richarddmorey

library(dplyr)
library(ggplot2)
library(ggdist)

url = 'https://osf.io/846cb/download'
httr::GET(url, httr::write_disk(tf <- tempfile(fileext = ".xlsx")))

readxl::read_xlsx(tf) %>%
  mutate(
    DIRECTION = factor(DIRECTION, levels = c("WEST","EAST"), 

richarddmorey

---
title: "Test SurveyJS and Rmarkdown"
output: 
  html_document:
    css: "https://unpkg.com/survey-jquery/survey.min.css"
editor_options: 
  chunk_output_type: console
---

A test javascript survey by building a JSON object for [SurveyJS](https://surveyjs.io/).

richarddmorey

richarddmorey

# Test answer to https://stackoverflow.com/questions/68473292/prevent-double-execution-of-shiny-reactive-with-two-dependency-pathways/68475515#68475515
library(shiny)
library(digest)

ui <- fluidPage(
  
  sidebarLayout(
    sidebarPanel(
      textInput("M", "M:", ""), 
      uiOutput("S_UI"),

richarddmorey

## Define MVTN parameters
mu = c(-1,0,1)
Sigma = matrix(
  c(1,  0,  0,
    0,  1, .5,
    0, .5,  1 
  ), 3,3)
p = length(mu)

library(mvtnorm)

richarddmorey

N = 20

library(mvtnorm)

# No shared variance between a and (b,c)
Sigma = matrix(c(1,0,0,
          0,1,.5,
          0,.5,1), 3, 3)

# Simulation

richarddmorey

https://www.theguardian.com/science/2021/feb/08/can-you-solve-it-think-of-a-number

The suggested answer using "yes" and "no" amounts to encoding in base 2, which is inefficient because there are
three possible response options ("yes", "no", "I don't know"). We want to find a way of encoding in base 3, which 
amounts to assigning 1/3 of the numbers to "yes", 1/3 to "no", and 1/3 to "I don't know". We can do that by telling
Johnny on the first step (and analogously afterward):

"I have two numbers in mind. One of them is 334. The other is less than 667 but greater than 333, but I will not tell you what it is. Is your
number less than both of my numbers?" 

richarddmorey

x = rlogis(10000,0,10)

mod = "model{
y ~ dlogis(0, 1/10)
}
"
library(rjags)
m = rjags::jags.model(file = textConnection(mod))
s = rjags::coda.samples(m,n.iter = 10000,variable.names = "y")

richarddmorey

library(tictoc)
df = data.frame(y = rnorm(10000000), x = rbinom(10000000,1,.5))

tic()
z = df[df$x == 1,]$y
toc()

rm(z)

tic()

richarddmorey

N = c(16, 16)
y = c(3, 9)
actual_result = prop.test(y, N, correct = FALSE)


compute_p_value = function(p0){
  ## Probability of all outcomes,
  ## assuming independence
  pr_X2s = 
    outer( dbinom(0:N[1], N[1], p0),