Richard Morey richarddmorey

## Bailey_Geary_2009.csv

          
            No
            homo
            CC
            Latitude
            north
            lat_from_o
            tpar
            meantemp
            hitemp
            lotemp
            diftemp
            Isosd
            dpop_30

            
              1
              1
              595
              27
              0
              35
              1
              17.8
              30
              1
              29
              0.218223437
              13

            
              2
              1
              475
              27
              0
              35
              1
              17.8
              30
              1
              29
              0.218223437
              13

            
              3
              1
              507
              27
              0
              35
              1
              17.8
              30
              1
              29
              0.218223437
              13

            
              4
              1
              570
              27
              0
              35
              1
              17.8
              30
              1
              29
              0.218223437
              13

            
              6
              1
              776
              4
              1
              4
              7
              24.7
              26
              11
              15
              0.212807364
              13

            
              7
              1
              506.333
              4
              1
              4
              7
              24.7
              26
              11
              15
              0.212807364
              13

            
              8
              1
              622.5
              4
              1
              4
              7
              24.7
              26
              11
              15
              0.212807364
              13

            
              5
              1
              500
              7
              1
              1
              7
              22.2
              25
              6
              19
              0.212807364
              13

            
              15
              1
              775
              40
              1
              32
              4
              5.8
              31
              -1
              32
              0.25577606
              16

## app2.R
# 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"),

## sample_constraints.R
## 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)

## corre_sim.R
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

## guess_number.txt
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?"

## rlogis.R
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")

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

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

rm(z)

tic()

## X2.R
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),

## eqt_pvalue.R

data(attitude)

library(cocor)
## EQtest at alpha = .05 needs a 90% CI
cocor(~ rating + complaints | rating + learning, data = attitude, alternative = "less",
      null.value = .2, test = "zou2007", conf.level = .9)

zou_ci_func = function(conf.level, ...)
{

## repro_subways.R
## Get file off OSF
tf = tempfile(fileext = ".xlsx")
download.file(url = "https://osf.io/846cb/download", destfile = tf)

## Create summaries
library(tidyverse)

readxl::read_xlsx(tf) %>%
  group_by(STN_NUMBER, DIRECTION) %>%
  summarise(
No	homo	CC	Latitude	north	lat_from_o	tpar	meantemp	hitemp	lotemp	diftemp	Isosd	dpop_30
1	1	595	27	0	35	1	17.8	30	1	29	0.218223437	13
2	1	475	27	0	35	1	17.8	30	1	29	0.218223437	13
3	1	507	27	0	35	1	17.8	30	1	29	0.218223437	13
4	1	570	27	0	35	1	17.8	30	1	29	0.218223437	13
6	1	776	4	1	4	7	24.7	26	11	15	0.212807364	13
7	1	506.333	4	1	4	7	24.7	26	11	15	0.212807364	13
8	1	622.5	4	1	4	7	24.7	26	11	15	0.212807364	13
5	1	500	7	1	1	7	22.2	25	6	19	0.212807364	13
15	1	775	40	1	32	4	5.8	31	-1	32	0.25577606	16
	# 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"),
	## 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)
	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
	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?"
	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")
	library(tictoc)
	df = data.frame(y = rnorm(10000000), x = rbinom(10000000,1,.5))

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

	rm(z)

	tic()
	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),

	data(attitude)

	library(cocor)
	## EQtest at alpha = .05 needs a 90% CI
	cocor(~ rating + complaints \| rating + learning, data = attitude, alternative = "less",
	null.value = .2, test = "zou2007", conf.level = .9)

	zou_ci_func = function(conf.level, ...)
	{
	## Get file off OSF
	tf = tempfile(fileext = ".xlsx")
	download.file(url = "https://osf.io/846cb/download", destfile = tf)

	## Create summaries
	library(tidyverse)

	readxl::read_xlsx(tf) %>%
	group_by(STN_NUMBER, DIRECTION) %>%
	summarise(