Converting Epoch to Date in Matlab - matlab

I have an array of Epoch milliseconds (array of numbers) in Matlab. I would like to convert these into UTC date-time format, such as DD-MM-YYYY HH:MM.
Is there a pre-defined Matlab way to do this or will I have to write my own function?

Suppose, you start with a vector time_unix, then:
>> time_unix = 1339116554872; % example time
>> time_reference = datenum('1970', 'yyyy');
>> time_matlab = time_reference + time_unix / 8.64e7;
>> time_matlab_string = datestr(time_matlab, 'yyyymmdd HH:MM:SS.FFF')
time_matlab_string =
20120608 00:49:14.872
Notes:
1) See the definition of matlab's time.
2) 8.64e7 is number of milliseconds in a day.
3) Matlab does not apply any time-zone shifts, so the result is the same UTC time.
4) Example for backward transformation:
>> matlab_time = now;
>> unix_time = round(8.64e7 * (matlab_time - datenum('1970', 'yyyy')))
unix_time =
1339118367664
To summarize, here are two functions:
function tm = unix2matlab(tu)
tm = datenum('1970', 'yyyy') + tu / 864e5;
end
function tu = matlab2unix(tm)
tu = round(864e5 * (tm - datenum('1970', 'yyyy')));
end
The matlab time here is numeric. You can always convert it to string using datestr()
Update for nanoseconds
time_unix_nanos = 1339116554872666666;
millis = round(time_unix_nanos / 1e6);
nanos = time_unix_nanos - 1e6 * millis;
time_matlab = unix2matlab(millis);
s = [datestr(time_matlab, 'yyyymmdd HH:MM:SS.FFF'), num2str(nanos)];
s =
20120608 00:49:14.872666666

I tried the above code, but the results were wrong. I realised the main error is related to the awkward definition of the Unix time (epoch time). Unix time (epoch time) is defined as the number of seconds after 1-1-1970, 00h:00, not the number of **milli**seconds (http://en.wikipedia.org/wiki/Unix_time). With this definition, the Unix time should therefore be divided by 8.64e5 instead of 8.64e7.
In addition, datenum('1970', 'yyyy') does not seem to result in the desired reference time of 1-1-1970, 00h:00.
Here's my improved code:
tMatlab = datenum (1970,1,1,0,0) + tUnix / 86400;

Serg's answer is what I normally use, when I'm working in MATLAB. Today I found myself wanting to do the conversion to date in MATLAB as the title says - without the datestring conversion  specified in the question body - and output the date number from the shell.
Here is what I settled on for the rounded date number:
TODAY_MATLAB="$[719529 + $[`date +%s` / 24/60/60]]"
This is really just the bash equivalent of what you would expect: 719529 is the datenum of the epoch (1970-01-01 or datenum(1970,1,1) in MATLAB). I'm also fumbling through ksh lately and it seems this can be done there with:
TODAY_EPOCH=`date +%s`
TODAY_MATLAB=`expr $TODAY_EPOCH / 24 / 60 / 60 + 719529`
As a side exercise, I added the decimal portion back onto the date in bash - I didn't bother in ksh, but it's only arithmetic and goes similarly:
N_DIGITS=7
FORMAT=$(printf "%%d.%%0%dd" $N_DIGITS)
NOW_EP_SEC=`date +%s`
SEC_PER_DAY=$(( 24*60*60 ))
NOW_EP_DAY=$(( $NOW_EP /$SEC_PER_DAY ))
SEC_TODAY=$(( $NOW_EP_SEC - $NOW_EP_DAY*$SEC_PER_DAY ))
TODAY_MATLAB="$(( NOW_EP_DAY+719529 ))"
FRACTION_MATLAB="$( printf '%07d' $(( ($SEC_TODAY*10**$N_DIGITS)/SEC_PER_DAY )) )"
MATLAB_DATENUM=$( printf $FORMAT $TODAY_MATLAB $FRACTION_MATLAB )
echo $MATLAB_DATENUM

Related

How to calculate Duration of Time between two date time through VBscript

How to calculate Duration of Time between two date time through VBscript
Date1 = 2021-01-22 11:43:38.000
Date2 = 2021-01-22 14:32:38.000
result should be HH:MM:SS
TimeSerial and FormatDateTime return a date or time that will take the regional settings of the computer into account. On my European computer no AM extension is shown because we use the 24h time format.
An additional problem with TimeSerial is that it will overflow once there are more than 32767 seconds.
A different approach could be to calculate the values for hours, minutes and seconds separately. A possible solution could be:
secValue = DateDiff("s",Date1,Date2)
hours = secValue \ 3600
hh = hours
if hours < 10 then
hh = Right("0" & hours, 2)
end if
mm = Right("0" & (secValue - hours * 3600) \ 60, 2)
ss = Right("0" & secValue mod 60, 2)
diff = hh & ":" & mm & ":" & ss
wscript.echo diff
Finally i answered to my Question Feeling great
Date1 = alA_Filling(0)
Date2 = alA_Filling(1)
secValue = DateDiff("s",Date1,Date2)
ts = TimeSerial(0, 0, secValue)
Duration = FormatDateTime(ts, vbLongTime)
But i got output like 2:49:00 AM why added AM to this may be this vbLongTime
how can i remove that AM

How to convert seconds into hh:mm format in Power Bi?

In Power Bi, I have a table that contains Name and TimeSpent by user in seconds.
I want to convert total seconds spent by all users into duration format (hh:mm)
When I am getting seconds in hh:mm format for each user from database query, the values are coming up like these 12:63 etc. After importing these values into power bi, I tried to set its datatype to DateTime format but power bi shows an error saying that it is not a valid value. If I set the datatype of the column as string then strings dont add up.
What can be the ideal way to do it?
you can solve this in one line:
measure = FORMAT(TIME(0;0;tableNAME[your_column]);"HH:mm:ss")
You can try the following DAX:
HHMMSS =
INT(Table[TimeSpent] / 3600) & ":" &
RIGHT("0" & INT((Table[TimeSpent] - INT(Table[TimeSpent] / 3600) * 3600) / 60), 2) & ":" &
RIGHT("0" & MOD(Table[TimeSpent], 3600), 2)
Source
Had a similar question but for D:HH:MM:SS, code below if it's of use.
DurTime (meas) =
VAR vDur = <<<duration in CALCULATE(SUM(seconds)) >>>
RETURN INT(vDur/86400) & ":" & //Days
RIGHT("0" & INT(MOD(vDur/3600,24)),2) & ":" & //Hours
RIGHT("0" & INT(MOD(vDur/60,60)),2) & ":" & //Minutes
RIGHT("0" & INT(MOD(vDur,60)),2) //Seconds
DAX code:
= TIME(0,0,SUM('Table'[Timespent]))
Then click the modelling tab and choose Format - Date Time and choose the appropriate format.
That's a better formula, which I'm using in PBI:
HHMMSS = FORMAT(TIME(int(Table[TimeSpent] / 3600); int(mod(Table[TimeSpent]; 3600) / 60);int(mod(mod(Table[TimeSpent]; 3600); 60))); "HH:mm:ss")
I wanted a Power BI Measure wich is easy to read for this problem, code below if it's of use.
HH:MM =
VAR TotalDuration = SUM(tableNAME[your_column] ) //if you use a measure just leave the SUM part out
VAR TotalHours = TRUNC (TotalDuration/3600)
VAR Min_ = FORMAT(TRUNC(TotalDuration - TotalHours * 3600),"00")
RETURN
TotalHours & ":" & Min_
The solution is adopted from the top answer of this question PowerBi Duration calculation in hh:mm:ss

Matlab Code for weekdays and weekends [duplicate]

This question already has an answer here:
Weekend extraction in Matlab
(1 answer)
Closed 6 years ago.
I were able to successfully made a schedule in which the output is 1 if time is between 7 AM-5PM and otherwise 0, time is based on my computer. However the day Monday-Sunday is based on my computer as well.. I cant find the solution to put an output 1 on Monday-Saturday and output 0 on Sunday. The code I have is below
function y = IsBetween5AMand7PM
coder.extrinsic('clock');
time = zeros(1,6);
time = clock;
current = 3600*time(4) + 60*time(5) + time(6); %seconds passed from the beginning of day until now
morning = 3600*7; %seconds passed from the beginning of day until 7AM
evening = 3600*17; %seconds passed from the beginning of day until 5PM
y = current > morning && current < evening;
end
Now the time here is correct already what I need is for the day (Monday-Sunday) to have my needed output. Also this matlab code is inside a matlab function on Simulink block.
If you use weekday like this, you can generate a 0/1 value as you specified for today's date:
if (weekday(now) > 1)
day_of_week_flag = 1;
else
day_of_week_flag = 0;
or if you like, this one-liner does the same thing, but may not be as easy to read if you're not familiar with the syntax:
day_of_week_flag = ( weekday(now) > 1);
You can also use date-strings like this to convert other dates:
day_of_week_flag = ( weekday('01-Mar-2016') > 1 )
Finally, if you have a numeric array of date/time values, like [2016 3 3 12 0 0], you first need to convert to a serial date using datenum, then use weekday:
time = clock;
day_of_week_flag = ( weekday(datenum(time)) > 1);
An alternate way to check without using weekday is the following:
time = clock;
day_of_week = datestr(time, 8);
if (day_of_week == 'Sun')
day_of_week_flag = 0;
else
day_of_week_flag = 1;

Compare dates in Lua

I have a variable with a date table that looks like this
* table:
[day]
* number: 15
[year]
* number: 2015
[month]
* number: 2
How do I get the days between the current date and the date above? Many thanks!
You can use os.time() to convert your table to seconds and get the current time and then use os.difftime() to compute the difference. see Lua Wiki for more details.
reference = os.time{day=15, year=2015, month=2}
daysfrom = os.difftime(os.time(), reference) / (24 * 60 * 60) -- seconds in a day
wholedays = math.floor(daysfrom)
print(wholedays) -- today it prints "1"
as #barnes53 pointed out could be off by one day for a few seconds so it's not ideal, but it may be good enough for your needs.
You can use the algorithms gathered here:
chrono-Compatible Low-Level Date Algorithms
The algorithms are shown using C++, but they can be easily implemented in Lua if you like, or you can implement them in C or C++ and then just provide Lua bindings.
The basic idea using these algorithms is to compute a day number for the two dates and then just subtract them to give you the number of days.
--[[
http://howardhinnant.github.io/date_algorithms.html
Returns number of days since civil 1970-01-01. Negative values indicate
days prior to 1970-01-01.
Preconditions: y-m-d represents a date in the civil (Gregorian) calendar
m is in [1, 12]
d is in [1, last_day_of_month(y, m)]
y is "approximately" in
[numeric_limits<Int>::min()/366, numeric_limits<Int>::max()/366]
Exact range of validity is:
[civil_from_days(numeric_limits<Int>::min()),
civil_from_days(numeric_limits<Int>::max()-719468)]
]]
function days_from_civil(y, m, d)
if m <= 2 then
y = y - 1
m = m + 9
else
m = m - 3
end
local era = math.floor(y/400)
local yoe = y - era * 400 -- [0, 399]
local doy = math.modf((153*m + 2)/5) + d-1 -- [0, 365]
local doe = yoe * 365 + math.modf(yoe/4) - math.modf(yoe/100) + doy -- [0, 146096]
return era * 146097 + doe - 719468
end
local reference_date = {year=2001, month = 1, day = 1}
local date = os.date("*t")
local reference_days = days_from_civil(reference_date.year, reference_date.month, reference_date.day)
local days = days_from_civil(date.year, date.month, date.day)
print(string.format("Today is %d days into the 21st century.",days-reference_days))
os.time (under Windows, at least) is limited to years from 1970 and up. If, for example, you need a general solution to also find ages in days for people born before 1970, this won't work. You can use a julian date conversion and subtract between the two numbers (today and your target date).
A sample julian date function that will work for practically any date AD is given below (Lua v5.3 because of // but you could adapt to earlier versions):
local
function div(n,d)
local a, b = 1, 1
if n < 0 then a = -1 end
if d < 0 then b = -1 end
return a * b * (math.abs(n) // math.abs(d))
end
--------------------------------------------------------------------------------
-- Convert a YYMMDD date to Julian since 1/1/1900 (negative answer possible)
--------------------------------------------------------------------------------
function julian(year, month, day)
local temp
if (year < 0) or (month < 1) or (month > 12)
or (day < 1) or (day > 31) then
return
end
temp = div(month - 14, 12)
return (
day - 32075 +
div(1461 * (year + 4800 + temp), 4) +
div(367 * (month - 2 - temp * 12), 12) -
div(3 * div(year + 4900 + temp, 100), 4)
) - 2415021
end

Google Bookmark Export date format?

I been working on parsing out bookmarks from an export file generated by google bookmarks. This file contains the following date attributes:
ADD_DATE="1231721701079000"
ADD_DATE="1227217588219000"
These are not standard unix style timestamps. Can someone point me in the right direction here? I'll be parsing them using c# if you are feeling like really helping me out.
Chrome uses a modified form of the Windows Time format (“Windows epoch”) for its timestamps, both in the Bookmarks file and the history files. The Windows Time format is the number of 100ns-es since January 1, 1601. The Chrome format is the number of microseconds since the same date, and thus 1/10 as large.
To convert a Chrome timestamp to and from the Unix epoch, you must convert to seconds and compensate for the difference between the two base date-times (11644473600).
Here’s the conversion formulas for Unix, JavaScript (Unix in milliseconds), Windows, and Chrome timestamps (you can rearrange the +/× and -/÷, but you’ll lose a little precision):
u : Unix timestamp eg: 1378615325
j : JavaScript timestamp eg: 1378615325177
c : Chrome timestamp eg: 13902597987770000
w : Windows timestamp eg: 139025979877700000
u = (j / 1000)
u = (c - 116444736000000) / 10000000
u = (w - 1164447360000000) / 100000000
j = (u * 1000)
j = (c - 116444736000000) / 10000
j = (w - 1164447360000000) / 100000
c = (u * 10000000) + 116444736000000
c = (j * 10000) + 116444736000000
c = (w / 10)
w = (u * 100000000) + 1164447360000000
w = (j * 100000) + 1164447360000000
w = (c * 10)
Note that these are pretty big numbers, so you’ll need to use 64-bit numbers or else handle them as strings like with PHP’s BC-math module.
In Javascript the code will look like this
function chromeDtToDate(st_dt) {
var microseconds = parseInt(st_dt, 10);
var millis = microseconds / 1000;
var past = new Date(1601, 0, 1).getTime();
return new Date(past + millis);
}
1231721701079000 looks suspiciously like time since Jan 1st, 1970 in microseconds.
perl -wle 'print scalar gmtime(1231721701079000/1_000_000)'
Mon Jan 12 00:55:01 2009
I'd make some bookmarks at known times and try it out to confirm.
Eureka! I remembered having read the ADD_DATE’s meaning at some website, but until today, I could not find it again.
http://MSDN.Microsoft.com/en-us/library/aa753582(v=vs.85).aspx
offers this explanation as a “Note” just before the heading “Exports and Imports”:
“Throughout this file[-]format definition, {date} is a decimal integer that represents the number of seconds elapsed since midnight January 1, 1970.”
Before that, examples of {date} were shown:
<DT><H3 FOLDED ADD_DATE="{date}">{title}</H3>
…
and
<DT>{title}
…
Someday, I will write a VBA macro to convert these to recognizable dates, but not today!
If someone else writes a conversion script first, please share it. Thanks.
As of the newest Chrome Version 73.0.3683.86 (Official Build) (64-bit):
When I export bookmark, I got an html file like "bookmarks_3_22_19.html".
And each item has an 'add_date' field which contains date string. like this:
Stack Overflow
This timestamp is actually seconds (not microseconds) since Jan 1st, 1970. So we can parse it with Javascript like following code:
function ChromeTimeToDate(timestamp) {
var seconds = parseInt(timestamp, 10);
var dt = new Date();
dt.setTime(seconds * 1000);
return dt;
}
For the upper example link, we can call ChromeTimeToDate('1553220774') to get Date.
ChromeTimeToDate('1553220774')
12:09:03.263 Fri Mar 22 2019 10:12:54 GMT+0800 (Australian Western Standard Time)
Initially looking at it, it almost looks like if you chopped off the last 6 digits you'd get a reasonable Unix Date using the online converter
1231721701 = Mon, 12 Jan 2009 00:55:01 GMT
1227217588 = Thu, 20 Nov 2008 21:46:28 GMT
The extra 6 digits could be formatting related or some kind of extended attributes.
There is some sample code for the conversion of Unix Timestamps if that is in fact what it is.
look here for code samples: http://www.epochconverter.com/#code
// my groovy (java) code finally came out as:
def convertDate(def epoch)
{
long dv = epoch / 1000; // divide by 1,000 to avoid milliseconds
String dt = new java.text.SimpleDateFormat("dd/MMM/yyyy HH:mm:ss").format(new java.util.Date (dv));
// to get epoch date:
//long epoch = new java.text.SimpleDateFormat("MM/dd/yyyy HH:mm:ss").parse("01/01/1970 01:00:00").getTime() * 1000;
return dt;
} // end of def
So firefox bookmark date exported as json gave me:
json.lastModified :1366313580447014
convert from epoch date:18/Apr/2013 21:33:00
from :
println "convert from epoch date:"+convertDate(json.lastModified)
function ConvertToDateTime(srcChromeBookmarkDate) {
//Hp --> The base date which google chrome considers while adding bookmarks
var baseDate = new Date(1601, 0, 1);
//Hp --> Total number of seconds in a day.
var totalSecondsPerDay = 86400;
//Hp --> Read total number of days and seconds from source chrome bookmark date.
var quotient = Math.floor(srcChromeBookmarkDate / 1000000);
var totalNoOfDays = Math.floor(quotient / totalSecondsPerDay);
var totalNoOfSeconds = quotient % totalSecondsPerDay;
//Hp --> Add total number of days to base google chrome date.
var targetDate = new Date(baseDate.setDate(baseDate.getDate() + totalNoOfDays));
//Hp --> Add total number of seconds to target date.
return new Date(targetDate.setSeconds(targetDate.getSeconds() + totalNoOfSeconds));
}
var myDate = ConvertToDateTime(13236951113528894);
var alert(myDate);
//Thu Jun 18 2020 10:51:53 GMT+0100 (Irish Standard Time)
#Python program
import time
d = 1630352263 #for example put here, if (ADD_DATE="1630352263")
print(time.ctime(d)) #Mon Aug 30 22:37:43 2021 - you will see