Logical result false when comparing 2 identical times
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Tiago Dias
am 7 Feb. 2018
Kommentiert: Walter Roberson
am 27 Feb. 2019
Hello, i want to compare times and something is wrong and i dont know why. Perhaps something is wrong in the Excel format of the data. but in matlab i think i placed the format of all the time equal.
clear;
clc;
close all;
tabela1 = readtable('galp.xlsx');
tabela2 = tabela1;
[n,m] = size(tabela1);
t_1 = datetime('01-01-2018 02:00','InputFormat','dd-MM-uuuu HH:mm');
t_2 = datetime('01-01-2018 06:00','InputFormat','dd-MM-uuuu HH:mm');
tempo_a_comparar = tabela2{121,1};
t_1
tempo_a_comparar
C = isequal(t_1,tempo_a_comparar)
6 Kommentare
bandari shravya
am 27 Feb. 2019
Getting error on comparing both datetimes
file2 = dir('matlabfiles');
filetest = struct2table(file2);
datemodified = datetime(filetest.date,'Format','dd/MMM/uuuu');
t1 = '08-Feb-2019 13:10:15';
to = datetime('26-Feb-2019','Format','dd/MMM/uuuu');
t2 = to ;
numfiles = length(file2);
for k = 1:numfiles
fprintf('%s ..',datemodified(k));
t2 = to ;
d = datemodified(k);
if isequal(d,t2)
fprintf('Done.\n');
else
fprintf('NOT .\n');
end
end
Walter Roberson
am 27 Feb. 2019
When you called datetime(filetest.date,'Format', 'dd/MMM/uuuu') then it converts the entire time and records it, and sets the format so that what is displayed for the time is in the format dd/MMM/uuuu . 'Format' of a datetime has to do with display, not with how it is used in computing or how the input is parsed. The parameter to control how a character vector is parsed is 'InputFormat', or sometimes 'ConvertFrom'.
You can use ymd() to extract the year, month, and day of a datetime, or you can use dateshift(d,'start','day') to refer to the beginning of the day.
Akzeptierte Antwort
Peter Perkins
am 7 Feb. 2018
The extension of the file you've posted does not match your code. The file you've posted appears to be a .xlsx file.
Add :ss.SSSSSSSSS to the end of all your formats. What you will see is this:
t_1 =
datetime
01/01/2018 02:00:00.000000000
tempo_a_comparar =
datetime
01/01/2018 02:00:00.000022949
ans =
-2.2949e-05
My initial reaction was to point out that Excel stores time in units of days since 1900. So in general, timestamps read from Excel are accurate only to about
>> eps(seconds(datetime - datetime(1900,0,0)))
ans =
4.7684e-07
seconds. But since the actual difference is 50 larger than that, I can only assume that what's in the .xlsx file is not what you think it is.
3 Kommentare
Walter Roberson
am 8 Feb. 2018
No, you need to read the SSSSS as well if you are dealing with text. However, you can
DateTimeArray = dateshift(DateTimeArray, 'start', 'second')
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