How do I quantify the uniformity of sampling time? 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: 

And some will look like this:


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).
 A: 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:

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:

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.

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.
A: 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 :



A: 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.
