#Scatter plot matplotlib example code#

# FIX #2: if numvars is odd, the bottom right corner plot doesn't have the # FIX #1: this needed to be changed from. Each row of 'data' is plottedįig.subplots_adjust(hspace=0.0, wspace=0.0) import itertoolsĭef scatterplot_matrix(data, names=, **kwargs):

#Scatter plot matplotlib example code#

Not a fix, but I made it optional to explicitly input names, so that it puts a default xi for variable i in the diagonal positions.īelow you'll find an updated version of your code that addresses these two points, otherwise preserving the beauty of your code. It just leaves it as the default 0.1 ticks. If you have an odd number of variables you are plotting with, the bottom right corner axes doesn't pull the correct xtics or ytics.

The axis tics weren't lining up like I would expect (i.e., in your example above, you should be able to draw a vertical and horizontal line through any point across all plots and the lines should cross through the corresponding point in the other plots, but as it sits now this doesn't occur. As I was working with it, I noticed a few little things that didn't look quite right. Thanks for sharing your code! You figured out all the hard stuff for us. # Set up ticks only on one side for the 'edge' subplots.įor i, j in zip(*np.triu_indices_from(axes, k=1)):Īxes.plot(data, data, **kwargs)Īxes.annotate(label, (0.5, 0.5), xycoords='axes fraction',įor i, j in zip(range(numvars), itertools.cycle((-1, 0))): Returns the matplotlib figureįig, axes = plt.subplots(nrows=numvars, ncols=numvars, figsize=(8,8))įig.subplots_adjust(hspace=0.05, wspace=0.05) Passed on to matplotlib's 'plot' command. Each row of 'data' is plottedĪgainst other rows, resulting in a nrows by nrows grid of subplots with theĭiagonal subplots labeled with 'names'. '''Plots a scatterplot matrix of subplots. Linestyle='none', marker='o', color='black', mfc='none')įig.suptitle('Simple Scatterplot Matrix')ĭef scatterplot_matrix(data, names, **kwargs): There's always a name associated with each data series, so you can omit having to specify names.)ĭata = 10 * np.random.random((numvars, numdata))įig = scatterplot_matrix(data, , If you're always going to be working with structured or rec arrays, then you can simplify this a touch. I'm not quite sure what your data looks like, but it's quite simple to just build a function to do this from scratch. The expectation is that you'd write a simple function to string things together however you'd like.

# Draw these two points with big triangles to make it clearĪx.Generally speaking, matplotlib doesn't usually contain plotting functions that operate on more than one axes object (subplot, in this case). # And we'll have the line stop at x = 5000 # anywhere so we're going to have it go through 0,0 # A line with a fixed slope can intercept the axis # Now add on a line with a fixed slope of 0.03 import numpy as npĭf = pd.DataFrame( )ĭf = df * 0.88 + np.random.rand(25) * 5000 In the example below I've put in a line with slope = 0.03. If you know you want the slope to be 0.88 then you can calculate the y position of the other point: y_1 = 0.88 * (5000 - 0) + 0įor the data you've provided in the question a line with slope 0.88 will fly off the top of the y axis very quickly ( y_1 = 4400 in the example above). Now you can rearrange the formula for y_1 as: y_1 = slope * (x_1 - x_0) + y_0 For this example let's assume you want the line to go through the origin (0,0) x_0 = 0 # We already know this as it was set earlier There are an infinite number of parallel lines with this slope so we'll have to set one of the points to start off with. You define the slope of a line as increase in y / increase in x which would be: slope = (y_1 - y_0) / (x_1 - x_0)Īnd this can rearrange to: (y_1 - y_0) = slope * (x_1 - x_0) Let's assume you want the x coordinates of those two points to span the x axis so you're going to set x_0 and x_1 manually: x_0 = 0Īlternatively you can just take the minimum and maximum values from the axis: x_min, x_max = ax.get_xlim() These two points have x and y coordinates so for the two points you'll have four numbers: x_0, y_0, x_1, y_1. Y_min, y_max = c, c + slope*(x_max-x_min)īuilding on answer you just want to add in a line connecting two points. # need a slope and c to fix the position of line If you would like to plot scatter with regression line, I would recommend to do it in seaborn with a minimum lines of codes.ĭf = pd.DataFrame(np.random.multivariate_normal(,, ], size=100), columns=) The correlation coefficient won't give the slope of the regression line, because your data are in different scales.