PA1_template.md

Reproducible Research: Peer Assessment 1

First, I load necessary libraries with messages supressed. Doing this once so future messages aren't silently supressed

library(dplyr)
library(ggplot2)

Loading and preprocessing the data

First I load the data and transform the date column into a more usable format

activity <- read.csv("activity.csv", header=TRUE)
#Convert the dates
Date <- strptime(activity$date, "%Y-%m-%d")

#Create a logical vectors of everything other than the original date
#then bind those columns with the new Date back to our data frame
lv <- names(activity) != "date"
activity <- cbind(activity[,lv], Date)

#Does everything look ok?
summary(activity)

##      steps          interval         Date                    
##  Min.   :  0.0   Min.   :   0   Min.   :2012-10-01 00:00:00  
##  1st Qu.:  0.0   1st Qu.: 589   1st Qu.:2012-10-16 00:00:00  
##  Median :  0.0   Median :1178   Median :2012-10-31 00:00:00  
##  Mean   : 37.4   Mean   :1178   Mean   :2012-10-31 00:25:34  
##  3rd Qu.: 12.0   3rd Qu.:1766   3rd Qu.:2012-11-15 00:00:00  
##  Max.   :806.0   Max.   :2355   Max.   :2012-11-30 00:00:00  
##  NA's   :2304

What is mean total number of steps taken per day?

totalByDay = activity %>%
              group_by(Date) %>%
              summarise(sum=sum(steps))

hist(totalByDay$sum, breaks=10, main = "Histogram of total daily steps",
     xlab = NULL)

Calculate mean and median of the daily totals, ignoring days that totalled to NA values. Based on the histogram I am expecting values between 10,000 and 11,000

mean(totalByDay$sum, na.rm=TRUE)

## [1] 10766

median(totalByDay$sum, na.rm=TRUE)

## [1] 10765

What is the average daily activity pattern?

averageByInterval <- activity %>%
                      group_by(interval) %>%
                      summarise(avg=mean(steps, na.rm = TRUE))
plot(averageByInterval, type="l", ylab ="Average number of steps")

Which interval has the maximum average number of steps?

averageByInterval <- activity %>%
                      group_by(interval) %>%
                      summarise(avg=mean(steps, na.rm = TRUE))

averageByInterval[averageByInterval$avg==max(averageByInterval$avg),]

## Source: local data frame [1 x 2]
## 
##     interval   avg
## 104      835 206.2

Imputing missing values

How many missing values are we dealing with? Is it a large proportion?

sum(is.na(activity$steps))

## [1] 2304

sum(is.na(activity$steps)) / nrow(activity)

## [1] 0.1311

I will replace the missing values with the average number of steps taken during that interval for which we have values

# Get just the rows with missing steps
missing <- activity[is.na(activity$steps),]

# Add a column for imputed values. This combination of mutate()
# and sapply() is a little messy -- there is probably a nicer way
missing <- mutate(missing, 
                  steps=sapply(interval, 
                               function(y) averageByInterval[averageByInterval$interval == y,]$avg))

# Create a new data frame with 1) original rows with non-null steps 2) newly created data points

newActivity = rbind(activity[!is.na(activity$steps),],
                    missing)

Lets check that we have the same number of rows we started with and that there are no null values in steps any more

nrow(activity) == nrow(newActivity)

## [1] TRUE

sum(is.na(newActivity$steps))

## [1] 0

Now I recreate our original histogram and recacluate the mean and median with our new data set.

newTotalByDay = newActivity %>%
                group_by(Date) %>%
                summarise(sum=sum(steps))

hist(newTotalByDay$sum, breaks=10, 
     main = "Histogram of total daily steps",
     xlab = NULL)

mean(newTotalByDay$sum)

## [1] 10766

median(newTotalByDay$sum)

## [1] 10766

Imputing missing values has virtually no impact on my estimates of total number of steps taken each day.

format(mean(newTotalByDay$sum) / mean(totalByDay$sum, na.rm = TRUE), nsmall=5)

## [1] "1.00000"

format(median(newTotalByDay$sum) / median(totalByDay$sum, na.rm = TRUE),nsmall=5)

## [1] "1.00011"

Are there differences in activity patterns between weekdays and weekends?

First I add a new factor variable indicating weekday or weekend

newActivity <- newActivity %>%
               mutate(dayType = "weekday")
newActivity[weekdays(newActivity$Date) %in% c("Saturday", "Sunday"),"dayType"] <- "weekend"
newActivity$dayType <- as.factor(newActivity$dayType)

newAverageByInterval <- newActivity %>%
                        group_by(interval, dayType) %>%
                        summarise(avg=mean(steps))

Is the average by interval different for weekdays and weekend? In general yes:

• Weekday has a higher peak in the morning
• Weekends are a little more dispersed during the day

ggplot(newAverageByInterval) + 
  geom_line(aes(x=interval,y=avg)) + 
  facet_wrap(~dayType)