How to split numeric values and store them?
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I have the array below with activities 1, 2 and 3 and the times the system has spent in these activities. Not every row is a new date. For example, the first value (24) is 1/1/2015, however, the two following values (22.3 + 1.7 = 24) is on 1/2/2015, and so on.
Modes2 =
0 0 24.0000
0 0 22.3000
0 0 1.7000
0 0 13.0000
0 11.0000 0
0 24.0000 0
0 24.0000 0
0 24.0000 0
0 24.0000 0
0 17.2833 0
0 3.7667 0
2.9500 0 0
7.1167 0 0
16.8833 0 0
24.0000 0 0
To further analyse these, the need to be stored per 6 hours in something like the following.
Time_periods1 =
1/1/2015 6 hr 0 0 0
1/1/2015 12 hr 0 0 0
1/1/2015 18 hr 0 0 0
1/1/2015 24 hr 0 0 0
1/2/2015 6 hr 0 0 0
1/2/2015 12 hr 0 0 0
1/2/2015 18 hr 0 0 0
1/2/2015 24 hr 0 0 0
1/3/2015 6 hr 0 0 0
1/3/2015 12 hr 0 0 0
1/3/2015 18 hr 0 0 0
1/3/2015 24 hr 0 0 0
1/4/2015 6 hr 0 0 0
1/4/2015 12 hr 0 0 0
I wrote the following, does not work, now I'm lost.
for i=1:size(Modes2,1)
for j=1:size(Time_periods2,1)
for k=1:7
if (Modes2(i,k)>0 && Modes2(i,k)<=6)
Time_periods2(j,k)=Modes2(i,k);
elseif (Modes2(i,k)>6 && Modes2(i,k)<=12)
Time_periods2(j+1,k)=Modes2(i,k)-6;
Time_periods2(j,k)=6;
elseif (Modes2(i,k)>12 && Modes2(i,k)<=18)
Time_periods2(j+2,k)=Modes2(i,k)-12;
Time_periods2(j+1,k)=6;
Time_periods2(j,k)=6;I
elseif (Modes2(i,k)>18 && Modes2(i,k)<=24)
Time_periods2(j+3,k)=Modes2(i,k)-18;
Time_periods2(j+2,k)=6;
Time_periods2(j+1,k)=6;
Time_periods2(j,k)=6;
end
end
end
end
Any ideas?
댓글 수: 3
Showing the expected output helps us much more than showing non-functioning code.
Please show the complete expected output for the sample data you have given.
Note that not everyone on this forum is from the USA, and readers might interpret "1/2/2015" as the 1st of February 2015. Using ISO 8601 dates is unambiguous.
Rik
2019년 11월 19일
It looks like the brunt of your code is trying to determine when 24 hours have passed. To find the sequential day, you could also use the line of code below.
day_ID=ceil(cumsum(sum(Modes2,2))/24);
If you use that to generate the first column, wouldn't that be fairly close to what you mean?
Ymkje Lize Neuteboom
2019년 11월 19일
채택된 답변
Rik
2019년 11월 19일
The code below should do the trick. You will also notice there are much more comments explaining what the code is attempting to do.
Modes2 =[
0 0 24.0000
0 0 22.3000
0 0 1.7000
0 0 13.0000
0 11.0000 0
0 24.0000 0
0 24.0000 0
0 24.0000 0
0 24.0000 0
0 17.2833 0
0 3.7667 0
2.9500 0 0
7.1167 0 0
16.8833 0 0
24.0000 0 0];
len=6;%set the boundary so we can change it later if needed
in=Modes2;%make a copy so we can change the data
out=zeros(ceil(sum(in(:))/len),size(in,2));%pre-allocate the output
ind_out=1;%start at row 1
for ind_in=1:size(in,1)
while sum(in(ind_in,:))>0
ExtractAtMost=len-sum(out(ind_out,:));
tmp=min(ExtractAtMost,in(ind_in,:));
%this min() operation works element by element, so it relies on
%every row having only 1 non-zero element
%store to output
out(ind_out,:)=tmp+out(ind_out,:);
%reduce input to prepare for next iteration
in(ind_in,:)=in(ind_in,:)-tmp;
if len-sum(out(ind_out,:)) <= 2*eps
%test if output row is already full, if so, go to next row
ind_out=ind_out+1;
end
end
end
댓글 수: 2
Ymkje Lize Neuteboom
2019년 11월 19일
Rik
2019년 11월 19일
You're welcome. If this solved your question, feel free to mark this as accepted answer. If not, feel free to comment with your remaining issues.
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