从给出错误日期的 csv 文件中读取日期

Question

I have a csv file with a column of dates in the format dd/mm/yyyy HH:MM. The first entry has '28/06/2017 17:10'. Excel for some reason displays this as 28/06/2017 5:10:00 PM. I want to read this column in matlab and output it in the format dd-mmmm-yyyy HH:MM:SS. I get a strange date for the ouptut in each column, with the first entry being: '12-August-0033 17:10:00', but the time is fine.

The dates begin in column 2, row 2 of my csv file.

[num,txt,raw] = xlsread('winter_cruise_2017_10min_avg.csv');
formatIn = 'dd/mm/yyyy HH:MM';
A = datestr(raw(2:end,2),formatIn);
formatOut = 'dd-mmmm-yyyy HH:MM:SS';
B = datestr(A,formatOut);

Answer 1

For a winter_cruise_2017_10min_avg.csv containing:

28/06/2017 17:10,

You can use datetime to interpret it correctly:

%% Load data
[num,txt,raw] = xlsread('winter_cruise_2017_10min_avg.csv');

%% Interpret text:
formatIn = 'dd/MM/yyyy HH:mm:ss'; % Corrected!
A = datetime(raw, 'InputFormat', formatIn);
%{
A = 
  datetime
   28-Jun-2017 17:10:00
%}

%% Reformat datetime
formatOut = 'dd-mmmm-yyyy HH:MM:SS';
B = datestr(A,formatOut);
%{
B =
    '28-June-2017 17:10:00'
%}