| title | output | ||||
|---|---|---|---|---|---|
Reproducible Research: Peer Assessment 1 |
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Start by loading and preprocessing already downloaded and unpacked csv file into data table
library(data.table)
setwd("\\\\nash/mtec-home/vdenis/My Documents/coursera")
data <- fread("activity.csv")
summary(data)## 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 :2304
Further prepare data: Convert date into proper date format
data[,date := as.Date(date)]## 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
## ---
## 17564: NA 2012-11-30 2335
## 17565: NA 2012-11-30 2340
## 17566: NA 2012-11-30 2345
## 17567: NA 2012-11-30 2350
## 17568: NA 2012-11-30 2355
1.1 Calculate the total number of steps taken per day
total_steps <- data[,sum(steps), by = date]
setnames(total_steps, "V1", "sum_steps")
head(total_steps, 10)## date sum_steps
## 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
## 7: 2012-10-07 11015
## 8: 2012-10-08 NA
## 9: 2012-10-09 12811
## 10: 2012-10-10 9900
1.2 If you do not understand the difference between a histogram and a barplot, research the difference between them. Make a histogram of the total number of steps taken each day
library(ggplot2)
qplot(sum_steps, data = total_steps, geom = "histogram")
1.3 Calculate and report the mean and median of the total number of steps taken per day
mm_steps <- total_steps[,list(mean(sum_steps, na.rm = TRUE),median(sum_steps, na.rm = TRUE))]
setnames(mm_steps, "V1", "mean_steps_per_day")
setnames(mm_steps, "V2", "median_steps_per_day")
mm_steps## mean_steps_per_day median_steps_per_day
## 1: 10766.19 10765
2.1 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)
i_steps <- data[,mean(steps, na.rm = TRUE), by = interval]
setnames(i_steps, "V1", "mean_steps_per_interval")
ggplot(data = i_steps, aes(x = interval, y = mean_steps_per_interval)) + geom_line() + geom_smooth(method = "loess")
2.2 Which 5-minute interval, on average across all the days in the dataset, contains the maximum number of steps?
i_steps[which(i_steps[,mean_steps_per_interval] == max(i_steps[,mean_steps_per_interval])), ]## interval mean_steps_per_interval
## 1: 835 206.1698
3.1 Calculate and report the total number of missing values in the dataset (i.e. the total number of rows with NAs)
sum(is.na(data[,steps]) | is.na(data[,date]) | is.na(data[,interval]))## [1] 2304
3.2 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.
help_interval <- data[!is.na(steps), as.integer(round(mean(steps))), by = interval]
setnames(help_interval, "interval", "testinterval")
setnames(help_interval, "V1", "newvalue")3.3 Create a new dataset that is equal to the original dataset but with the missing data filled in.
new_data <- data
new_data[is.na(steps),steps:=help_interval[testinterval == interval,newvalue]]## steps date interval
## 1: 2 2012-10-01 0
## 2: 0 2012-10-01 5
## 3: 0 2012-10-01 10
## 4: 0 2012-10-01 15
## 5: 0 2012-10-01 20
## ---
## 17564: 5 2012-11-30 2335
## 17565: 3 2012-11-30 2340
## 17566: 1 2012-11-30 2345
## 17567: 0 2012-11-30 2350
## 17568: 1 2012-11-30 2355
3.4 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?
total_steps_new <- new_data[,sum(steps), by = date]
setnames(total_steps_new, "V1", "sum_steps")
qplot(sum_steps, data = total_steps_new, geom = "histogram")
Both mean and median slightly decrease
mm_steps <- total_steps_new[,list(mean(sum_steps, na.rm = TRUE),median(sum_steps, na.rm = TRUE))]
setnames(mm_steps, "V1", "mean_steps_per_day")
setnames(mm_steps, "V2", "median_steps_per_day")
mm_steps## mean_steps_per_day median_steps_per_day
## 1: 10765.64 10762
4.1 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.
new_data[, day_of_week := as.factor(ifelse(weekdays(date, abbreviate = TRUE) == "Sa" | weekdays(date, abbreviate = TRUE) == "So", "Weekend", "Weekday"))]## steps date interval day_of_week
## 1: 2 2012-10-01 0 Weekday
## 2: 0 2012-10-01 5 Weekday
## 3: 0 2012-10-01 10 Weekday
## 4: 0 2012-10-01 15 Weekday
## 5: 0 2012-10-01 20 Weekday
## ---
## 17564: 5 2012-11-30 2335 Weekday
## 17565: 3 2012-11-30 2340 Weekday
## 17566: 1 2012-11-30 2345 Weekday
## 17567: 0 2012-11-30 2350 Weekday
## 17568: 1 2012-11-30 2355 Weekday
4.2 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).
i_steps <- new_data[,mean(steps, na.rm = TRUE), by = list(day_of_week, interval)]
setnames(i_steps, "V1", "mean_steps_per_interval")
ggplot(data = i_steps, aes(x = interval, y = mean_steps_per_interval)) + geom_line() + geom_smooth(method = "loess") + facet_wrap(~day_of_week)
Weekdays seem on average more spiky while during weekends activities are more spread during the day and start and end later.