I want to get the number of days between two dates based on specific condition, here is the image illustration of what I am talking about:
I need to devise a formula to calculate number of days from dates (column C) it takes ID = 1 to reach from L1 to L2 , so ideally the output for ID = 1 should be:
L1 : 0
L2 : 2022-07-14 - 2022-07-06 = 8
Same for other ids (2,3). I am just a beginner trying to learn, so I apologize for my ordinary question. Thank you
DAYS will give you the day count between dates. try:
=DAYS(SINGLE(FILTER(C:C; B:B="L2"; A:A=1));
SINGLE(FILTER(C:C; B:B="L1"; A:A=1)))
=DAYS(SINGLE(FILTER(C:C; B:B="L2"; A:A=2));
SINGLE(FILTER(C:C; B:B="L1"; A:A=2)))
=DAYS(SINGLE(FILTER(C:C; B:B="L2"; A:A=3));
SINGLE(FILTER(C:C; B:B="L1"; A:A=3)))
Related
I have a list of date ranges for the past 8 quarters given by the below function
q) findLastYQuarters:{reverse("d"$(-3*til y)+m),'-1+"d"$(-3*-1+til y)+m:3 bar"m"$x}[currentDate;8]
q) findLastYQuarters
2020.01.01 2020.03.31
2020.04.01 2020.06.30
2020.07.01 2020.09.30
2020.10.01 2020.12.31
2021.01.01 2021.03.31
2021.04.01 2021.06.30
2021.07.01 2021.09.30
2021.10.01 2021.12.31
I need to produce a separate list that labels each item in this list by a specific format; the second list would need to be
1Q20,2Q20,3Q20,4Q20,1Q21,2Q21,3Q21,4Q21
This code needs to be able to run on it's own, so how can I take the first list as an input and produce the second list? I thought about casting the latter date in the range as a month and dividing it by 3 to get the quarter and extracting the year, but I couldn't figure out how to actually implement that. Any advice would be much appreciated!
I'm sure there are many ways to solve this, a function like f defined below would do the trick:
q)f:{`$string[1+mod[`month$d;12]%3],'"Q",/:string[`year$d:x[;0]][;2 3]}
q)lyq
2020.01.01 2020.03.31
2020.04.01 2020.06.30
2020.07.01 2020.09.30
2020.10.01 2020.12.31
2021.01.01 2021.03.31
2021.04.01 2021.06.30
2021.07.01 2021.09.30
2021.10.01 2021.12.31
q)f lyq
`1Q20`2Q20`3Q20`4Q20`1Q21`2Q21`3Q21`4Q21
Figured it out.
crop:findLastYQuarters;
crop[0]:crop[0][1];
crop[1]:crop[1][1];
crop[2]:crop[2][1];
crop[3]:crop[3][1];
crop[4]:crop[4][1];
crop[5]:crop[5][1];
crop[6]:crop[6][1];
crop[7]:crop[7][1];
labels:()
labelingFunc:{[r] temp:("." vs string["m"$r]); labels,((string(("J"$temp[1])%3)),"Q",(temp[0][2,3])};
leblingFunc each crop;
labels
I'm looking forward to assign a specific count of persons to a specific shift. For example I got six persons and three different shifts. Now I have to assign exact two persons to every shift. I tried something like this but..
NOTE: this won't work, so please edit as fast as possible to misslead people, I even removed the "." after it so nobody is copying it:
person(a)
person(b)
person(c)
person(d)
person(e)
person(f)
shift("mor")
shift("aft")
shift("nig")
shiftCount(2).
{ assign(P,S) : shift(S)} = 1 :- person(P).
% DO NOT COPY THIS! SEE RIGHT ANSWER DOWN BELOW
:- #count{P : assign(P,"mor")} = K, shiftCount(K).
:- #count{P : assign(P,"aft")} = K, shiftCount(K).
:- #count{P : assign(P,"nig")} = K, shiftCount(K).
#show assign/2.
Is this possible to count the number of assigned shifts, so I can assign exactly as many people as a given number?
The output of the code above (when the "." are inserted) is:
clingo version 5.5.0
Reading from stdin
Solving...
Answer: 1
assign(a,"nig") assign(b,"aft") assign(c,"mor") assign(d,"mor")
assign(e,"mor") assign(f,"mor")
SATISFIABLE
Models : 1+
Calls : 1
Time : 0.021s (Solving: 0.00s 1st Model: 0.00s Unsat: 0.00s)
CPU Time : 0.000s
Here you can defently see, that the morning ("mor") shift is used more than two times, as difined in the shiftCount. What do I need to change to get the wanted result?
Replace your 3 lines constraints with
{assign(P,S): person(P)} == K :- shift(S), shiftCount(K).
or alternatively if you want to use the constraint writing:
:- {assign(P,S): person(P)} != K, shift(S), shiftCount(K).
First line states: For a given shiftCount K and for every shift S: the number of assignments over all people P for this shift S is K.
The constraint reads: it can not be the case for a shiftCount K and a shift S that the number of assignments over all people P to the shift S is not K.
Please do not alter your question / sample code dramatically since this may leads to the case that this answer won't work anymore.
In SSRS,how to count the number of dates present in a column?
I am developing a report where I need to display the total number of dates where Date_of_Delivery.Value is updated in a specific month & also I need to display the same for where Date_of_Delivery.Value is Not updated.
Please insist me.
If you want a count of the number of times a date is in a certain time period, you would use the IFF to perform the check and then SUM the results.
=SUM(IIF(Fields!Date_of_Delivery.Value >= CDATE("01/01/2016") AND Fields!Date_of_Delivery.Value <= CDATE("01/31/2016"), 1, 0)
The IFF will check to see if the Date of Delivery is between two dates and return 1 if true otherwise 0. The SUM then sums up all the results.
You should probably use some Parameters for your date so you can just change the parameters instead of the code in the report.
=SUM(IIF(Fields!Date_of_Delivery.Value >= Parameters!START_DATE.Value AND Fields!Date_of_Delivery.Value <= Parameters!END_DATE.Value, 1, 0)
I have a .mat file that contains data from the years 2006-2100. Each year, there is a different number of lines. I need to count how many lines are 2006, how many are 2007, etc.
The set up, by column, is: Year, Month, Day, Lat, Long
I just want to count the number of rows containing the same Year entry and get an array back with an array containing that info.
I'm thinking a for or while loop should work, but I don't know how to right it.
If we assume your data are in a numeric matrix, you can just do:
num_lines2006 = sum(data(:,1)==2006);
data2006 = data(data(:,1)==2006),:);
If you want to add a column with number of rows for corresponding year, here is a solution with a loop:
for k=size(data,1):-1:1
num_year(k,1) = sum(data(:,1)==data(k,1));
end
data = [data num_year];
Here is a solution without loop:
[unq_year,~,idx] = unique(data(:,1),'stable');
num_year = grpstats(data(:,1),unq_year,#numel);
data = [data num_year(idx)];
To count numeric entries, you may want to use histc
years = unique(data(:,1);
counts = histc(data(:,1),years);
Since you just want to count the number of rows you could just write something simple like:
years = unique(data(:, 1));
counts = arrayfun(#(year) nnz(data(:, 1) == year), years);
years contains the unique years, and numRows the number of times they are found.
You could also use a one-liner inspired by Jonas' answer:
[counts, years] = hist(data(:,1), unique(data(:,1))');
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.