Regular Expression to timestamp

2019 7 24
2 个回答
回答已采纳
更新时间：2019 7 25
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I have a log file that gives me a timestamp with two formats:
when time is between [0h,1h[:
11:32.456 - MM:SS.zzz
when time is between [1h,1000h[:
22:45:45.234 - HH:MM:SS.zzz
So i created this regular expression:
(?<hour>\d+)?:?(?<minute>\d+):(?<second>\d+)\.(?<fraction>\d+)
The problem is that if the time is less than one hour ( [0h,1h[ - 11:32.456) the regular expression gives me this:
Hour:1
Minute: 1
seconds: 32
fraction: 456
I need the next result:
Hour: 0
Minute: 11
seconds: 32
fraction: 456
If the time is greater than 1 hour all goes fine.
How can i solve this problem?
Thanks.

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 采纳的回答

Star Strider
Star Strider 2019-7-24

0 个投票

I would just use the hour, minute and second functions.
Note that second will give you the fractions as well.
If you want to separate the fractions:
t1 = datetime('now','Format','dd-MMM-yyyy HH:mm:ss.SSS');
s = second(t1)
fraction = rem(s,1)

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Marco Silva
Marco Silva 2019-7-24
Thank you for your advice, however, when my timestamp has the format 11:32.456, I receive the error "Unable to convert '11:32.456' to datetime using the format 'HH:mm:ss.SSS'."
And i cant modify the timestamp because my log file has a several entries
Star Strider
Star Strider 2019-7-24
My pleasure.
Use a different input format for those times:
t = datetime('11:32.456', 'InputFormat','mm:ss.SSS');
or add hours:
intime = '11:32.456';
newtime = sprintf('00:%s', intime);
t = datetime(newtime, 'InputFormat','HH:mm:ss.SSS');
You can check to see what the format is with the number lf elements returned by strfind:
clns = strfind(timestr, ':')
where ‘timestr’ is the string you are using to create your datetime object. If there is only one element in the result, add the '00:' using the sprintf call first.
Marco Silva
Marco Silva 2019-7-25
Good trick, thank you!

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Stephen23
Stephen23 2019-7-25
编辑：Stephen23 2019-7-25

0 个投票

>> rgx = '(?<H>\d*)(?(1):)(?<M>\d+):(?<S>\d+)\.(?<F>\d+)';
>> regexp('11:32.456',rgx,'names')
ans =
H: ''
M: '11'
S: '32'
F: '456'
>> regexp('22:45:45.234',rgx,'names')
ans =
H: '22'
M: '45'
S: '45'
F: '234'

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