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My datasets cover a long time, say 20 years, and the data are sampled at different time. For example, some are obtaiend at 1:00(stored as 100), some are 23:00(2300). If I plot the histogram of the sampling time, it will look like this:

enter image description here

And some will look like this:

enter image description here

enter image description here

Is there a metric that I can use to quantify the uniformity of the histogram? So I can know that for the 1st plot, the data are sampled equally, 2nd plot's overall histogram has a little bit un-uniformity, and 3rd plot lacks great uniformity (it lacks data sampled at 1600 - 2000).

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  • $\begingroup$ I'm also after an answer to this question - I've started by getting a normalised value for each bin representing the amount of data it has in proportion to the total amount of data. I haven't yet come up with a way to reduce this to a single value representing the eveness of the readings. $\endgroup$ – greenglass Jun 3 '17 at 13:58
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    $\begingroup$ @greenglass I'm working on some similar problem. I think maybe (after bin the data), 1. each bin - average, 2. calculate the std. This might be a simple metrics to quantify the uniformity. $\endgroup$ – cqcn1991 Jun 8 '17 at 1:01
  • $\begingroup$ that is simple and could work - it doesn't take into consideration that lots missing in neighbouring bins is worse than far apart bins, at least in my case. But your comment has definitely got my mind going on this again. $\endgroup$ – greenglass Jun 8 '17 at 7:27
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There are many metrics. They are best used in conjunction with visualizing the data appropriately.

Among the solutions worth considering are to compare the distributions of the frequencies (regardless of time) to your reference distribution, the uniform one. Theory suggests that the deviations from perfect uniformity--the residuals--should be about the size of the square root of the average frequency. You can exploit that to compare datasets with different absolute frequencies: standardize the residuals (by dividing them by their expected deviations).

This has a close mathematical relationship to chi-squared tests. Indeed, we can use the standard Normal distribution as a reference for the standardized residuals, whence the sum of their squares is the usual chi-squared statistic. When it's small--around the number of distinct times or less--you have near-perfect uniformity. That gives you a good reference value for comparison.

Let's look at your data from this point of view. Here are versions of your three datasets:

Figure 1: the data.

We can order these residuals and plot them against the expected values of the first, second, ..., twenty fourth order statistics of the standard Normal distribution. The horizontal deviations of these plots around a diagonal line signal non-uniformity:

Figure 2: qq plots on a common scale

Notice the chi-squared statistics posted in each plot. The value of $15.8$ at the left isn't even as great as $24$ (the number of data values), perfectly consistent with a uniform distribution. The middle value of $563$ is large. What it means is that although the residuals line up in the plot, their values are too spread out: this is an over-dispersed dataset. Finally, the right hand value of $28000$ is huge. It signals major variations in this dataset.

Even more insight can be had by redrawing these plots, each on its own axis, so we can see the details of the variation.

Figure 3: QQ plots on their own axes.

Now you can see clearly how uniformly dispersed the first two datasets are. But by inspecting their vertical scales, you can see that the "dispersed" data are spread out around seven times more than the "uniform" data: that measures the over-dispersion.

Just about all statistical software produces plots like these: they are called "QQ" (quantile-quantile) plots.

This method works well for any dataset. Interpreting the chi-squared statistic becomes a little delicate when the average frequency drops below $5$ or so, but for almost any exploratory application that's no problem.

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    $\begingroup$ I've come to the conclusion that my specific question is more different to the original poster's question than I initially thought. There is more than one answer here that I believe gives me somewhere to continue working from, but I think this one best addresses my needs and the original question, as well as giving some good background information. That is why I am giving it the bounty. $\endgroup$ – greenglass Jun 11 '17 at 10:42
  • $\begingroup$ @whuber: Could be the source code of this analysis be shared? Many thanks. $\endgroup$ – Maximilian Jul 3 '18 at 12:28
  • $\begingroup$ @Maximilian Ordinarily I keep code used for answers. I am sorry that I did not retain any copy of this, likely because it was quick and straightforward to create. $\endgroup$ – whuber Jul 3 '18 at 12:50
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You could construct a Monte carlo-esque test with the given parameters and test how similar they are. That is generate a large number of uniform data points and measure how much of them overlap with your data. Since your data is already frequency you can just count how many points there are in your simulation in a time period versus how much there are in your data. for example your generate 240000 points (since the total sum of the frequencies is roughly 240000) on a uniform distribution with parameters (0,2400) and measure how much points there are in each 100 interval. Then $\dfrac{\sum_{i=1}^{24} frequency_{ith hour}-numberofobservations_{ith interval}}{totalsumoffrequencies}$

this will give you the fraction of data points that behave different then you would expect. You could (applying CLT) test this against a normal distribution. You have to re multiply with the sum of frequency though.

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The uniform distribution has the highest entropy. Entropy can be used as a measure of uniformity.

$$S=-\sum_{i=1}^np(x_i)\log(p(x_i))$$

Minimum is $0$. Maximum is $\log(n)$. The exponential version is more intuitive : it is somehow the percentage of values covered :

$$p=e^S/n$$

Examples :

Unifrom

Half

Some distribution

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I am not sure what you are after. Whether you need some test as well (ChiSquare might do it)?

Anyway. To explore this kind of data I would plot a line graph of multiple frequency distributions as a function of time (for different periods).

So 'frequency' instead of 'absolute numbers' such that the variation in amount of events per period does not have an effect.

Numeric example:

say you start from some table of period x time (I use dummy data, but you could make such a thing with your data)

rand_table <- matrix(qnorm(runif(24*20,0,1),200,5),24)
day_effect <- sin(c(1:24)/12*3.14)*10
year_effect <- runif(20,-100,100)

table <- rand_table + matrix(rep(day_effect,20),24,byrow=0) + matrix(rep(year_effect,24),24,byrow=1)
colnames(table) <- (1997:2016)
table

          1997     1998     1999     2000     2001     2002     2003     2004     2005     2006     2007     2008     2009     2010     2011     2012     2013     2014     2015     2016
 [1,] 260.0667 233.5953 120.2019 173.8588 256.3401 204.9747 177.7684 151.1975 138.7639 177.9552 240.7447 205.3817 265.8322 283.5408 262.5569 270.4157 174.4056 128.5618 290.2204 211.4510
 [2,] 249.5244 239.8960 120.5107 192.3158 253.0286 213.0315 169.0203 151.1792 148.4686 189.3689 243.1242 208.7573 262.9110 284.1739 258.1356 277.2521 169.0257 126.3684 294.5213 215.3311
 [3,] 247.9368 241.0808 122.3087 187.8790 254.8316 221.1269 171.4014 153.3314 141.9503 190.8465 245.8284 215.3114 267.7581 293.8443 261.3726 280.1178 177.0729 132.3668 297.9602 204.8105
 [4,] 259.6248 236.9574 131.4456 181.1347 254.8909 215.5816 167.8606 162.4480 143.0204 197.6665 248.4771 207.5274 261.8768 288.2050 262.8231 279.2299 173.8911 128.8646 304.1465 211.6245
 [5,] 258.5498 235.3907 124.2587 199.1903 256.9186 201.3183 175.2693 158.9103 150.4161 199.6419 245.8176 201.3408 267.2395 296.9404 271.8963 281.5610 170.9048 135.5900 307.9319 225.0224
 [6,] 254.2567 238.7180 125.5673 206.0158 258.8102 218.9670 173.4169 156.2924 145.0003 194.5099 252.9973 217.6832 264.4040 290.2716 267.9381 276.0918 166.2304 133.1175 307.3783 223.6769
 [7,] 263.8015 240.6280 128.1640 189.2185 260.7813 213.8075 179.2625 153.9522 150.5207 192.3161 245.8665 216.3049 275.7316 299.2184 271.1639 273.7122 179.7725 132.4717 304.0137 216.3689
 [8,] 251.1121 229.2925 125.1061 199.1912 251.5387 213.5779 173.9617 161.8092 143.6716 191.9641 245.3583 219.0199 266.3403 297.8517 271.8039 277.2551 184.1632 138.4491 303.5132 212.9774
 [9,] 262.7223 228.7230 123.4121 188.1462 256.0832 214.9447 168.3505 160.6458 141.1269 189.5807 252.1686 211.2347 258.2162 286.4550 257.3694 277.0731 164.2448 142.6826 297.7478 214.7545
[10,] 249.4369 237.1748 121.9888 187.5911 247.8664 205.6249 164.0456 158.7430 146.7741 194.0744 234.6995 208.7744 267.5273 277.2283 269.1293 276.5450 172.7910 141.0752 290.0011 206.9505
[11,] 251.5613 232.9232 122.3697 181.4821 239.0499 212.5056 166.7117 148.2396 146.7692 184.4266 242.7564 213.7748 259.5819 274.8612 259.7164 279.9896 167.4115 128.5881 295.3626 212.8600
[12,] 254.9230 220.9191 114.5943 171.9050 242.3551 209.4729 168.1985 131.6664 131.7506 182.0200 243.0256 204.5998 256.6656 282.8597 265.3859 272.2841 161.0621 122.4714 301.9434 207.6163
[13,] 244.3254 220.8979 109.9595 174.6536 247.1941 207.7537 165.4744 145.6344 133.9057 185.2712 242.3778 198.1467 255.9428 271.8196 257.6499 269.8779 156.8241 124.0382 294.1261 207.0789
[14,] 244.6379 222.5761 111.4693 174.1807 247.1866 194.8534 168.4106 143.9778 137.0152 179.4115 233.5407 201.8264 260.9790 279.9378 249.5447 267.6091 152.3627 117.3667 285.6467 202.8995
[15,] 241.6316 224.1021 113.3249 164.2898 242.0647 203.1320 152.5925 139.5406 143.6150 182.4157 229.9307 207.7646 253.3011 273.1239 254.5200 267.1596 152.2835 113.4475 290.8282 193.2609
[16,] 236.1205 217.1519 103.7728 177.2118 235.3146 196.6978 165.7939 138.1489 135.7740 179.4173 240.3620 196.5932 251.1842 276.0529 242.9755 259.4906 157.0868 113.2675 277.5009 198.8784
[17,] 244.0012 224.7148 106.9760 172.4818 237.0167 197.9934 152.9706 140.7621 118.6633 176.8517 227.0027 191.5034 248.9712 265.6904 242.9542 271.0733 150.5976 112.6630 284.8537 193.8020
[18,] 241.4985 213.2320 111.5817 170.3487 237.5654 188.5293 155.9515 143.3172 121.2416 176.1494 231.1264 198.5020 253.3602 269.5122 251.1248 269.7529 149.3875 114.2572 271.4105 202.0268
[19,] 235.2160 221.5018 108.2831 170.1034 238.3747 200.2492 156.0825 135.2035 124.5123 171.6472 233.6136 204.2768 247.2021 278.7483 247.4591 263.8084 157.1870 121.6157 284.1122 202.8897
[20,] 251.9713 226.7143 102.0953 170.3932 236.9628 199.0248 172.7265 137.3076 134.7663 171.9151 238.1566 188.2712 252.4209 275.7938 262.8806 259.6602 148.8275 114.1513 284.5481 204.1796
[21,] 245.5455 223.6492 109.9065 177.2015 241.7890 193.4855 156.2468 138.8739 138.1672 179.0033 233.4918 197.4053 247.6878 274.7670 251.1480 260.8251 148.7647 118.0140 284.0358 201.4156
[22,] 245.4394 217.7951 115.2334 178.6744 249.4580 204.2626 152.8850 137.1021 130.8307 182.5549 232.3583 200.7744 256.4592 277.6917 248.8487 262.9982 153.6351 122.6311 287.4040 195.8459
[23,] 252.1241 225.9910 113.6773 186.8966 243.3909 208.2465 161.2245 143.8986 133.0586 186.5644 224.3530 205.1357 263.8198 283.8469 255.0983 265.6281 161.3566 128.1783 294.0537 199.1899
[24,] 238.2365 229.5241 117.0020 187.2129 250.2044 208.5918 174.3084 142.9330 133.0209 187.7178 242.7608 209.7624 259.7822 277.7875 256.6742 276.9512 157.4175 125.0456 298.1174 210.7647

and then plot the frequency for each year separately

norm_table <- t(t(table)/colSums(table))

#plot
plot(-100,-100,xlim=c(0,24),ylim=c(0,0.1),xlab="hour",ylab="frequency")
plotcolor <- hsv(seq(0.1,0.8,length.out=20),1,1)
for (i_year in 1:20) {
  lines(x<-c(1:24),y<-norm_table[,i_year],col=plotcolor[i_year])
}

frequency distribution for each different year

If a reduction to categorical data is fine four your purpose and you'd wish to have a quick test, then you can do a chi-squared-test on the cross table. To see which values are strongest outliers you could plot their variation from the predicted values.

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