清空筛选条件
清空筛选条件

Inverse of log formula

24 次查看（过去 30 天）
Lisa
Lisa 2017-5-1
回答： Image Analyst 2017-5-1
Hi, I have the following formula:
y = -log(x / mean(x))
y and x are vector’s.
However, in my case I only know y, but not x. Therefore, I would like to solve the inverse of the formula for x. I am aware of that exp is the inverse of log:
x=-exp(y)
but I am not sure how to consider the term mean(x)?
Thank you in advance.
Lisa

请先登录，再回答此问题。

采纳的回答

Star Strider
Star Strider 2017-5-1
You cannot determine ‘mean(x)’ unless you have access to the original ‘x’ vector. (If you had ‘x’, you would not need to do the inversion.) You can only recover the normalised ratio ‘x/mean(x)’.
  3 个评论
显示 1更早的评论
Lisa
Lisa 2017-5-1
Thank you very much.
Star Strider
Star Strider 2017-5-1
@Lisa — As always, my pleasure!
@John D’Errico — Thank you!

请先登录，再进行评论。

更多回答（2 个）

Image Analyst
Image Analyst 2017-5-1
Try fzero().
Image Analyst
Image Analyst 2017-5-1
Lisa, is this what you are after - to find the x for a y that is a specified value? You can do it numerically like this, where I find the x where y = 1:
% Define a range of x values.
x = sort(rand(1, 500), 'ascend');
% Compute the y values over that range.
fprintf('mean(x) = %f\n', mean(x));
y = -log(x / mean(x));
% Plot curve.
plot(x, y, 'b-', 'LineWidth', 2);
grid on;
xlabel('x', 'FontSize', 20);
ylabel('y', 'FontSize', 20);
% Find the x closest to y = 1 (for example).
[minValue, index] = min(abs(y-1))
% Get x and y values
x1 = x(index);
y1 = y(index);
fprintf('Closest point: y = %f at x = %f\n', y1, x1);
% Draw lines
hold on;
line([0, x1], [y1, y1], 'Color', 'r', 'LineWidth', 2);
line([x1, x1], [0, y1], 'Color', 'r', 'LineWidth', 2);
ax = gca;
ax.XAxisLocation = 'origin';
In the command window:
mean(x) = 0.514679
minValue =
0.0017406261319084
index =
93
Closest point: y = 1.001741 at x = 0.189011
And the plot of the results:

类别

在 Help Center 和 File Exchange 中查找有关 Time Series 的更多信息

标签

Community Treasure Hunt

Find the treasures in MATLAB Central and discover how the community can help you!

Start Hunting!

Translated by