How can I delete some specific rows from a matrix?

I have an easy question but somehow I am not being able to solve it.

Suppose I have a datetime matrix, A where

'2003-06-24'
'2003-07-10'
'2003-07-18'
'2003-07-26'
'2003-08-03'
'2003-08-11'
'2003-08-19'
'2003-08-27'

And I have another matrix B that contains every day/date from '2003-06-01' to '2003-09-01'.

Can anyone please tell me how can I filter out the rest of the dates from Matrix B so that only the dates from matrix A remain in B?

In another word, how can I make B just like A? (by removing all the unnecessary dates)

Any feedback from you will be greatly appreciated!

Answer 1

[Steven Lord]

Do you have a character matrix or do you have a datetime matrix?

C = ['2003-06-24'
'2003-07-10'
'2003-07-18'
'2003-07-26'
'2003-08-03'
'2003-08-11'
'2003-08-19'
'2003-08-27']

C = 8×10 char array
    '2003-06-24'
    '2003-07-10'
    '2003-07-18'
    '2003-07-26'
    '2003-08-03'
    '2003-08-11'
    '2003-08-19'
    '2003-08-27'

dt = datetime(C)

dt = 8×1 datetime array
   24-Jun-2003
   10-Jul-2003
   18-Jul-2003
   26-Jul-2003
   03-Aug-2003
   11-Aug-2003
   19-Aug-2003
   27-Aug-2003

I'll assume you're using the datetime matrix dt. Let's create B:

B = datetime('2003-06-01'):datetime('2003-09-01');

You can use the set functions like intersect, union, setxor, setdiff, etc. on datetime arrays.

BnotDt = setdiff(B, dt);
whos dt B BnotDt

  Name        Size            Bytes  Class       Attributes

  B           1x93              744  datetime              
  BnotDt      1x85              680  datetime              
  dt          8x1                64  datetime              

Element 24 of B is in dt:

B(24)

ans = datetime
   24-Jun-2003

dt(1)

ans = datetime
   24-Jun-2003

Element 24 of B is not in BnotDt. [It's not in a different location in BnotDt either, since BnotDt is sorted.]

BnotDt(23:25)

ans = 1×3 datetime array
   23-Jun-2003   25-Jun-2003   26-Jun-2003

issorted(BnotDt)

ans = logical
   1

[Ashfaq Ahmed]

Hi!

Thank you so much!

While this is a very intersting approach to solve the problem, my question is how do I know that the 24th element of B is in dt? Because I just posted an example case. In my original problem, there are more than 10,000 rows from 2001 to 2021. Can you kindly give me an idea on how to automate this finding?

Thank you again!

[Steven Lord]

Use the set functions.

C = ['2003-06-24'
'2003-07-10'
'2003-07-18'
'2003-07-26'
'2003-08-03'
'2003-08-11'
'2003-08-19'
'2003-08-27'];
dt = datetime(C);
B = datetime('2003-06-01'):datetime('2003-09-01');
[isItAMember, whereIsIt] = ismember(dt, B)

isItAMember = 8×1 logical array
   1
   1
   1
   1
   1
   1
   1
   1

whereIsIt = 8×1

24 40 48 56 64 72 80 88

Answer 2

[Peter Perkins]

Surely you don't just have a list of dates. Presumably you have data at each date. Use a timetable to store all that.

At that point, it becomes one line:

ttB = ttB(ttA.Time,:)