Project 1
Steve Scicluna
11 March 2019
Background
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 includes the number of steps taken in 5 minute intervals each day.
The data is available to be downloaded from the course website at https://d396qusza40orc.cloudfront.net/repdata%2Fdata%2Factivity.zip as repdata_data_activity.zip. A prior check of this file confirms that the zipped contents comprise a single activity.csv file.
Loading and pre-processing the data
The first step is to download the .zip file from the internet, unzip the zip file, and read the zipped activity.csv into R.
download.file("https://d396qusza40orc.cloudfront.net/repdata%2Fdata%2Factivity.zip",
"repdata_data_activity.zip")
unzip("repdata_data_activity.zip")
activity <- read.csv("activity.csv", stringsAsFactors = FALSE)Initial look at the data
The results displayed below show the data table’s dimensions (rows x columns), describe the table’s variables, and summarise the tables variables.
dim(activity)## [1] 17568 3 str(activity)## 'data.frame': 17568 obs. of 3 variables:
## $ steps : int NA NA NA NA NA NA NA NA NA NA ...
## $ date : chr "2012-10-01" "2012-10-01" "2012-10-01" "2012-10-01" ...
## $ interval: int 0 5 10 15 20 25 30 35 40 45 ... summary(activity)## steps date interval
## Min. : 0.00 Length:17568 Min. : 0.0
## 1st Qu.: 0.00 Class :character 1st Qu.: 588.8
## Median : 0.00 Mode :character Median :1177.5
## Mean : 37.38 Mean :1177.5
## 3rd Qu.: 12.00 3rd Qu.:1766.2
## Max. :806.00 Max. :2355.0
## NA's :2304Convert the date field from a character string to a date string
The str(activity) results show that the date field is formatted as a character string. In order that this field may be used for calculations, the format needs to be changed to a POSIXct date field. A ‘day’ field shall also be added as it shall be required later on.
The str(activity) and summary(activity) commands shall then be repeated to verify this has been done.
library(lubridate)
activity$date <- ymd(activity$date)
activity$day <- weekdays(activity$date)
str(activity)## 'data.frame': 17568 obs. of 4 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 ...
## $ day : chr "Monday" "Monday" "Monday" "Monday" ... summary(activity)## steps date interval day
## Min. : 0.00 Min. :2012-10-01 Min. : 0.0 Length:17568
## 1st Qu.: 0.00 1st Qu.:2012-10-16 1st Qu.: 588.8 Class :character
## Median : 0.00 Median :2012-10-31 Median :1177.5 Mode :character
## 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 :2304What is mean total number of steps taken per day?
For this part of the assignment, missing values in the dataset are going to be ignored. The following blocks of R code shall:
- Calculate the total number of steps taken per day
- Make a histogram of the total number of steps taken each day
- Calculate and report the mean and median of the total number of steps taken per day
1. Calculate the total number of steps taken per day
library(dplyr)
totalsteps <- activity %>% group_by(date) %>% summarise(stepsperday = sum(steps, na.rm = TRUE))
totalsteps$stepsperday## [1] 0 126 11352 12116 13294 15420 11015 0 12811 9900 10304
## [12] 17382 12426 15098 10139 15084 13452 10056 11829 10395 8821 13460
## [23] 8918 8355 2492 6778 10119 11458 5018 9819 15414 0 10600
## [34] 10571 0 10439 8334 12883 3219 0 0 12608 10765 7336
## [45] 0 41 5441 14339 15110 8841 4472 12787 20427 21194 14478
## [56] 11834 11162 13646 10183 7047 02. Make a histogram of the total number of steps taken each day
hist(totalsteps$stepsperday,
breaks = 5,
xlab = "Steps per day",
main = "Daily activity data",
ylim = c(0, 30),
col = "wheat")
3. Calculate and report the mean and median of the total number of steps taken per day
options("scipen" = 100, "digits" = 2)
oldmean <- mean(totalsteps$stepsperday)
oldmedian <- median(totalsteps$stepsperday)The mean and median number of steps taken each day are 9354.23 and 10395 respectively.
What is the average daily activity pattern?
ASSUMPTION: Intervals with NA are excluded from this analysis.
1. Time series plot
This next block of R code shall make a time series plot of the number of steps taken each day at five minute intervals starting at midnight, averaged across all days.
library(dplyr)
dailypattern <- activity %>%
group_by(interval) %>%
summarise(stepsperinterval = mean(steps, na.rm = TRUE))
# create time series plot
plot(x = dailypattern$interval,
y = dailypattern$stepsperinterval,
type = "l",
xlab = "Interval (time, 24 hour format)",
ylab = "Average steps per interval (all days)",
main = "Daily activity pattern")
2. Maximum steps per interval
This next block of R code shall calculate the time of day (in 24 hour format) for the five minute interval that had the highest number of steps averaged across all days, and the average number of steps for that interval.
options("scipen" = 100, "digits" = 0)
maxstepsinterval <- filter(dailypattern, stepsperinterval == max(stepsperinterval))
maxstepsinterval$interval## [1] 835 maxstepsinterval$stepsperinterval## [1] 206The maximum number of steps taken in any five minute interval during the day, averaged across all days, was 206 steps taken at 0835 (24 hour time format).
Imputing missing values
1. Calculate and report the total number of missing values in the dataset (i.e. the total number of rows with NAs)
nrow(activity[is.na(activity$steps),])## [1] 23042304 rows are missing ‘steps per interval’ data.
2. Strategy for imputing missing values
The 2,304 rows with missing values shall have their NAs substituted with the average for the relevant 5-minute interval.
3. Create a new dataset that is equal to the original dataset but with the missing data filled in.
# create new table to hold populated NA values
popactivity <- activity
# identify the rows with NA values
narows <- is.na(popactivity$steps)
# calculate the average steps per interval across all days
avinterval <- tapply(popactivity$steps, popactivity$interval, mean, na.rm = TRUE, simplify = TRUE)
# apply average steps per interval to rows with NA values
popactivity$steps[narows] <- avinterval[as.character(popactivity$interval[narows])]
# create new table with imputed steps per day
imputedsteps <- popactivity %>% group_by(date) %>% summarise(stepsperday = sum(steps, na.rm = TRUE))
# confirm NA rows have been populated
summary(activity)## steps date interval day
## Min. : 0 Min. :2012-10-01 Min. : 0 Length:17568
## 1st Qu.: 0 1st Qu.:2012-10-16 1st Qu.: 589 Class :character
## Median : 0 Median :2012-10-31 Median :1178 Mode :character
## Mean : 37 Mean :2012-10-31 Mean :1178
## 3rd Qu.: 12 3rd Qu.:2012-11-15 3rd Qu.:1766
## Max. :806 Max. :2012-11-30 Max. :2355
## NA's :2304 summary(popactivity)## steps date interval day
## Min. : 0 Min. :2012-10-01 Min. : 0 Length:17568
## 1st Qu.: 0 1st Qu.:2012-10-16 1st Qu.: 589 Class :character
## Median : 0 Median :2012-10-31 Median :1178 Mode :character
## Mean : 37 Mean :2012-10-31 Mean :1178
## 3rd Qu.: 27 3rd Qu.:2012-11-15 3rd Qu.:1766
## Max. :806 Max. :2012-11-30 Max. :2355 # minimum steps, mean, median, and quartiles should all increase
summary(totalsteps)## date stepsperday
## Min. :2012-10-01 Min. : 0
## 1st Qu.:2012-10-16 1st Qu.: 6778
## Median :2012-10-31 Median :10395
## Mean :2012-10-31 Mean : 9354
## 3rd Qu.:2012-11-15 3rd Qu.:12811
## Max. :2012-11-30 Max. :21194 summary(imputedsteps)## date stepsperday
## Min. :2012-10-01 Min. : 41
## 1st Qu.:2012-10-16 1st Qu.: 9819
## Median :2012-10-31 Median :10766
## Mean :2012-10-31 Mean :10766
## 3rd Qu.:2012-11-15 3rd Qu.:12811
## Max. :2012-11-30 Max. :211944. Histogram
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?
hist(imputedsteps$stepsperday,
breaks = 5,
xlab = "Imputed steps per day",
main = "Daily activity data",
sub = "(NA values re-populated with average interval values)",
ylim = c(0, 35),
col = "cyan")
options("scipen" = 100, "digits" = 0)
newmean <- mean(imputedsteps$stepsperday)
newmedian <- median(imputedsteps$stepsperday)
diffmean <- newmean - oldmean
diffmedian <- newmedian - oldmedianBefore the missing ‘steps per interval’ data was re-populated with values imputed from the average for each relevant interval, the mean and median number of steps taken per day were 9354 and 10395 respectively.
After the missing ‘steps per interval’ data was re-populated with values imputed from the average for each relevant interval, the mean and median number of steps taken per day were 10766 and 10766 respectively.
Imputing data increased the mean and median number of steps per day by 1412 and 371 respectively.
Are there differences in activity patterns between weekdays and weekends?
1. New factor variable
Create a new factor variable in the dataset with two levels - “weekday” and “weekend” indicating whether a given date is a weekday or weekend day.
popactivity <- popactivity %>%
mutate(typeofday = ifelse(popactivity$day == "Saturday" | popactivity$day == "Sunday",
"Weekend", "Weekday"))
head(popactivity)## steps date interval day typeofday
## 1 2 2012-10-01 0 Monday Weekday
## 2 0 2012-10-01 5 Monday Weekday
## 3 0 2012-10-01 10 Monday Weekday
## 4 0 2012-10-01 15 Monday Weekday
## 5 0 2012-10-01 20 Monday Weekday
## 6 2 2012-10-01 25 Monday Weekday2. Panel time series plot
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.
# create table for the two time series plots
popdailypattern <- popactivity %>% group_by(interval, typeofday) %>% summarise(stepsperinterval = mean(steps, na.rm = TRUE))
# create panel time series plots using ggplot2
library(ggplot2)
qplot(interval,
stepsperinterval,
data = popdailypattern,
geom = "line",
ylim = c(0, 250),
xlab = "Interval (time, 24 hour format)",
ylab = "Average steps per interval",
main = "Imputed daily activity pattern (by type of day",
facets = typeofday~.)