I'm trying to take out two separate years from a date table.
% Date table
Datez = [2001 2;2001 5;2001 9;2001 11;2002 3;2002 5;2002 7;2002 9;2002 11;...
2003 2;2003 4;2003 6;2003 8;2003 10;2003 12;2004 3;2004 5;2004 7;...
2004 9;2004 11; 2005 10;2005 12]
I want to take out all values as 1 or 0. I want the dates from 2001-11 to 2002-11 plus all values from 2004-11 to 2005-11.
In total I should get a new vector, called test:
test = [0;0;0;1;1;1;1;1;1;0;0;0;0;0;0;0;0;0;0;1;1;0] % final result
I tried these combinations, but I don't know how to combine these four statements into a vector that looks like "test" or if there are any better solutions?
xjcr = 1:length(Datez)
(Datez(xjcr,1) >= 2001 & Datez(xjcr,2) >= 11) % greater than 2001-11
(Datez(xjcr,1) <= 2002 & Datez(xjcr,2) <= 11) % smaller than 2002-11
(Datez(xjcr,1) >= 2004 & Datez(xjcr,2) >= 11) % greater than 2004-11
(Datez(xjcr,1) <= 2005 & Datez(xjcr,2) <= 11) % smaller than 2005-11
Any ideas are much appreciated, thanks in advance!
Your issue is that you do not want to filter on two items independently, years greater than 2001 and months greater than November. This would give you December 2001 but not January 2002. The solution I believe is to treat your two composite numbers as a single number so that the comparison operator can operate on them as a pair. Here is an easy method:
Datez2 = Datez(:,1)*100 + Datez(:,2);
test = (Datez2>=200111 & Datez2<=200211) | (Datez2>=200411 & Datez2<=200511)
Maybe multiplying by 12 and adding (month - 1) would be best depending on if you are building something that needs to be very robust or if you are just hacking something together.
Related
For a lecture exercise, I have to represent in Answer Set Programming (we use Clingo as interpreter) a the following integrity constraint:
"You have to plan the calendar of a Masterclass. Normally, the lectures are on Fridays (8 hours) and Saturday(4 or 5 hours). And the 7th and 16th week are full, which means the lectures goes from Monday to Friday, with 8 hours per day, and on Saturday, with 4 or 5 hours of lecture."
The basic settings for the problem are the following:
#const n_weeks = 2. % for now we limit the problem size to 2 weeks
#const n_days = 6. % days in a fullweek
week(1..n_weeks).
day(1..n_days).
hour(1..8). % from the 1st to the 8th hour of the day
% the second week is a fullweek (lectures from 1st to 8th hour from Monday to Friday)
fullweek(2).
% We number all the weekdays (mon-fri) (we need it for the saturday)
fullday(1..5).
% some professors just for test
prof("prof1").
prof("prof2").
prof("prof3").
prof("prof4").
% subj, total hours, prof
subject("subj1", 8, "prof1").
subject("subj2", 14, "prof2").
subject("subj3", 24, "prof3").
subject("subj4", 11, "prof1").
% The main predicate, to print out at the end.
0 {calendar(W, D, H, W*100+D*10+H, lecture(S, P))} 1 :- week(W), day(D), hour(H), subject(S, _, P).
Now, as mentioned above (the final line in bold), we have some problems with the following constraint:
"In this masterclass the hours of a lecture on Saturday can be 4 or 5."
For now, me and my colleagues represented this constraint like this:
% The Saturday has 4 or 5 hours of lecture
:- #count{I : calendar(W, D, _, I, lecture(_, _))} > 5, week(W), day(D), not fullday(D).
:- #count{I : calendar(W, D, _, I, lecture(_, _))} < 4, week(W), day(D), not fullday(D).
Is it the right way to represent constraint like this? There is a better approach?
I do not believe there is the "right way" to represent the constraint as long as it is technically correct. I suggest to consider the following points:
The way on how to express Saturday is complicated, i.e. you can replace the variable D by 6 and eliminate the predicates day and fullday.
I do not understand why you use "lecture(_, _)" instead of the underscore.
I am not sure why you use the variable I for counting and think you like to count the hours instead.
Maybe it make sense to use disjunction explicitly, i.e. use a predicate like "hours_on_sunday(H)" and write a rule that H must be 4 or 5.
I have some ECG data for a number of subjects. For each subject, I can export an excel file with the RR interval, Heart Rate and other measures. The problem is that I have a timestamp starting at the time of recording (in this case 11:22:3:00).
I need to compare the date with other subjects and I want to automate the procedure in Matlab.
I need to flexibly compare, for instance, the first 3 minutes of subjects in condition 1 with those of sbj in condition 2. Or minutes 4 to 8 of condition 1 and 2 and so forth. To do this, I am thinking that the best way is to shift the time vector for each subject so that it starts from 0.
There are a couple of problems to note: I CANNOT create just one vector for all subjects. This would be inaccurate because the heart measures are variable for each individual.
So, IN SHORT I need to shift the time vector for each participant so that it starts at 0 and increases exactly like the original one. So, in this example:
H: M: S: MS RR HR
11:22:03:000 0.809 74.1
11:22:03:092 0.803 74.7
11:22:03:895 0.768 78.1
11:22:04:663 0.732 81.9
11:22:05:395 0.715 83.9
11:22:06:110 0.693 86.5
11:22:06:803 0.705 85.1
11:22:07:508 0.706 84.9
11:22:08:214 0.749 80.1
11:22:08:963 0.762 78.7
11:22:09:725 0.766 78.3
would become:
00:00:00:0000
00:00:00:092
00:00:00:895
00:00:01:663
and so forth...
I would like to do it in Matlab...
P.S.
I was working around the idea of extracting the info in 4 different variables.
Then, I could subtract the values for each cell from the first cell.
For instance:
11-11 = 0; 22-22=0; 03-03=0; ms: keep the same value
Maybe this could kind of work, except that it wouldn't if I have a subject that started, say, at 11:55:05:00
Thank you all for any help.
Gluce
Basic timestamp normalization just subtracts the minimum (or first, assuming they're properly ordered) time from the rest.
With MATLAB's datetime object, this is just subtraction, which yields a duration object:
ts = ["11:22:03:000", "11:22:03:092", "11:22:03:895", "11:22:04:663"];
% Convert to datetime & normalize
t = datetime(ts, 'InputFormat', 'HH:mm:ss:SSS');
t.Format = 'HH:mm:ss:SSS';
nt = t - t(1);
% Reformat & display
nt.Format = 'hh:mm:ss.SSS';
Which returns:
>> nt
nt =
1×4 duration array
00:00:00.000 00:00:00.092 00:00:00.895 00:00:01.663
Alternatively, you can normalize the datetime array itself:
ts = ["11:22:03:000", "11:22:03:092", "11:22:03:895", "11:22:04:663"];
t = datetime(ts, 'InputFormat', 'HH:mm:ss:SSS');
t.Format = 'HH:mm:ss:SSS';
[h, m, s] = hms(t);
[t.Hour, t.Minute, t.Second] = deal(h - h(1), m - m(1), s - s(1));
Which returns the same:
>> t
t =
1×4 datetime array
00:00:00:000 00:00:00:092 00:00:00:895 00:00:01:663
I simply want to generate a series of dates 1 year apart from today.
I tried this
CurveLength=30;
t=zeros(CurveLength);
t(1)=datestr(today);
x=2:CurveLength-1;
t=addtodate(t(1),x,'year');
I am getting two errors so far?
??? In an assignment A(I) = B, the number of elements in B and
Which I am guessing is related to the fact that the date is a string, but when I modified the string to be the same length as the date dd-mmm-yyyy i.e. 11 letters I still get the same error.
Lsstly I get the error
??? Error using ==> addtodate at 45
Quantity must be a numeric scalar.
Which seems to suggest that the function can't be vectorised? If this is true is there anyway to tell in advance which functions can be vectorised and which can not?
To add n years to a date x, you do this:
y = addtodate(x, n, 'year');
However, addtodate requires the following:
x must be a scalar number, not a string.
n must be a scalar number, not a vector.
Hence the errors you get.
I suggest you use a loop to do this:
CurveLength = 30;
t = zeros(CurveLength, 1);
t(1) = today; % # Whatever today equals to...
for ii = 2:CurveLength
t(ii) = addtodate(t(1), ii - 1, 'year');
end
Now that you have all your date values, you can convert it to strings with:
datestr(t);
And here's a neat one-liner using arrayfun;
datestr(arrayfun(#(n)addtodate(today, n, 'year'), 0:CurveLength))
If you're sequence has a constant known start, you can use datenum in the following way:
t = datenum( startYear:endYear, 1, 1)
This works fine also with months, days, hours etc. as long as the sequence doesn't run into negative numbers (like 1:-1:-10). Then months and days behave in a non-standard way.
Here a solution without a loop (possibly faster):
CurveLength=30;
t=datevec(repmat(now(),CurveLength,1));
x=[0:CurveLength-1]';
t(:,1)=t(:,1)+x;
t=datestr(t)
datevec splits the date into six columns [year, month, day, hour, min, sec]. So if you want to change e.g. the year you can just add or subtract from it.
If you want to change the month just add to t(:,2). You can even add numbers > 12 to the month and it will increase the year and month correctly if you transfer it back to a datenum or datestr.
I have a dataset for which I have extracted the date at which an event occurred. The date is in the format of MMDDYY although MatLab does not show leading zeros so often it's MDDYY.
Is there a method to find the mean or median (I could use either) date? median works fine when there is an odd number of days but for even numbers I believe it is averaging the two middle ones which doesn't produce sensible values. I've been trying to convert the dates to a MatLab format with regexp and put it back together but I haven't gotten it to work. Thanks
dates=[32381 41081 40581 32381 32981 41081 40981 40581];
You can use datenum to convert dates to a serial date number (1 at 01/01/0000, 2 at 02/01/0000, 367 at 01/01/0001, etc.):
strDate='27112011';
numDate = datenum(strDate,'ddmmyyyy')
Any arithmetic operation can then be performed on these date numbers, like taking a mean or median:
mean(numDates)
median(numDates)
The only problem here, is that you don't have your dates in a string type, but as numbers. Luckily datenum also accepts numeric input, but you'll have to give the day, month and year separated in a vector:
numDate = datenum([year month day])
or as rows in a matrix if you have multiple timestamps.
So for your specified example data:
dates=[32381 41081 40581 32381 32981 41081 40981 40581];
years = mod(dates,100);
dates = (dates-years)./100;
days = mod(dates,100);
months = (dates-days)./100;
years = years + 1900; % set the years to the 20th century
numDates = datenum([years(:) months(:) days(:)]);
fprintf('The mean date is %s\n', datestr(mean(numDates)));
fprintf('The median date is %s\n', datestr(median(numDates)));
In this example I converted the resulting mean and median back to a readable date format using datestr, which takes the serial date number as input.
Try this:
dates=[32381 41081 40581 32381 32981 41081 40981 40581];
d=zeros(1,length(dates));
for i=1:length(dates)
d(i)=datenum(num2str(dates(i)),'ddmmyy');
end
m=mean(d);
m_str=datestr(m,'dd.mm.yy')
I hope this info to be useful, regards
Store the dates as YYMMDD, rather than as MMDDYY. This has the useful side effect that the numeric order of the dates is also the chronological order.
Here is the pseudo-code for a function that you could write.
foreach date:
year = date % 100
date = (date - year) / 100
day = date % 100
date = (date - day) / 100
month = date
newdate = year * 100 * 100 + month * 100 + day
end for
Once you have the dates in YYMMDD format, then find the median (numerically), and this is also the median chronologically.
You see above how to present dates as numbers.
I will add no your issue of finding median of the list. The default matlab median function will average the two middle values when there are an even number of values.
But you can do it yourself! Try this:
dates; % is your array of dates in numeric form
sdates = sort(dates);
mediandate = sdates(round((length(sdates)+1)/2));
the thing is that, the 1st number is already ORACLE LONG,
second one a Date (SQL DATE, no timestamp info extra), the last one being a Short value in the range 1000-100'000.
how can I create sort of hash value that will be unique for each combination optimally?
string concatenation and converting to long later:
I don't want this, for example.
Day Month
12 1 --> 121
1 12 --> 121
When you have a few numeric values and need to have a single "unique" (that is, statistically improbable duplicate) value out of them you can usually use a formula like:
h = (a*P1 + b)*P2 + c
where P1 and P2 are either well-chosen numbers (e.g. if you know 'a' is always in the 1-31 range, you can use P1=32) or, when you know nothing particular about the allowable ranges of a,b,c best approach is to have P1 and P2 as big prime numbers (they have the least chance to generate values that collide).
For an optimal solution the math is a bit more complex than that, but using prime numbers you can usually have a decent solution.
For example, Java implementation for .hashCode() for an array (or a String) is something like:
h = 0;
for (int i = 0; i < a.length; ++i)
h = h * 31 + a[i];
Even though personally, I would have chosen a prime bigger than 31 as values inside a String can easily collide, since a delta of 31 places can be quite common, e.g.:
"BB".hashCode() == "Aa".hashCode() == 2122
Your
12 1 --> 121
1 12 --> 121
problem is easily fixed by zero-padding your input numbers to the maximum width expected for each input field.
For example, if the first field can range from 0 to 10000 and the second field can range from 0 to 100, your example becomes:
00012 001 --> 00012001
00001 012 --> 00001012
In python, you can use this:
#pip install pairing
import pairing as pf
n = [12,6,20,19]
print(n)
key = pf.pair(pf.pair(n[0],n[1]),
pf.pair(n[2], n[3]))
print(key)
m = [pf.depair(pf.depair(key)[0]),
pf.depair(pf.depair(key)[1])]
print(m)
Output is:
[12, 6, 20, 19]
477575
[(12, 6), (20, 19)]