Datenum with a.m. and p.m.

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Brenton Hirao
Brenton Hirao 2019 年 11 月 15 日
コメント済み: Adam Danz 2019 年 11 月 15 日
I have a data set with timestamps:
'11/5/2019 12 a.m. EST'
'11/10/2019 1 p.m. EST'
How can I convert these timestamps into a vector of date numbers?
thanks for the help.
  4 件のコメント
Brenton Hirao
Brenton Hirao 2019 年 11 月 15 日
Walter Robinson, the periods are part of the text. The data only contains EST.
Adam Danz
Adam Danz 2019 年 11 月 15 日
So you're converting from strings to datenum?
It might be necessary to remove the dots first using strrep but maybe not (my machine has been shut down for the night).



Shubham Gupta
Shubham Gupta 2019 年 11 月 15 日
Assuming there is ".m. EST" attached to each "timestamp" at the end. You can follow the steps:
Step-1 Replace ".m. EST" to just "m" (Hint : regexprep)
Step-2 Convert the output into the date numbers (Hint: datenum)
You should get the output with date numbers in a array. (Not sure what do mean by "continuous"?)
  2 件のコメント
Walter Roberson
Walter Roberson 2019 年 11 月 15 日
S = '11/5/2019 12 a.m. EST';
temp = regexprep(S, '\.m\.', 'm');
t = datetime(temp, 'Inputformat', 'd/M/yyyy h a z', 'TimeZone', 'America/Chicago');
The above will have "eastern time" as the attached timezone. You can switch it to EST but you will get a warning,
Warning: 'EST' specifies a time zone with a fixed offset from UTC, -05:00. This zone does not follow daylight saving time, and so may give unexpected results. See the datetime.TimeZone property for details about specifying time zones.
Adam Danz
Adam Danz 2019 年 11 月 15 日
Instead of the regexprep() you could also do
temp = strrep(S,'.','');


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