = zeros(CurveLength, 1);
t(1) = today; % # Whatever today equals to...
for ii = 2:CurveLength
t(ii) = addtodate(t(1), ii - 1, 'year');
end
Now that you have all your date values, you can convert it to strings with:
datestr(t);
And here's a neat one-liner using arrayfun;
datestr(arrayfun(#(n)addtodate(today, n, 'year'), 0:CurveLength))
If you're sequence has a constant known start, you can use datenum in the following way:
t = datenum( startYear:endYear, 1, 1)
This works fine also with months, days, hours etc. as long as the sequence doesn't run into negative numbers (like 1:-1:-10). Then months and days behave in a non-standard way.
Here a solution without a loop (possibly faster):
CurveLength=30;
t=datevec(repmat(now(),CurveLength,1));
x=[0:CurveLength-1]';
t(:,1)=t(:,1)+x;
t=datestr(t)
datevec splits the date into six columns [year, month, day, hour, min, sec]. So if you want to change e.g. the year you can just add or subtract from it.
If you want to change the month just add to t(:,2). You can even add numbers > 12 to the month and it will increase the year and month correctly if you transfer it back to a datenum or datestr.
I need to compute time difference in minutes with four input parameters, DATE_FROM, DATE_TO, TIME_FROM, TIME_TO. And one output parameter DIFF_TIME. I have created a function module, I need to write a formula which computes the time diff in minutes.
Any help would be great!
Thanks,
Sai.
Use CL_ABAP_TSTMP=>TD_SUBTRACT to get the number of seconds between two date/time pairs.
(then, to get the number of minutes, divide the number of seconds by 60).
Example:
DATA(today_date) = CONV d( '20190704' ).
DATA(today_time) = CONV t( '000010' ).
DATA(yesterday_date) = CONV d( '20190703' ).
DATA(yesterday_time) = CONV t( '235950' ).
cl_abap_tstmp=>td_subtract(
EXPORTING
date1 = today_date
time1 = today_time
date2 = yesterday_date
time2 = yesterday_time
IMPORTING
res_secs = DATA(diff) ).
ASSERT diff = 20. " verify expectation or short dump
If the values are guaranteed to be in the same time zone, it's easy enough that you don't need any special function module or utility method. Read this, then get the difference of the dates and multiply that by 24 * 60 and get the difference of the times (which is in seconds) and divide that by 60. Sum it up and there you are.