Logical result false when comparing 2 identical times
조회 수: 2 (최근 30일)
이전 댓글 표시
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)
댓글 수: 6
bandari shravya
2019년 2월 27일
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
2019년 2월 27일
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.
채택된 답변
Peter Perkins
2018년 2월 7일
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.
댓글 수: 3
Walter Roberson
2018년 2월 8일
No, you need to read the SSSSS as well if you are dealing with text. However, you can
DateTimeArray = dateshift(DateTimeArray, 'start', 'second')
추가 답변 (0개)
참고 항목
카테고리
Help Center 및 File Exchange에서 Logical에 대해 자세히 알아보기
Community Treasure Hunt
Find the treasures in MATLAB Central and discover how the community can help you!
Start Hunting!