Reproducible Research: Peer Assessment 1
Loading and preprocessing the data
Here, we shall analyze the activity (number of steps taken during every 5-minute interval of the day) of an individual. We shall find out at what time of the day s/he is most active and try to find out the pattern of activity, if any. For convenience, we assume that the input-file is located in the same file-directory from where this Rmarkup file is being run.
f <- read.csv("activity.csv")
g <- aggregate(f$steps, list(f$date), sum, na.rm=FALSE)
colnames(g) <- c("day", "steps")What is mean total number of steps taken per day?
a. Histogram of the total number of steps taken each day :-
hist(g$steps, col = "RED", main = "Steps taken each day", xlab = "Number of Steps")
mn <- mean(g$steps, na.rm=TRUE)
md <- median(g$steps, na.rm=TRUE)b. The mean and median (when calculated ignoring rows with missing-number) of total number of steps taken each day are 1.076618910^{4} and 10765, respectively.
What is the average daily activity pattern?
a <- aggregate(f$steps, list(f$interval), mean, na.rm=TRUE)
colnames(a) <- c("interval", "steps")
plot(a$interval, a$steps, type = "l", xlab = "Interval", ylab = "Average no. of steps", main = "Average no. of steps taken daily in 5 minutes intervals")
maxavg <- a[a$steps == max(a$steps),"interval"]The 835 -th 5-minute interval, on average across all the days in the dataset, contains the maximum number of steps.
Imputing missing values
x <- nrow(f)
y <- nrow(f[!is.na(f$steps),])1. There are total (x - y) = 2304 rows in the dataset with missing values (i.e. the total number of rows with NAs).
2. The strategy I take for filling up these NAs, is to use the mean for that 5-minute interval.
3. To create a new dataset ‘e’, similar to the original one, ‘f’, with NAs filled up as per the strategy, I first identify the row-numbers, for which the ‘steps’-column is NA. Then fill-up the ‘steps’-value of those rows, with the ‘steps’-value for that ‘interval’ from the data.frame ‘a’ (as defined above). Then we calclate the mean and median of the average number of steps taken each day, and plot a histogram of the same.
e <- f
t <- which(is.na(f$steps))
for (i in 1:length(t)) {
e[t[i],"steps"] <- a[(a$interval == e[t[i],"interval"]),"steps"]
}
b <- aggregate(e$steps, list(e$date), sum, na.rm=FALSE)
## Keeping na.rm=FALSE here, to keep it syntactically same as 'g'-calculation, though it does not mean anything as 'e' does not contain any NA.
colnames(b) <- c("day_b", "steps_b")
mn_b <- mean(b$steps_b)
md_b <- median(b$steps_b)
hist(b$steps_b, col = "BLUE", main = "Steps taken each day", xlab = "Number of Steps")
The mean and median of the new data set are 1.076618910^{4} and 1.076618910^{4}, which are same as the estimates from the first part of the assignment.
- Important Note :- Had we calculated ‘g’ (day-wise step-count) with ‘na.rm=TRUE’ option, then mean and medians of g$steps would have been a smaller number. Because, in that case step-count would have been 0 for 2012-10-01, which then would have been taken for calculating the mean/ median.
Are there differences in activity patterns between weekdays and weekends?
I am creating a ‘day’ factor-variable - with two levels ‘weekday’ and ‘weekend’. Then attaching it to the dataset with filled-in missing values. Then through a for-loop, appropriate values are assigned to this variable.
day <- factor(c("weekday","weekend"))
e <- data.frame(day, e)
for (i in 1:nrow(e)) {
if ((weekdays(as.Date(e[i,"date"])) == "Saturday") | (weekdays(as.Date(e[i,"date"])) == "Sunday")) {
e[i,"day"] <- "weekend" }
else {
e[i,"day"] <- "weekday" }
}- The dataset-‘p’ contains the mean number of steps in the ’interval’s during the ’weekday’s and ’weekend’s. Then I use the lattice-system (xyplot) to generate the panel plot, as instructed.
library(lattice)
p <- aggregate(e$steps, list(e$day, e$interval),mean)
colnames(p) <- c("day", "interval", "steps")
xyplot(steps ~ interval | day, data = p, type = "l", xlab = "Interval", ylab = "Number of steps", layout = c(1,2))