julia

is there any possibility to change the RHS of non-linear constraints in julia?

由 邮差的信 提交于 2021-02-10 12:50:10
问题 Is there any possibility to change the RHS of non-linear constraints? using JuMP, Ipopt, Juniper,Gurobi,CPUTime T=0; ZT=zeros(2,1); ZB=zeros(2,1); #-----Model parameters-------------------------------------------------------- sig=0.86; landa=50; E=T0=T1=.0833; T2=0.75; gam2=1; gam1=0; a1=5; a2=4.22; a3=977.4; ap=977.4; C1=949.2; c0=114.24; f(x) = cdf(Normal(0, 1), x); #--------------------------------------------------------------------------- ALT= Model(optimizer_with_attributes(Juniper

How to properly configure Julia portable or standalone

由 守給你的承諾、 提交于 2021-02-10 12:44:35
问题 How to correctly configure Julia portable or stanalone, For everything to happen from the external storage USB drives: Add, update packages ? In Julia's helps is there any procedure? This is what happens to me after running the .bat file. From the USB flash I still have reference to the installation of Julia in C: 回答1: All you need to do is to configure your JULIA_DEPOT_PATH . This is the place where Julia is installing it's packages and holding pre-compiled codes. Assume your installation

outlier异常值检验原理和处理方法

由 别等时光非礼了梦想. 提交于 2021-02-10 09:32:42
https://study.163.com/provider/400000000398149/index.htm?share=2&shareId=400000000398149 （ 欢迎关注博主主页，学习python视频资源，还有大量免费python经典文章） Before we tackle how to handle them, let’s quickly define what an outlier is. An outlier is any data point that is distinctly different from the rest of your data points. When you’re looking at a variable that is relatively normally distributed, you can think of outliers as anything that falls 3 or more standard deviations from its mean. While this will suffice as a working definition, keep in mind that there’s no golden rule for defining what an outlier is.

Combinations that add up to a number - Julia lang

由 南笙酒味 提交于 2021-02-10 07:14:42
问题 I'm new to Julia. Is there a way to add up elements from a list that add up to a certain value target? I have done this with Python's itertools library like in the example below but I find it extremely slow for larger datasets. import itertools numbers = [1, 2, 3, 7, 7, 9, 10] result = [seq for i in range(len(numbers), 0, -1) for seq in itertools.combinations(numbers, i) if sum(seq) == 10] print result 回答1: While like mentioned by Kermit the problem is NP-hard, it is still worth knowing how

Combinations that add up to a number - Julia lang

由 人盡茶涼 提交于 2021-02-10 07:13:38
问题 I'm new to Julia. Is there a way to add up elements from a list that add up to a certain value target? I have done this with Python's itertools library like in the example below but I find it extremely slow for larger datasets. import itertools numbers = [1, 2, 3, 7, 7, 9, 10] result = [seq for i in range(len(numbers), 0, -1) for seq in itertools.combinations(numbers, i) if sum(seq) == 10] print result 回答1: While like mentioned by Kermit the problem is NP-hard, it is still worth knowing how

Combinations that add up to a number - Julia lang

由 做~自己de王妃 提交于 2021-02-10 07:13:18
问题 I'm new to Julia. Is there a way to add up elements from a list that add up to a certain value target? I have done this with Python's itertools library like in the example below but I find it extremely slow for larger datasets. import itertools numbers = [1, 2, 3, 7, 7, 9, 10] result = [seq for i in range(len(numbers), 0, -1) for seq in itertools.combinations(numbers, i) if sum(seq) == 10] print result 回答1: While like mentioned by Kermit the problem is NP-hard, it is still worth knowing how

Combinations that add up to a number - Julia lang

由 筅森魡賤 提交于 2021-02-10 07:11:27
问题 I'm new to Julia. Is there a way to add up elements from a list that add up to a certain value target? I have done this with Python's itertools library like in the example below but I find it extremely slow for larger datasets. import itertools numbers = [1, 2, 3, 7, 7, 9, 10] result = [seq for i in range(len(numbers), 0, -1) for seq in itertools.combinations(numbers, i) if sum(seq) == 10] print result 回答1: While like mentioned by Kermit the problem is NP-hard, it is still worth knowing how

Julia FixedEffectsModels IV regression doesn't match R IV regression

由 怎甘沉沦 提交于 2021-02-10 05:48:18
问题 When I run (what I think is the same regression) in R and in Julia I get very different results. I think this is because the IV regression is using an indicator variable to instrument for another indicator variable but I can't figure out if I am doing something else wrong. I have tried several different methods, but the data starts like this: rc D tau May_ret Jun_ret Jul_ret Aug_ret Sep_ret 1 -43 0 0 0.04529617 0.02106667 0.009868421 0.032573290 0.010473186 2 -19 0 0 0.01973333 0.05752213 -0

Using an array for in array indexing

由 徘徊边缘 提交于 2021-02-10 05:35:30
问题 Is there a nicer way to do this? a = rand(10,10) a[[CartesianIndex(i,i) for i in 1:10]] .= something I.e. using setindex! without a CartesianIndex or something like a[i,i for i in 1:10] .= something Or in general using setindex! with an array for syntax? 回答1: I like the CartesianIndex solution, but one other way would be to create a boolean mask: julia> a = rand(5, 5); julia> a[[i == j for i ∈ 1:5, j ∈ 1:5]] .= 0; julia> a 5×5 Array{Float64,2}: 0.0 0.376169 0.0248078 0.957535 0.522565 0

Plotly + Julia + Latex

由 六眼飞鱼酱① 提交于 2021-02-10 05:09:27
问题 I would like to use latex with the Plotly backend. Look in this example: Even the x-axis and y-axis log scales are not formatted properly! Do you know how to do that? Best, v. using Plots, LaTeXStrings plotlyjs() x = 10 .^ LinRange(-5,5,10) plot(x, 1 ./ x, yaxis=:log, xaxis=:log, m = 2, xlabel="x", label="1/x",ylabel="f(x)") plot!(x, 10 ./ x, m=2,label="10/x") plot!(x, 1 ./ (x.^2), m=2, label=L"f(x)=$1/x^2$") savefig("./test.png") 回答1: Seems to work with the gr() backend, so I expect
订阅 julia