Datevec problem - not able to convert string to numeric array

Hi I have problem with datevec. I have vector like this. I am trying to parse data for obtaining second. But it always gives an error for me. I have my time vector in String Format in 2017a.

"2017-10-16T14:39:00.221Z"
"2017-10-16T14:39:02.221Z"
"2017-10-16T14:39:03.221Z"
"2017-10-16T14:39:04.221Z"
"2017-10-16T14:39:06.221Z"
"2017-10-16T14:39:08.221Z"
"2017-10-16T14:39:09.221Z"
"2017-10-16T14:39:10.221Z"
"2017-10-16T14:39:11.221Z"

The code that I write is

Time= strrep(Time, 'Z', ''); %find .XXXXZ replace to''
Time= strrep(Time, 'T', ' '); %find T replace to' '
formatIn = 'yyyy-mm-dd HH:MM:SS'
datevec(Time,formatIn)

The answer of matlab is

formatIn =

    'yyyy-mm-dd HH:MM:SS'

Error using datevec (line 103) The input to DATEVEC was not an array of character vectors. What Can I do???

Answer 1

[Andrei Bobrov]

A = {'2017-10-16T14:39:00.221Z'
'2017-10-16T14:39:02.221Z'
'2017-10-16T14:39:03.221Z'
'2017-10-16T14:39:04.221Z'
'2017-10-16T14:39:06.221Z'
'2017-10-16T14:39:08.221Z'
'2017-10-16T14:39:09.221Z'
'2017-10-16T14:39:10.221Z'
'2017-10-16T14:39:11.221Z'};
Time = strung(A);

dt = datetime(regexprep(A,{'T','Z'},{' ',''}),'I','uuuu-MM-dd HH:mm:ss.SSS');
out = datevec(dt);

[Stephen23]

datetime accepts literals, so regexprep is not required. Perhaps something like this (untested):

'uuuu-MM-dd''T''HH:mm:ss.SSS''Z'''

[Andrei Bobrov]

Thank you Stephen!

Answer 2

[Orkun OZENER]

Thank you for your answer It works. But what is I is doing at there..I did not find the meaning of it..Is it a format?

dt = datetime(regexprep(A,{'T','Z'},{' ',''}),'I','uuuu-MM-dd HH:mm:ss.SSS');

[Stephen23]

"I did not find the meaning of it..Is it a format?"

Did you read the datetime help? It explains what that is.

Answer 3

[Peter Perkins]

Stephen's mod of Andrei's answer looks right, but I would add that there is likely no need to call datevec. In other words, you probably don't want to convert text to a numeric array, you want to convert it to datetimes.