Change value based on the values of another column

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Krishna
Krishna el 5 de Ag. de 2022
Comentada: Star Strider el 8 de Ag. de 2022
I have a table with 2 columns and 6000 rows each. Sample values in the 1st column is like:
-0.00554
-0.00503
-0.00406
-0.00406
-0.00316
-0.00274
-0.00274
0
0
0
0
0
0
0.00233
0.00452
0.00552
0.00715
0.00831
0
0
-9.00000000000000e-05
-9.00000000000000e-05
-9.00000000000000e-05
-9.00000000000000e-05
-9.00000000000000e-05
-9.00000000000000e-05
-9.00000000000000e-05
0
0
0
0.00552
0.00715
0.00831
0.00715
0.00831
0
0
0
0
0
-0.00406
-0.00316
The corresponding values in the 2nd column will be starting from 1,2,3, etc. until the positive value changes again to negative value in the 1st column. To be more specific, 1st column will be having values starting with -ve sign followed by 0s and +ve values. So, the -ve value will be changing to +ve and again back to -ve. Once the -ve value show up again, the count on the 2nd column should start from 1 again. Please help me wih this.

Respuesta aceptada

Star Strider
Star Strider el 5 de Ag. de 2022
Another approach —
v = [-0.00554
-0.00503
-0.00406
-0.00406
-0.00316
-0.00274
-0.00274
0
0
0
0
0
0
0.00233
0.00452
0.00552
0.00715
0.00831
0
0
-9.00000000000000e-05
-9.00000000000000e-05
-9.00000000000000e-05
-9.00000000000000e-05
-9.00000000000000e-05
-9.00000000000000e-05
-9.00000000000000e-05
0
0
0
0.00552
0.00715
0.00831
0.00715
0.00831
0
0
0
0
0
-0.00406
-0.00316];
sv = sign(v);
sv(sv>=0) = 1; % Adjust 'sign' Vector
pn = [0 strfind(sv.', [1 -1]) numel(v)]; % +vn To -ve Transition Indices
for k = 1:numel(pn)-1
idx = [pn(k) pn(k+1)-1]+1; % Index Range
Col2(idx(1):idx(2)) = (idx(1) : idx(2)) - (idx(1)-1); % Create Column #2 (As Row Vector)
end
Result = [v, Col2(:)] % Full Matrix
Result = 42×2
-0.0055 1.0000 -0.0050 2.0000 -0.0041 3.0000 -0.0041 4.0000 -0.0032 5.0000 -0.0027 6.0000 -0.0027 7.0000 0 8.0000 0 9.0000 0 10.0000
ResultMtx = [Result(1:10,:) Result(11:20,:) Result(21:30,:) Result(31:40,:)] % Display (Remove Later)
ResultMtx = 10×8
-0.0055 1.0000 0 11.0000 -0.0001 1.0000 0.0055 11.0000 -0.0050 2.0000 0 12.0000 -0.0001 2.0000 0.0072 12.0000 -0.0041 3.0000 0 13.0000 -0.0001 3.0000 0.0083 13.0000 -0.0041 4.0000 0.0023 14.0000 -0.0001 4.0000 0.0072 14.0000 -0.0032 5.0000 0.0045 15.0000 -0.0001 5.0000 0.0083 15.0000 -0.0027 6.0000 0.0055 16.0000 -0.0001 6.0000 0 16.0000 -0.0027 7.0000 0.0072 17.0000 -0.0001 7.0000 0 17.0000 0 8.0000 0.0083 18.0000 0 8.0000 0 18.0000 0 9.0000 0 19.0000 0 9.0000 0 19.0000 0 10.0000 0 20.0000 0 10.0000 0 20.0000
ResultEnd = Result(41:end,:) % Display (Remove Later)
ResultEnd = 2×2
-0.0041 1.0000 -0.0032 2.0000
.
  10 comentarios
Krishna
Krishna el 8 de Ag. de 2022
Editada: Krishna el 8 de Ag. de 2022
@Star Strider Consider this as my table. The previous one was wrong. Sorry. The relationship between input2 and output is, when input2 value changes to a -ve from +ve value within the subset, output will again count from 0. The difference is found from input1 alone.
Input1 input2 Output
1 -0.0103 0
3 -0.0105 2
5 -0.0091 4
7 -0.0082 6
10 -0.0073 9
11 -0.0059 10
15 -0.0054 14
20 -0.0027 19
21 0 20
22 0 21
23 0 22
24 0 23
25 -0.0059 24
26 -0.0054 25
27 -0.002723 26
35 0 34
36 0.0023 35
38 0.0068 37
40 0.0100 39
41 0 40
42 -0.0030 0
43 -0.0026 1
44 -0.0062 2
......
Star Strider
Star Strider el 8 de Ag. de 2022
The values for ‘Input1’ are different, and I do not understand how either it or ‘Output’ are incremented.
I just do not see any sort of pattern here that lends itself to being coded.

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Más respuestas (2)

dpb
dpb el 5 de Ag. de 2022
Editada: dpb el 5 de Ag. de 2022
If there's always at least one zero before the new -ive value excepting the initial element, then
ix=find(diff(sign([0;x]))==-1);
will locate the beginng line of each section.
Or, actually, on reflection,
ix=find(diff(sign([0;x]))<0);
will find a transition from either 0 to -ive or +ive to -ive
  2 comentarios
Krishna
Krishna el 5 de Ag. de 2022
There is no 0 at some cases. new -ve value is also present immediately after a +ve value.
Krishna
Krishna el 5 de Ag. de 2022
I'm running this code inside a loop as depicted below:
for k = 1:rows
ix = find(diff(table.first_column([0;k]))<0);
if(ix < 0)
table.second_column(k) = 1;
end
end
But, while running this code, I'm getting an error like:
"Array indices must be positive integers or logical values."

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Andrei Bobrov
Andrei Bobrov el 5 de Ag. de 2022
v = [-0.00554
-0.00503
-0.00406
-0.00406
-0.00316
-0.00274
-0.00274
0
0
0
0
0
0
0.00233
0.00452
0.00552
0.00715
0.00831
0
0
-9.00000000000000e-05
-9.00000000000000e-05
-9.00000000000000e-05
-9.00000000000000e-05
-9.00000000000000e-05
-9.00000000000000e-05
-9.00000000000000e-05
0
0
0
0.00552
0.00715
0.00831
0.00715
0.00831
0
0
0
0
0
-0.00406
-0.00316];
s = sign(v);
p = diff([0;s]) == -1 & s == -1;
i = accumarray(cumsum(p),1);
x = ones(size(p));
x(p) = x(p) - [ 0;i(1:end-1)];
out = cumsum(x);

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