Reproducible Research: Peer Assessment 1
Loading and preprocessing the data
Show any code that is needed to
1) Load the data (i.e. read.csv())
2) Process/transform the data (if necessary) into a format suitable for your analysis
## Get necessary packages loaded
install.packages('knitr', dependencies = T, repos="http://cran.rstudio.com/")## Installing package into 'C:/Users/Andrea/Documents/R/win-library/3.1'
## (as 'lib' is unspecified)
## also installing the dependency 'evaluate'## package 'evaluate' successfully unpacked and MD5 sums checked
## package 'knitr' successfully unpacked and MD5 sums checked
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## The downloaded binary packages are in
## C:\Users\Andrea\AppData\Local\Temp\Rtmp2Fwu5g\downloaded_packageslibrary(knitr)## Warning: package 'knitr' was built under R version 3.1.3if(!require(tidyr)){install.packages("tidyr", dependencies = T, repos="http://cran.rstudio.com/")}## Loading required package: tidyr library(tidyr)
if(!require(dplyr)){install.packages("dplyr", dependencies = T, repos="http://cran.rstudio.com/")}## Loading required package: dplyr
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## Attaching package: 'dplyr'
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## The following object is masked from 'package:stats':
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## filter
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## The following objects are masked from 'package:base':
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## intersect, setdiff, setequal, union library(dplyr)
if(!require(lubridate)){install.packages("lubridate", dependencies = T, repos="http://cran.rstudio.com/")}## Loading required package: lubridate library(lubridate)
if(!require(la)){install.packages("lattice", dependencies = T, repos="http://cran.rstudio.com/")}## Loading required package: la## Warning in library(package, lib.loc = lib.loc, character.only = TRUE,
## logical.return = TRUE, : there is no package called 'la'## Installing package into 'C:/Users/Andrea/Documents/R/win-library/3.1'
## (as 'lib' is unspecified)## package 'lattice' successfully unpacked and MD5 sums checked
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## The downloaded binary packages are in
## C:\Users\Andrea\AppData\Local\Temp\Rtmp2Fwu5g\downloaded_packages library(lattice)## Warning: package 'lattice' was built under R version 3.1.3## little functions annoying but necessary b/c I haven't been able to find a better way
as.numeric.factor <- function(x) {as.numeric(levels(x))[x]}
## pull down a fresh dataset if it doesn't already exist in your working directory
## Load the data & process
dataset_url <-"https://d396qusza40orc.cloudfront.net/repdata%2Fdata%2Factivity.zip"
if(!file.exists("Dataset.zip")){
download.file(dataset_url, destfile="Dataset.zip")}
if(!file.exists("activity.csv")){
unzip("Dataset.zip", overwrite= TRUE)}
if(!exists("activity")){
activity <- read.csv("activity.csv", sep = ",")}
dateDownloaded <- date()
cat("Data file downloaded & evaluated ", dateDownloaded," from ", dataset_url)## Data file downloaded & evaluated Mon May 25 15:03:43 2015 from https://d396qusza40orc.cloudfront.net/repdata%2Fdata%2Factivity.zipWhat is mean total number of steps taken per day?
For this part of the assignment, you can ignore the missing values in the dataset.
1) Calculate the total number of steps taken per day
TotalbyDay<-aggregate(activity$steps, list(Day=activity$date), FUN = sum, na.rm=TRUE)
names(TotalbyDay)<-c("Day", "TotalSteps")
head(TotalbyDay)## Day TotalSteps
## 1 2012-10-01 0
## 2 2012-10-02 126
## 3 2012-10-03 11352
## 4 2012-10-04 12116
## 5 2012-10-05 13294
## 6 2012-10-06 154202) If you do not understand the difference between a histogram and a barplot, research the difference between them. Make a histogram of the total number of steps taken each day
hist(TotalbyDay$TotalSteps, col = "green", breaks = 100
,xlab = c("Total Number of Steps per Day")
,xlim = c(0,25000)
, ylab = c("Frequency of Number of Daily Steps")
,main = "Histogram of Number of Daily Steps"
)
3) Calculate and report the mean and median of the total number of steps taken per day
sum1<- as.data.frame(rbind(
c("Total Number of Steps", sum(TotalbyDay$TotalSteps)),
c("Mean Number of Steps", trunc(mean(TotalbyDay$TotalSteps))),
c("Median Number of Steps", median(TotalbyDay$TotalSteps))
))
colnames(sum1)<-c("Description", "Number")
print(sum1)## Description Number
## 1 Total Number of Steps 570608
## 2 Mean Number of Steps 9354
## 3 Median Number of Steps 10395What is the average daily activity pattern?
1)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)
AvgbyInterval<-aggregate(activity$steps, list(interval=activity$interval)
, FUN = mean, na.rm=TRUE)
names(AvgbyInterval)<-c("interval", "AvgSteps")
x<- which.max(AvgbyInterval$AvgSteps)
maxAvg<-AvgbyInterval[x,]
label <- paste("Peak Avg Steps:",trunc(maxAvg$AvgSteps),"@ Time:"
,trunc(maxAvg$interval))
plot(AvgbyInterval$interval,AvgbyInterval$AvgSteps, type = "l"
, col = "blue", lwd = 2
,xlab = "Time Stamp (24HRMM)"
,ylab = "Average Number of Steps"
)
title("Average Number of Steps Across All Days per 5 Minute Interval")
text (maxAvg$interval+500, maxAvg$AvgSteps, label, cex = 0.8)
# dev.copy(png, file = "Plot1.png", width = 480, height = 480)2)Which 5-minute interval, on average across all the days in the dataset, contains the maximum number of steps?
print(paste("Time:",maxAvg$interval))## [1] "Time: 835"Imputing missing values
Note that there are a number of days/intervals where there are missing values (coded as NA). The presence of missing days may introduce bias into some calculations or summaries of the data.
1) Calculate and report the total number of missing values in the dataset (i.e. the total number of rows with NAs)
nrow(activity)## [1] 17568 colSums(is.na(activity))## steps date interval
## 2304 0 02)Devise a strategy for filling in all of the missing values in the dataset. The strategy does not need to be sophisticated. For example, you could use the mean/median for that day, or the mean for that 5-minute interval, etc.
Since I've already calculated the mean for the 5-minute intervals, I'll
recycle that data for the strategy by merging the AvgbyInterval.AvgSteps
with the data from activity.3)Create a new dataset that is equal to the original dataset but with the missing data filled in.
activity2<-merge(activity, AvgbyInterval, by.x = "interval")
remove<-subset(activity2, is.na(steps))
remove<-remove[,c(1,4,3)]
colnames(remove)<-c("interval", "steps","date")
keep<-subset(activity2, !is.na(steps))
keep<-keep[,1:3]
activity.imputed<-rbind(keep, remove)
nrow(activity.imputed)## [1] 17568 colSums(is.na(activity.imputed))## interval steps date
## 0 0 04a)Make a histogram of the total number of steps taken each day and Calculate and report the mean and median total number of steps taken per day.
TotalbyDay.Imputed<-aggregate(activity.imputed$steps
,list(Day=activity.imputed$date)
, FUN = sum, na.rm=TRUE)
names(TotalbyDay.Imputed)<-c("Day", "TotalSteps")
hist(TotalbyDay.Imputed$TotalSteps, col = "blue", breaks = 100
,xlab = c("Total Number of Steps per Day")
,xlim = c(0,25000)
, ylab = c("Frequency of Number of Daily Steps")
,main = "Histogram of Number of Daily Steps"
)
sum2<- as.data.frame(rbind(
c("Total Number of Steps", trunc(sum(TotalbyDay.Imputed$TotalSteps))),
c("Mean Number of Steps", trunc(mean(TotalbyDay.Imputed$TotalSteps))),
c("Median Number of Steps", trunc(median(TotalbyDay.Imputed$TotalSteps)))
))
colnames(sum2)<-c("Description", "Number")
print(sum2)## Description Number
## 1 Total Number of Steps 656737
## 2 Mean Number of Steps 10766
## 3 Median Number of Steps 107664b)Do these values differ from the estimates from the first part of the assignment?
yes
4c)What is the impact of imputing missing data on the estimates of the total daily number of steps?
summary<- merge(sum1, sum2, by = "Description")
summary$Number.x<-as.numeric.factor(summary$Number.x)
summary$Number.y<-as.numeric.factor(summary$Number.y)
colnames(summary)<-c("Description", "UnadjustedSteps", "AdjustedSteps")
summary<-mutate(summary,
PctChange.From.UnAdjusted =
trunc(100*((UnadjustedSteps - AdjustedSteps)/UnadjustedSteps)))
print(summary[,c(1,4)] )## Description PctChange.From.UnAdjusted
## 1 Mean Number of Steps -15
## 2 Median Number of Steps -3
## 3 Total Number of Steps -15Are there differences in activity patterns between weekdays and weekends?
For this part the weekdays() function may be of some help here. Use the dataset with the filled-in missing values for this part.
1) Create a new factor variable in the dataset with two levels - “weekday” and “weekend” indicating whether a given date is a weekday or weekend day.
activity3<-mutate(activity.imputed, DOW = wday(activity$date, label = TRUE))
day.id<-as.data.frame(cbind(
DOW = c("Sun","Mon","Wed","Fri","Sat","Thurs","Tues"),
End = c("weekend","weekday","weekday","weekday"
,"weekend","weekday","weekday")
))
activity3<-merge(activity3, day.id, by.x = "DOW")
print(str(activity3))## 'data.frame': 17568 obs. of 5 variables:
## $ DOW : Ord.factor w/ 7 levels "Sun"<"Mon"<"Tues"<..: 6 6 6 6 6 6 6 6 6 6 ...
## $ interval: int 210 2050 210 2055 215 2055 210 210 210 520 ...
## $ steps : num 0 0 56 0 0 0 0 0 0 6 ...
## $ date : Factor w/ 61 levels "2012-10-01","2012-10-02",..: 57 2 9 33 5 17 27 44 21 27 ...
## $ End : Factor w/ 2 levels "weekday","weekend": 1 1 1 1 1 1 1 1 1 1 ...
## NULL2) Make a panel plot containing 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 weekday days or weekend days (y-axis). See the README file in the GitHub repository to see an example of what this plot should look like using simulated data.
AvgByEnd.Imputed<-aggregate(activity3$steps,
list(TimeStamp=activity3$interval
,WeekEnd.WeekDay = activity3$End)
,FUN = mean, na.rm=TRUE)
names(AvgByEnd.Imputed)<-c("TimeStamp", "WeekEnd.WeekDay", "AvgSteps")
xyplot(AvgByEnd.Imputed$AvgSteps ~ AvgByEnd.Imputed$TimeStamp
| AvgByEnd.Imputed$WeekEnd.WeekDay
,layout = c(1, 2)
,type = "l"
,xlab = "5 minute Time Stamp"
,ylab = "Average Number of Steps")