In Python, I am attempting to find a way to plot/rescale kde's so that they match up with the histograms of the data that they are fitted to:

enter image description here

The above is a nice example of what I am going for, but for some data sources , the scaling gets completely screwed up, and you get the following results, coming from the following code:

import numpy as np
import matplotlib as plot
import seaborn as sns

x1 = np.array([0.0, 0.0, 0.0, 0.0, 0.5, -0.12500000000000003, 0.0, -0.4, 0.0, 0.25])

## Simple histogram, weighted to reveal probabilities
plt.hist(x1, weights=np.ones(len(x1))/len(x1));

## Histogram + kde, but clearly something has gone wrong
sns.distplot(x1, hist=True, kde=True) 

Correct histogram:

enter image description here

Incorrect histogram + kde scaling:

enter image description here

Now, I am aware that kernel density estimators are "meant" to integrate over 1, or have the area beneath them equal to one (as recounted here, and many other answers on stack exchange), so they do not necessarily have to line up with a weighted histogram.

However, I find that a kde that does line up with a histogram would be much more informative, revealing a best estimate as to how the histogram really looks (if more data were simply available).

Is there a way to do this? To consistently get a kde image that looks like the top one? I will be extremely appreciative of any help on this.

  • $\begingroup$ Tell your KDE to use an appropriate range (aka kernel half-width) to match the histogram bin width. At a guess, your problem is that your software automatically chooses that range and all you need to do is look up the help page for the argument to override it. $\endgroup$ – whuber May 17 '19 at 21:23

Small discrete sample. Here is a 'standard graphics' plot in R of your observations.

 x = c(0.0, 0.0, 0.0, 0.0, 0.5, -0.12500000000000003, 0.0, -0.4, 0.0, 0.25)

hist(x, br=10, prob=T, col = "skyblue2")
  • The argument prob=T makes a histogram on a density scale in which the total area of the histogram bars sums to unity.
  • The argument br=10 'suggests' approximately ten bins, to provide a more reasonable match to the values in this extremely small and discrete sample; the default would give five bins.

  • The command rug puts small tick marks along the horizontal axis to show locations of observations within bins.

enter image description here

 lines(density(x), lwd=2, col="brown")
  • The lines statement overlays the default kernel density estimator (KDE) of the density procedure onto the histogram. One can change the bandwidth of the KDE with an appropriate argument.

In my experience, the area under KDE curves, made with the default density in R, is very nearly unity. Thus KDE's are calibrated to facilitate easy comparison with density histogram.

Of course, such comparisons are more fruitful with relatively large samples from continuous distributions, than with small discrete samples.

Large, continuous sample. For example, here is a histogram of a sample of size $n = 5000$ from $\mathsf{Gamma}(6, 1).$ The broken brown curve is the default KDE and the thin black curve is the density of the population.

y = rgamma(5000, 6, 1)
hist(y, prob=T, col="skyblue2", main="Histogram of sample of 5000 from GAMMA(6, 1)")
curve(dgamma(x, 6, 1), 0, 20, add=T) # `curve` requres fcn of `x`
lines(density(y), lwd=2, col="brown", lty="dashed")

enter image description here

Summary: The point is not to show that 'R is better than Python', but my knowledge of R is greater than my knowledge of Python, so I am more likely to make useful statements about R.

The point is to highlight characteristics of samples, and specifications for histograms and KDE's that can lead to fruitful comparisons. (Presumably there are ways to control the appearance of histograms and KDE's in Python roughly to match what I have shown in R.)

Addendum: With enough adjustments of bin widths and centers of the histogram and the kernel shapes and bandwidth of the KDE, I suppose you can get the KDE to emulate the histogram of almost any data as you seem to be suggesting.

x = c(0.0, 0.0, 0.0, 0.0, 0.5, -0.12500000000000003, 0.0, -0.4, 0.0, 0.25)
cutp = seq(-.42, .6, by=.05)
hist(x, br=cutp, prob=T, ylim=c(0,17), col = "skyblue2");  rug(x)
lines(density(x, adjust=.1))

enter image description here

| cite | improve this answer | |
  • $\begingroup$ Thanks for answering, BruceET. I think you caught unto the weird thing I found, and this is true in Python as well as R it seems, that for large datasets (or more continuous ones), the default kde's seems to match the histograms pretty well, whereas for small datasets (like the one I posted), the kde's go out of wack. Would you know the specific reason why this happens and, if so, what would be done in R to combat it? $\endgroup$ – Coolio2654 May 18 '19 at 17:39
  • $\begingroup$ I have not found KDEs to be useful for very small datasets or discrete distributions. I doubt that is their intended use. $\endgroup$ – BruceET May 18 '19 at 17:47
  • $\begingroup$ That may well be true, but there must be a way to scale the resulting kde to the heights of a normalized histogram. If that is not possible, I will think more on your post. $\endgroup$ – Coolio2654 May 18 '19 at 19:03
  • 1
    $\begingroup$ See Addendum, documentation on density, and perhaps Wichkam's paper. I didn't say it can't be done, only gently suggested maybe it shouldn't. $\endgroup$ – BruceET May 18 '19 at 20:41

View the histogram as a density estimator, so that the area under the curve is one. I, don't know about Python, but it must be possible. Then, there is one thing that can still make the plots different, and that is the bin size of histogram/kernel width of kde, choose them to be comparable. There must be some arguments to your Python code that can do it.

| cite | improve this answer | |

Apparently the kernel bandwidth is too small in your example. According to the documentation of scipy.stats.gaussian_kde, the scipy implementation of kde only supports simple "rules of thumb" for guessing the bandwidth. I would start with "Silverman's rule of thumb", which is $$h_{SROT} = 0.9\cdot\min\left\{\hat{\sigma},\hat{\mbox{IQR}}/1.35\right\}$$ where $\hat{\sigma}$ is the stdev of your data, and $\hat{\mbox{IQR}}$ its interquartile range.

A much better method, however, is the rule by Sheather and Jones, that is available in the R function density with the option bw="SJ". This method tries to minimize the asymptotic mean integrated square error (AMISE):

Sheather, Jones: "A reliable data-based bandwidth selection method for kernel density estimation." Journal of the Royal Statistical Society series B 53, pp. 683-690, 1991

| cite | improve this answer | |

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.