SLB stats & predictions

Risorse Utili

Nel nostro impegno a fornire un servizio completo agli appassionati del basket SLB Great Britain, abbiamo compilato una lista delle risorse utilizzabili online. Queste includono link a guide pratiche, articoli informativi suggeriti dagli esperti del settore sportivo ed eventuale materiale educativo riguardante l'analisi statistica applicata alle scommesse sportive professionalmente organizzate.

• Guide Pratichedonnagoodman/Software-Development-for-Business-Analytics<|file_sep|>/R-Scripts/Project-1.Rmd --- title: "Project 1" author: "Donna Goodman" date: "10/8/2018" output: html_document --- {r setup, include=FALSE} knitr::opts_chunk$set(echo = TRUE) ## Project 1 #### Please see the attached pdf for all the code and outputs. {r} #install.packages("dplyr") library(dplyr) {r} #reading in the file projects <- read.csv(file="Projects.csv", header=TRUE) head(projects) {r} #showing the dimensions of the data set dim(projects) {r} #counting how many NA's there are sum(is.na(projects)) {r} #looking at the structure of the data set str(projects) {r} #finding out what unique values exist for the Target column unique(projects$Target) {r} #finding out what unique values exist for the Activity column unique(projects$Activity) #### Question 1 {r} #using select to extract columns for Target and Activity question1 <- projects %>% select(Target, Activity) head(question1) #### Question 2 {r} #using filter to extract rows with Target = "New Business" question2 <- projects %>% filter(Target == "New Business") head(question2) #### Question 3 {r} #using group_by to find out how many rows we have for each value of Target question3 <- projects %>% group_by(Target) %>% summarise(count = n()) question3 #### Question 4 {r} #using group_by and summarise to find out how many rows we have for each value of Target and Activity question4 <- projects %>% group_by(Target, Activity) %>% summarise(count = n()) question4 #### Question 5 {r} #using group_by and summarise to find out how many rows we have for each value of Target and Activity and then filtering by those that have count > 50 question5 <- projects %>% group_by(Target, Activity) %>% summarise(count = n()) %>% filter(count > 50) question5 #### Question 6 {r} #using group_by and summarise to find out how many rows we have for each value of Target and Activity and then filtering by those that have count > 50 question6 <- projects %>% group_by(Target) %>% summarise(count = n()) %>% filter(count > 50) question6 #### Question 7 {r} #using mutate to add new column called AverageAge that contains the average age of members in each project. question7 <- projects %>% mutate(AverageAge = mean(Age)) head(question7) Note that the above code also includes some NA values. {r} #looking at how many NA's are in Age column sum(is.na(projects$Age)) So there are NA values in Age. Using na.rm=TRUE removes the NA's from the calculation. {r} #using mutate to add new column called AverageAge that contains the average age of members in each project without NA values. question7b <- projects %>% mutate(AverageAge = mean(Age, na.rm=TRUE)) head(question7b) Note that this time there are no NA values. #### Question 8 {r} #using arrange to order rows by descending order of Start.Date question8 <- projects %>% arrange(desc(Start.Date)) head(question8) tail(question8) Note that this does not work because Start.Date is an object of type factor not Date. {r} #checking the structure of Start.Date str(projects$Start.Date) As we can see Start.Date is an object of type factor so it needs to be converted into type Date first before it can be ordered by arrange. First converting Start.Date into type Date using as.Date() function. Then using arrange to order rows by descending order of Start.Date {r} question8b <- projects %>% mutate(DateStart = as.Date(Start.Date)) %>% arrange(desc(DateStart)) head(question8b) tail(question8b) As we can see this works fine now. #### Question 9 We need to calculate the length of time each project lasts. To do this we need to subtract Start.Date from End.Date. However Start.Date and End.Date are objects of type factor not Date. So we need to convert both these columns into type Date first before we can subtract them from one another. First converting Start.Date and End.Date into type Date using as.Date() function. Then using mutate to create new column called Length that contains number of days between Start.Date and End.Date. Note that this calculation will give us negative numbers because we are subtracting Start.Date from End.Date. To correct this we will take absolute value of this calculation using abs() function. This will give us positive numbers only which will represent number of days between Start.Date and End.Date. Then using arrange to order rows by descending order of Length. {r} question9 <- projects %>% mutate(DateStart = as.Date(Start.Date), DateEnd = as.Date(End.Date), Length = abs(DateEnd - DateStart)) %>% arrange(desc(Length)) head(question9) tail(question9) As we can see this works fine now. #### Project Questions #### Q1: Which activity has been performed on most number of projects? Using group_by and summarise functions first grouped by Activity then counted number of times each activity appears in Projects dataset. Then sorted data frame in descending order so activity with most number of projects appears on top row. As we can see project activity with most number of projects is Campaign followed by Workshop followed by Seminar. {r} q1 <- projects %>% group_by(Activity) %>% summarise(count = n()) %>% arrange(desc(count)) q1 #### Q2: Which region has most number of new business activities? Using group_by and summarise functions first grouped by Region then counted number of times each region appears in Projects dataset where target equals