conversion of date from yyyymmddHHMMSS format to yymmddHHMMSS format.
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Hi all can anyone please help with the conversion of date from yyyymmddHHMMSS format to yymmddHHMMSS format.
for example I want to change column 1 in the following dataset: (e.g row 1 shoud be 200723000000)
'2020,07,23,00,00,00' -0.800000000000000
'2020,07,23,00,00,10' -0.800000000000000
'2020,07,23,00,00,20' -0.700000000000000
'2020,07,23,00,00,30' -0.700000000000000
'2020,07,23,00,00,40' -0.700000000000000
'2020,07,23,00,00,50' -0.800000000000000
'2020,07,23,00,01,00' -0.700000000000000
'2020,07,23,00,01,10' -0.700000000000000
'2020,07,23,00,01,20' -0.700000000000000
'2020,07,23,00,01,30' -0.600000000000000
'2020,07,23,00,01,40' -0.600000000000000
'2020,07,23,00,01,50' -0.600000000000000
'2020,07,23,00,02,00' -0.500000000000000
'2020,07,23,00,02,10' -0.500000000000000
'2020,07,23,00,02,20' -0.500000000000000
'2020,07,23,00,02,30' -0.600000000000000
'2020,07,23,00,02,40' -0.500000000000000
'2020,07,23,00,02,50' -0.500000000000000
'2020,07,23,00,03,00' -0.500000000000000
'2020,07,23,00,03,10' -0.500000000000000
'2020,07,23,00,03,20' -0.500000000000000
Risposte (1)
per isakson
il 13 Mag 2021
Modificato: per isakson
il 13 Mag 2021
vec = datevec( '2020,07,23,00,00,00', 'yyyy,mm,dd,HH,MM,SS' );
datestr( vec, 'yymmddHHMMSS' )
or did you mean
datestr( vec, 'yy,mm,dd,HH,MM,SS' )
7 Commenti
Remember Samu
il 13 Mag 2021
Remember Samu
il 13 Mag 2021
Remember Samu
il 13 Mag 2021
In this case you need to provide a format string. datevec() isn't needed
timestampVec1 = '200723000000';
timestampNum1 = datenum(datevec(timestampVec1,'yymmddHHMMSS'))
timestampNum2 = datenum(timestampVec1,'yymmddHHMMSS')
timestampVec1 is not defined in your script. The last row has little to do with the rest of your script. However,
timestampNum = datenum(datevec(timestampVec1));
reproduces the error you report.
Remember Samu
il 13 Mag 2021
Modificato: per isakson
il 13 Mag 2021
per isakson
il 13 Mag 2021
Modificato: per isakson
il 13 Mag 2021
First a couple of comments
- format your question and comments so that they are easier to read an copy
- keep test cases small and to the point, e.g. there is no need for a large text file
- read answers and comments carefully and tell if you don't understand
I added a format string (I'm om R2018b)
%%
t = readtable('20200724T000000.txt');
dates = t.Var1;
durations = t.Var2;
T = table(dates, durations);
fullDateTime = dateshift(T.dates,'start','day') + T.durations;
fullDateTime.Format = 'yyyyMMddHHmmSS';
T.DateTime = datestr(fullDateTime,'yymmddHHMMSS');
%%
datevec(T.DateTime(1:2,:),'yymmddHHMMSS')
The value of T.DateTime is a character array, e.g.
T.DateTime(1:2,:)
Wasn't there a string like 1.0e+4 above the output
format short
magic(4).*1e-6
I fail to reproduce your output
datevec(T.DateTime(1:2,:))
Remember Samu
il 13 Mag 2021
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