I have succesfully imported Scipy Optimize's function minimize to Julia, but when I try to use it so that I specify constraints to it:
julia> #pyimport scipy.optimize.minimize as so
julia> so.minimize(f, x0,
constraints={"type": "ineq", "fun": g},
options={"maxiter": 1000})
Julia throws an error about the curly braces:
ERROR: LoadError: syntax: { } vector syntax is discontinued
Anyone having an idea of how the minimize function could be used properly in Julia?
I now realized that the curly brackets present dicts in Python and the Julia-version should therefore look like:
julia> so.minimize(objective_function, coords,
constraints = Dict("type" => "ineq", "fun" => g),
options = Dict("maxiter" => 1000))
So my guess is now that even though the function comes from Scipy, the input arguments need to be Julia syntax.
I am trying to plot a function using PyPlot on an IJulia notebook, but I keep obtaining error messages.
When I ran this code:
function gtest2(x)
6.34*(log2(1+exp(10.0*(x+0.5))))^0.8
end
using PyPlot
x = -1.0:0.1:1.0;
plot(x, gtest2(x));
I got errors like these:
MethodError: no method matching ^(::Array{Float64,1}, ::Float64)
Closest candidates are: ^(::Float64, ::Float64) at math.jl:355 ...
I tried to defined a different type of variable while defining my function using gtest2(x::Number) or gtest2(x::Float64) but I have the same errors.
It does the same using linespace instead of -1.0:0.1:1.0. I understand that the format the function sees in input does not match the definition but I don't get what I'm doing wrong because simple functions work:
function f(x)
x
end
plot(x,f(x))
Why am I getting those errors in the first case?
I am using IJulia notebook 0.5.1 on safari.
Your code doesn't properly handle vectors thus you need to either change gtest
using the . vectorization syntax
function gtest2(x)
6.34*(log2.(1 + exp.(10.0*(x + 0.5)))).^0.8
end
or even easier use the dot vectorization as follows
plot(x, gtest2.(x));
To learn more about dot vectorization please see the following in the docs: https://docs.julialang.org/en/latest/manual/functions.html#man-vectorized-1
The first definition works also with:
map(gtest2, x)
or
gtest2.(x)
I'm trying to make a grid in ggplot to plot 4 graphs, as if it were a basic pair (mfrow = c (2,2)). However, I can not execute the code. I have already tried with gridExtra and cowplot with the functions plot_grid, grid.arrange, ggplot2.multiplot and also tried with the multiplot function. The error that appears is as follows:
Error: Aesthetics must be either length 1 or the same as the data (8598): alpha, x, y, group
gridExtra::grid.arrange(ggplot(),ggplot(),ggplot(),ggplot(), nrow=2)
produces
you may want to debug your code for each individual plot first.
I'm having trouble solving a discrepancy between something breaking at runtime, but using the exact same data and operations in the python console, having it work fine.
# f_err - currently has value 1.11819388872025
# l_scales - currently a numpy array [1.17840183376334 1.13456764589809]
sq_euc_dists = self.se_term(x1, x2, l_scales) # this is fine. It calls cdists on x1/l_scales, x2/l_scales vectors
return (f_err**2) * np.exp(-0.5 * sq_euc_dists) # <-- errors on this line
The error that I get is
AttributeError: 'Zero' object has no attribute 'exp'
However, calling those exact same lines, with the same f_err, l_scales, and x1, x2 in the console right after it errors out, somehow does not produce errors.
I was not able to find a post referring to the 'Zero' object error specifically, and the non-'Zero' ones I found didn't seem to apply to my case here.
EDIT: It was a bit lacking in info, so here's an actual (extracted) runnable example with sample data I took straight out of a failed run, which when run in isolation works fine/I can't reproduce the error except in runtime.
Note that the sqeucld_dist function below is quite bad and I should be using scipy's cdist instead. However, because I'm using sympy's symbols for matrix elementwise gradients with over 15 partial derivatives in my real data, cdist is not an option as it doesn't deal with arbitrary objects.
import numpy as np
def se_term(x1, x2, l):
return sqeucl_dist(x1/l, x2/l)
def sqeucl_dist(x, xs):
return np.sum([(i-j)**2 for i in x for j in xs], axis=1).reshape(x.shape[0], xs.shape[0])
x = np.array([[-0.29932052, 0.40997373], [0.40203481, 2.19895326], [-0.37679417, -1.11028267], [-2.53012051, 1.09819485], [0.59390005, 0.9735], [0.78276777, -1.18787904], [-0.9300892, 1.18802775], [0.44852545, -1.57954101], [1.33285028, -0.58594779], [0.7401607, 2.69842268], [-2.04258086, 0.43581565], [0.17353396, -1.34430191], [0.97214259, -1.29342284], [-0.11103534, -0.15112815], [0.41541759, -1.51803154], [-0.59852383, 0.78442389], [2.01323359, -0.85283772], [-0.14074266, -0.63457529], [-0.49504797, -1.06690869], [-0.18028754, -0.70835799], [-1.3794126, 0.20592016], [-0.49685373, -1.46109525], [-1.41276934, -0.66472598], [-1.44173868, 0.42678815], [0.64623684, 1.19927771], [-0.5945761, -0.10417961]])
f_err = 1.11466725760716
l = [1.18388412685279, 1.02290811104357]
result = (f_err**2) * np.exp(-0.5 * se_term(x, x, l)) # This runs fine, but fails with the exact same calls and data during runtime
Any help greatly appreciated!
Here is how to reproduce the error you are seeing:
import sympy
import numpy
zero = sympy.sympify('0')
numpy.exp(zero)
You will see the same exception you are seeing.
You can fix this (inefficiently) by changing your code to the following to make things floating point.
def sqeucl_dist(x, xs):
return np.sum([np.vectorize(float)(i-j)**2 for i in x for j in xs],
axis=1).reshape(x.shape[0], xs.shape[0])
It will be better to fix your gradient function using lambdify.
Here's an example of how lambdify can be used on partial d
from sympy.abc import x, y, z
expression = x**2 + sympy.sin(y) + z
derivatives = [expression.diff(var, 1) for var in [x, y, z]]
derivatives is now [2*x, cos(y), 1], a list of Sympy expressions. To create a function which will evaluate this numerically at a particular set of values, we use lambdify as follows (passing 'numpy' as an argument like that means to use numpy.cos rather than sympy.cos):
derivative_calc = sympy.lambdify((x, y, z), derivatives, 'numpy')
Now derivative_calc(1, 2, 3) will return [2, -0.41614683654714241, 1]. These are ints and numpy.float64s.
A side note: np.exp(M) will calculate the element-wise exponent of each of the elements of M. If you are trying to do a matrix exponential, you need np.linalg.exmp.
I am programming in Julia but using PyPloy library. I want to plot an histogram with log y-axis. But when I use the following code:
using PyPlot
List = [rand() for i = 1:100]
plt.hist(List)
plt.gca().set_yscale("log")
I get the following error:
type PyObject has no field set_yscale
while loading In[45], in expression starting on line 3
in getindex at /home/rm/.julia/v0.4/PyCall/src/PyCall.jl:642
in pysequence_query at /home/rm/.julia/v0.4/PyCall/src/conversions.jl:743
in pytype_query at /home/rm/.julia/v0.4/PyCall/src/conversions.jl:759
in convert at /home/rm/.julia/v0.4/PyCall/src/conversions.jl:808
in pycall at /home/rm/.julia/v0.4/PyCall/src/PyCall.jl:812
in fn at /home/rm/.julia/v0.4/PyCall/src/conversions.jl:181
in close_queued_figs at /home/rm/.julia/v0.4/PyPlot/src/PyPlot.jl:295
Is this a path error? If so, is there a simpler way to do a log-log plot with a different command?
Thanks in advance.
I feel like this should be more prominently explained in the documentation, but if you scroll down to the bottom of the Readme for PyCall (which PyPlot uses) it says:
Important: The biggest difference from Python is that object attributes/members are accessed with o[:attribute] rather than o.attribute, so that o.method(...) in Python is replaced by o[:method](...)
So, as #jverzani mentioned, after you call any module-level function from PyPlot that returns an object, that object is a PyObject and all of the attributes and methods have to be called using the bracket notation with a symbol.