Introduction

This assignment makes use of data from a personal activity monitoring device. This device collects data at 5 minute intervals throughout the day. The data consists of two months of data from an anonymous individual collected during the months of October and November, 2012 and include the number of steps taken in 5 minute intervals each day.

This document presents the results from Project Assignment 1 in the Coursera course Reproducible Research, written in a single R markdown document that can be processed by knitr and transformed into an HTML file.

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
rm(list = ls()); cat("\014")


# setwd("C:/Users/mXXXX/OneDrive/061 Coursera/spec_DataScience/datascienceCoursera_5ReproR/week2_assignment")  #for Surface PC
setwd("C:/Users/marco/OneDrive/061 Coursera/spec_DataScience/datascienceCoursera_5ReproR/week2_assignment")  #for Home PC

url_data <- "https://d396qusza40orc.cloudfront.net/repdata%2Fdata%2Factivity.zip"
download.file(url_data, destfile = "./activity.zip",method = "auto")


# unzipp file
unzip(zipfile="./activity.zip")


dat <- read.csv(file = "activity.csv")


require(lubridate)
require(dplyr)
require(ggplot2)

dat$date <- ymd(dat$date)

What 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
  2. 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
  3. Calculate and report the mean and median of the total number of steps taken per day

dat.temp <- group_by(dat, date)
dat.summarize <- summarize(dat.temp
                           , sumSteps = sum(steps, na.rm = TRUE)
                           , meanSteps = mean(steps, na.rm =TRUE)
                           , sumInterval = sum(interval))

# let's get rid of the NaN values
dat.summarize_good <- dat.summarize[!is.na(dat.summarize$meanSteps), ]

step_mean <- mean(dat.summarize_good$sumSteps)
step_median <- median(dat.summarize_good$sumSteps)


ggplot(dat.summarize_good, aes(x = sumSteps)) +
        geom_histogram(aes(fill = ..count..), colour = "grey4", binwidth = 1000) + 
        labs(title = "Histogram of Steps per day", x = "Steps per day", y = "Frequency")

We have a mean value of 1.076618910^{4}
and a median value of 10765

What 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)
  2. Which 5-minute interval, on average across all the days in the dataset, contains the maximum number of steps?
dat.temp2 <- group_by(dat, interval)
dat.summarize2 <- summarize(dat.temp2
                           , meanSteps = mean(steps, na.rm =TRUE)
                           )
# let's get rid of the NaN values
dat.summarize2_good <- dat.summarize2[!is.na(dat.summarize2$meanSteps), ]

ggplot(dat.summarize2_good, aes(x = interval, y = meanSteps)) +
        geom_line(color = "firebrick")



dat.summarize2_good[which.max(dat.summarize2_good$meanSteps), ]
#> # A tibble: 1 x 2
#>   interval meanSteps
#>      <int>     <dbl>
#> 1      835  206.1698

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)
  2. 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.
  3. Create a new dataset that is equal to the original dataset but with the missing data filled in.
  4. 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. Do these values differ from the estimates from the first part of the assignment? What is the impact of imputing missing data on the estimates of the total daily number of steps?

sum(is.na(dat$steps))
#> [1] 2304


# this time, with tapply
dat.filled <- dat
ndx <- is.na(dat.filled$steps)
int_avg <- tapply(dat.filled$steps, dat.filled$interval, mean, na.rm=TRUE, simplify=T)
dat.filled$steps[ndx] <- int_avg[as.character(dat.filled$interval[ndx])]
sum(is.na(dat.filled$steps))
#> [1] 0


dat.temp3 <- group_by(dat.filled, date)
dat.summarize3 <- summarize(dat.temp3
                           , sumSteps = sum(steps, na.rm = TRUE)
                           , meanSteps = mean(steps, na.rm =TRUE))
head(dat.summarize3)
#> # A tibble: 6 x 3
#>         date sumSteps meanSteps
#>       <date>    <dbl>     <dbl>
#> 1 2012-10-01 10766.19  37.38260
#> 2 2012-10-02   126.00   0.43750
#> 3 2012-10-03 11352.00  39.41667
#> 4 2012-10-04 12116.00  42.06944
#> 5 2012-10-05 13294.00  46.15972
#> 6 2012-10-06 15420.00  53.54167

step_mean_filled <- mean(dat.summarize3$sumSteps)
step_median_filled <- median(dat.summarize3$sumSteps)


ggplot(dat.summarize3, aes(x = sumSteps)) +
        geom_histogram(aes(fill = ..count..), colour = "grey4", binwidth = 1000) + 
        labs(title = "Histogram of Steps per day", x = "Steps per day", y = "Frequency")

Are 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.
  2. 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.

dat.filled <- mutate(dat.filled, weektype = ifelse(weekdays(dat.filled$date, abbreviate = FALSE) == "Samstag" | weekdays(dat.filled$date, abbreviate = FALSE) == "Sonntag", "weekend", "weekday"))
#> Warning: package 'bindrcpp' was built under R version 3.3.3
dat.filled$weektype <- as.factor(dat.filled$weektype)
head(dat.filled)
#>       steps       date interval weektype
#> 1 1.7169811 2012-10-01        0  weekday
#> 2 0.3396226 2012-10-01        5  weekday
#> 3 0.1320755 2012-10-01       10  weekday
#> 4 0.1509434 2012-10-01       15  weekday
#> 5 0.0754717 2012-10-01       20  weekday
#> 6 2.0943396 2012-10-01       25  weekday


dat.temp4 <- group_by(dat.filled, interval, weektype)
dat.summarize4 <- summarize(dat.temp4
                            , sumSteps = sum(steps, na.rm = TRUE)
                            , meanSteps = mean(steps, na.rm =TRUE))
head(dat.summarize4)
#> # A tibble: 6 x 4
#> # Groups:   interval [3]
#>   interval weektype    sumSteps  meanSteps
#>      <int>   <fctr>       <dbl>      <dbl>
#> 1        0  weekday 101.3018868 2.25115304
#> 2        0  weekend   3.4339623 0.21462264
#> 3        5  weekday  20.0377358 0.44528302
#> 4        5  weekend   0.6792453 0.04245283
#> 5       10  weekday   7.7924528 0.17316562
#> 6       10  weekend   0.2641509 0.01650943


s <- ggplot(dat.summarize4, aes(x=interval, y=meanSteps, color = weektype)) +
        geom_line() +
        facet_grid(. ~ weektype)

print(s)