Loading and preprocessing the data

We extract the .csv file from the zip and adjust the column “date” withe a Date format:

DF <- read.csv(unz("activity.zip", "activity.csv"), colClasses=c("integer", "Date", "integer"))

head(DF)
##   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
summary(DF)
##      steps             date               interval     
##  Min.   :  0.00   Min.   :2012-10-01   Min.   :   0.0  
##  1st Qu.:  0.00   1st Qu.:2012-10-16   1st Qu.: 588.8  
##  Median :  0.00   Median :2012-10-31   Median :1177.5  
##  Mean   : 37.38   Mean   :2012-10-31   Mean   :1177.5  
##  3rd Qu.: 12.00   3rd Qu.:2012-11-15   3rd Qu.:1766.2  
##  Max.   :806.00   Max.   :2012-11-30   Max.   :2355.0  
##  NA's   :2304
str(DF)
## 'data.frame':    17568 obs. of  3 variables:
##  $ steps   : int  NA NA NA NA NA NA NA NA NA NA ...
##  $ date    : Date, format: "2012-10-01" "2012-10-01" ...
##  $ interval: int  0 5 10 15 20 25 30 35 40 45 ...

What is mean total number of steps taken per day?

First we select the data we need and group it per date. Then we sum the steps for each day:

library(dplyr)
stepsDay<-DF %>%
        select(steps, date) %>%
        group_by(date) %>%
        summarise(sumSteps=sum(steps))

head(stepsDay)
## Source: local data frame [6 x 2]
## 
##         date sumSteps
## 1 2012-10-01       NA
## 2 2012-10-02      126
## 3 2012-10-03    11352
## 4 2012-10-04    12116
## 5 2012-10-05    13294
## 6 2012-10-06    15420

Now we prepare the histogram of the total number of steps taken each day: (we adjust the bindwith to 1.000 steps)

library(ggplot2)
ggplot(data=stepsDay, aes(sumSteps)) + 
        geom_histogram(binwidth = 1000) +
        labs(title="Steps taken each day", 
             x = "Number of steps each day", y = "Number of times in a day")

Now we calculate the mean and median total number of steps taken per day with this code

mean(stepsDay$sumSteps, na.rm=TRUE)
## [1] 10766.19
median(stepsDay$sumSteps, na.rm=TRUE)
## [1] 10765

What is the average daily activity pattern?

Now we make a time series plot of the 5-minute interval (x-axis) and the average number of steps taken, averaged across all days (y-axis):

#To plot average daily activity
DF %>%  group_by(interval) %>% 
        summarize(mean=mean(steps, na.rm=TRUE)) %>% 
        plot(., type="l", main="Average daily activity", 
             xlab="5-minute interval", ylab="Average steps across all days")

#To display the interval with the max. number of steps in average on the plot
max<-DF %>%  group_by(interval) %>% 
        summarize(mean=mean(steps, na.rm=TRUE)) %>% 
        slice(which.max(mean)) %>% select(interval)

abline(v=max, col="purple")
axis(1, at=max, labels = max, pos=0, col.axis="purple")

In purple we see the interval with the maximum number of steps. And here the exact
interval and his mean:

DF %>%  group_by(interval) %>% 
        summarize(mean=mean(steps, na.rm=TRUE)) %>% 
        slice(which.max(mean)) #slice selects rows by position
## Source: local data frame [1 x 2]
## 
##   interval     mean
## 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.

summary(DF)
##      steps             date               interval     
##  Min.   :  0.00   Min.   :2012-10-01   Min.   :   0.0  
##  1st Qu.:  0.00   1st Qu.:2012-10-16   1st Qu.: 588.8  
##  Median :  0.00   Median :2012-10-31   Median :1177.5  
##  Mean   : 37.38   Mean   :2012-10-31   Mean   :1177.5  
##  3rd Qu.: 12.00   3rd Qu.:2012-11-15   3rd Qu.:1766.2  
##  Max.   :806.00   Max.   :2012-11-30   Max.   :2355.0  
##  NA's   :2304

  1. There are 2304 missing values in the dataset

  2. We will fill the NA’s with 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.

# Split main DF in two DF, one with NAs and one without NAs
DF_noNA <-DF %>% filter(complete.cases(.))
DF_NA <-DF %>% filter(!complete.cases(.))

# Create DF with intervals and their step means
meanSteps <- DF %>% group_by(interval) %>% summarize(mean=mean(steps, na.rm=TRUE))

# Merge means with DF with NAs by "interval"
DF_NA <- merge(meanSteps, DF_NA, by = "interval", all.y=TRUE)
DF_NA <- DF_NA[,c(2,4,1)] #remove column not needed and reorder columns
colnames(DF_NA) <-c("steps", "date", "interval") #rename columns

# Merge DF with no NA's and with the DF with means
DF_new <- rbind(DF_noNA, DF_NA)

Here the histogram of the new DF

library(dplyr)
library(ggplot2)
stepsDay_new<- DF_new %>%
        select(steps, date) %>%
        group_by(date) %>%
        summarise(sumSteps=sum(steps)) 


ggplot(data=stepsDay_new, aes(sumSteps)) + 
        geom_histogram(binwidth = 1000) +
        labs(title="Steps taken each day (new DF)", 
             x = "Number of steps each day", y = "Number of times in a day")

Here the mean and median of the new DF

mean(stepsDay_new$sumSteps, na.rm=TRUE)
## [1] 10766.19
median(stepsDay_new$sumSteps, na.rm=TRUE)
## [1] 10766.19
summary(stepsDay$sumSteps) # before
##    Min. 1st Qu.  Median    Mean 3rd Qu.    Max.    NA's 
##      41    8841   10760   10770   13290   21190       8
summary(stepsDay_new$sumSteps) # now
##    Min. 1st Qu.  Median    Mean 3rd Qu.    Max. 
##      41    9819   10770   10770   12810   21190

Now the estimates have changed as follows:
- We have now more observations as shown in the histogram (we added data to the NAs)
- Mean stays the same (the strategy was to use the mean to fulfil the missing values)
- Now the median is the same as the mean

Are there differences in activity patterns between weekdays and weekends?

For this part we are going to use the data set with the filled-in missing values.

We will add a new column with the information, Weekday or Weekend.

library(dplyr)

DF_new$weekday <- as.factor(
        ifelse(weekdays(DF_new$date) %in% c("Samstag","Sonntag"), "Weekend", "Weekday")
        ) 

str(DF_new)
## 'data.frame':    17568 obs. of  4 variables:
##  $ steps   : num  0 0 0 0 0 0 0 0 0 0 ...
##  $ date    : Date, format: "2012-10-02" "2012-10-02" ...
##  $ interval: int  0 5 10 15 20 25 30 35 40 45 ...
##  $ weekday : Factor w/ 2 levels "Weekday","Weekend": 1 1 1 1 1 1 1 1 1 1 ...

Now we make a plot containing a time series plot of the 5-minute interval (x-axis) and the average number of steps taken, averaged across all weekday days or weekend days (y-axis).

library(dplyr)
library(lattice)
DF_new_weekday <- DF_new %>% 
        group_by(interval, weekday) %>%
        summarise(steps = mean( steps, na.rm=TRUE ))

xyplot(steps~interval | weekday, data = DF_new_weekday,
       type="l", ylab="Number of steps", 
       xlab="Interval",
       layout=c(1,2))