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# Discussion Figure 2 - Pressure audiograms for Elasmobranchs
setwd("H:/1. Shark Project/9. Data analysis/Behavioral Audiogram Analysis")
#getwd()

library(ggplot2)
library(ggpubr)
library(magrittr)


d <- read.table("Elasmo_audiogramms.txt",header=T,na.strings=c("NA","NaN",""),sep="\t")

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FBlikeNumber <-
  function(url) {
    require(magrittr)
    require(rvest)
    x <- read_html(url, encoding = "UTF-8") %>% as.character
    re <- regexpr("[,0-9]+(?= 人說這讚)", x, perl = T)
    substr(x, re, re + attr(re, "match.length")) %>%
      gsub(",", "", .) %>%
      as.integer
  }

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TEST_GUI = function() {
  library(tcltk2)
  win1 <- tktoplevel()
  win1$env$rb1 <- tk2radiobutton(win1)
  win1$env$rb2 <- tk2radiobutton(win1)
  win1$env$rb3 <- tk2radiobutton(win1)
  rbValue <- tclVar("PANDA")
  
  tkconfigure(win1$env$rb1, variable = rbValue, value = 1)
  tkconfigure(win1$env$rb2, variable = rbValue, value = 2)

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# see https://www.ptt.cc/bbs/R_Language/M.1540859231.A.848.html

theta <- 8 # 本例以 theta = 8 為例
n.delta <- 500 # 設定方格在 x 和 y 的格數
x.sim <- seq(-theta, theta, length = n.delta)
y.sim <- seq(-14, 14, length = n.delta) # 太窄或太寬可以調整
f.xy <-
  function(x, y, th) {
    1 / (2 * th * sqrt(2 * pi)) * exp(-1 / 2 * (y - x) ^ 2)
  }

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library(data.table)
library(ggplot2)
library(magrittr)

# raw data
set.seed(123)
d <-
  data.table(
    Week = gl(2, 12, labels = c("48", "49")),
    Factory = gl(3, 4, labels = c("Factory1", "Factory2", "Factory3")),

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rm(dat)
library(httr)
library(xml2)
library(RCurl)
library(XML)
library(dplyr)
library(data.table)


## 抓八卦板的資料

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# dat 為輸入的二因子資料
set.seed(1234)
dat <- data.frame(
    y = rnorm(80, 5, 2),
    x1 = gl(2, 40, labels = c("A","B")),
    x2 = gl(4, 10, labels = c("1","2","3","4"))
)

# 利用 tapply 求出各組之 mean 和 sd
y.mean <- with(dat, tapply(y, paste(x1, x2, sep="."), mean))

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# 第一題
# 已知 x 如下
x <- c(23,52,63,2,34,6,45,67,73,24,63,75,35,23,7,45,7,34)
# 以抽出且放回的方法隨機抽出 x 中的數，
# 抽出的量和 x 一樣多（也就是 18 個）。
# 求出這 18 個數的平均。
# 把上述的過程重複 10000 次，並把每次的平均記錄在一個 vector 「假的平均」裡。
# 最後，試著看懂 ?quantile，求出「假的平均」的 2.5％ 和 97.5% 分位數。
# 以下是提示，試著把 ??? 填上適當的 R code。
重覆次數 <- 10000

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## string to class Date 
dates <- c("01/27/92", "03/27/92")
x <- as.Date(dates, "%m/%d/%y") # see ?strptime to learn more

## class Date to string
format(x, "%x")
format(x, "%F")
format(x, "%c")
format(x, "%b %D %Y")