When i compare 2 time period, if there is decrease - does the % have to be red and negative no?
example
1.389 users in the selected time interval compared to 1.260 users of the previous period.
An increase of 10.24% compared to the previous period. There is nothing strange.
Related
I want to know when a time series stabilizes after its peak.
This time series shows a peak, and then it goes down (see images below (1); (2)).
I would like to calculate the moment at which this time-series stabilizes (becomes flatter) after its peak.
In an ideal world, the data should go down to 0 and stay there. But as you see, it does not reach 0, nor stay 100% stable.
I thought of various different ways/possibilities:
-Calculate a tangent point to see the slope change. But the data has many small ups & downs even after smoothing it.
-Calculate the average at the tail (end of the series /e.g. if time = 8000; calculate last 2000 values mean average), then calculate an interval (margin +- this value), then calculate the time at which the first value appears in this interval.
-Calculate pronounced changes in the trend or slope.
*Maybe you have a better idea I did not consider. Feel free to share it if you've already dealt with this in the past.
I need to know the time at which this stabilization happened.
Ideally, you could mute/ignore all values before the peak value, but without deleting these rows (time should stay). Calculating the peak is easy (max value).
I also standardized the data so it starts at y=0 (I have various time series, I make them all start at y=0 to compare them later*).
I do not know how to provide the data because it is about 8k values.
I would really appreciate your help.
Thank you very much.
I'm using Graphite and Grafana and I'm trying to plot a series against a time shifted version of itself for comparison.
(I.e. is the current value similar to this time last week?)
What I'd like to do is plot;
the 5 minute moving average of the series
a band consisting of the 5 minute moving average of the series timeshifted by 7 days, bounded above and below by the standard deviation of itself
That way I can see if the current moving average falls within a band limited by the standard deviation of the moving average from a week ago.
I have managed to produce a band based on the timeshifted moving average, but only by offsetting either side by a constant amount. I can't work out any way of offsetting by the standard deviation (or indeed by any dynamic value).
I've copied a screenshot of the sort of thing I'm trying to achieve. The yellow line is the current moving average, the green area is bounded by the historical moving average offset either side by the standard deviation.
Is this possible at all in Grafana using Graphite as the backend?
I'm not quite on the latest version, but can easily upgrade (and will do so shortly anyway).
Incidentally, I'm not a statistician, if what I'm doing actually makes no sense mathematically, I'd love to know! ;-) My overall goal is to explore better alternatives, instead of using static thresholds, for highlighting anomalous or problematic server performance metrics - e.g. CPU load, disk IOPS, etc.
I am trying to create a simple bouncy ball simulator, and I want the balls to move at approximately the same speed as they would in real life. However, currently, after being dropped from 5 feet, the ball hits the ground almost instantly and bounces a few times before stopping in than a second. I know I could just experiment with my gravity value until it's semi-realistic, but I'm confused as to why my current gravity value doesn't work. Here's how I got the current one:
Gravity in real life = 9.8 m/sec^2
= 32.152 ft/sec^2
= 1.072 ft per 1/30th of a sec^2 (my frame rate is set to 30 in my program)
= 102.887 pixels per 1/30th of a sec^2 (a foot is 96 pixels in my program)
Here's my code for moving the ball (using Processing 3.2.1):
void move() {
dy+=102.88704; //gravity
x+=dx;
y+=dy;
z+=dz;
if(y+size*8>480) {
dy*=-0.85;
}
y=constrain(y,-100000.0,480-(size*8));
}
Currently, x and z just stay at 0. Since it's being dropped from 5 feet, it hits the ground when it gets to 480-size*8 (size is in inches). The 0.85 value is temporary and I might tweak it later, but it shouldn't have any impact on this issue. Any and all help is greatly appreciated. Thanks!
Your mistake is in converting your measurements from seconds to 1/30 seconds. Note that the unit is ft/sec^2: that is seconds squared. So to convert the time unit from seconds to 1/30 seconds you must also square the 1/30. Therefore
32.152 ft/sec^2
= 32.152/30^2 ft/(1/30 sec)^2
= 0.035724 ft/(1/30 sec)^2
= 0.035724 * 96 pixels/(1/30 sec)^2
= 3.4295 pixels/(1/30 sec)^2
So try replacing the number 102.88704 with 3.4295 and see if that fixes the problem.
I'm trying to get some information out of a couple of waveforms, which I currently have in the format of a CSV table with the columns: Time, Volume, Pressure, Flow. (The data is the flow/pressure/volume data obtained from a healthcare ventilator.)
I want to get excel / R / another-programme-that-I've-not-yet-thought-of to do the following with the waveform:
Break it up into individual breaths - starting with when the pressure starts to rise above a baseline (shortly followed by the flow rising above 0)
For each breath, extract:The highest pressure that occurs, the pressure just before the start of the next breath, the lowest pressure that occurs
Does anyone know how to do that, without doing it manually? (It'd mean trawling through a few hours-worth of ventilator data for the project I'm trying to do)
I've attached a copy of the shapes of the waves I'm dealing with, to try to help make more sense.Pressure & Volume against time
My intention is to work out the breath-to-breath variability in maximum, minimum, and baseline pressures.
Ultimately, I've come up with my own fix using excel.
For each pressure value, I've taken the mean of that value and the two either side of it. I've then compared this to the previous mean. For my data, a change of this mean of more than 1% in either direction identifies the beginning and end of the up- and down-stroke of the pressure curve, and I can identify where in the curve I am based on the absolute value for pressure (the peak should never be lower than the highest baseline for the data I've collected), and for the slopes, the first direction of change of the mean (this smooths out occasionally inconsistent slopes).
I'm then graphing the data, along with the transition points, the calculated phase of the breath, and the breath count, so I can sense-check it visually.
Ged
Subject: Finding One Coin of 13 in 3 Steps
There is a pile of thirteen coins, all of equal size. Twelve are of
equal weight. One is of a different weight. In three weighings(using scales) find
the unequal coin and determine if it is heavier or lighter.
I scratched my head on this one. I have found an answer but about 12.
Is it possible to do for 13 ?
So if it is possible can we end up with a method that can calculate the number of steps that are needed to find the unequal coin in pile of N. Pseudocode is just fair enough.
NOTE: Do not forget we do not know if the coin is lighter or heavier.
PS: Solution for 12 and some interesting thoughts here.
No, we cannot find a method that is guaranteed to determine which coin is not equal to the others and if it is heavier or lighter then the others, not with the restrictions you lay out.
One weighing of coins has three possible results: left pan down and right pan up (so the total of weights on the left is greater than the total of weights on the right), left pan up and right pan down (so the total of weights on the left is less than the total of weights on the right), or the pans balance (so the total of weights on the left is equal to the total of weights on the right). If we want to distinguish between four or more possibilities with just one weighing, we may fail since we can guarantee only three. Similarly, two weighings can distinguish between at most nine possibilities, and three weighings can handle at most 27 possibilities. The problem has 13 coins, each of which may be light or heavy, so there are 26 possibilities to begin. It looks like we may be able to handle them.
However, the problem comes at the first weighing. What happens if we place four or fewer weights on each pan? If one side goes up, all we know is that the special coin is among the five or more coins we did not use. However, that is 10 possibilities: light or heavy, for five coins. Therefore two more weighings is not guaranteed to distinguish between them.
Now, what happens if we place five or more weights on each pan for the first weighing? If the left pan rises, either one of the five or more weights on the left is light or one of the five or more weights on the right is heavy. That is at least 10 possibilities, so two more weighings is not guaranteed to distinguish between them.
Either way we may end up with 10 or more possibilities to solve in two weighings, which spoils any solution. Any method that has only three possible results at each step will need to be more sophisticated than the weighing pans.