InClassEX_4/9

EX1

# 本題欲了解下列function之功能，先執行
lapply(lapply(search(), ls), length)

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# 從結果觀察，第一層lapply應為顯示出search()所找尋到物件的數量，第二層應為該物件數量之長度

EX2

functionP <- function(rate, loan, year){
  month <- 12*year
  pay <- outer(loan, (rate/(1-outer((1+rate),(-month), "^"))), "*")
  return(pay)
}

functionP(year = seq(10, 30, 5),
        loan = c(5000000, 10000000, 15000000, 20000000),
        rate = c(0.02, 0.05, 0.07, 0.09))

## , , 1
## 
##          [,1]      [,2]      [,3]      [,4]
## [1,] 110240.5  250718.6  350104.3  450014.5
## [2,] 220481.0  501437.1  700208.5  900029.0
## [3,] 330721.5  752155.7 1050312.8 1350043.6
## [4,] 440961.9 1002874.3 1400417.1 1800058.1
## 
## , , 2
## 
##          [,1]      [,2]      [,3]      [,4]
## [1,] 102913.7  250038.4  350001.8  450000.1
## [2,] 205827.4  500076.7  700003.6  900000.2
## [3,] 308741.0  750115.1 1050005.4 1350000.2
## [4,] 411654.7 1000153.5 1400007.2 1800000.3
## 
## , , 3
## 
##          [,1]      [,2]      [,3]    [,4]
## [1,] 100870.4  250002.1  350000.0  450000
## [2,] 201740.8  500004.1  700000.1  900000
## [3,] 302611.2  750006.2 1050000.1 1350000
## [4,] 403481.6 1000008.2 1400000.1 1800000
## 
## , , 4
## 
##          [,1]      [,2]    [,3]    [,4]
## [1,] 100263.7  250000.1  350000  450000
## [2,] 200527.4  500000.2  700000  900000
## [3,] 300791.1  750000.3 1050000 1350000
## [4,] 401054.8 1000000.4 1400000 1800000
## 
## , , 5
## 
##          [,1]    [,2]    [,3]    [,4]
## [1,] 100080.2  250000  350000  450000
## [2,] 200160.4  500000  700000  900000
## [3,] 300240.7  750000 1050000 1350000
## [4,] 400320.9 1000000 1400000 1800000

EX3

library(magrittr)

## Warning: package 'magrittr' was built under R version 3.4.3

## 
## Attaching package: 'magrittr'

## The following object is masked from 'package:purrr':
## 
##     set_names

## The following object is masked from 'package:tidyr':
## 
##     extract

dta3 <- read.table("hs0.txt", header = TRUE) 

# (a)
mapply(function(x,y){
  t.test(dta3[, x], dta3[ ,y])$p.value
  }, x = rep(7:11, each = 5), y = 7:11) %>%
  matrix(5, 5)

##           [,1]      [,2]      [,3]      [,4]      [,5]
## [1,] 1.0000000 0.5812665 0.6728327 0.7572410 0.8676815
## [2,] 0.5812665 1.0000000 0.8903494 0.3775863 0.7150322
## [3,] 0.6728327 0.8903494 1.0000000 0.4515844 0.8118467
## [4,] 0.7572410 0.3775863 0.4515844 1.0000000 0.6377587
## [5,] 0.8676815 0.7150322 0.8118467 0.6377587 1.0000000

# (b)
yvar <- names(dta3)[7:11]
lapply(yvar, function(y) {
  oneway.test(substitute(i ~ race, list(i = as.name(y))), data = dta3)$p.value}) %>%
  unlist()

## [1] 8.794587e-04 3.045127e-04 6.490966e-05 1.217540e-05 4.209418e-02

# (c)
lm(math ~ -1 + ses, data = dta3)

## 
## Call:
## lm(formula = math ~ -1 + ses, data = dta3)
## 
## Coefficients:
##   seshigh     seslow  sesmiddle  
##     56.17      49.17      52.21

ggplot(dta3, aes(ses, math))+
  stat_summary(fun.data = mean_cl_boot, na.rm = TRUE)+
  scale_x_discrete(limits = c("low", "middle", "high"))+
  scale_y_continuous(breaks = seq(40, 70, by = 1))+
  labs(x = "SES", y = "Average Math Score")

EX4

fcircle <- function(x) {sqrt(1^2 - x^2)}
N <- 10000
sum(fcircle(runif(N, 0, 1)) > runif(N, 0, 1)) / N * (1 * 1)

## [1] 0.7856