Introduction

It is now possible to collect a large amount of data about personal movement using activity monitoring devices such as a Fitbit, Nike Fuelband, or Jawbone Up. These type of devices are part of the “quantified self” movement - a group of enthusiasts who take measurements about themselves regularly to improve their health, to find patterns in their behavior, or because they are tech geeks. But these data remain under-utilized both because the raw data are hard to obtain and there is a lack of statistical methods and software for processing and interpreting the data.
This assignment makes use of data from a personal activity monitoring device. This device collects data at 5 minute intervals through out the day. The data consists of two months of data from an anonymous individual collected during the months of October and November, 2012 and include the number of steps taken in 5 minute intervals each day. #### Requirements to run

Make a directory named as "data" in your working directory to store the raw data.

Loading the libraries

library(data.table)
library(knitr)
library(dplyr)
library(ggplot2)
library(xtable)
opts_chunk$set(echo = TRUE, results = 'hold')       
#Define Text results. hold output pieces and push them to end of chunk

Specifying the URL and name of the file to be stored

fileURL<-"https://d396qusza40orc.cloudfront.net/repdata%2Fdata%2Factivity.zip"
fileStoredName<-"ActivityMonitorData"

Defining a function to load the data with try cases

getData<-function(fileOpen="ActivityMonitorData",direct="R_programming"){
      print(paste("The present working directory is:",workingDir<-getwd()))
      date<-format(Sys.Date(),"%Y")
      newFileName<-paste(fileStoredName,date,sep = "-")
      fileList<-list.files()
      if(sum("data"==fileList)>0){
            if(!file.exists(paste("./data/",newFileName,".zip",sep=""))){
                  download.file(fileURL,method="curl",destfile = paste("./data/",newFileName,".zip",sep=""))
                  if(!file.exists(paste("./data/",newFileName,sep="")))
                        unzip(zipfile=paste("./data/",newFileName,
                                            ".zip",sep=""), exdir=paste("./data/",newFileName,sep=""))
            }
      }
      else
            print("You seem to be in a different directory")
      actualDoc<-list.files(paste("./data/",newFileName,sep=""),full.names = T)
      
      output<- tryCatch(
            {
                  fread(actualDoc,sep=",",header=T,data.table = TRUE)
            },
            error=function(cond){
                  message(paste("file does not seem to exist:", actualDoc))
                  message("Here's the original error message:")
                  message(cond)
                  return(2)
            },
            warning=function(cond) {
                  message(paste("file caused a warning:", actualDoc))
                  message("Here's the original warning message:")
                  message(cond)
                  return(3)
            },
            finally={
                  message(paste("Processed file: ", actualDoc," was opened successfully"))
            }
      )
      output
}

Calling the function to load the data

activityData<-getData()
## Processed file:  ./data/ActivityMonitorData-2015/activity.csv  was opened successfully

Preprocessing the data

activityData$recordDate<-as.Date(strptime(activityData$date,"%Y-%m-%d"))
activityData<-mutate(activityData, steps=as.numeric(steps))

str() view of the preprocessed data

## Classes 'data.table' and 'data.frame':   17568 obs. of  4 variables:
##  $ steps     : num  NA NA NA NA NA NA NA NA NA NA ...
##  $ date      : chr  "2012-10-01" "2012-10-01" "2012-10-01" "2012-10-01" ...
##  $ interval  : int  0 5 10 15 20 25 30 35 40 45 ...
##  $ recordDate: Date, format: "2012-10-01" "2012-10-01" ...
##  - attr(*, ".internal.selfref")=<externalptr>

Processing the data to get the mean ,median and total steps taken in a day

groupedActivityData<- group_by(activityData, recordDate)
summary<-summarize(groupedActivityData, meanSteps=mean(steps,na.rm = T), totalSteps=sum(steps,na.rm =T),
                   medianSteps=median(steps,na.rm=T))
maximumSteps=max(summary$totalSteps)

The Total steps in a day are:

##  [1]     0   126 11352 12116 13294 15420 11015     0 12811  9900 10304
## [12] 17382 12426 15098 10139 15084 13452 10056 11829 10395  8821 13460
## [23]  8918  8355  2492  6778 10119 11458  5018  9819 15414     0 10600
## [34] 10571     0 10439  8334 12883  3219     0     0 12608 10765  7336
## [45]     0    41  5441 14339 15110  8841  4472 12787 20427 21194 14478
## [56] 11834 11162 13646 10183  7047     0

A histogram of the Total number of steps per day.

par(mfrow=c(1,1),bg="beige")
hist(summary$totalSteps, main = "Histogram: No. Of Steps", xlab="Total No. of steps per day",
     angle=120, col="brown", density=20,xlim=c( 0 , maximumSteps+5000),ylim=c(0,20),breaks=10)

plot of chunk unnamed-chunk-9

#<--!barplot(summary$total,names.arg = summary$recordDate, angle= 60, density= 20, col="gold", main="Bar Plot: No. Of Steps", ylab="Total No. of steps", xlab="days", axisnames = T, cex.names = 0.5, axes=T,axis.lty = 1,xpd = T,las=2,pos=1,beside = F)-->

The mean and median of the total steps per day are below:

mean(summary$totalSteps)
median(summary$totalSteps)
## [1] 9354.23
## [1] 10395

Make a time series plot (i.e. type = “l”) of the 5-minute interval (x-axis) and the average number of steps taken, averaged across all days (y-axis)

A graph of the average number of steps per 5-Minute-Interval.

par(mfrow=c(1,1),bg="beige")
plot(k$Group.1,k$x, type ="l",main="Time Series Plot of Average No. of Steps",
     xlab="5 minute Intervals", ylab="average steps taken", col="blue",xaxt="n")
axis(1,labels=T, las=2)
abline(h=maximum<-max(k$x),col="magenta", lwd=1)
abline(v=maximumInterval<-k$Group.1[k$x==max(k$x)],col="magenta",lwd=1)
text(y=150,x=round(k$Group.1[k$x==max(k$x)],3)-90,paste("max. Interval is=",round(maximumInterval,3)),
     col="magenta1", srt=90)

plot of chunk unnamed-chunk-12

Report which 5-minute interval has the max average of steps?

The 5-Minute-Interval with the highest number of steps is:

print(maximumInterval)
## [1] 835

Calculate and report the total number of missing values in the dataset

The total number of NAs in the original data is:

numberOfNA<-sum(is.na(activityData$steps))
numberOfNA
## [1] 2304

adding a new column to the data table, containing information about the missing values

newActivityData<-mutate(activityData,IsNA=is.na(activityData$steps))

Filling in the missing values by replacing the NA’s with the mean of the steps taken at the same time interval

answer<-c()
for(i in 1:nrow(newActivityData)){
      if(newActivityData$IsNA[i])
           answer[i]<- filter(k,Group.1==newActivityData$interval[i])$x
      else
            answer[i]<-newActivityData$steps[i]
}

This new dataframe has the estimated values of the NAs imputed in.

newActivityData<-newActivityData[,steps:=answer]

Processing the data to get the mean ,median and total steps taken in a day

newGroupedActivityData<-group_by(newActivityData, recordDate)
newSummary<-summarize(newGroupedActivityData, meanSteps=mean(steps,na.rm = T), totalSteps=sum(steps,na.rm = T),
                      medianSteps=median(steps,na.rm=T))

A histogram of the Total number of steps per day.

par(mfrow=c(1,1),bg="beige")
newMaximumSteps=max(newSummary$totalSteps)
hist(newSummary$totalSteps, main = "Histogram: No. Of Steps", xlab="Total No. of steps per day",
     angle= 120, col="brown", density=20,xlim=c( 0 , newMaximumSteps+5000),ylim=c(0,25),breaks=10)

plot of chunk unnamed-chunk-19

What is the impact of imputing missing data on the estimates of the total daily number of steps?

The Total steps in a day are:

print((newSummary$totalSteps))
##  [1] 10766.19   126.00 11352.00 12116.00 13294.00 15420.00 11015.00
##  [8] 10766.19 12811.00  9900.00 10304.00 17382.00 12426.00 15098.00
## [15] 10139.00 15084.00 13452.00 10056.00 11829.00 10395.00  8821.00
## [22] 13460.00  8918.00  8355.00  2492.00  6778.00 10119.00 11458.00
## [29]  5018.00  9819.00 15414.00 10766.19 10600.00 10571.00 10766.19
## [36] 10439.00  8334.00 12883.00  3219.00 10766.19 10766.19 12608.00
## [43] 10765.00  7336.00 10766.19    41.00  5441.00 14339.00 15110.00
## [50]  8841.00  4472.00 12787.00 20427.00 21194.00 14478.00 11834.00
## [57] 11162.00 13646.00 10183.00  7047.00 10766.19

The mean and median of the total steps per day are below:

mean(newSummary$totalSteps)
median(newSummary$totalSteps)
## [1] 10766.19
## [1] 10766.19

The new mean value is exactly the same as the original, however, the median is very slightly different. This shows that imputing values based on means does not create a large impact in the results. Because these new values were put in based on the other steps values, the data isn’t too different.

Given below is the data table divided between weekdays and weekends:

weekdayData<-newActivityData[,weekday:= (weekdays(newActivityData$recordDate)=="Sunday" |
                                               weekdays(newActivityData$recordDate)=="Saturday") ]
weekdayData<-mutate(weekdayData,weekday=as.factor(weekday))
levels(weekdayData$weekday)<- c("weekday", "weekend")
newK <- aggregate(weekdayData$steps, by= list(weekdayData$interval,weekdayData$weekday),
                  FUN = mean, na.rm=T)
colnames(newK) <- c("interval","day_of_week", "steps")

panel plot containing a time series of the average number of steps taken across all weekdays and weekend days

qplot(interval, steps, data=newK, facets = day_of_week~.,
      main = "Activity Pattern On Weekdays vs Weekends" ) + geom_line(col="black")

plot of chunk unnamed-chunk-23