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2019-05-01 10:23:32 +0200 commented question Legend attached to curve in plot ("PlotLabels")

Having read the solution ("add labels manually") I now see why a more concrete example than f(x) = n*x is useful. I will add a simplified version of one of my real plots.

2019-05-01 10:17:36 +0200 commented answer Legend attached to curve in plot ("PlotLabels")

This is very useful: it's simple and works out of the box in many cases.

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2019-04-30 12:48:44 +0200 asked a question Legend attached to curve in plot ("PlotLabels")

Hello,

I have a plot with several curves (say, 5) and I need to clearly mark them.

The plot should be grayscale friendly (I am giving curves different shades, but it is not enough), and line style (dotted...) already has a different meaning. I am using a legend, but it goes in a separate box and does not really help a lot.

Since the different curves never cross in a given plot, just attaching the legend label to the curve tail would be great: how can I achieve this? To be clear, I am looking for an equivalent of Mathematica PlotLabels.

The following example gives an idea of the 'most overlapping' case I can be interested in (actually an exaggeration, a real plot would stop at nexp=2.5):

f(x,n) = e^(-1/(n*x))
listexp = srange(0.5, 3+0.1, 0.5)
lenght = float(len(listexp))
xMinPlot = 0.1

p = Graphics()
for i,nexp in enumerate(listexp) :
    c = 10^nexp
    p += plot_semilogx(f(x,c), (x, xMinPlot, 10), hue=i/lenght)

p.show()

Using slelievre answer (and imitating the PlotLabels idea, putting gray lines between plot and label), I came up with this:

text_options = {...}
offsets = [0.01, 0.01, 0.01, -0.02, -0.02, 0.02]
xlabel = xMinPlot*0.85
xline  = xMinPlot*0.97

for i,nexp in enumerate(listexp) :
    c = 10^nexp
    yline = f(xMinPlot,c)
    p += line(((xline,yline),(xlabel,yline+offsets[i])), color='gray', thickness=0.5)
    p += text('$c = 10^{{ {:.1f} }}$'.format(float(nexp)),
        (xlabel, yline+offsets[i]), **text_options)

p.show(xmin=5*10^-2, xmax=10)

This approach requires to analyse each plot case by case and playing around to find a solution, but I am afraid there is nothing one can do to partially automate it. Also, I am not sure that I am doing this the best way, or whether the result could look better/more professional.