I'd like to make "SORTKEY" like the below. It's not the same observations for each one.
Basically, each one is 3 obs but if flg=1 then "SORTKEY" includes that observation.
In this example, it means SORTKEY = 2 is 4 obs, SORTKEY ^=2 is 3 obs.
Is there the way to make the SORTKEY manually?. If you have a good idea, please give me some advice.
I want the following dataset, using the "test" dataset.
/*
SORTKEY NO FLG
1 1 0
1 2 0
1 3 0
2 4 0
2 5 0
2 6 0
2 7 1
3 8 0
3 9 0
3 10 0
*/
data test;
input no flg;
cards;
1 0
2 0
3 0
4 0
5 0
6 0
7 1
8 0
9 0
10 0
;
run;
Use a sequence counter to track the 3-rows-per-sortkey requirement.
Example:
data want;
set have;
retain sortkey 1;
seq+1;
if seq > 3 and flag ne 1 then do;
seq = 1;
sortkey+1;
end;
run;
Related
I recently worked on a task where I needed to identify new clients.
I managed to find something similar on google and the final result was this measure that I don't understand
and maybe you can help me understand the logic behind this measure. I obviously thought wrongly that it should be >=MIN(Sheet1[Data])))
not <MIN(Sheet1[Data])))
I improvised some data along with the formula.
new_cust =
CALCULATE(
DISTINCTCOUNT(Sheet1[Cust_id])
,FILTER(
ALL(Sheet1[Data])
,Sheet1[Data]<=MAX(Sheet1[Data])
)
)
-
CALCULATE(
DISTINCTCOUNT(Sheet1[Cust_id])
,FILTER(
ALL(Sheet1[Data])
,Sheet1[Data]<MIN(Sheet1[Data])
)
)
Cust_id Data New_Cust
1 1/1/2023 1
1 1/2/2023 0
2 1/3/2023 1
2 1/4/2023 0
2 1/5/2023 0
3 1/6/2023 1
3 1/7/2023 0
1 2/1/2023 0
1 2/2/2023 0
3 2/3/2023 0
3 2/4/2023 0
3 2/5/2023 0
4 2/6/2023 1
4 2/7/2023 0
4 2/8/2023 0
1 3/1/2023 0
1 3/2/2023 0
2 3/3/2023 0
2 3/4/2023 0
3 3/5/2023 0
3 3/6/2023 0
4 3/7/2023 0
4 3/8/2023 0
6 3/9/2023 1
6 3/10/2023 0
Thank you in advance for your understanding and help
I'm using SAS Base/ SAS Enterprise Guide and I'm stuck.
I need to create the Association Rules matrix to calculate the support, lift, confidence and other association metrics.
My table is similar to: (one line per client with all the products associated to)
data test;
input item1 item2 item3 item4 ;
datalines ;
1 0 1 0
1 1 1 0
1 0 1 0
1 0 1 1;
As you know for the market basket analysis I want something like that:
item1 item2 item3 item4
item1 4 1 4 1
item2 1 1 1 0
item3 4 1 4 1
item4 1 0 4 4
Any idea?
Thank you for your help.
I think your example WANT is incorrect.
data test;
input item1-item4;
datalines;
1 0 1 0
1 1 1 0
1 0 1 0
1 0 1 1
;;;;
proc print;
run;
proc corr noprint sscp out=sscp(drop=int: where=(_type_ eq 'SSCP' and _NAME_ ne: 'Int'));
var item:;
run;
proc print;
run;
I have a vector
A = [ 1 1 1 2 2 3 6 8 9 9 ]
I would like to write a loop that counts the frequencies of values in my vector within a range I choose, this would include values that have 0 frequencies
For example, if I chose the range of 1:9 my results would be
3 2 1 0 0 1 0 1 2
If I picked 1:11 the result would be
3 2 1 0 0 1 0 1 2 0 0
Is this possible? Also ideally I would have to do this for giant matrices and vectors, so the fasted way to calculate this would be appreciated.
Here's an alternative suggestion to histcounts, which appears to be ~8x faster on Matlab 2015b:
A = [ 1 1 1 2 2 3 6 8 9 9 ];
maxRange = 11;
N = accumarray(A(:), 1, [maxRange,1])';
N =
3 2 1 0 0 1 0 1 2 0 0
Comparing the speed:
K>> tic; for i = 1:100000, N1 = accumarray(A(:), 1, [maxRange,1])'; end; toc;
Elapsed time is 0.537597 seconds.
K>> tic; for i = 1:100000, N2 = histcounts(A,1:maxRange+1); end; toc;
Elapsed time is 4.333394 seconds.
K>> isequal(N1, N2)
ans =
1
As per the loop request, here's a looped version, which should not be too slow since the latest engine overhaul:
A = [ 1 1 1 2 2 3 6 8 9 9 ];
maxRange = 11; %// your range
output = zeros(1,maxRange); %// initialise output
for ii = 1:maxRange
tmp = A==ii; %// temporary storage
output(ii) = sum(tmp(:)); %// find the number of occurences
end
which would result in
output =
3 2 1 0 0 1 0 1 2 0 0
Faster and not-looping would be #beaker's suggestion to use histcounts:
[N,edges] = histcounts(A,1:maxRange+1);
N =
3 2 1 0 0 1 0 1 2 0
where the +1 makes sure the last entry is included as well.
Assuming the input A to be a sorted array and the range starts from 1 and goes until some value greater than or equal to the largest element in A, here's an approach using diff and find -
%// Inputs
A = [2 4 4 4 8 9 11 11 11 12]; %// Modified for variety
maxN = 13;
idx = [0 find(diff(A)>0) numel(A)]+1;
out = zeros(1,maxN); %// OR for better performance : out(maxN) = 0;
out(A(idx(1:end-1))) = diff(idx);
Output -
out =
0 1 0 3 0 0 0 1 1 0 3 1 0
This can be done very easily with bsxfun.
Let the data be
A = [ 1 1 1 2 2 3 6 8 9 9 ]; %// data
B = 1:9; %// possible values
Then
result = sum(bsxfun(#eq, A(:), B(:).'), 1);
gives
result =
3 2 1 0 0 1 0 1 2
Matlab - Hello, I want to combine two binary images with same size (111x111), but first i want to divide the image into 3 x 3 matrix patch (37 sub matrix), with the two conditions:
1.If the 3 x 3 patches from image 2 matrix values is all white (1) then the result matrix = image 1 matrix , example:
image 1 patch: image 2 patch: result:
1 1 0 1 1 1 1 1 0
1 0 1 1 1 1 1 0 1
1 1 1 1 1 1 1 1 1
2. Else, i want to keep the center value of 3 x 3 patches (index (2,2)) from image 1, but the other value from image 2
image 1 patch: Image 2 patch : result:
0 0 0 1 0 1 1 0 1
0 0 0 1 1 0 1 0 0
0 0 0 1 0 1 1 0 1
And do the whole image and combine the whole 3 x 3 patches into result image (111x111 again)
My Code so far (Using mat2cell):
clear;
clc;
I1 = imread('image1.bmp');
I2 = imread('image2.bmp');
TI1 = im2bw(I1); %Thresholding I1
TI2 = im2bw(I2); %Thresholding I2
%Mat2cell patch
cellTI1 = mat2cell(TI1, 3*ones(size(TI1,1)/3,1), 3*ones(size(TI1,2)/3,1))
cellTI2= mat2cell(TI2, 3*ones(size(TI2,1)/3,1), 3*ones(size(TI2,2)/3,1))
% Im Confused with the loop
result1 = ones(37,37);
for i=1:3
for j=1:3
for m=1:37
for n=1:37
if TI2{m,n} == [1 1 1;
1 1 1;
1 1 1]
result1 = TI1(m,n);
else
result1 = [TI2{1,1}(1,1) TI2{1,1}(1,2) TI2{1,1}(1,3);
TI2{1,1}(2,1) TI1{1,1}(2,2) TI2{1,1}(3,2);
TI2{1,1}(3,1) TI2{1,1}(3,2) TI2{1,1}(3,3)];
end
end
end
Sorry for my bad English,
Thanks
This little code snippet is supposed to loop through a sorted data frame. It keeps a count of how many successive rows have the same information in columns aIndex and cIndex and also bIndex and dIndex. If these are the same, it deposits the count and increments it for the next time around, and if they differ, it deposits the count and resets it to 1 for the next time around.
for (i in 1:nrow(myFrame)) {
if (myFrame[i, aIndex] == myFrame[i, cIndex] &
myFrame[i, bIndex] == myFrame[i, dIndex]) {
myFrame[i, eIndex] <- count
count <- (count + 1)
} else {
myFrame[i, eIndex] <- count
count <- 1
}
}
It's been running for a long time now. I understand that I'm supposed to vectorize whenever possible, but I'm not really seeing it here. What am I supposed to do to make this faster?
Here's what an example few rows should look like after running:
aIndex bIndex cIndex dIndex eIndex
1 2 1 2 1
1 2 1 2 2
1 2 4 8 3
4 8 1 4 1
1 4 1 4 1
I think this will do what you want; the tricky part is that the count resets after the difference, which effectively puts a shift on the eIndex.
There (hopefully) is an easier way to do this, but this is what I came up with.
tmprle <- rle(((myFrame$aIndex == myFrame$cIndex) &
(myFrame$bIndex == myFrame$dIndex)))
myFrame$eIndex <- c(1,
unlist(ifelse(tmprle$values,
Vectorize(seq.default)(from = 2,
length = tmprle$lengths),
lapply(tmprle$lengths,
function(x) {rep(1, each = x)})))
)[-(nrow(myFrame)+1)]
which gives
> myFrame
aIndex bIndex cIndex dIndex eIndex
1 1 2 1 2 1
2 1 2 1 2 2
3 1 2 4 8 3
4 4 8 1 4 1
5 1 4 1 4 1
Maybe this will work. I have reworked the rle and sequence bits.
dat <- read.table(text="aIndex bIndex cIndex dIndex
1 2 1 2
1 2 1 2
1 2 4 8
4 8 1 4
1 4 1 4", header=TRUE, as.is=TRUE,sep = " ")
dat$eIndex <-NA
#identify rows where a=c and b=d, multiply by 1 to get a numeric vector
dat$id<-(dat$aIndex==dat$cIndex & dat$bIndex==dat$dIndex)*1
#identify sequence
runs <- rle(dat$id)
#create sequence, multiply by id to keep only identicals, +1 at the end
count <-sequence(runs$lengths)*dat$id+1
#shift sequence down one notch, start with 1
dat$eIndex <-c(1,count[-length(count)])
dat
aIndex bIndex cIndex dIndex eIndex id
1 1 2 1 2 1 1
2 1 2 1 2 2 1
3 1 2 4 8 3 0
4 4 8 1 4 1 0
5 1 4 1 4 1 1