Histogram fill from a specified x/y value

Question

I have the following histogram plot:

BPT Diagnostic Diagram

Now, I'm sure many of you will point me in the direction of the matplotlib libraries, and other resources, but for some reason what I have in mind, and subsequently, read from various different sources doesn't quite work with my axes histogram. (Correct me if I am wrong!)

My question is this:

How do I hatch fill the histogram from the position on the line splits overlaid on the histograms in the (2,2,1) and (2,2,4) diagrams.

The x histogram (position 2,2,1) has a vline plotted at x = -0.222, and the y histogram has a hline plotted at y = 0.49.

An example working code, modified for just any array from the original, is as follows:

import numpy as np
import scipy
import matplotlib as mpl
import matplotlib.pyplot as plt
import pylab

X = 0.32, 0.41, 0.45, 0.53, -0.23,  0.34,  0.35, 0.47, 0.48, 0.33, 0.49, -0.10, -0.23,       0.45, 0.19
Y = 0.56, 0.67, 0.49, 0.61,  0.00, -0.02, -0.12, 0.12, 0.23, 0.44, 0.56,  0.13,  0.56, 0.67, 0.28

binsize = 0.1

min_x_data, max_x_data   = np.min(X), np.max(X)
num_x_bins               = np.floor((max_x_data - min_x_data) / binsize)

min_y_data, max_y_data   = np.min(Y), np.max(Y)
num_y_bins               = np.floor((max_y_data - min_y_data) / bin size)

fig = plt.figure(221)

axScatter = fig.add_subplot(223)
axScatter.scatter(X, Y)
axScatter.set_xlim(-2.0, 1.5)
axScatter.set_ylim(-2.0, 2.5)

axHistX = fig.add_subplot(221)
axHistX.set_xlim(-2.0, 1.5)
axHistX.set_ylim(0, 10)

axHistY = fig.add_subplot(224)
axHistY.set_xlim(0, 10)
axHistY.set_ylim(-2.0, 2.5)

axHistX.hist(X, num_x_bins, ec='0.3', fc='none', histtype='step')
axHistY.hist(Y, num_y_bins, ec='0.3', fc='none', histtype='step', orientation='horizontal')

axScatter.axhline(y=0.49, xmin=0, xmax=1, linestyle='-.',c='k')
axScatter.axvline(x=-0.222, ymin=0, ymax=1, linestyle='-.',c='k')
axHistX.axvline(x=-0.222, ymin=0, ymax=1, linestyle='-.',c='k')
axHistY.axhline(y=0.49, xmin=0, xmax=1, linestyle='-.',c='k')

plt.show()

A little difficult as I have my data as a csv file, and the BPT plot itself is quite complex. Any general/generic/adaptable example to get me started would be fine. — GCien, Oct 30 '14 at 16:25
If you can simply make random data (I assume you're using numpy so `np.random` will do) and provide the code. It makes it **much** easier to help you, as otherwise people would have to spend 20 minutes just getting the subplots looking roughly like yours look before they even write the shading part. — Ffisegydd, Oct 30 '14 at 16:26
[how-to-draw-a-rectangle-over-a-specific-region-in-a-matplotlib-graph](http://stackoverflow.com/questions/13013781/how-to-draw-a-rectangle-over-a-specific-region-in-a-matplotlib-graph) — gboffi, Oct 30 '14 at 16:34
Thank you for your response gboffi, but I don't really want to resort to that just yet! I could just as well use ms paint. What happens if I change my dataset? — GCien, Oct 30 '14 at 16:39
Uh, possible misunderstanding... you want to shade UNDER the histogram, don't you? while I wanted to shade the UPPER HALF of the subplot... my bad, I'm very sorry for bothering and for your lost time. — gboffi, Oct 30 '14 at 17:09

tillsten · Accepted Answer · 2014-10-30T17:34:41.250

6

So here my last try, but strongly suggest you to learn at the basics of python and matplotlib.

def step_hist(ax, X, num_x_bins=10,               
              hatch_from= -2, hatch_till=0.5,
              orientation='h'):
    #make histogram by hand.
    hist, edges = np.histogram(X, bins=num_x_bins)
    #generate (x,y) points for a step-plot
    edges = np.repeat(edges, 2)
    hist = np.hstack((0, np.repeat(hist, 2), 0))
    #plot step_hist
    #indices where we want the  plot hatached
    fill_region  =(hatch_from<edges)&(edges<hatch_till) 
    #apply hatching my using fill_between.
    if orientation == 'h':    
        ax.fill_between(edges[fill_region], 
                        hist[fill_region], 0, 
                        color='none', edgecolor='k',
                        hatch='xxx')
        ax.plot(edges, hist, 'k')
    elif orientation == 'v':
        ax.fill_betweenx(edges[fill_region], 
                        hist[fill_region], 0, 
                        color='none', edgecolor='k',
                        hatch='xxx')
        ax.plot(hist, edges, 'k')

a, b = np.random.randn(500), np.random.randn(500)

ax_top = plt.subplot(221)
ax_right = plt.subplot(224)
ax_cent = plt.subplot(223)

ax_cent.scatter(a, b, c='k')
step_hist(ax_top, a)
step_hist(ax_right, a, orientation='v')

resulting picture

edited Oct 30 '14 at 17:34

answered Oct 30 '14 at 16:57

tillsten

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Could I ask just how you have been able to specify where the hatching starts/ends, and also if I wanted hatch="//////" how would this be implemented? Thank you very much for this answer. – GCien Oct 30 '14 at 17:09
1

Seriously just read the code, it is very simple. The hatch is argument of the fill_between function and hatch region is defined by fill_region. – tillsten Oct 30 '14 at 17:10
Apologies, I see the hatch arg there at the end. – GCien Oct 30 '14 at 17:12
Plus I am also having problems implementing it. i.e., AttributeError: 'AxesSubplot' object has no attribute 'step_hist' – GCien Oct 30 '14 at 17:13
Don't call ax.step_hist, that doesn't make sense, because it is a new function. I added the axes as an argument, now you can call step_hist(ax, data) – tillsten Oct 30 '14 at 17:18
Let us [continue this discussion in chat](http://chat.stackoverflow.com/rooms/63948/discussion-between-michael-roberts-and-tillsten). – GCien Oct 30 '14 at 17:20
Sorry, i don't want to help you anymore, if you are not willing to spend a few minutes to understand the code. Only thing left for you is to do the vertical orientation. For that switch X and Y and use fill_between_x. – tillsten Oct 30 '14 at 17:23
It doesn't work my friend, so thank you for your effort but you need to work with me. – GCien Oct 30 '14 at 17:26
1

@MichaelRoberts This works fine for me using your test data set, I don't understand why you think it doesn't. tillsten has provided a fully working answer, they could have added a few sentences of explanation of what it does, but your expectation that they run through the entire code to you is wrong. – Ffisegydd Oct 31 '14 at 09:13
@Ffisegydd The answer has been heavily edited from what we see here. I had many issues but he has now provided a very valid answer. – GCien Oct 31 '14 at 19:40

Histogram fill from a specified x/y value

1 Answers1