+
+
+
+
+
+
+Reproducible Research Project 1
+Dyutit Mohanty
+2024-03-11
+ +
+
+Loading and preprocessing the data
+filename <- "activity.csv"
+data <- read.csv(filename, header=TRUE)
+head(data)## 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
+
+What is mean total number of steps taken per day?
+
+
+Calculate the total number of steps taken per day
+data2 <- data[!(is.na(data$steps)), ]
+steps_per_day <- aggregate(steps ~ date, data=data2, FUN = sum)
+
+Histogram of the total number of steps taken each day
+hist(steps_per_day$steps, main="Histogram of the total number of steps taken each day", xlab="Steps", ylab="Frequency")
+
+Calculate and report the mean and median of the total number of +steps taken per day
+mean_steps_per_day <- mean(steps_per_day$steps)
+median_steps_per_day <- median(steps_per_day$steps)
+mean_steps_per_day## [1] 10766.19median_steps_per_day## [1] 10765
+
+What is the average daily activity pattern?
+
+
+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)
+my_mean <- function(x) mean(x, na.rm = TRUE) # R kept throwing an error when "mean" was used as arg so I defined my own mean function
+data3 <- aggregate(steps ~ interval, data, FUN=my_mean)
+head(data3)## interval steps
+## 1 0 1.7169811
+## 2 5 0.3396226
+## 3 10 0.1320755
+## 4 15 0.1509434
+## 5 20 0.0754717
+## 6 25 2.0943396plot(y=data3$steps,x=data3$interval, type="l", xlab="Intervals", ylab="Steps")
+
+Which 5-minute interval, on average across all the days in the +dataset, contains the maximum number of steps?
+index_max_steps <- which.max(data3$steps)
+interval_max_steps <- data3$interval[index_max_steps]
+interval_max_steps## [1] 835max(data3$steps)## [1] 206.1698
+
+Imputing missing values
+
+
+Calculate and report the total number of missing values in the +dataset (total number of rows with NAs)
+data4 <- subset(data, is.na(data$steps))
+dims <- dim(data4)
+nrows_na <- dims[1]
+nrows_na## [1] 2304
+
+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.
+# I am using the mean for the interval
+data_nona <- data
+
+for (i in 1:nrow(data_nona)){
+ if (is.na(data_nona$steps[i])){
+ interval <- data_nona$interval[i]
+ index <- which(data3$interval == interval)
+ mean_val <- data3$steps[index]
+ data_nona$steps[i] <- mean_val
+ }
+}
+
+head(data_nona)## steps date interval
+## 1 1.7169811 2012-10-01 0
+## 2 0.3396226 2012-10-01 5
+## 3 0.1320755 2012-10-01 10
+## 4 0.1509434 2012-10-01 15
+## 5 0.0754717 2012-10-01 20
+## 6 2.0943396 2012-10-01 25
+
+Create a new dataset that is equal to the original dataset but with +the missing data filled in.
+head(data_nona)## steps date interval
+## 1 1.7169811 2012-10-01 0
+## 2 0.3396226 2012-10-01 5
+## 3 0.1320755 2012-10-01 10
+## 4 0.1509434 2012-10-01 15
+## 5 0.0754717 2012-10-01 20
+## 6 2.0943396 2012-10-01 25
+
+Make a histogram of the total number of steps taken each day and +Calculate and report the mean and median total number of steps taken per +day. Do these values differ from the estimates from the first part of +the assignment? What is the impact of imputing missing data on the +estimates of the total daily number of steps?
+data4 <- aggregate(steps ~ date, data_nona, FUN=sum)
+hist(data4$steps, main="Histogram of the total number of steps taken each day", xlab="Steps", ylab="Frequency")
mean_steps <- mean(data4$steps)
+median_steps <- median(data4$steps)
+mean_steps## [1] 10766.19median_steps## [1] 10766.19
+
+Slight increase in average
+
+
+
+
+
+
+Are there differences in activity patterns between weekdays and +weekends?
+
+
+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.
+data5 <- data_nona
+data5$date <- as.Date(data5$date)
+data5$day <- weekdays(data5$date)
+
+for (i in 1:nrow(data5)){
+ if (data5$day[i] %in% c("Saturday", "Sunday")){
+ data5$weekday[i] <- "weekend"
+ }
+ else {
+ data5$weekday[i] <- "weekday"
+ }
+}
+
+fct_days <- factor(data5$weekday)
+summary(fct_days)## weekday weekend
+## 12960 4608
+
+Make a panel 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). See the README file in the +GitHub repository to see an example of what this plot should look like +using simulated data.
+wknd_data <- data5[data5$weekday == "weekend",]
+wkday_data <- data5[data5$weekday == "weekday",]
+
+wknd_data <- aggregate(steps ~ interval, wknd_data, mean)
+wkday_data <- aggregate(steps ~ interval, wkday_data, mean)
+
+par(mfrow=c(2, 1))
+plot(wknd_data$interval, wknd_data$steps, type="l")
+plot(wkday_data$interval, wkday_data$steps, type="l")