Convert UTC timestamp to Simple binary encoded message - python-2.7

I am trying to convert UTC timestamp to simple binary encoded message.
Would like to achieve what is mentioned in example here.
Binary Encoding Example
The following timestamp:
UTC timestamp 14:17:22 Friday, October 4, 2024
is expressed in binary code (nanoseconds since Unix epoch) this way:
007420bf8838fb17 (8 bytes in nanoseconds since Unix epoch synced to a master clock to microsecond accuracy.
What I have done so far is,
import struct
from datetime import datetime
dt = datetime(2024, 10, 4, 14, 17, 22, 0)
timestamp = (dt - datetime(1970, 1, 1)).total_seconds() * 1000000000
utc_timestamp = struct.pack('d', timestamp)
Output I see on CLI is '\xf4\x02\x82\xa1E\xfb\xb7C' but, as per example in shared link, expected is 007420bf8838fb17

TL;DR: I think there's either an error in linked example, or the description of the representation is incorrect or incomplete.
import struct
from datetime import datetime, timedelta, timezone
utc_tz = timezone(timedelta(0))
t = datetime(2024, 10, 4, 14, 17, 22, 0, utc_tz)
nanos = int(t.timestamp()) * 1_000_000_000
print(hex(timestamp))
This gives the result of 0x17fb45a18202f400, not the quoted 0x17fb3888bf297400 (as presented, I think it's a little-endian byte string, so this value has the order of bytes reversed).
The quoted answer, converted to decimal seconds (dividing by 1_000_000_000) is 1728037042.0005898. That value obviously has a sub-second quantity, which the source datetime does NOT have.
Decoding the seconds component of the quoted answer:
import time
answer = 0x17fb3888bf297400
secs = int(answer / 1_000_000_000)
print(time.gmtime(secs))
gives:
time.struct_time(tm_year=2024, tm_mon=10, tm_mday=4, tm_hour=10, tm_min=17, tm_sec=22, tm_wday=4, tm_yday=278, tm_isdst=0)
which looks almost correct, save that the hour component is 4 hours earlier. So ... my guess is that this is actually a US/Eastern timestamp (not UTC), and contains a sub-second component of 5898ns, and the example is misleading.
If the schema you're using is encoding timestamps as a 64-bit number of nanoseconds since the Unix epoch (midnight 1 January 1970), then I think your example code has only one error: use format <q (or <Q) rather than d, and you'll get a little-endian 64-bit integer as required. You will also need to convert the timestamp to an integer to have struct.pack() accept it.

Related

Convert Unix Timestamp to Datetime in Power BI

Good morning Stack Overflow,
I have a question, I do have a list of Timestamp, I believe there are Unix Time Stamp, I'd like to convert to datetime on PBI, but I'm getting an error, using the following formula :' #datetime(1957,1,1,0,0,0) + #duration(0,0,0,[Timestamp]) ' , Timestamp is the column that I'd like to convert.
Below a sample list of the data :
Is there a way to fix this issue?
You're close but you need to remember that Unix timestamps are recorded in milliseconds since 1970, not seconds since 1957. Try this instead
#datetime(1970, 1, 1, 0, 0, 0) + #duration(0, 0, 0, [Timestamp]/1000)
Make sure the column is a datetime data type.

How to get Current Date in Ballerina

Is there a way to get the current date in ballerina?
As I was browsing through some code examples I came across the syntax to get the current time. Shown below is how to get the current date in Ballerina:
Note: first you have to import the time package given below for this to work.
import ballerina/time;
Then put the following lines of code:
time: Time currentTime = time:[currentTime][2]();
string customTimeString = currentTime.format("dd-MM-yyyy");
This will give the following output:
08-07-2018
This is work for ballerina 0.991 and 1.0 first you have to import the time package
Then it will give the current date if you want to get in a format it will included the code
import ballerina/time;
To get current time
time:Time time = time:currentTime();
string standardTimeString = time:toString(time);
io:println("Current system time in ISO format: ", standardTimeString);
To format the time
string|error customTimeString = time:format(time, "yyyy-MM-dd-E");
if (customTimeString is string) {
io:println("Current system time in custom format: ", customTimeString);
}
y -Years
M -months
d -date
E -day
h -hour
m -Minuit
s -seconds
For Swan Lake Update 3 they seem to have removed the time:currentTime() function.
It seems they have replaced it with time:utcNow().
According to the ballerina documentation,
"The time:Utc is the tuple representation of the UTC. The UTC represents the number of seconds from a specified epoch. Here, the epoch is the UNIX epoch of 1970-01-01T00:00:00Z."
So you can convert this above tuple representation to RFC 3339 timestamp by using,
time:Utc currTime = time:utcNow();
string date = time:utcToString(currTime);
io:println(date);
Then you will get a result like below,
2023-01-14T17:04:15.639510400Z
Using ballerina time library you can convert to other different representations as well.

Python - Timestamp Conversion

I have converted some sensor data into CSV format. The data contains a timestamp attribute that is divided into two parts as follows:
measurement_timestamp_begin:
seconds: 3733032665
fractions: 3056174174
I need to convert this into standard UNIX timestamp format. I saw several posts for doing the conversion but each method receives a single argument. I don't understand the fraction part. Does it mean seconds.fraction? e.g. 3733032665.3056174174.
Following code converts above mentioned timestamp format to standard UNIX timestamp format.
#!/usr/bin/env python
import datetime, time
unix_epoch = datetime.date(*time.gmtime(0)[0:3])
ntp_epoch = datetime.date(1900, 1, 1)
ntp_delta = (unix_epoch - ntp_epoch).days * 24 * 3600
def ntp_to_unix_time(date):
return (date - ntp_delta)
print datetime.datetime.fromtimestamp(int(ntp_to_unix_time(3733032665.3056174174))).strftime('%Y-%m-%d %H:%M:%S')

How do I convert an aware datetime to a POSIX timestamp?

I'd like to specify 12 PM on a particular date in the central timezone and adjust for daylight savings time (CDT). I'd then like to convert this to a POSIX timestamp.
The first thing I reached for is:
d = datetime.datetime(2017, 6, 27, 12, tzinfo=???)
But I don't have a concrete CDT class. pytz does however:
z = pytz.timezone('EST5EDT')
d = datetime.datetime(2017, 6, 27, 12, tzinfo=z)
But this does not work (pytz documentation says as much but I don't understand why the constructor cannot use the timezone information). If I use November (fall back) or June (spring forward) I get still get 12:00:00-05:00 as the time portion.
Even if this did work the method to convert to a POSIX timestamp assumes a naive datetime:
posix = time.mktime(d.timetuple())
This timestamp represents 12PM in my local time zone.
Then there is normalize() with code and examples that I find very difficult to follow:
au_dt = au_tz.normalize(utc_dt.astimezone(au_tz))
I also tried to subtract my aware time from an epoch defined at January 1, 1970 but that doesn't work unless the datetime gets the UTC offset correct (see my second point above).
Can anyone help me with a mental model of how this stuff works in general and a solution to this problem in particular?
This solution seems to work:
import pytz
import datetime
# Naive datetime (no timezone).
d = datetime.datetime(2017, 11, 27, 12)
cdt = pytz.timezone('US/Central')
# Give it a central time zone context.
d = cdt.localize(d)
# Determine the time in UTC.
d = d.astimezone(pytz.utc)
# Create a naive POSIX epoch.
epoch = datetime.datetime(1970, 1, 1)
# Give it context in the UTC time zone.
epoch = pytz.utc.localize(epoch)
# Number of seconds have elapsed since the epoch.
print (d - epoch).total_seconds()

Cannot parse datetime in pm format with booster, C++, graphlab

I've tried to convert datetime string into datetime of an SArray (uses C++ booster library), but it does not seem to understand the %p format specifier. http://www.boost.org/doc/libs/1_43_0/doc/html/date_time/date_time_io.html
This documentation says specifiers marked with ! do not currently work for input.
Does that mean that you cannot parse anything with pm or PM?
I was able to get the string-to-datetime conversion to work by making two small changes:
Use %I for the hour on a 12-hour clock (%H is for a 24 hour clock).
Use %P (upper case) for the AM/PM flag.
Here's what works for me:
sf = gl.SFrame({'date': ['2015-11-06 02:12:42 pm',
'2015-11-05 03:43:11 pm']})
sf['date2'] = sf['date'].str_to_datetime('%Y-%m-%d %I:%M:%S %p')