calculate total percentage on either side of histogram

Hi, is it possible to calculate the total percentage on either side of histogram - like in the digram below? NB: I dont need to place the value on the chart - or get the percentage for the individual bars, just the percent for either side.

[dpb]

What is the divider, mean, origin, ... ? Not clear, precisely.

The sum of the bars shown isn't 37% but 36.1% so that doesn't match.

Possibly what you're looking for is

T=TargetValue;
pct=length(val<=T)/length(val)*100;

where val is the data vector and TargetValue is the value against which to compare--could be

TargetValue=mean(val(:));

for example, or

TargetValue=0;

[aine gormley]

Hi, thanks for your reply. My data is also temperature, mean linear regressions over time. Its a huge dataset, some of the slopes are postive and some are negative (as in the diagram above). I will give this a try and get back to you

[aine gormley]

hi, does this make sense? pct is coming up at 3%, from the histogram it looks much larger than that so i do not think it is calculating it right?

>> slopes = slope_TSA(:);
h=histogram(slopes(slopes>0));
h.FaceColor='red';
hold on
h1=histogram(slopes(slopes<0));
h1.FaceColor='blue';
xlim([-0.0002 0.00034]);
ylim([0 4400]);
title('Distribution of combined slopes - TSA IRR');
xlabel('Rate of change');
ylabel('Rate of change frequency');
pct=length(h<=0)/length(x)*100;
>> 

[aine gormley]

Hi, I have also tried this - yielding 4%

percantages = 100 * length(slopes(slopes<0)) / length(slopes);

[aine gormley]

oh I think I got it - does this code make sense (the answer is yielding 34% which looks logical):

slopes = slope_TSA(:);
nn = numel(slopes);
n2 = numel(slopes(slopes>0));
percantages = (n2 / nn) * 100;

[dpb]

If slopes is an array rather than a vector, then length() returns max(size(x)) while numel(x) is the total number in the array. For a vector they're the same result which is what I was assuming.

If you have more than one variable in the array, probably what you need to do is to do the calculation on each variable rather than the entire population???

It's not clear what the data really are so I can't really tell just what it is that is best thing to do for the actual case--as noted, my original comment was based on the assumption of one variable in a vector in which case numel() and length() should have (and would have) provided the same result; that they don't means something else is going on and need to know what that "something" is.

[aine gormley]

hi - the only variable is temperature change (ie the slope from a linear regression over time), so I think that is ok?

[dpb]

Ok, yes, that should work just fine.

So then slope_TSA was an array and hence the expression

slopes=slope_TSA(:);

maybe? If that were so and the previous was done using the array, that could be the cause of the unexpected answer, perhaps.

I was just trying to figure out why you got a nonsensical answer the one time...it's not terribly important if you now do have the correct result, just "inquiring minds" and all... :)

Answer 1

[the cyclist]

Taking a very different tack ... maybe you'd find the cdfplot command useful?