Assignment 1_Answer key.R

######### RUN THE CODE BELOW IN R. R-STUDIO IS THE RECOMMENDED IDE. BOTH R AND R-STUDIO ARE FREE.
######### QUESTIONS SHOULD BE POSTED TO PIAZZA
######### THE ACTUAL ASSIGNMENT BEGINS ON LINE 71


### Multilateral Development Institution Data
foo <- read.csv("https://tinyurl.com/yb4phxx8") # read in the data

# column names
names(foo)

# dimensions of the data set
dim(foo)

# quick look at the data structure
head(foo)

# one thing to be very careful with (in this data set) is the use of dates. 8 columns involve dates.

# take note of the columns representing calendar dates
date.columns <- c(11, 12, 14, 15, 16, 17, 18, 25)

# these columns need some tweaking--I want to address missing values, calling the blank (empty) 
# elements "NA" instead of leaving them blank, and I wish to tell R these are "Date" objects.

for(i in date.columns)  # this "for loop" only loops through the "date.columns" -- no other columns.
  
{
  
  # identify which values are missing in the "i"th column of the foo data set
  which_values_are_missing <- which(as.character(foo[, i]) == "")
  
  # those values that are missing (blank) in the "i"th column are replaced by <NA>
  # because R knows how to handle "NA" -- NA means something special in R--blanks are handled 
  # more unpredictably (which is bad).
  foo[which_values_are_missing, i] <- NA
  
  # last step--replace each of these columns (which is structured as a column of "factor" values)
  # as a column of dates--i.e., convert them to an object of "class" = Date. They are dates, after all.
  # And if you convert them to the Date class, R will know they are dates and you can manipulate 
  # dates in a simple, straightforward way. Otherwise, you won't be able to easily manipulate them
  # arithmetically.  E.g., for simple Date operations, see lines 48-58 below...
  # **By the way, if you don't understand what a "factor" is in R, you should Google it.** 
  foo[, i] <- as.Date(as.character(foo[, i]))
  
}

# Now R knows that these columns are comprised of dates
# for example...  Replicate this yourself...

foo[3,12]
# [1] "1968-03-13"

foo[4,12]
# [1] "1968-07-03"

foo[3,12] - foo[4,12]
# Time difference of -112 days

# Also, one additional helpful hint... How to eliminate rows with NAs...
# The "is.na" function--for more info, Google it or type ?is.na at the R command prompt in the console.
which.have.NAs <- which(is.na(foo$Rating == TRUE)) # for which rows is the claim "is.na" a TRUE claim?

# Then, if you wanted to, e.g., remove all those rows, retaining only the rows with ratings...
new_foo <- foo[-which.have.NAs, ]
# Notice I called this tweaked data set "new_foo" instead of rewriting over the original data set...
# It's a bit safer to do this, in case I decide I want to quickly revert back to the original data set.

###########################################################################

### ASSIGNMENT 1 -- You may want to read ALL the questions before you begin. 
### NOTE: FOR ALL QUESTIONS BELOW, ONLY CONSIDER PROJECTS WITH 
### non-missing "Circulation.Date" >= 2009-01-01. 
### EXCLUDE ALL OTHER PROJECTS FROM YOUR ANALYSIS.
### YOU MUST provide a link to your R code. ------ DON'T FORGET TO DO THIS!!!!!!!!!!!!
# Take note of the column names: i.e., you can type: names(foo)
# fyi: the column called "Rating" is the success rating at completion. 0 = lowest, 3 = highest.


noNA_foo <- new_foo[!is.na(new_foo$CirculationDate), ]
df <- noNA_foo[which(noNA_foo$CirculationDate >= as.Date("2009-01-01")), ]

# Check:
summary(df$CirculationDate)



# (1) ... (see prompt in the assignment)
# (a) Is this claim true? Explain. 

# Some projects have no original completion dates.
# We need to disregard them.

mask <- which(is.na(df$OriginalCompletionDate))
df_withDates <- df[-mask,]
expected_duration <- mean(df_withDates$OriginalCompletionDate) - mean(df_withDates$ApprovalDate) 
expected_duration
# Time difference of 650.9313 days, which is slightly less than two full years.
# The claim is partially true depending on your tolerance.

# (b)

# Create a circulation year column
df_withDates$CirculationYear <- format(df_withDates$CirculationDate, "%Y")

# Are there rows with NAs in the Revised Completion Date column?
sum(is.na(df_withDates$RevisedCompletionDate))  # No

# And an delay column
df_withDates$Delay <- df_withDates$RevisedCompletionDate - df_withDates$OriginalCompletionDate 

# The simplest way to do it probably involves a for loop in which you iterate through
# circulation years. I'm providing a fancier solution so that you can get used to 
# reading more complex R code. The dplyr library provides a data processing pipeline
# that is very similar to data analysis in SQL. DataCamp has tons of resources on it.

library(dplyr)  
delayByYear <- df_withDates %>%  # %>% means "pass what is to the left to the function that follows"
  group_by(CirculationYear) %>%    
  summarise(mean.delay = mean(Delay), 
            median.delay = median(Delay),
            IQR.delay = quantile(Delay, 0.75) - quantile(Delay, 0.25))
delayByYear

# Not the prettiest plot, but you get the point.
plot(delayByYear$CirculationYear, delayByYear$mean.delay, 
     pch=15, col="red", ylim=c(100, 800), 
     xlab="Circulation Year", ylab="Delay (days)", main="Project Delay")
points(delayByYear$CirculationYear, delayByYear$median.delay, pch=16, col="blue")
points(delayByYear$CirculationYear, delayByYear$IQR.delay, pch=17, col="green")
legend("bottomleft", pch=c(15,16,17), 
       col=c("red","blue","green"), 
       legend=c("Mean Delay", "Median Delay", "IQR of Delays"))
grid(nx=NA, ny=NULL)

# RESPONSE:
# Project delays have fluctuated over the years. Starting with an average of 658 days in
# 2009, it decreased to 503 in 2013 before increasing again to 627 in 2015. The average went
# down to 566 in 2018. The median delay also fluctuates but tends to be much lower than mean delays.
# The IQR follows a similar trend to the mean.

# (c)

# I did a year-by-year analysis, but this was not necessary.

# Create an actual duration column
df_withDates$ActualDuration <- df_withDates$RevisedCompletionDate - df_withDates$ApprovalDate

# Actual duration statistics
mean(df_withDates$ActualDuration) 
median(df_withDates$ActualDuration)
quantile(df_withDates$ActualDuration)
IQR(df_withDates$ActualDuration) 

df_withDates$ExpectedDuration <- df_withDates$OriginalCompletionDate - df_withDates$ApprovalDate

mean(df_withDates$ExpectedDuration)
median(df_withDates$ExpectedDuration) 
quantile(df_withDates$ExpectedDuration) 
IQR(df_withDates$ExpectedDuration)
range(df_withDates$ExpectedDuration)

ActualDurationByYear <- df_withDates %>%  
  group_by(CirculationYear) %>%    
  summarise(mean.duration = mean(ActualDuration), 
            median.duration = median(ActualDuration),
            IQR.duration = quantile(ActualDuration, 0.75) - quantile(ActualDuration, 0.25))
ActualDurationByYear

plot(ActualDurationByYear$CirculationYear, 
     ActualDurationByYear$mean.duration, pch=15, col="red", ylim=c(300, 1300),
     xlab="Circulation Year", ylab="Duration (days)", main="Actual Project Duration")
points(ActualDurationByYear$CirculationYear, 
       ActualDurationByYear$median.duration, pch=16, col="blue")
points(ActualDurationByYear$CirculationYear, 
       ActualDurationByYear$IQR.duration, pch=17, col="green")
legend("bottomleft", pch=c(15,16,17), 
       col=c("red","blue","green"), 
       legend=c("Mean", "Median", "IQR"))
grid(nx=NA, ny=NULL)

# RESPONSE:

# A similar analysis as shown in 1(b)

# (2) 

# We want project between 2010 and now
df2010 <- noNA_foo[which(noNA_foo$CirculationDate >= as.Date("2010-01-01")), ]

# The most straightforward way is...
print("Distribution of Project Ratings")
prop.table(table(df2010$Rating)) * 100  
# 2.4% of projects received a rating of 0, 12.9% of 1, 71% of 2, and 13% of 3. Hence, the 
# majority of projects receive a rating of 2.


# (3) 

# Only PATA projects
df_PATA <- df2010[which(df2010$Type == "PATA"),]

print("Distribution of Project Ratings (PATA only)")
table(df_PATA$Rating) / length(df_PATA$Rating) * 100  
# 1% of projects received a rating of 0, 8% of 1, 72% of 2, and 19% of 3. 
# We see that PATA have less projects rated as 0 or 1 and more rated as 3.


# (4) 


# There are several ways to do it. Here's a way:
df_TopQuant <- df[which(df$RevisedAmount >= quantile(df$RevisedAmount, 0.9)),]
df_BottomQuant <- df[which(df$RevisedAmount <= quantile(df$RevisedAmount, 0.1)),]

# Compare the ratings
(table(df_TopQuant$Rating) / length(df_TopQuant$Rating) * 100) - 
  (table(df_BottomQuant$Rating) / length(df_BottomQuant$Rating) * 100)
# The differences seem to be quite small (i.e., about 2% difference max)

# Compare other features
# I will look at count instead of percentages because, given that there are 
# more categories, the percentages might be misleading (e.g., a difference of 100%
# because there is one project of a given type )
# Dept
table(df_TopQuant$Dept) - table(df_BottomQuant$Dept)
# More projects in the top quantile are within the SDCC, whereas
# more projects in the bottom  quantile are in the SERD, ERCD and SARD departments.

# Cluster
(table(df_TopQuant$Cluster) / length(df_TopQuant$Cluster) * 100) - 
  (table(df_BottomQuant$Cluster) / length(df_BottomQuant$Cluster) * 100)
# A few significant differences percentage-wise, but if you look at the counts you will
# see that there's a lot of missing data here, so not very conclusive.

# Country
(table(df_TopQuant$Country) / length(df_TopQuant$Country) * 100) - 
  (table(df_BottomQuant$Country) / length(df_BottomQuant$Country) * 100)
# Several differences here.

# Hence, the budget seems to be decorrelated with the rating. However,
# it might be associated with some other project characteristics, such as country
# department. Nevertheless, we would not be able to draw causal conclusions given that
# the data is observational and neither unobservable nor observable characteristics
# are balanced across the "treatment" and control group.



# (5)
# (a) decision problem or objective?
# R: Minimize project completion delays
# (b) lever or levers?
# R: Budget alocation (we may want to consider levers which describe how the budget is spent as well)
# (c) ideal RCT design?
# R: Allocate budget randomly, which will in expectation make projects with more and less
# budget comparable because their other characteristics will overlap, and then measure
# the difference in ratings. 
# (d) dependent variable(s) and independent variable(s) in the modeler
# R: The independent variable will be the budget allocated, whereas the dependent variable
# will be the project delay.
# (e) And---Why would running RCTs and modeling/optimizing over RCT results be preferable 
# to using (observational, non-RCT) "foo" data?
# R: Because in observational data, projects with higher and lower budget allocations
# are not guaranteed to be comparable in expectation. For example, such projects may be deployed
# in different regions of the world, which might affect the difference in ratings. Hence,
# We cannot infer that the difference in ratings is due to the difference in budget given that
# there are confounding variables.