PA1_template.md

title	author	date	output
Reproducible Research: Peer Assessment 1	Xavier Musy	Novermber 16, 2014	html_document keep_md true

library(plyr)
library(timeDate)

Loading and preprocessing the data

 ## unzip file if we need to
fileName <- "activity.csv"
if (!file.exists(fileName)){
  message("unzipping data...")
  unzip("activity.zip")
}

## read data
activity <- read.csv(fileName)

## explore data
head(activity)

##   steps       date interval
## 1    NA 2012-10-01        0
## 2    NA 2012-10-01        5
## 3    NA 2012-10-01       10
## 4    NA 2012-10-01       15
## 5    NA 2012-10-01       20
## 6    NA 2012-10-01       25

## lots of NA steps, so let's see where there's data
head( activity[ which(activity$steps > 0,),] )

##     steps       date interval
## 555   117 2012-10-02     2210
## 556     9 2012-10-02     2215
## 627     4 2012-10-03      410
## 631    36 2012-10-03      430
## 644    25 2012-10-03      535
## 647    90 2012-10-03      550

What is mean total number of steps taken per day?

sumDaySteps <- aggregate(steps ~ date, data=activity, FUN=sum, na.rm=TRUE)
hist(sumDaySteps$steps, xlab="Steps", ylab="Frequency (days)", main="Steps per day histogram", breaks=25)
abline(v=mean(sumDaySteps$steps, na.rm=T), col = "blue", lwd = 2, lty="dashed")

Mean number of steps per day:

mean(sumDaySteps$steps)

## [1] 10766

Median number of steps per day:

median(sumDaySteps$steps)

## [1] 10765

What is the average daily activity pattern?

meanPerInterval = aggregate(steps ~ interval, activity, mean, rm=TRUE)
plot(meanPerInterval, type="l", main="Average number of steps per 5-min interval")

Accross all days, the interval with the maximum number of steps is:

meanPerInterval[which(meanPerInterval$steps == max(meanPerInterval$steps)), "interval"]

## [1] 835

Imputing missing values

Number of rows with missing values:

length(which(is.na(activity)))

## [1] 2304

As an imputing data strategy, we impute NA steps by replacing these with the mean number of steps for that interval. We impute the data in a seperate data set.

names(meanPerInterval)[names(meanPerInterval)=="steps"] <- "mean.steps" # rename steps before merging in
imputedActivity <- merge(x=activity, y=meanPerInterval, by="interval", all.x=TRUE) # merge mean into activity set
imputedActivity <- arrange(imputedActivity,date, interval) # sort back to what it was, by date, by interval
imputedIndex <- which(is.na(imputedActivity$steps)) # find rows of NA steps
imputedActivity[imputedIndex, "steps"] <- round(imputedActivity[imputedIndex, "mean.steps"]) # impute data

We plot imputed data:

sumDayStepsImputed <- aggregate(steps ~ date, data=imputedActivity, FUN=sum)
hist(sumDayStepsImputed$steps, xlab="Steps", ylab="Frequency (days)", main="Imputed steps per day histogram", breaks=25)
abline(v=mean(sumDayStepsImputed$steps), col = "blue", lwd = 2, lty="dashed")

Mean number of steps per day:

mean(sumDayStepsImputed$steps)

## [1] 10766

Median number of steps per day:

median(sumDayStepsImputed$steps)

## [1] 10762

For imputed data, the mean number of steps remained the same, but the median changed slightly.

Are there differences in activity patterns between weekdays and weekends?

Let's mark which data is weekday or weekend

activity$isweekday <- as.factor(ifelse(isWeekday(activity$date), "weekday", "weekend"))

We observe that weekends have an increase in steps per 5-minute interval throughout the day, albeit starting a little later in the day than on weekdays. While weekdays have a higher number of steps as a single peak in the day.

par(mfrow=c(2,1))
meanPerIntervalWeekend = aggregate(steps ~ interval, activity[activity$isweekday == "weekend",], mean) 
plot(meanPerIntervalWeekend, type="l", main="Average number of steps per 5-min interval on weekends")
abline(h=mean(meanPerIntervalWeekend$steps), col = "blue", lwd = 2, lty="dashed")
meanPerIntervalWeekday = aggregate(steps ~ interval, activity[activity$isweekday == "weekday",], mean) 
plot(meanPerIntervalWeekday, type="l", main="Average number of steps per 5-min interval on weekdays")
abline(h=mean(meanPerIntervalWeekday$steps), col = "blue", lwd = 2, lty="dashed")