Change value based on the values of another column

Question

Krishna 2022-8-5

0
链接

此问题的直接链接

https://ww2.mathworks.cn/matlabcentral/answers/1774720-change-value-based-on-the-values-of-another-column

评论： Star Strider 2022-8-8

采纳的回答： Star Strider

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

-0.00316

-0.00274

0

0.00233

0.00452

0.00552

0.00715

0.00831

0

-9.00000000000000e-05

0

0.00552

0.00715

0.00831

0.00715

0.00831

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.

0 个评论
显示 -2更早的评论隐藏 -2更早的评论

请先登录，再进行评论。

请先登录，再回答此问题。

Answer 1

Star Strider 2022-8-5

0
链接

此回答的直接链接

https://ww2.mathworks.cn/matlabcentral/answers/1774720-change-value-based-on-the-values-of-another-column#answer_1021785

在 MATLAB Online 中打开

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 个评论
显示 8更早的评论隐藏 8更早的评论

Star Strider 2022-8-8

在 MATLAB Online 中打开

I do not understand.

Taking a wild guess —

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)
    Col2(idx(1):idx(2)) = diff([v(idx(1)); v(idx(1) : idx(2))]);    % Create Column #2 (As Row Vector)
end
Result = [v, Col2(:)]                                           % Full Matrix
Result = 42×2
   -0.0055         0
   -0.0050    0.0005
   -0.0041    0.0010
   -0.0041         0
   -0.0032    0.0009
   -0.0027    0.0004
   -0.0027         0
         0    0.0027
         0         0
         0         0
ResultMtx = [Result(1:10,:) Result(11:20,:) Result(21:30,:) Result(31:40,:)]    % Display (Remove Later)
ResultMtx = 10×8
   -0.0055         0         0         0   -0.0001         0    0.0055    0.0055
   -0.0050    0.0005         0         0   -0.0001         0    0.0072    0.0016
   -0.0041    0.0010         0         0   -0.0001         0    0.0083    0.0012
   -0.0041         0    0.0023    0.0023   -0.0001         0    0.0072   -0.0012
   -0.0032    0.0009    0.0045    0.0022   -0.0001         0    0.0083    0.0012
   -0.0027    0.0004    0.0055    0.0010   -0.0001         0         0   -0.0083
   -0.0027         0    0.0072    0.0016   -0.0001         0         0         0
         0    0.0027    0.0083    0.0012         0    0.0001         0         0
         0         0         0   -0.0083         0         0         0         0
         0         0         0         0         0         0         0         0
ResultEnd = Result(41:end,:)                                                    % Display (Remove Later)
ResultEnd = 2×2
   -0.0041         0
   -0.0032    0.0009

.

Star Strider 2022-8-8

I’mn getting infinitesimally closer to understanding it, however my comment to your other post (copied in part here) describes some of my confusion:

I’ve been trying to understand this, however much needs to be clarified.

Input Output

-0.0055 0

-0.0050 -0.0055

-0.0041 -0.0091

-0.0041 -0.0082

-0.0032 -0.0073 % <— This seems to me to be anomalous - Should it be -0.0114? If not, why not?

-0.0027 -0.0059 % And similarly here ...

-0.0027 -0.0054 % And similarly here ...

0 -0.0027 % And similarly here.

0 0

0.0023 0.0023

0.0045 0.0068

0.0055 0.0100

-0.0025 0

-0.0005 -0.0030

-0.0021 -0.0026

-0.0041 -0.0062

Unfortunately, I don’t understand the description of the problem, either, although the earlier problem was straightforward.

.

Krishna 2022-8-8

编辑：Krishna 2022-8-8

@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 2022-8-8

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.

请先登录，再进行评论。

Answer 2

dpb 2022-8-5

0
链接

此回答的直接链接

https://ww2.mathworks.cn/matlabcentral/answers/1774720-change-value-based-on-the-values-of-another-column#answer_1021745

编辑：dpb 2022-8-5

在 MATLAB Online 中打开

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 个评论
显示无隐藏无

Krishna 2022-8-5

There is no 0 at some cases. new -ve value is also present immediately after a +ve value.

Krishna 2022-8-5

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

But, while running this code, I'm getting an error like:

"Array indices must be positive integers or logical values."

请先登录，再进行评论。

Answer 3

Andrei Bobrov 2022-8-5

0
链接

此回答的直接链接

https://ww2.mathworks.cn/matlabcentral/answers/1774720-change-value-based-on-the-values-of-another-column#answer_1021905

在 MATLAB Online 中打开

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);

0 个评论
显示 -2更早的评论隐藏 -2更早的评论

请先登录，再进行评论。

Change value based on the values of another column

0 个评论
显示 -2更早的评论隐藏 -2更早的评论

采纳的回答

10 个评论
显示 8更早的评论隐藏 8更早的评论

更多回答（2 个）

2 个评论
显示无隐藏无

0 个评论
显示 -2更早的评论隐藏 -2更早的评论

另请参阅

类别

标签

产品

版本

Community Treasure Hunt

Change value based on the values of another column

0 个评论 显示 -2更早的评论隐藏 -2更早的评论

采纳的回答

10 个评论 显示 8更早的评论隐藏 8更早的评论

更多回答（2 个）

2 个评论 显示 无隐藏 无

0 个评论 显示 -2更早的评论隐藏 -2更早的评论

另请参阅

类别

标签

产品

版本

Community Treasure Hunt

0 个评论
显示 -2更早的评论隐藏 -2更早的评论

10 个评论
显示 8更早的评论隐藏 8更早的评论

2 个评论
显示无隐藏无

0 个评论
显示 -2更早的评论隐藏 -2更早的评论