## How do I create a vector with the first day of each month?

on 27 Apr 2013

### John Doe (view profile)

How do I create a column vector with the first day of each month? Considering I have a list of days...

Jan Simon

### Jan Simon (view profile)

on 27 Apr 2013

Please explain what "I have a list of dates" exactly means. Guessing the type and size of the inputs would be an inefficient approach.

cfjunior

### cfjunior (view profile)

on 27 Apr 2013

a list of days would be a column vector that goes from 10/01/1998 till 06/31/1999 in numbers

Jan Simon

### Jan Simon (view profile)

on 27 Apr 2013

And example in valid Matlab syntax would be more helpful. The input is still not clear. Something like this would be much easier to understand:

```data = [datenum('10/01/1998'), datenum('06/31/1999')]
```

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### John Doe (view profile)

on 27 Apr 2013
Edited by John Doe

### John Doe (view profile)

on 27 Apr 2013

Not quite sure what you are asking for here.

I assume you have a vector with all days from Monday-Sunday, and you know which day the first day of the year is:

```weekdays = [{'Monday'},{'Tuesday'},{'Wednesday'},{'Thursday'},{'Friday'},{'Saturday'},{'Sunday'}];
```
```dayNumber = zeros(1,12);
dayNumber(1) = 2;       % 2013 started on a Tuesday.
numDays = [31 28 31 30 31 30 31 31 30 31 30 31];  % Not a leap year
```
```for i = 1:11
dayNumber(i+1) = mod((dayNumber(i)+days(i)-1),7)+1;
end
```
```for i = 1:12
dayName(i) = weekdays(dayNumber(i));
end
```

If you need to account for leap years: It is a leap year every fourth year, except years that can be divided by 100, except years that can be divided by 400.

A general approach would be something like this:

```year = 2013;
```
```year0 = 1900;
startDay = 1;  % Year 1900 started on a monday.
```
```if year > year0
for i = 1:(year-year0)
if mod((year0+i),4)~=0
startDay = startDay + 365;
elseif mod((year0+i),4)==0 && mod((year0+i),100)~=0
startDay = startDay + 366;
elseif mod((year0+i),4)==0 && mod((year0+i),100)==0 && mod((year0+i),400)~=0
startDay = startDay + 365;
elseif mod((year0+i),4)==1 && mod((year0+i),400)==0
startDay = startDay + 366;
end
end
end
dayNumber = mod(startDay,7)+1;
```

This gives dayNumber = 2, for year = 2013. Then continue with the code above. (numDays(2)=29, if leap year)

cfjunior

on 27 Apr 2013

thanks a lot!

Jan Simon

### Jan Simon (view profile)

on 27 Apr 2013

@Robert: Nicer and more efficient:

```weekdays = {'Monday','Tuesday','Wednesday','Thursday','Friday','Saturday','Sunday'};
```

### Jan Simon (view profile)

on 27 Apr 2013
```indata = datenum('10/01/1998'):datenum('06/31/1999');
dvec   = datevec(data);
duniq  = unique(dvec(:, 1:2), 'rows');
result = datenum(v(:,1), v(:,2), 1);
```

Now result contains a row vector of the serial date numbers of the first day of each month contained in the input data.

cfjunior

### cfjunior (view profile)

on 29 Apr 2013

Thanks a lot man!!

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