How to identify the hourly/daily missing data points?

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Andi
Andi 2021-12-7
评论： Chunru 2021-12-7
Hi everyone,
My dataset consists of long time series (about 20 years, excel sheet) with years, month, day, hour, mint, second. Each row represent one event. Some time there is no occurance on an hour or a day or even a month. Manually, its very hard to check with hour has no event or which day has no event.
May someone suggest me how can i handle this problem.
output shuld be like in 7 column year, month, day, hour mint,. second, events
data=load('data.txt');
year=data(:,1);
month=data(:,2);
day=data(:,3);
hour=data(:,4);
minute=data(:,5);
second=data(:,6);
observation = datenum(year,month,day,hour,minute,second);

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Chunru
Chunru 2021-12-7
Ran in:
% Assume you have the observation as a datenum vector
% Here we generate an example datenum vector
observation = datenum(2021, 1, 1, [0:.5:3 5:1.2:8 8.1:0.1:9]', 0, 0); % YMDHMS
observation = sort(observation); % sort it if necessary
idx = find(diff(observation) > 1/24) % difference is greater than 1h
idx = 3×1
7 8 9
datestr(observation(idx))
ans = 3×20 char array
'01-Jan-2021 03:00:00' '01-Jan-2021 05:00:00' '01-Jan-2021 06:12:00'
datestr(observation(idx+1))
ans = 3×20 char array
'01-Jan-2021 05:00:00' '01-Jan-2021 06:12:00' '01-Jan-2021 07:24:00'
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Chunru
Chunru 2021-12-7
编辑：Chunru 2021-12-7
idx = find(diff(observation) > 1) % gaps more than 1 day
Chunru
Chunru 2021-12-7
find the missing day: convert the datenum to integer day by using floor. Then if difference between 2 adjacent integer day is greater than 1, there is data missing for that day.
idx = find(diff(floor(observation)) > 1)

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