How to convert hours, minutes, seconds to seconds?

Hi,

How can I return an array with

hh:min:sec

into

seconds?

For example

VarName2(1:5)
ans = 
    '14:54:25'
    '14:54:25'
    '14:54:25'
    '14:54:26'
    '14:54:26'

into a array with seconds?

Answer 1

[Daniel]

Assuming you mean seconds in the day

    % initialize variables
    NUM_SECONDS_PER_DAY = 86400.0;
    timeStrings = {'14:54:25';'14:54:25';'14:54:25';'14:54:26';'14:54:26'};
    % convert times to fractional days using datenum
    timeFractionalDays = datenum(timeStrings);
    % leave only the part with the most recent day fraction
    timeDayFraction = mod(timeFractionalDays,1);
    % multiply by number of seconds in a day
    timeInSeconds = timeDayFraction .* NUM_SECONDS_PER_DAY

This produces the result

    timeInSeconds =

       1.0e+04 *

        5.3665
        5.3665
        5.3665
        5.3666
        5.3666

[Lizan]

The time string is actually stored in a big vector valled "VameName2". I just gave some numbers for example.

Obtained following error:

Error using datenum (line 179)
DATENUM failed.
Caused by:
    Error using datevec (line 104)
    The input to DATEVEC was not an array of strings.

When I wrote,

timeStrings = {VarName2};

[Daniel]

if VarName2 is a cell array of strings, try substituting VarName2 in for timeStrings like this:

    % initialize variables
    NUM_SECONDS_PER_DAY = 86400.0;
    % convert times to fractional days using datenum
    timeFractionalDays = datenum(VarName2);
    % leave only the part with the most recent day fraction
    timeDayFraction = mod(timeFractionalDays,1);
    % multiply by number of seconds in a day
    timeInSeconds = timeDayFraction .* NUM_SECONDS_PER_DAY

[Lizan]

I accepted your answer a bit to early, ..

I plotted the data and I am noticing that the time is starting from 980 min or so.. I expected it to start from zero.

[Daniel]

Is the first string in VarName2 00:00:00? That is the only time that would return a 0 value from this algorithm, since it is returns seconds in the day. What were you looking to get your answer in terms of?

[Lizan]

Well, the data is for example in format 14:54:25. I guess I would like to have the time in seconds from the first time entry.

so

    '14:54:25' - 0 sec
    '14:54:25' - 0 sec
    '14:54:25' - 1 sec
    '14:54:26' - 2 sec
    '14:54:26' - 2 sec
etc

[Mohammad Abouali]

14:54:25 is 53665 seconds

Subtract this number, i.e. 53665 from the calculation of Daniel or my method.

[Daniel]

Simply subtract the first value of the answer vector from itself.

    % initialize variables
    NUM_SECONDS_PER_DAY = 86400.0;
    % convert times to fractional days using datenum
    timeFractionalDays = datenum(VarName2);
    % leave only the part with the most recent day fraction
    timeDayFraction = mod(timeFractionalDays,1);
    % multiply by number of seconds in a day
    timeInSeconds = timeDayFraction .* NUM_SECONDS_PER_DAY;
    % find answer starting from first value
    timeInSeconds = timeInSeconds - timeInSeconds(1)

[Lizan]

Thanks!

Answer 2

[Stephen23]

The approach for MATLAB >=R2014b is to use the DURATION class, e.g.:

C = {'14:54:25'; '14:54:25'; '14:54:25'; '14:54:26'; '14:54:26'};
S = seconds(duration(C))

S = 5×1

53665 53665 53665 53666 53666