conversion of date from yyyymmddHHMMSS format to yymmddHHMMSS format.

Question

Remember Samu 2021-5-13

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此问题的直接链接

https://ww2.mathworks.cn/matlabcentral/answers/828983-conversion-of-date-from-yyyymmddhhmmss-format-to-yymmddhhmmss-format

编辑： per isakson 2021-5-13

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

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Answer 1

per isakson 2021-5-13

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此回答的直接链接

https://ww2.mathworks.cn/matlabcentral/answers/828983-conversion-of-date-from-yyyymmddhhmmss-format-to-yymmddhhmmss-format#answer_698793

编辑：per isakson 2021-5-13

在 MATLAB Online 中打开

vec = datevec( '2020,07,23,00,00,00', 'yyyy,mm,dd,HH,MM,SS' );
datestr( vec, 'yymmddHHMMSS' )
ans = '200723000000'

or did you mean

datestr( vec, 'yy,mm,dd,HH,MM,SS' )
ans = '20,07,23,00,00,00'

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Remember Samu 2021-5-13

hi Per,

so I ran the following code but my date vector is not correct, please see the following:

t = readtable('E:\Data\2020\B220_DT80\20200724T000000.txt');

t(1:8642,:);

height(t);

rng('default')

dates = t.Var1;

durations = t.Var2;

T = table(dates, durations);

fullDateTime = dateshift(T.dates,'start','day') + T.durations;

fullDateTime.Format = 'MM/dd/yyyy HH:mm:ss';

T.DateTime = fullDateTime;

T.DNI = t.Var4;

refdata = T(:,{'DateTime','DNI'});

timestampVec1 = datenum(datevec(T.DateTime))

and when i ran,

datevec(T.DateTime), the following came up it looks like its adding zeros, insted of (2020 07 23 08 52 20), its giving(2.0200 0.0070 0.0230 0.0080 0.0520 0.0202).

2.0200 0.0070 0.0230 0.0080 0.0520 0.0202

2.0200 0.0070 0.0230 0.0080 0.0520 0.0302

2.0200 0.0070 0.0230 0.0080 0.0520 0.0402

2.0200 0.0070 0.0230 0.0080 0.0520 0.0502

2.0200 0.0070 0.0230 0.0080 0.0530 0.0002

2.0200 0.0070 0.0230 0.0080 0.0530 0.0102

2.0200 0.0070 0.0230 0.0080 0.0530 0.0202

2.0200 0.0070 0.0230 0.0080 0.0530 0.0302

2.0200 0.0070 0.0230 0.0080 0.0530 0.0402

2.0200 0.0070 0.0230 0.0080 0.0530 0.0502

2.0200 0.0070 0.0230 0.0080 0.0540 0.0002

2.0200 0.0070 0.0230 0.0080 0.0540 0.0102

2.0200 0.0070 0.0230 0.0080 0.0540 0.0202

2.0200 0.0070 0.0230 0.0080 0.0540 0.0302

2.0200 0.0070 0.0230 0.0080 0.0540 0.0402

2.0200 0.0070 0.0230 0.0080 0.0540 0.0502

2.0200 0.0070 0.0230 0.0080 0.0550 0.0002

2.0200 0.0070 0.0230 0.0080 0.0550 0.0102

2.0200 0.0070 0.0230 0.0080 0.0550 0.0202

2.0200 0.0070 0.0230 0.0080 0.0550 0.0302

2.0200 0.0070 0.0230 0.0080 0.0550 0.0402

2.0200 0.0070 0.0230 0.0080 0.0550 0.0502

2.0200 0.0070 0.0230 0.0080 0.0560 0.0002

2.0200 0.0070 0.0230 0.0080 0.0560 0.0102

Remember Samu 2021-5-13

编辑：per isakson 2021-5-13

在 MATLAB Online 中打开

hi Per,

so I ran the following code but my date vector is not correct, please see the following:

t = readtable('E:\Data\2020\B220_DT80\20200724T000000.txt');
t(1:8642,:);
height(t);
rng('default') 
dates = t.Var1; 
durations = t.Var2; 
T = table(dates, durations);
fullDateTime = dateshift(T.dates,'start','day') + T.durations;
fullDateTime.Format = 'MM/dd/yyyy HH:mm:ss';
T.DateTime = fullDateTime;
T.DNI = t.Var4;
refdata = T(:,{'DateTime','DNI'});
timestampVec1 = datenum(datevec(T.DateTime))

and when i ran,

datevec(T.DateTime)

the following came up it looks like its adding zeros, insted of (2020 07 23 08 52 20), its giving(2.0200 0.0070 0.0230 0.0080 0.0520 0.0202).

0200    0.0070    0.0230    0.0080    0.0530    0.0002
0200    0.0070    0.0230    0.0080    0.0530    0.0102
0200    0.0070    0.0230    0.0080    0.0530    0.0202
0200    0.0070    0.0230    0.0080    0.0530    0.0302
0200    0.0070    0.0230    0.0080    0.0530    0.0402
0200    0.0070    0.0230    0.0080    0.0530    0.0502
0200    0.0070    0.0230    0.0080    0.0540    0.0002
0200    0.0070    0.0230    0.0080    0.0540    0.0102
0200    0.0070    0.0230    0.0080    0.0540    0.0202
0200    0.0070    0.0230    0.0080    0.0540    0.0302
0200    0.0070    0.0230    0.0080    0.0540    0.0402
0200    0.0070    0.0230    0.0080    0.0540    0.0502
0200    0.0070    0.0230    0.0080    0.0550    0.0002
0200    0.0070    0.0230    0.0080    0.0550    0.0102
0200    0.0070    0.0230    0.0080    0.0550    0.0202
0200    0.0070    0.0230    0.0080    0.0550    0.0302
0200    0.0070    0.0230    0.0080    0.0550    0.0402
0200    0.0070    0.0230    0.0080    0.0550    0.0502
0200    0.0070    0.0230    0.0080    0.0560    0.0002
0200    0.0070    0.0230    0.0080    0.0560    0.0102

per isakson 2021-5-13

编辑：per isakson 2021-5-13

在 MATLAB Online 中打开

20200724T000000.txt

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')
ans = 2×6
        2020           7          23           0           0           0
        2020           7          23           0           0          10

The value of T.DateTime is a character array, e.g.

T.DateTime(1:2,:)
ans = 2×12 char array
    '200723000000'
    '200723000010'

Wasn't there a string like 1.0e+4 above the output

format short
magic(4).*1e-6
ans = 4×4
	1.0e+-4 *

    0.1600    0.0200    0.0300    0.1300
    0.0500    0.1100    0.1000    0.0800
    0.0900    0.0700    0.0600    0.1200
    0.0400    0.1400    0.1500    0.0100

I fail to reproduce your output

datevec(T.DateTime(1:2,:))
Error using datevec (line 289)
Cannot parse date 200723000000.

Remember Samu 2021-5-13

Thank you so much Per, i needed to pass it in this format (datevec(T.DateTime(1:2,:),'yymmddHHMMSS')) and it worked!

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conversion of date from yyyymmddHHMMSS format to yymmddHHMMSS format.

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conversion of date from yyyymmddHHMMSS format to yymmddHHMMSS format.

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