I have worked with a simple C program to find the Day for Given Date. For it, I have written a lot of lines to calculate the day and month and to find the kind of the given year. While Surfing I came to know about a single line code to find the day for the given date. The code is as below
( d += m < 3 ? y --: y- 2, 23 * m / 9 + d + 4 + y / 4 - y / 100 + y / 400) % 7 ;
// 0 - Sunday, 6 - saturday
It gave the correct answer for all inputs, but I couldn't understand the values used in this expression.
Why the sum of day and month is checked for less than 3.
Why the year is reduced by one and the condition fails it decreases the year by 2.
Why the numbers 3, 23 and 9 are used in this expression.
I have confused about the operator precedence on this statement. Can anyone explain how this works?
What I've found so far:
23 * m / 9 results in
1 2 3
2 5 2
3 7 3
4 10 2
5 12 3
6 15 2
7 17 3
8 20 3
9 23 2
10 25 3
11 28 2
12 30 3
This expression adds the days over 28 days of a month.
The expression y / 4 - y / 100 + y / 400 results in:
1995 483 0
1996 484 1
1997 484 1
1998 484 1
1999 484 1
2000 485 2
2001 485 2
with the result, adding one day every 4 years (except leap years)
Because every year with 365 days (mod 7 == 1) increments the weekday by 1, the years are added to the days.
The expression d + (m < 3 ? y --: y- 2) is for correcting the leap year calculation. If we have a leap year, we can correct by one day only if we have a month >= march.
Related
I've got a dataset that has id, start date and a claim value (in dollars) in each row - most ids have more than one row - some span over 50 rows. The earliest date for each ID/claim varies, and the claim values are mostly different.
I'd like to do a rolling sum of the value of IDs that have claims within 365 days of each other, to report each ID that has claims that have exceeded a limiting value across each period. So for an ID that had a claim date on 1 January, I'd sum all claims to 31 December (inclusive). Most IDs have several years of data so for the example above, I'd also need to check that if they had a claim on 1 May that they hadn't exceeded the limit by 30 April the following year and so on. I normally see this referred to as a 'rolling sum'. My site has many SAS products including base, stat, ets, and others.
I'm currently testing code on a small mock dataet and so far I've converted a thin file to a fat file with one column for each claim value and each date of the claim. The mock dataset is similar to the client dataset that I'll be using. Here's what I've done so far (noting that the mock data uses days rather than dates - I'm not at the stage where I want to test on real data yet).
data original_data;
input ppt $1. day claim;
datalines;
a 1 7
a 2 12
a 4 12
a 6 18
a 7 11
a 8 10
a 9 14
a 10 17
b 1 27
b 2 12
b 3 14
b 4 12
b 6 18
b 7 11
b 8 10
b 9 14
b 10 17
c 4 2
c 6 4
c 8 8
;
run;
proc sql;
create table ppt_counts as
select ppt, count(*) as ppts
from work.original_data
group by ppt;
select cats('value_', max(ppts) ) into :cats
from work.ppt_counts;
select cats('dates_',max(ppts)) into :cnts
from work.ppt_counts;
quit;
%put &cats;
%put &cnts;
data flipped;
set original_data;
by ppt;
array vars(*) value_1 -&cats.;
array dates(*) dates_1 - &cnts.;
array m_vars value_1 - &cats.;
array m_dates dates_1 - &cnts.;
if first.ppt then do;
i=1;
do over m_vars;
m_vars="";
end;
do over m_dates;
m_dates="";
end;
end;
if first.ppt then do:
i=1;
vars(i) = claim;
dates(i)=day;
if last.ppt then output;
i+1;
retain value_1 - &cats dates_1 - &cnts. 0.;
run;
data output;
set work.flipped;
max_date =max(of dates_1 - &cnts.);
max_value =max(of value_1 - &cats.);
run;
This doesn't give me even close to what I need - not sure how to structure code to make this correct.
What I need to end up with is one row per time that an ID exceeds the yearly limit of claim value (say in the mock data if a claim exceeds 75 across a seven day period), and to include the sum of the claims. So it's likely that there may be multiple lines per ID and the claims from one row may also be included in the claims for the same ID on another row.
type of output:
ID sum of claims
a $85
a $90
b $80
On separate rows.
Any help appreciated.
Thanks
If you need to perform a rolling sum, you can do this with proc expand. The code below will perform a rolling sum of 5 days for each group. First, expand your data to fill in any missing gaps:
proc expand data = original_data
out = original_data_expanded
from = day;
by ppt;
id day;
convert claim / method=none;
run;
Any days with gaps will have missing value of claim. Now we can calculate a moving sum and ignore those missing days when performing the moving sum:
proc expand data = original_data
out = want(where=(NOT missing(claim)));
by ppt;
id day;
convert claim = rolling_sum / transform=(movsum 5) method=none;
run;
Output:
ppt day rolling_sum claim
a 1 7 7
a 2 19 12
a 4 31 12
a 6 42 18
a 7 41 11
...
b 9 53 14
b 10 70 17
c 4 2 2
c 6 6 4
c 8 14 8
The reason we use two proc expand statements is because the rolling sum is calculated before the days are expanded. We need the rolling sum to occur after the expansion. You can test this by running the above code all in a single statement:
/* Performs moving sum, then expands */
proc expand data = original_data
out = test
from = day;
by ppt;
id day;
convert claim = rolling_sum / transform=(movsum 5) method=none;
run;
Use a SQL self join with the dates being within 365 days of itself. This is time/resource intensive if you have a very large data set.
Assuming you have a date variable, the intnx is probably the better way to calculate the date interval than 365 depending on how you want to account for leap years.
If you have a claim id to group on, that would also be better than using the group by clause in this example.
data have;
input ppt $1. day claim;
datalines;
a 1 7
a 2 12
a 4 12
a 6 18
a 7 11
a 8 10
a 9 14
a 10 17
b 1 27
b 2 12
b 3 14
b 4 12
b 6 18
b 7 11
b 8 10
b 9 14
b 10 17
c 4 2
c 6 4
c 8 8
;
run;
proc sql;
create table want as
select a.*, sum(b.claim) as total_claim
from have as a
left join have as b
on a.ppt=b.ppt and
b.day between a.day and a.day+365
group by 1, 2, 3;
/*b.day between a.day and intnx('year', a.day, 1, 's')*/;
quit;
Assuming that you have only one claim per day you could just use a circular array to keep track of the pervious N days of claims to generate the rolling sum. By circular array I mean one where the indexes wrap around back to the beginning when you increment past the end. You can use the MOD() function to convert any integer into an index into the array.
Then to get the running sum just add all of the elements in the array.
Add an extra DO loop to zero out the days skipped when there are days with no claims.
%let N=5;
data want;
set original_data;
by ppt ;
array claims[0:%eval(&n-1)] _temporary_;
lagday=lag(day);
if first.ppt then call missing(of lagday claims[*]);
do index=max(sum(lagday,1),day-&n+1) to day-1;
claims[mod(index,&n)]=0;
end;
claims[mod(day,&n)]=claim;
running_sum=sum(of claims[*]);
drop index lagday ;
run;
Results:
running_
OBS ppt day claim sum
1 a 1 7 7
2 a 2 12 19
3 a 4 12 31
4 a 6 18 42
5 a 7 11 41
6 a 8 10 51
7 a 9 14 53
8 a 10 17 70
9 b 1 27 27
10 b 2 12 39
11 b 3 14 53
12 b 4 12 65
13 b 6 18 56
14 b 7 11 55
15 b 8 10 51
16 b 9 14 53
17 b 10 17 70
18 c 4 2 2
19 c 6 4 6
20 c 8 8 14
Working in a known domain of date integers, you can use a single large array to store the claims at each date and slice out the 365 days to be summed. The bookkeeping needed for the modular approach is not needed.
Example:
data have;
call streaminit(20230202);
do id = 1 to 10;
do date = '01jan2012'd to '02feb2023'd;
date + rand('integer', 25);
claim = rand('integer', 5, 100);
output;
end;
end;
format date yymmdd10.;
run;
options fullstimer;
data want;
set have;
by id;
array claims(100000) _temporary_;
array slice (365) _temporary_;
if first.id then call missing(of claims(*));
claims(date) = claim;
call pokelong(
peekclong(
addrlong (claims(date-365))
, 8*365)
,
addrlong(slice(1))
);
rolling_sum_365 = sum(of slice(*));
if dif1(claim) < 365 then
claims_out_365 = lag(claim) - dif1(rolling_sum_365);
if first.id then claims_out_365 = .;
run;
Note: SAS Date 100,000 is 16OCT2233
I am trying to group by dataset in three month groups, or quarters, but as I'm starting from an arbitrary date, I cannot use the quarter function in sas.
Example data below of what I have and quarter is the column I need to create in SAS.
The start date is always the same, so my initial quarter will be 3rd Sep 2018 - 3rd Dec 2018 and any active date falling in that quarter will be 1, then quarter 2 will be 3rd Dec 2018 - 3rd Mar 2019 and so on. This cannot be coded manually as the start date will change depending on the data, and the number of quarters could be up to 20+.
The code I have attempted so far is below
data test_Data_op;
set test_data end=eof;
%let j = 0;
%let start_date = start_Date;
if &start_Date. <= effective_dt < (&start_date. + 90) then quarter = &j.+1;
run;
This works and gives the first quarter correctly, but I can't figure out how to loop this for every following quarter? Any help will be greatly appreciated!
No need for a DO loop if you already have the start_date and actual event dates. Just count the number of months and divide by three. Use the continuous method of the INTCK() function to handle start dates that are not the first day of a month.
month_number=intck('month',&start_date,mydate,'cont')+1;
qtr_number=floor((month_number-1)/3)+1;
Based on the comment by #Lee. Edited to match the data from the screenshot.
The example shows that May 11 would be in the 3rd quarter since the seed date is September 3.
data have;
input mydate :yymmdd10.;
format mydate yymmddd10.;
datalines;
2018-09-13
2018-12-12
2019-05-11
;
run;
%let start_date='03sep2018'd;
data want;
set have;
quarter=floor(mod((yrdif(&start_date,mydate)*4),4))+1;
run;
If you want the number of quarters to extend beyond 4 (e.g. September 4, 2019 would be in quarter 5 rather than cycle back to 1), then remove the "mod" from the function:
quarter=floor(yrdif(&start_date,mydate)*4)+1;
The traditional use of quarter means a 3 month time period relative to Jan 1. Make sure your audience understands the phrase quarter in your data presentation actually means 3 months relative to some arbitrary starting point.
The funky quarter can be functionally computed from a months apart derived using a mix of INTCK for the baseline months computation and a logical expression for adjusting with relation to the day of the month of the start date. No loops required.
For example:
data have;
do startDate = '11feb2019'd ;
do effectiveDate = startDate to startDate + 21*90;
output;
end;
end;
format startDate effectiveDate yymmdd10.;
run;
data want;
set have;
qtr = 1
+ floor(
( intck ('month', startDate, effectiveDate)
-
(day(effectiveDate) < day(startDate))
)
/ 3
);
format qtr 4.;
run;
Extra
Comparing my method (qtr) to #Tom (qtr_number) for a range of startDates:
data have;
retain seq 0;
do startDate = '01jan1999'd to '15jan2001'd;
seq + 1;
do effectiveDate = startDate to startDate + 21*90;
output;
end;
end;
format startDate effectiveDate yymmdd10.;
run;
data want;
set have;
qtr = 1
+ floor( ( intck ('month', startDate, effectiveDate)
- (day(effectiveDate) < day(startDate))
) / 3 );
month_number=intck('month',startDate,effectiveDate,'cont')+1;
qtr_number=floor((month_number-1)/3)+1;
format qtr: month: 4.;
run;
options nocenter nodate nonumber;title;
ods listing;
proc print data=want;
where qtr ne qtr_number;
run;
dm 'output';
-------- OUTPUT ---------
effective month_ qtr_
Obs seq startDate Date qtr number number
56820 31 1999-01-31 1999-04-30 1 4 2
57186 31 1999-01-31 2000-04-30 5 16 6
57551 31 1999-01-31 2001-04-30 9 28 10
57916 31 1999-01-31 2002-04-30 13 40 14
58281 31 1999-01-31 2003-04-30 17 52 18
168391 90 1999-03-31 1999-06-30 1 4 2
168483 90 1999-03-31 1999-09-30 2 7 3
168757 90 1999-03-31 2000-06-30 5 16 6
168849 90 1999-03-31 2000-09-30 6 19 7
169122 90 1999-03-31 2001-06-30 9 28 10
169214 90 1999-03-31 2001-09-30 10 31 11
169487 90 1999-03-31 2002-06-30 13 40 14
169579 90 1999-03-31 2002-09-30 14 43 15
169852 90 1999-03-31 2003-06-30 17 52 18
169944 90 1999-03-31 2003-09-30 18 55 19
280510 149 1999-05-29 2001-02-28 7 22 8
280875 149 1999-05-29 2002-02-28 11 34 12
281240 149 1999-05-29 2003-02-28 15 46 16
282035 150 1999-05-30 2000-02-29 3 10 4
282400 150 1999-05-30 2001-02-28 7 22 8
282765 150 1999-05-30 2002-02-28 11 34 12
I was working on a homework problem regarding using arrays and looping to create a new variable to identify the date of when the maximum blood lead value was obtained but got stuck. For context, here is the homework problem:
In 1990 a study was done on the blood lead levels of children in Boston. The following variables for twenty-five children from the study have been entered on multiple lines per subject in the file lead_sum2018.txt in a list format:
Line 1
ID Number (numeric, values 1-25)
Date of Birth (mmddyy8. format)
Day of Blood Sample 1 (numeric, initial possible range: -9 to 31)
Month of Blood Sample 1 (numeric, initial possible range: -9 to 12)
Line 2
ID Number (numeric, values 1-25)
Day of Blood Sample 2 (numeric, initial possible range: -9 to 31)
Month of Blood Sample 2 (numeric, initial possible range: -9 to 12)
Line 3
ID Number (numeric, values 1-25)
Day of Blood Sample 3 (numeric, initial possible range: -9 to 31)
Month of Blood Sample 3 (numeric, initial possible range: -9 to 12)
Line 4
ID Number (numeric, values 1-25)
Blood Lead Level Sample 1 (numeric, possible range: 0.01 – 20.00)
Blood Lead Level Sample 2 (numeric, possible range: 0.01 – 20.00)
Blood Lead Level Sample 3 (numeric, possible range: 0.01 – 20.00)
Sex (character, ‘M’ or ‘F’)
All blood samples were drawn in 1990. However, during data entry the order of blood samples was scrambled so that the first blood sample in the data file (blood sample 1) may not correspond to the first blood sample taken on a subject, it could be the first, second or third. In addition, some of the months and days and days of blood sampling were not written on the forms. At data entry, missing month and missing day values were each coded as -9.
The team of investigators for this project has made the following decisions regarding the missing values. Any missing days are to set equal to 15, any missing months are to be set equal to 6. Any analyses that are done on this data set need to follow those decisions. Be sure to implement the SAS syntax as indicated for each question. For example, use SAS arrays and loops if the item states that these must be used.
Here is the data that the HW references (it is in list format and was contained in a separate file called lead_sum2018.txt):
1 04/30/78 6 10
1 -9 7
1 14 1
1 1.62 1.35 1.47 F
2 05/19/79 27 11
2 20 -9
2 5 6
2 1.71 1.31 1.76 F
3 01/03/80 11 7
3 6 6
3 27 2
3 3.24 3.4 3.83 M
4 08/01/80 5 12
4 28 -9
4 3 4
4 3.1 3.69 3.27 M
5 12/26/80 21 5
5 3 7
5 -9 12
5 4.35 4.79 5.14 M
6 06/20/81 7 10
6 11 3
6 22 1
6 1.24 1.16 0.71 F
7 06/22/81 19 6
7 3 12
7 29 8
7 3.1 3.21 3.58 F
8 05/24/82 26 7
8 31 1
8 9 10
8 2.99 2.37 2.4 M
9 10/11/82 2 7
9 25 5
9 28 3
9 2.4 1.96 2.71 F
10 . 10 8
10 30 12
10 28 2
10 2.72 2.87 1.97 F
11 11/16/83 19 4
11 15 11
11 7 -9
11 4.8 4.5 4.96 M
12 03/02/84 17 6
12 11 2
12 17 11
12 2.38 2.6 2.88 F
13 04/19/84 2 12
13 -9 6
13 1 7
13 1.99 1.20 1.21 M
14 02/07/85 4 5
14 17 5
14 21 11
14 1.61 1.93 2.32 F
15 07/06/85 5 2
15 16 1
15 14 6
15 3.93 4 4.08 M
16 09/10/85 12 10
16 11 -9
16 23 6
16 3.29 2.88 2.97 M
17 11/05/85 12 7
17 18 1
17 11 11
17 1.31 0.98 1.04 F
18 12/07/85 16 2
18 18 4
18 -9 6
18 2.56 2.78 2.88 M
19 03/02/86 19 4
19 11 3
19 19 2
19 0.79 0.68 0.72 M
20 08/19/86 21 5
20 15 12
20 -9 4
20 0.66 1.15 1.42 F
21 02/22/87 16 12
21 17 9
21 13 4
21 2.92 3.27 3.23 M
22 10/11/87 7 6
22 1 12
22 -9 3
22 1.43 1.42 1.78 F
23 05/12/88 12 2
23 21 4
23 17 12
23 0.55 0.89 1.38 M
24 08/07/88 17 6
24 27 11
24 6 2
24 0.31 0.42 0.15 F
25 01/12/89 4 7
25 15 -9
25 23 1
25 1.69 1.58 1.53 M
A) Input the data and in the data step:
1) make sure that Date of Birth variable is recorded as a SAS date;
2) use SAS arrays and looping to create a SAS date variable for each of the three blood samples and to address the missing data in accordance to the decisions of the investigators. Hint: use a single array and do loop to recode the missing values for day and month, separately, and an array/do loop for creating the SAS date variable;
3) use a SAS function to create a variable for the highest, i.e., maximum, blood lead value for each child;
4) use SAS arrays and looping to identify the date on which this largest value was obtained and create a new variable for the date of the largest blood lead value;
5) determine the age of the child in years when the largest blood lead value was obtained (rounded to two decimal places);
6) create a new variable based on the age of the child in years when the largest lead value was obtained (call it, “agecat”) that takes on three levels: for children less than 4 years old, agecat should equal 1; for children at least 4 years old, but less than 8, agecat should equal 2; and for children at least 8 years of age, agecat should be 3.;
7) print out the variables for the date of birth, date of the largest lead level, age at blood sample for the largest blood lead level, agecat, sex, and the largest blood lead level (Only print out these requested variables). All dates should be formatted to use the mmddyy10. format on the output.
The code I used in response to this was:
libname HW3 'C:\Users\johns\Desktop\SAS';
filename HW3new 'C:\Users\johns\Desktop\SAS\lead_sum2018.txt';
data one;
infile HW3new;
informat dob mmddyy8.;
input #1 id dob dbs1 mbs1
#2 dbs2 mbs2
#3 dbs3 mbs3
#4 bls1 bls2 bls3 sex;
array dbs{3} dbs1 dbs2 dbs3;
array mbs{3} mbs1 mbs2 mbs3;
do i=1 to 3;
if dbs{i}=-9 then dbs{i}=15;
end;
do i=4 to 6;
if mbs{i}=-9 then mbs{i}=6;
end;
array date{3} mdy1 mdy2 mdy3;
do i=1 to 3;
date{i}=mdy(mbs{i}, dbs{i}, 1990);
end;
maxbls=max(of bls1-bls3);
array bls{3} bls1 bls2 bls3;
array maxdte{3} maxdte1 maxdte2 maxdte3;
do i=1 to i=3;
if bls{i}=maxbls then maxdte=i;
end;
agemax=maxdte-dob;
ageest=round(agemax/365.25,2);
if agemax=. then agecat=.;
else if agemax < 4 then agecat=1;
else if 4 <= agemax < 8 then agecat=2;
else if agemax ge 8 then agecat=3;
run;
I received this error:
22 maxbls=max(of bls1-bls3);
23 array bls{3} bls1 bls2 bls3;
24 array maxdte{3} maxdte1 maxdte2 maxdte3;
25 do i=1 to i=3;
26 if bls{i}=maxbls then maxdte=i;
ERROR: Illegal reference to the array maxdte.
27 end;
Does anyone have any tip is regards to this issue? What did I do wrong? Was I supposed to create an additional array for the date of when the maximum blood lead sample value was collected? Thanks!
**I'm stuck on #4 of Part A, but I included the other parts for context. Thanks!
**Edits: I included the data that I had to read into SAS and the file name of the file it came from
Just from looking at the code immediately prior to the error, you have a problem on this line:
26 if bls{i}=maxbls then maxdte=i;
You are getting the error because you are attempting to assign a value to the array maxdte. Arrays cannot be assigned values like that (unless you are using the deprecated do over syntax...) Instead, choose an element of the array and assign the value to the element. E.g. you could do:
26 if bls{i}=maxbls then maxdte{1}=i;
Or instead of a literal 1, you could use a variable containing the relevant array index.
You are not properly handling ID field from lines #2-4
input #1 id dob dbs1 mbs1
#2 dbs2 mbs2
#3 dbs3 mbs3
#4 bls1 bls2 bls3 sex;
For example you need to skip field 1 on line 2-3 or read the ids into array perhaps to check they are all the same.
input #1 id dob dbs1 mbs1
#2 id2 dbs2 mbs2
#3 id3 dbs3 mbs3
#4 id4 bls1 bls2 bls3 sex;
This example show how to check that you have 4 lines with the same ID and if you do read the rest of the variables or execute LOSTCARD. ID 3 has a missing record;
353 data ex;
354 infile cards n=4 stopover;
355 input #1 id #2 id2 #3 id3 #4 id4 #;
356 if id eq id2 eq id3 eq id4
357 then input #1 id dob:mmddyy. dbs1 mbs1
358 #2 id2 dbs2 mbs2
359 #3 id3 dbs3 mbs3
360 #4 id4 bls1 bls2 bls3 sex :$1.;
361 else lostcard;
362 format dob mmddyy.;
363 cards;
NOTE: LOST CARD.
RULE: ----+----1----+----2----+----3----+----4----+----5----+----6----+----7----+----8----+----9----+----0
372 3 01/03/80 11 7
373 3 27 2
374 3 3.24 3.4 3.83 M
375 4 08/01/80 5 12
NOTE: LOST CARD.
376 4 28 -9
NOTE: LOST CARD.
377 4 3 4
NOTE: The data set WORK.EX has 3 observations and 15 variables.
data ex;
infile cards n=4 stopover;
input #1 id #2 id2 #3 id3 #4 id4 #;
if id eq id2 eq id3 eq id4
then input #1 id dob:mmddyy. dbs1 mbs1
#2 id2 dbs2 mbs2
#3 id3 dbs3 mbs3
#4 id4 bls1 bls2 bls3 sex :$1.;
else lostcard;
format dob mmddyy.;
cards;
1 04/30/78 6 10
1 -9 7
1 14 1
1 1.62 1.35 1.47 F
2 05/19/79 27 11
2 20 -9
2 5 6
2 1.71 1.31 1.76 F
3 01/03/80 11 7
3 27 2
3 3.24 3.4 3.83 M
4 08/01/80 5 12
4 28 -9
4 3 4
4 3.1 3.69 3.27 M
;;;;
run;
proc print;
run;
I have the following example dataset which consists of the # of fish caught per check of a net. The nets are not checked at uniform intervals. The day of the check is denoted in julian days as well as the number of days the net had been fishing since last checked (or since it's deployment in the case of the first check)
http://textuploader.com/9ybp
Site_Number Check_Day_Julian Set_Duration_Days Fish_Caught
2 5 3 100
2 10 5 70
2 12 2 65
2 15 3 22
100 4 3 45
100 10 6 20
100 18 8 8
450 10 10 10
450 14 4 4
In any case, I would like to turn the raw data above into the following format:
http://textuploader.com/9y3t
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18
2 0 0 100 100 100 70 70 70 70 70 65 65 22 22 22 0 0 0
100 0 45 45 45 20 20 20 20 20 20 8 8 8 8 8 8 8 8
450 10 10 10 10 10 10 10 10 10 10 4 4 4 4 0 0 0 0
This is a matrix which assigns the # of fish caught during the period to EACH of the days that were within that period. The columns of the matrix are Julian days, the rows are site numbers.
I have tried to do this with some matrix functions but I have had much difficulty trying to populate all the fields that are within the time period, but I do not necessarily have a row of data for?
I had posted my small bit of code here, but upon reflection, my approach is quite archaic and a bit off point. Can anyone suggest a method to convert the data into the matrix provided? I've been scratching my head and googling all day but now I am stumped.
Cheers,
C
Two answers, the second one is faster but a bit low level.
Solution #1:
library(IRanges)
with(d, {
ir <- IRanges(end=Check_Day_Julian, width=Set_Duration_Days)
cov <- coverage(split(ir, Site_Number),
weight=split(Fish_Caught, Site_Number),
width=max(end(ir)))
do.call(rbind, lapply(cov, as.vector))
})
Solution #2:
with(d, {
ir <- IRanges(end=Check_Day_Julian, width=Set_Duration_Days)
site <- factor(Site_Number, unique(Site_Number))
m <- matrix(0, length(levels(site)), max(end(ir)))
ind <- cbind(rep(site, width(ir)), as.integer(ir))
m[ind] <- rep(Fish_Caught, width(ir))
m
})
I don't see a super obvious matrix transformation here. This is all i've got assuming the raw data is in a data.frame called dd
dd$Site_Number<-factor(dd$Site_Number)
mm<-matrix(0, nrow=nlevels(dd$Site_Number), ncol=18)
for(i in 1:nrow(dd)) {
mm[as.numeric(dd[i,1]), (dd[i,2]-dd[i,3]):dd[i,2] ] <- dd[i,4]
}
mm
Let's say I have maximum temperature data for the last 20 years. My data frame has a column for month, day, year and MAX_C (temperature data). I want to calculate the mean (and standard deviation, and range) maximum temperature from June 31 of one year to July 1 of the preceding year (i.e. mean max daily temp from July 1, 1991 to June 31, 1992). Is there an efficient way to do this?
My approach, thus far, has been to create an array:
maxt.prev12<-tapply(maxt$MAX_C,INDEX=list(maxt$month,maxt$day,maxt$year),mean)
I put mean in as the function as tapply was not producing an array without a function after the INDEX, but mean is not actually calculating anything here. Then I was thinking about trying to take January through June from one the matrices (i.e. 1992), and July through December from the preceding matrix (i.e. 1991), and then computing the mean. I'm not entirely sure how to do that part, however, there must be a more efficient way of performing these calculations in R
EDIT
Here is a simple sample set of data
maxt
day month year MAX_C
1 1 1990 29
1 2 1990 28
1 3 1990 32
1 4 1990 26
1 5 1990 24
1 6 1990 32
1 7 1990 30
1 8 1990 28
1 9 1990 28
1 10 1990 24
1 11 1990 30
1 12 1990 30
1 1 1991 25
1 2 1991 26
1 3 1991 28
1 4 1991 25
1 5 1991 24
1 6 1991 32
1 7 1991 26
1 8 1991 32
1 9 1991 26
1 10 1991 26
1 11 1991 27
1 12 1991 26
1 1 1992 27
1 2 1992 25
1 3 1992 29
1 4 1992 32
1 5 1992 27
1 6 1992 27
1 7 1992 24
1 8 1992 25
1 9 1992 28
1 10 1992 26
1 11 1992 31
1 12 1992 27
I would create an "indicator year" column which was equal to the year if month in July-Dec but equal to year-1 when month in Jan-June.
EDITED month reference in light of the fact it was numeric rather than character:
> maxt$year2 <- maxt$year
> maxt[ maxt$month %in% 1:6, "year2"] <-
+ maxt[ maxt$month %in% 1:6, "year"] -1
> # month.name is a 12 element constant vector in all versions of R
> # check that it matches the spellings of your months
>
> mean_by_year <- tapply(maxt$MAX_C, maxt$year2, mean, na.rm=TRUE)
> mean_by_year
1989 1990 1991 1992
28.50000 27.50000 27.50000 26.83333
If you wanted to change the labels so they reflected the non-calendar year derivation:
> names(mean_by_year) <- paste(substr(names(mean_by_year),3,4),
+ as.character( as.numeric(substr(names(mean_by_year),3,4))+1),
sep="_")
> mean_by_year
89_90 90_91 91_92 92_93
28.50000 27.50000 27.50000 26.83333
Although I don't think it will be quite right at the millennial turn.