Read HDI (.csv) file

library(knitr)
file_url <- "https://raw.githubusercontent.com/amarnathbose/Datafiles/master/hdi2017_2.csv"
h <- read.csv(file_url)
kable(head(h))

Query: India’s position in HDI ranking

# What is India's rank in the 4 development attributes
which(h$country=='India')

## [1] 130

h[h$country=='India','hdi']

## [1] 0.64

rank(-h$hdi)[which(h$country=='India')]

## [1] 130

India’s rank is 130. This can also be seen by 130 and the HDI score is 0.64.

Query: India’s rank for 4 parameters, separately

paste('Life Exp rank = ',rank(-h$lifeexp)[which(h$country=='India')])

## [1] "Life Exp rank =  134"

paste('Exp Years Schooling rank = ',rank(-h$eyrsschool)[which(h$country=='India')])

## [1] "Exp Years Schooling rank =  128"

paste('Avg Years Schooling rank = ',rank(-h$ayrsschool)[which(h$country=='India')])

## [1] "Avg Years Schooling rank =  137"

paste('Per Capita GNI rank = ',rank(-h$pcgni)[which(h$country=='India')])

## [1] "Per Capita GNI rank =  124"

Investigation into Balanced Development

using 1. Life Expectancy 2. Expected and Average Years of Schooling 3. Per Capita GNI

l <- rank(-h$lifeexp)
e <- rank(-h$eyrsschool)
a <- rank(-h$ayrsschool)
p <- rank(-h$pcgni)
h <- cbind(h,l,e,a,p)
kable(head(h))

for (i in 1:dim(h)[1])
  h[i,'d'] <- max(h[i,8:11])-min(h[i,8:11])
kable(head(h),caption='Top 6 countries')

# 10 countries with extreme unbalanced HDI indicators
h <- h[order(-h$d),]
kable(head(h[,c(2,3,8:12)],10))

# 10 countries with best balanced HDI indicators
h <- h[order(h$d),]
kable(head(h[,c(2,3,8:12)],10))

The algorithm

Consider a unit circle inscribed in a square. The ratio of the areas of the circle and the square is \(\frac{\pi}{4}\). Simulate a large number of points in the square. (With the centre of the square defined as (0,0) this is equivalent to generating random (x,y) where \(-1\leqslant x \leqslant +1\) and \(-1\leqslant y \leqslant +1\). The points inside the circle have the property \(x^2 + y^2 \leqslant 1\), since the circle has unit radius, with centre (0,0)

x <- runif(10000,-1,1)
y <- runif(10000,-1,1)

z <- x^2 + y^2
pi_approx <- length(z[z<=1])*4/length(z)

Effect of varying the number of simulated points on the convergence of \(\pi\)

f <- function(ntrials) {
  x <- runif(ntrials,-1,1)
  y <- runif(ntrials,-1,1)
  z <- x^2 + y^2
  pia <- length(z[z<=1])*4/length(z)
  return(pia)
}
ntrials <- 10^seq(3,6,0.05)
pis <- sapply(ntrials, f)

library(ggplot2)
qplot(ntrials,pis,main='Monte Carlo simulation of pi',xlab='Trials',ylab='pi',geom=c("point", "line"))

Read the GenderDiscrimination data set

file_url <- "https://raw.githubusercontent.com/amarnathbose/Datafiles/master/GenderDiscrimination.csv"
g <- read.csv(file_url)
#g <- read.csv('GenderDiscrimination.csv',header=T)
kable(g[sample(1:dim(g)[1],6,replace=F),])

summary(g)

##     Gender      Experience        Salary      
##  Female:140   Min.   : 2.00   Min.   : 53400  
##  Male  : 68   1st Qu.: 7.00   1st Qu.: 66000  
##               Median :10.00   Median : 74000  
##               Mean   :12.05   Mean   : 79844  
##               3rd Qu.:16.00   3rd Qu.: 88000  
##               Max.   :39.00   Max.   :194000

Plot to see how salary relates to experience

qplot(Experience, Salary, data=g, main="Salary versus Experience")

## Superimpose the line of best fit on this scatterplot

options(repr.plot.width=6, repr.plot.height=5)
plot(g$Experience,g$Salary, main="Salary versus Experience")
l1 <- lm(Salary ~ Experience, data=g)
abline(l1)

## Examine the best fit linear model

names(summary(l1))

##  [1] "call"          "terms"         "residuals"     "coefficients" 
##  [5] "aliased"       "sigma"         "df"            "r.squared"    
##  [9] "adj.r.squared" "fstatistic"    "cov.unscaled"

l1$coefficients

## (Intercept)  Experience 
##   59033.075    1727.311

summary(l1)$r.squared

## [1] 0.3149884

summary(l1)$coefficients

##              Estimate Std. Error   t value     Pr(>|t|)
## (Intercept) 59033.075  2499.8476 23.614670 1.592315e-60
## Experience   1727.311   177.4755  9.732669 1.168931e-18

Model Salary on Experience as well as Gender

qplot(Experience, Salary, data=g, geom='point', color=Gender, main="Salary versus Experience - Gender Differentiated")

g <- cbind(g,Female=ifelse(as.numeric(g$Gender)==1,1,0))
l2 <- lm(Salary ~ Experience + Experience*Female + Female, data=g)
summary(l2)

## 
## Call:
## lm(formula = Salary ~ Experience + Experience * Female + Female, 
##     data = g)
## 
## Residuals:
##    Min     1Q Median     3Q    Max 
## -71048  -9278  -1701   9166  47932 
## 
## Coefficients:
##                   Estimate Std. Error t value Pr(>|t|)    
## (Intercept)        58299.3     2998.6  19.442  < 2e-16 ***
## Experience          2753.0      199.8  13.781  < 2e-16 ***
## Female              8034.3     4110.6   1.955    0.052 .  
## Experience:Female  -2086.2      287.3  -7.261 7.95e-12 ***
## ---
## Signif. codes:  0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1
## 
## Residual standard error: 15110 on 204 degrees of freedom
## Multiple R-squared:  0.5561, Adjusted R-squared:  0.5495 
## F-statistic: 85.18 on 3 and 204 DF,  p-value: < 2.2e-16

options(repr.plot.width=6, repr.plot.height=4)
plot(g$Experience,g$Salary, col=ifelse(as.numeric(g$Gender)==1,"blue","red"), 
     #pch=ifelse(as.numeric(g$Gender)==1,1,19), 
     pch=19,
     main="Salary versus Experience - Gender Differentiated",xlab="Experience in years",ylab="Salary")
legend("topleft", c("Female", "Male"), col=c("blue", "red"),lty=1)
abline(l2$coefficients[1]+l2$coefficients[3],l2$coefficients[2]+l2$coefficients[4],col="blue")
abline(l2$coefficients[1],l2$coefficients[2],col="red")

summary(l2)

## 
## Call:
## lm(formula = Salary ~ Experience + Experience * Female + Female, 
##     data = g)
## 
## Residuals:
##    Min     1Q Median     3Q    Max 
## -71048  -9278  -1701   9166  47932 
## 
## Coefficients:
##                   Estimate Std. Error t value Pr(>|t|)    
## (Intercept)        58299.3     2998.6  19.442  < 2e-16 ***
## Experience          2753.0      199.8  13.781  < 2e-16 ***
## Female              8034.3     4110.6   1.955    0.052 .  
## Experience:Female  -2086.2      287.3  -7.261 7.95e-12 ***
## ---
## Signif. codes:  0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1
## 
## Residual standard error: 15110 on 204 degrees of freedom
## Multiple R-squared:  0.5561, Adjusted R-squared:  0.5495 
## F-statistic: 85.18 on 3 and 204 DF,  p-value: < 2.2e-16

options(repr.plot.width=6, repr.plot.height=3.5)
ggplot(g, aes(x=Experience, y=Salary, color=Gender)) + geom_point() + geom_smooth(method='lm')

A function to return multiples of 3

# function to return multiples of 3
m3 <- function(f=1, t) {
  mul <- c()
  for (i in f:t) {
    k <- i%%3
    if (k==0)
      mul <- c(mul,i)
  }
  return(mul)
}

Invoke the function

j <- m3(,15) # m3(to=15)
j

## [1]  3  6  9 12 15

Now, a function to return multiples of an arbitrary integer, passed as a function argument

# function to return multiples of an arbitrary integer, m
mm <- function(f, t, m) {
  mul <- c()
  for (i in f:t) {
    k <- i%%m
    if (k==0)
      mul <- c(mul,i)
  }
  return(mul)
}

Invoke the improved function

m3(13,29)

## [1] 15 18 21 24 27

mm(13,29,3)

## [1] 15 18 21 24 27

mm(12,200,17)

##  [1]  17  34  51  68  85 102 119 136 153 170 187