PA1_template.md

Reproducible Research: Peer Assessment 1

Loading and preprocessing the data

First of all, I force the locale to US english. Then I load the data and convert the dates from factors to actual dates.

Sys.setlocale("LC_TIME", "en_US")

## [1] "en_US"

setwd("~/datascience/5. Reproducible Research/RepData_PeerAssessment1")
data <- read.csv(unz("activity.zip", "activity.csv"))
data$date <- as.Date(data$date)

What is mean total number of steps taken per day?

steps_per_day = aggregate(steps ~ date, data, sum)
hist(steps_per_day$steps, xlab="Steps per day", main="Histogram of steps per day")

# Mean number of total steps per day
mean_per_day=mean(steps_per_day$steps)
# Median number of total steps per day
median_per_day=median(steps_per_day$steps)

The mean total steps per day is 1.0766 × 10⁴ and the median total number of steps per day is 10765.

What is the average daily activity pattern?

data$date = NULL
mean_steps_by_interval = aggregate(.~interval, FUN=mean, data=data)

plot(mean_steps_by_interval, type="l")

# The highest interval
mean_steps_by_interval[which.max(mean_steps_by_interval$steps),]

##     interval steps
## 104      835 206.2

Imputing missing values

data <- read.csv(unz("activity.zip", "activity.csv"))
data_imputed = data
data_imputed$date <- as.Date(data_imputed$date)

# Number of NA's
sum(is.na(data$steps))

## [1] 2304

Use the previously calculated means per 5-minute interval to fill empty spots in the dataset

My Strategy is to use the previously computed means per interval. In a loop I will fill out any interval that has a NA in it with the mean of that interval over all other days.

for (i in 1:nrow(data_imputed)) {
  if (is.na(data_imputed[i,]$steps)) {
    data_imputed[i,]$steps = mean_steps_by_interval[which(mean_steps_by_interval$interval==data_imputed[i,]$interval), ]$steps
  }
}

steps_per_day = aggregate(steps ~ date, data_imputed, sum)
hist(steps_per_day$steps, xlab="Steps per day", main="Histogram of steps per day")

mean(steps_per_day$steps)

## [1] 10766

median(steps_per_day$steps)

## [1] 10766

The mean value does not differ from the non-imputed data, which is not surprising since I added mean values in the missing spots which will not change anything to the overall mean. The median has changed only slightly.

Are there differences in activity patterns between weekdays and weekends?

weekends = weekdays(data_imputed$date) %in% c("Saturday", "Sunday")
data_imputed$daytype = factor(weekends, labels=c("weekday", "weekend"))

mean_steps_by_interval_weekends = aggregate(.~interval, FUN=mean, data=data_imputed[which(data_imputed$daytype=="weekend"),])
mean_steps_by_interval_weekdays = aggregate(.~interval, FUN=mean, data=data_imputed[which(data_imputed$daytype=="weekday"),])

mean_steps_by_interval_weekdays$date = NULL
mean_steps_by_interval_weekdays$daytype = NULL
mean_steps_by_interval_weekends$date = NULL
mean_steps_by_interval_weekends$daytype = NULL
par(mfrow=c(2,1))


with(data, { 
  plot(mean_steps_by_interval_weekends, type="l", ylab="Number of steps", main="Mean steps per interval in weekends")
  plot(mean_steps_by_interval_weekdays, type="l", ylab="Number of steps", main="Mean steps per interval on weekdays")
})

As can be seen in the graph above, there is more walking activity during the day in the weekends than during weekdays. This could be due to a day job where the person mostly sits.