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how to get serial date numbers?

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UPT
UPT 2012 年 9 月 16 日
hello.i have a structure like this: data = struct('Date', {cell(2208,1)}, 'Hour', {zeros(2208,1)}, 'DryBulb', {zeros(2208,1)}, 'DewPnt', {zeros(2208,1)}, 'SYSLoad', {zeros(2208,1)}, 'NumDate', {zeros(2208,1)});.. 'Date' has 92 days and 'Hour' has 24 hours,from 1 to 24.how can i express 'Date' and 'Hour' in serial date numbers and to present these serial date numbers in 'NumDate'?thank you

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per isakson
per isakson 2012 年 9 月 16 日
編集済み: per isakson 2012 年 9 月 20 日
Hint:
sdn = bsxfun( @plus, [0:91], transpose([0:23]/24) );
datestr( sdn(1 ), 'yyyy-mm-dd HH:MM:SS' )
datestr( sdn(end), 'yyyy-mm-dd HH:MM:SS' )
data.NumDate = sdn(:);
.
--- Working code in response to comment ---
Try this
sd1 = datenum( ' 1 Jan 2004', 'dd mmm yyyy' );
sd2 = datenum( '31 Mar 2004', 'dd mmm yyyy' );
sdn = bsxfun( @plus, [sd1:1:sd2], transpose([0:23]/24) );
datestr( sdn( 1), 'yyyy-mm-dd HH:MM:SS' )
datestr( sdn(end), 'yyyy-mm-dd HH:MM:SS' )
This code avoids round-off errors in sdn(1,:)
>> sdn( 1, 1:5 )
ans =
731947 731948 731949 731950 731951
>> all( sdn(1,1:5) == [ 731947, 731948, 731949, 731950, 731951 ] )
ans =
1
Whether it is wise to rely on this for comparisons is questionable
And to make sdn a column vector
sdn = sdn(:);
  7 件のコメント
per isakson
per isakson 2012 年 9 月 19 日
編集済み: per isakson 2012 年 9 月 20 日
A hint is a hint nothing more. I'm convinced one should not use code, which one doesn't understand. Seriously, there is nothing in my hint that could possibly account for your dates.
See my answer above
UPT
UPT 2012 年 9 月 19 日
thank you very much per isakson..:)..i did a mistake with the first code but anyway, thank you again..:)

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