create a array base on specific condition ?
8 Ansichten (letzte 30 Tage)
Ältere Kommentare anzeigen
MUKESH KUMAR
am 30 Aug. 2018
Kommentiert: MUKESH KUMAR
am 31 Aug. 2018
I had a array like this
A=[0 0 0 10 0 0 0 0 8 0 0 5 0 0 0 3 0 2 0 0 0 1 0 0 0];
and now I want to create a array B like this in which
B(4)=10-8=2;
[B(4)=A(4)-next upcoming non zero value ],
B(9)=8-5=3;[B(9)=A(9)-next non zero value]
and similarly for
B(12)=5-3=2;
B(16)=3-2=1;
B(18)=2-1=1;
B(22)=1;
and rest of the B values are zero. thanks
2 Kommentare
Akzeptierte Antwort
Stephen23
am 30 Aug. 2018
Bearbeitet: Stephen23
am 30 Aug. 2018
>> idx = A~=0;
>> A(idx) = [-diff(A(idx)),1]
A =
0 0 0 2 0 0 0 0 3 0 0 2 0 0 0 1 0 1 0 0 0 1 0 0 0
5 Kommentare
Stephen23
am 31 Aug. 2018
Bearbeitet: Stephen23
am 31 Aug. 2018
@MUKESH KUMAR: you get negative values because although in your question you gave values which decrease in magnitude (so their differences are all positive), in your real A data all of the values increase in magnitude (so their differences are all negative). Lets have a look at some of the values:
>> B(find(B))
ans =
-1
-5
-3
-2
... lots more here
-3
-2
1
>> A(find(A))
ans =
1
2
7
10
12
13
... lots more here
128
131
133
In your question you wrote: "I had a array like this"
A=[0 0 0 10 0 0 0 0 8 0 0 5 0 0 0 3 0 2 0 0 0 1 0 0 0];
"and now I want to create a array B like this in which"
B(4)=10-8=2;
[B(4)=A(4)-next upcoming non zero value ],
B(9)=8-5=3;[B(9)=A(9)-next non zero value]
Lets try your exact calculation method with the real A values:
B(6334) = A(6334) - A(6478) = -1
B(6478) = A(6478) - A(7487) = -5
B(7487) = A(7487) - A(7543) = -3
...etc
All are negative, all follow the method that you gave in your question, and all are exactly the values that are in B.
Weitere Antworten (1)
jonas
am 30 Aug. 2018
v=A(find(A~=0));
vid=find(A~=0);
B=A
B(vid)=B(vid)-[v(2:end) 0]
not the most elegant solution
0 Kommentare
Siehe auch
Kategorien
Mehr zu Loops and Conditional Statements finden Sie in Help Center und File Exchange
Community Treasure Hunt
Find the treasures in MATLAB Central and discover how the community can help you!
Start Hunting!