How can I convert the number 1 into a date value 20150101 (yyyyMMdd)?

How can I convert A:

A = [ 1 2 3 4 5 ... 731]

into

[ 20150101 20150102 20150103 20150104 20150105 ... 20161231]

Answer 1

[Jan]

This works but is slow (see comments):

A   = [ 1 2 3 4 5 731];
num = datenum('31-Dec-2014') + A;
vec = sscanf(datestr(num, 'yyyymmdd ').', '%d')

or faster:

mat = datevec(datenum('31-Dec-2014') + A);
vec = mat * [10000; 100; 1; 0; 0; 0]

[EDITED]

dt = datetime(2015, 1, A, 'Format', 'uuuuMMdd')

>> 1×6 datetime array
  20150101   20150102   20150103   20150104   20150105   20161231

This looks similar to the wanted output, but it cannot be converted to a double vector directly.

[Jan]

Some interesting timings (note: I had a severe bug in the first version of these timings and deleted it):

A = 1:1e6;
n = 10;

% Conversion by SSCANF:
tic; for k = 1:1
   num = datenum('31-Dec-2014') + A;
   out = sscanf(datestr(num, 'yyyymmdd ').', '%d');
end, toc * n    % Too slow to wait

% Numerical conversion:
tic; for k = 1:n
  mat = datevec(datenum('31-Dec-2014') + A);
  out = mat * [10000; 100; 1; 0; 0; 0];
end, toc

% Undocumented internal MEX functions:
tic; for k = 1:n
  mat = datevecmx(datenummx(2014, 12, 31) + A);
  out = mat * [10000; 100; 1; 0; 0; 0];
end, toc

% Datetime object and DATEVEC (Andrei):
tic; for k = 1:n
  x   = datevec(datetime([2015 01 01]) + A -1);
  out = x(:,1:3) * [10000;100;1];
end, toc

% Datetime and year/month/day:
tic; for k = 1:n
   dt  = datetime([2015 01 01]) + A - 1;
   out = year(dt) * 10000 + month(dt) * 100 + day(dt);
end, toc

% yyyymmdd (Peter):
tic; for k = 1:n
   dt  = datetime([2015 01 01]) + A - 1;
   out = yyyymmdd(dt);
end, toc

% Create a datetime object only:
tic; for k = 1:n
  dt = datetime([2015 01 01]) + A - 1;   % Faster
end, toc

tic; for k = 1:n
   dt = datetime(2015, 1, A);            % Nicer
end, toc

 R2016b/64, Win7:

               382.38 seconds.   % sscanf(datestr(datenum()))
 Elapsed time is 2.13 seconds.   % datevec(datenum())
 Elapsed time is 2.11 seconds.   % MEX
 Elapsed time is 4.26 seconds.   % datevec(datetime)
 Elapsed time is 7.67 seconds.   % datetime -> year/month/day
 Elapsed time is 3.22 seconds.   % yyyymmdd

 % Creation of the datetime array, no conversion to output:
 Elapsed time is 0.56 seconds.   % datetime() + A
 Elapsed time is 1.55 seconds.   % datetime(A)

[André Gadêlha]

what is 'tic' and 'toc'?

[Guillaume]

@Jan,

"This looks similar to the wanted output, but it cannot be converted to a double vector directly."

Of course, it can:

dt = datetime(2015, 1, A, 'Format', 'uuuuMMdd')
str2double(cellstr(dt))

By the way, I would recommend spelling fully parameter names so that newbies don't wonder what the 'F' option is.

@André

In Matlab, you can type doc something or help something to learn about something. It's the fastest way to learn about functions you don't know about. So, try:

doc tic

[Jan]

@Guillaume: I meant directly. The format 'uuuuMMdd' does not convert the internal storage of the values, but concerns the output to strings only. Then cellstr(dt) is a complicated conversion already. str2double means some work also, because the conversion from a string to a double is surprisingly complicated to consider exceptions and care for replying the double which is as near as possible to the value represented in the string.

I do not know, how the date and time values are store internally: As serial date numbers or as date vectors. For the last case year(dt)*10000+... would be a "direct" conversion. But if date numbers are used, even cellstr(dt) means 2 conversions already to obtain the numerical values for year, month and day, and to create a string in the uuuuMMdd format.

Answer 2

[Peter Perkins]

There's a datetime method for that:

>> A = (1:731)';
>> d = datetime(2015,1,A)
d = 
  6×1 datetime array
   01-Jan-2015
   02-Jan-2015
[snip]
   30-Dec-2016
   31-Dec-2016
>> ymd = yyyymmdd(d);
ymd =
    20150101
    20150102
[snip]
    20161230
    20161231

But ask yourself why you want that. You are most likely better off sticking with d, the datetime vector, unless you need text, in which case ...

>> string(d,'yyyMMdd')
ans = 
  731×1 string array
    "20150101"
    "20150102"
    "20150103"
[snip]

... or unless you reaaly need those double values to hand off to some other function that only accepts dates in that form.

[Jan]

+1. yyyymmdd()? Wow, this does really match the OP's needs. I will include it in the runtime comparison later.

[Andrei Bobrov]

+1

Answer 3

[Andrei Bobrov]

out = datetime([2015 01 01],'F','uuuuMMdd') + A -1

or

x = datevec(datetime([2015 01 01]) + A -1);
out = x(:,1:3)*[10000;100;1];