Reproducible Research: Peer Assessment 1

Jonathan Hill

Loading necessary R packages

library(dplyr)

## 
## Attaching package: 'dplyr'
## 
## The following object is masked from 'package:stats':
## 
##     filter
## 
## The following objects are masked from 'package:base':
## 
##     intersect, setdiff, setequal, union

library(ggplot2)

## Warning: package 'ggplot2' was built under R version 3.1.3

Loading and preprocessing the data

I’ll name my dataset “activity” and convert the date variable from a factor to a date.

activity <- read.csv("activity.csv",header=TRUE)
activity$date <- as.Date(activity$date)

Then process the data so that it is summarized by the day and calculate the mean and median.

activityDay <- activity %>%
        group_by(date) %>%
        summarize (stepsDay = sum(steps)
                   )
meanstepsperday <- mean(activityDay$stepsDay,na.rm=TRUE)
medianstepsperday <- median(activityDay$stepsDay,na.rm=TRUE)

What is mean total number of steps taken per day?

The mean total number of steps taken per day is 1.076618910^{4} and the median is 10765.

Histogram of steps per day

ggplot(data=activityDay,mapping = aes(date,stepsDay))+
        geom_bar(stat="identity", fill ="dark blue")+
        labs(y="Steps per Day", x="Date", title="Histogram of Steps per Day")

## Warning: Removed 8 rows containing missing values (position_stack).

What is the average daily activity pattern?

First I will reshape the data and summarize it by the day.

activityInt <- activity %>%
        group_by(interval) %>%
        summarize (stepsInt = sum(steps, na.rm=TRUE)/61
        )

Then graph it here:

ggplot(data =activityInt,mapping=aes(interval,stepsInt))+
        geom_line()+
        labs(y="Average steps per interval",x="5-Minute Interval",title="Time Series Graph")

Next, I calculate the 5-minute interval in which the maximum number of steps are taken.

maxInt <- activityInt$interval[which.max(x = activityInt$stepsInt)]

The 5-minute interval in which the maximum average steps are taken is 835.

Imputing missing values

missingval <- sum(is.na(activity$steps))

The dataset has 2304 missing values. Therfore, I will merge it with the dataset from the average daily activity pattern above. Then I will use an ifelse statement to replace the missing values with the average number of steps in a given 5-minute interval.

mergedactivity <- merge(activityInt,activity,by="interval")
mergedactivity$steps <- ifelse(is.na(mergedactivity$steps),
                         mergedactivity$stepsInt,
                         mergedactivity$steps)

Finally, in order to bring the data back to its original state, I drop the average calculation variable.

mergedactivity$stepsInt <- NULL

Here is the new histogram.

cleanhist <- mergedactivity %>%
        group_by(date) %>%
        summarize (stepsDay = sum(steps)
        )

ggplot(data=cleanhist,mapping = aes(date,stepsDay))+
        geom_bar(stat="identity", fill ="dark blue")+
        labs(y="Steps per Day", x="Date", title="Histogram of Steps per Day")

cleanmean <- mean(cleanhist$stepsDay)
cleanmedian <- median(cleanhist$stepsDay)

The new mean is 1.058101410^{4} and the new median is 1.039510^{4}. By imputing the missing values, the mean decreases by 185.1749733 and the median decreases by 370.

Are there differences in activity patterns between weekdays and weekends?

Creating a new factor variable to indicate weekend or weekday:

mergedactivity$dow <- weekdays(mergedactivity$date)
mergedactivity$weekend <- ifelse(mergedactivity$dow == "Saturday","weekend",ifelse(mergedactivity$dow == "Sunday","weekend","weekday"))
mergedactivity$weekend <- as.factor(mergedactivity$weekend)

Manipulating the data

cleanInt <- mergedactivity %>%
        group_by(interval,weekend) %>%
        summarize (stepsInt = mean(steps, na.rm=TRUE)
        )

Creating the faceted line graphs

ggplot(data =cleanInt,mapping=aes(interval,stepsInt))+
        geom_line()+
        facet_grid(weekend~.)+
        labs(y="Average steps per interval",x="5-Minute Interval",title="Faceted Time Series Graph")