This assignment will be described in multiple parts. You will need to write a report that answers the questions detailed below. Ultimately, you will need to complete the entire assignment in a single R markdown document that can be processed by knitr and be transformed into an HTML file.
Throughout your report make sure you always include the code that you used to generate the output you present. When writing code chunks in the R markdown document, always use echo = TRUE so that someone else will be able to read the code. This assignment will be evaluated via peer assessment so it is essential that your peer evaluators be able to review the code for your analysis.
For the plotting aspects of this assignment, feel free to use any plotting system in R (i.e., base, lattice, ggplot2)
Fork/clone the GitHub repository created for this assignment. You will submit this assignment by pushing your completed files into your forked repository on GitHub. The assignment submission will consist of the URL to your GitHub repository and the SHA-1 commit ID for your repository state.
NOTE: The GitHub repository also contains the dataset for the assignment so you do not have to download the data separately.
if (!file.exists("activity.csv")) {
unzip("activity.zip")
}
activity <- read.csv("activity.csv")
activity$date <- as.Date(activity$date)
activity$interval <- as.factor(activity$interval)
(For this part of the assignment, you can ignore the missing values in the dataset.)
library(plyr)
steps_per_day <- ddply(activity, .(date), summarise, sum = sum(steps, na.rm=T))
steps_per_day
## date sum
## 1 2012-10-01 0
## 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
## 7 2012-10-07 11015
## 8 2012-10-08 0
## 9 2012-10-09 12811
## 10 2012-10-10 9900
## 11 2012-10-11 10304
## 12 2012-10-12 17382
## 13 2012-10-13 12426
## 14 2012-10-14 15098
## 15 2012-10-15 10139
## 16 2012-10-16 15084
## 17 2012-10-17 13452
## 18 2012-10-18 10056
## 19 2012-10-19 11829
## 20 2012-10-20 10395
## 21 2012-10-21 8821
## 22 2012-10-22 13460
## 23 2012-10-23 8918
## 24 2012-10-24 8355
## 25 2012-10-25 2492
## 26 2012-10-26 6778
## 27 2012-10-27 10119
## 28 2012-10-28 11458
## 29 2012-10-29 5018
## 30 2012-10-30 9819
## 31 2012-10-31 15414
## 32 2012-11-01 0
## 33 2012-11-02 10600
## 34 2012-11-03 10571
## 35 2012-11-04 0
## 36 2012-11-05 10439
## 37 2012-11-06 8334
## 38 2012-11-07 12883
## 39 2012-11-08 3219
## 40 2012-11-09 0
## 41 2012-11-10 0
## 42 2012-11-11 12608
## 43 2012-11-12 10765
## 44 2012-11-13 7336
## 45 2012-11-14 0
## 46 2012-11-15 41
## 47 2012-11-16 5441
## 48 2012-11-17 14339
## 49 2012-11-18 15110
## 50 2012-11-19 8841
## 51 2012-11-20 4472
## 52 2012-11-21 12787
## 53 2012-11-22 20427
## 54 2012-11-23 21194
## 55 2012-11-24 14478
## 56 2012-11-25 11834
## 57 2012-11-26 11162
## 58 2012-11-27 13646
## 59 2012-11-28 10183
## 60 2012-11-29 7047
## 61 2012-11-30 0
hist(steps_per_day$sum, ylab="Number of Days", col="blue", xlab="Number of Steps", main="Histogram of Steps Per Day, Oct-Nov 2012")
mean(steps_per_day$sum)
## [1] 9354.23
median(steps_per_day$sum)
## [1] 10395
Make a time series plot (i.e. type = “l”) of the 5-minute interval (x-axis) and the average number of steps taken, averaged across all days (y-axis).
library(ggplot2)
steps_per_interval <- ddply(activity, .(interval), summarise, sum = sum(steps, na.rm=T))
p <- ggplot(steps_per_interval, aes(x=interval, y=sum, group=1))
p + geom_line() + labs(title = "Average Steps per Day by 5-min Intervals, Oct-Nov 2012") + labs(x = "5-minute Intervals", y = "Average Number of Steps")
steps_per_interval[ which(steps_per_interval$sum==(max(steps_per_interval$sum))), ]
## interval sum
## 104 835 10927
Note that there are a number of days/intervals where there are missing values (coded as NA). The presence of missing days may introduce bias into some calculations or summaries of the data. Calculate and report the total number of missing values in the dataset.
NA_values <- activity[!complete.cases(activity),]
nrow(NA_values)
## [1] 2304
The strategy does not need to be sophisticated. For example, you could use the mean/median for that day, or the mean for that 5-minute interval, etc. Create a new dataset that is equal to the original dataset but with the missing data filled in.
# Calculate the mean for each interval
interval_mean <- ddply(activity, .(interval), summarise, mean = mean(steps, na.rm=T))
# Add the interval mean as a new variable to the activity dataset
activity_with_interval_mean <- join(activity, interval_mean)
## Joining by: interval
# Write function that will replace NA values with the interval mean
replace_NA <- function(dataset, variable, replacement) {
for (i in 1:nrow(dataset)) {
if (is.na(dataset[i, variable])) {
dataset[i, variable] <- dataset[i, replacement]
}
}
dataset
}
# Run the function on the dataset
complete_activity <- replace_NA(activity_with_interval_mean, variable=1, replacement=4)
complete_activity <- complete_activity[, -4]
head(complete_activity)
## steps date interval
## 1 1.7169811 2012-10-01 0
## 2 0.3396226 2012-10-01 5
## 3 0.1320755 2012-10-01 10
## 4 0.1509434 2012-10-01 15
## 5 0.0754717 2012-10-01 20
## 6 2.0943396 2012-10-01 25
complete_steps_per_day <- ddply(complete_activity, .(date), summarise, sum = sum(steps))
complete_steps_per_day$sum <- round(complete_steps_per_day$sum)
hist(complete_steps_per_day$sum, ylab="Number of Days", col="red", xlab="Number of Steps", main="Histogram of Steps Per Day, Oct-Nov 2012")
Mean comparison
mean(steps_per_day$sum)
## [1] 9354.23
mean(complete_steps_per_day$sum)
## [1] 10766.16
Median comparison
median(steps_per_day$sum)
## [1] 10395
median(complete_steps_per_day$sum)
## [1] 10766
Separate histograms
par(mfrow=c(1,2))
hist(steps_per_day$sum, ylab="Number of Days", col=rgb(1,0,0,0.2), xlab="Number of Steps", main="Steps Per Day, Excl. NA")
hist(complete_steps_per_day$sum, ylab="Number of Days", col=rgb(0,0,1,0.2), xlab="Number of Steps", main="Steps Per Day, Impute NA")
Overlapping histograms
hist(steps_per_day$sum,ylim=c(0,25),breaks=10,col=rgb(1,0,0,0.2), ylab="Number of Days", xlab="Number of Steps", main="Steps Per Day, NA vs Imputed")
par(new=TRUE)
hist(complete_steps_per_day$sum,ylim=c(0,25),breaks=10,col=rgb(0,0,1,0.2),main="",xlab="",ylab="")
legend('topleft',c('NA','Imputed'),
fill = c(rgb(1,0,0,0.2), rgb(0,0,1,0.2)), bty = 'n',
border = NA)
For this part the weekdays() function may be of some help here. Use the dataset with the filled-in missing values for this part. 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.
library(timeDate)
complete_activity$day_of_week <- ifelse(isWeekday(complete_steps_per_day$date)==TRUE, "weekday", "weekend")
library(lattice)
xyplot(steps ~ interval | day_of_week, layout = c(1, 2), data=complete_activity, type="l")