How would I perform logical comparisons with times given in the form of a string vector?

I am given a vector of time values in "military time." I need to run through the entire vector and determine how many times are within a certain time window. How would I do this? Here's what I've tried so far:

temptime = IE322data{:,3};
time = datenum(temptime);
for j = 1:length(time)
    if time(j) > datenum(6:00) & time(j) < datenum(14:00)
        morningcount = morningcount + 1;
    end
end

[Jan]

Please post a small example for the input data. What is "military time"?

Answer 1

[Jan]

Are you sure that this is working:

temptime = IE322data{:,3};

What is the contents of IE322data ? The command works only, if this variable is a {1 x n} row cell. Then this would be more clear:

temptime = IE322data{1,3};

If is a cell array, you need round parentheses:

temptime = IE322data(:, 3);

The command datenum(6:00) replies an empty matrix, because 6:00 is same as 6:0, which is the empty matrix also. Even with quotes, I guess the command does not reply, what you want:

datenum('6:00')
%  738157.25

So this is 6 o'clock today. Now it would be useful to know, what the contents of your input data is. Are you searching for times at the current day? If not, maybe you want the fractional part only?

temptime = IE322data(:, 3);
time     = datenum(temptime);
morningcount = sum(time > rem(datenum('6:00'), 1) & ...
                   time < rem(datenum('14:00'), 1));

Do you have a reason to work with the old datenum function instead of modern datetime objects?

Are you sure that "> 6:00" is wanted and not ">= 6:00"?

Answer 2

[Steven Lord]

% time = datenum(temptime);

Instead of converting the dates and times to serial date numbers, I recommend converting them to a datetime array.

s = ["2021-04-17 12:15"; "2021-04-17 19:18"; "2021-04-18 02:01"; "2021-04-18 11:59"]

s = 4×1 string array
    "2021-04-17 12:15"
    "2021-04-17 19:18"
    "2021-04-18 02:01"
    "2021-04-18 11:59"

t = datetime(s)

t = 4×1 datetime array
   17-Apr-2021 12:15:00
   17-Apr-2021 19:18:00
   18-Apr-2021 02:01:00
   18-Apr-2021 11:59:00

Now you can use timeofday (if you don't care about the date information) and compare that with a duration.

start = hours(6);
finish = hours(19);
inrange = (start <= timeofday(t)) & (timeofday(t) <= finish)

inrange = 4×1 logical array
   1
   0
   0
   1

Or if you had a timetable with these dates as the row times you could try a timerange.

Finally, if you want to bin the dates and times in t, you could do that with histcounts or histogram.

bins = datetime(2021, 4, 17):hours(6):datetime(2021, 4, 19);
histogram(t, bins)

[counts, binedges, binnumber] = histcounts(t, bins)

counts = 1×8

0 0 1 1 1 1 0 0

binedges = 1×9 datetime array
   17-Apr-2021 00:00:00   17-Apr-2021 06:00:00   17-Apr-2021 12:00:00   17-Apr-2021 18:00:00   18-Apr-2021 00:00:00   18-Apr-2021 06:00:00   18-Apr-2021 12:00:00   18-Apr-2021 18:00:00   19-Apr-2021 00:00:00

binnumber = 4×1

3 4 5 6

Element 2 of t is in the bin whose left edge is binedges(binnumber(2)).

fprintf('%s is in the range [%s, %s)\n', ...
    string(t(2)), ...
    string(binedges(binnumber(2))), ...
    string(binedges(binnumber(2)+1)))

17-Apr-2021 19:18:00 is in the range [17-Apr-2021 18:00:00, 18-Apr-2021)

[Jack Ramsey]

This would work out perfectly, except my data is separated in a pretty weird way. My data looks like this:

Column 2 is a datetime vector. However, column 3 is technically a string vector. Is there some kind of way for me to combine the two in a single datetime vector? My biggest problem is the fact that the time is in string form.