Example fails due to MethodError for "linestyle" attribute

JuliaPy / PyPlot.jl

Plotting for Julia based on matplotlib.pyplot

https://github.com/JuliaPy/PyPlot.jl

MIT License

475 stars 87 forks source link

Example fails due to MethodError for "linestyle" attribute #493

Closed smonsays closed 4 years ago

smonsays commented 4 years ago

When trying to execute the example in the README:

using PyPlot
x = range(0; stop=2*pi, length=1000); y = sin.(3 * x + 4 * cos.(2 * x));
plot(x, y, color="red", linewidth=2.0, linestyle="--")
title("A sinusoidally modulated sinusoid")

I get the following error in Julia 1.4.2 and PyPlot v2.9.0

MethodError: no method matching py_linestyle(::Symbol, ::String)

smonsays commented 4 years ago

Sorry, I realised I still had PyPlot loaded through the Plots.jl backend