How to find the most common statistical value in an array?

11 次查看（过去 30 天）
ref
ref 2012-6-27
Assume that the array consists of real time recordings and appears to have five common values with unknown std (e.g. 5.1356 +/- 1.265, 13.2136 +- 1.564 and so on....). How could you find the most common values in that array with its limits included? One way to do that would be to use the mode function with predefined edges for your common values. But how could you do that without "cheating" ?

请先登录，再回答此问题。

采纳的回答

Star Strider
Star Strider 2012-6-27
I've had to deal with these sorts of problems, so after looking at the plot of your data (that do not appear to be too noisy at least w.r.t. the transition step magnitudes), as an initial approach I suggest:
datadif = diff([0; data]); % (for a column vector) note that starting with '0' results in 'datadif' being the same length as 'data' with essentially the same relative indices
then graph it (perhaps on the same axes as 'data') and see if the +ve spikes in 'datadif' will help you define where your data subset limits are. If it looks good, then use 'find' (with a threshold) or something similar to identify the x-axis indices for the transitions. After that. a 'for' loop might be the best way to calculate your summary statistics on each subset, especially if you only need them (since you data look flat between the transitions) and you apparently only need to deal with your data once.
I'm not sure this is the answer you're looking for, but it will help you define your data subsets if that is part of the solution to your problem.
  4 个评论
显示 2更早的评论
ref
ref 2012-6-28
Ok Strider, it seems that
datadif = diff([0; data]);
worked, but with a little data manipulation. My measurements had a frequency of 8-10 Hz, so diff function did not work in the first place. But after averaging with a factor equal to frequency it worked just fine. See pic plz. http://imageshack.us/photo/my-images/14/diffe.jpg/ I do not have time to check if the histogram works, but i will though.
Star Strider
Star Strider 2012-6-28
I'm glad it worked! It might be worthwhile for you to create a second data file, linearly resampling your original data uniformly to 10 Hz and using that dataset for convenience.
The histogram will create peaks corresponding to data frequencies. There are routines for finding and characterising chromatography peaks that would likely work on histogram peaks if you need to do this often.
Thank you for accepting my answer.

请先登录，再进行评论。

更多回答（2 个）

grapevine
grapevine 2012-6-27
Take a look at this function:
mode - Most frequent values in array
http://www.mathworks.nl/help/techdoc/ref/mode.html
  5 个评论
显示 3更早的评论
Walter Roberson
Walter Roberson 2012-6-27
What would be permissible to access in that situation? The data used to create that graph? That graph appears to just be sort() of the original data, and when one is doing statistical work, the order of the data is not relevant (unless it is a time-dependent process), so having access to the sorted data would be equivalent to having access to the original data.
ref
ref 2012-6-27
Hi Walter. Yes obviously the data are permissible to access. I just plotted them that way to prove that sorted from smallest to largest value, the mean values are easier to be observed from anyone. The original data file is time depended, but again the problem is to find a mean value with its std.

请先登录，再进行评论。

Image Analyst
Image Analyst 2012-6-28
If your data are not integers, but are instead floating point numbers ( single or double), then depending on how they were generated, you may find that none of them are equal, even though you think they should be. See the FAQ on this topic: http://matlab.wikia.com/wiki/FAQ#Why_is_0.3_-_0.2_-_0.1_.28or_similar.29_not_equal_to_zero.3F In that case you should use hist() or histc() which will group values that are close to each other into bins.

类别

在 Help Center 和 File Exchange 中查找有关 Data Preprocessing 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