| title | output | ||||
|---|---|---|---|---|---|
Reproducible Research: Peer Assessment 1 |
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First, loading the zip file and unziping it, then reading the csv data file:
if (!file.exists("raw_data.zip")){
print("loading the zip file............... ")
fileURL <- "https://d396qusza40orc.cloudfront.net/repdata%2Fdata%2Factivity.zip"
download.file(fileURL, "raw_data.zip", method="curl")
}
if (!file.exists("activity.csv")) {
print("unziping the data...................")
unzip("raw_data.zip")
}
library(readr)
activity_full <- read_csv("activity.csv")
library(knitr)
kable(head(activity_full), align='c')steps date interval
NA 2012-10-01 0
NA 2012-10-01 5
NA 2012-10-01 10
NA 2012-10-01 15
NA 2012-10-01 20
NA 2012-10-01 25
Second, processing the data; I delete the lines with NA values:
activity <- activity_full[ !is.na(activity_full$steps),]
kable( head(activity), align='c' )steps date interval
0 2012-10-02 0
0 2012-10-02 5
0 2012-10-02 10
0 2012-10-02 15
0 2012-10-02 20
0 2012-10-02 25
I calcute down below the total number of steps taken per day:
activity_sum <- aggregate(steps~date, data=activity, sum)
head(activity_sum)## date steps
## 1 2012-10-02 126
## 2 2012-10-03 11352
## 3 2012-10-04 12116
## 4 2012-10-05 13294
## 5 2012-10-06 15420
## 6 2012-10-07 11015
Then I plot the histogram of the total number of steps taken each day
library(ggplot2)
qplot(steps, data = activity_sum, bins=50)
The next step is to calculate the mean and median of the total number of steps taken per day
mean_steps <- mean(activity_sum$steps)
median_steps <- round(median(activity_sum$steps),2)The mean of total steps is 1.0766189\times 10^{4}, and their median is 1.0765\times 10^{4}.
averages1 <- c( round( mean(activity_sum$steps), 2), round( median(activity_sum$steps), 2))
averages1## [1] 10766.19 10765.00
The mean of total steps is 10766.19 and their median is 10765.
We need to plot the average daily activity pattern corresponding to each 5-minute interval: (the average calculated using the mean function)
interval_avrg <- aggregate(steps~interval, data=activity, mean)
p1 <- ggplot(data=interval_avrg, aes(x=interval, y=steps)) +
geom_line()+
ggtitle("The average daily activity pattern") +
xlab(" 5-minute intervals") + ylab("average steps")+
theme_minimal()
p1
Let's get the 5-minute interval which, on average across all the days in the dataset, contains the maximum number of steps:
int_max <- interval_avrg[which.max(interval_avrg$steps),]
kable(int_max, digits=2, align='c') interval steps
104 835 206.17
It's the 835 5-minute interval!
We'll calculate the number of rows that contains NA's values. To do so, we can use the summary function:
kable(summary(activity_full), align='c') steps date interval
Min. : 0.00 Min. :2012-10-01 Min. : 0.0
1st Qu.: 0.00 1st Qu.:2012-10-16 1st Qu.: 588.8
Median : 0.00 Median :2012-10-31 Median :1177.5
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 :2304 NA NA
It's only the steps column that has missing value and it has 2304 missing rows.
Since many rows from the first day are missing, my strategy to replace the NAs will be to use the mean of the corresponding 5-minutes-interval across all the days.
The dataframe activity_fill_na is a copy of the original full raw data, but is going to be modified in order to replace the NA values from the steps column:
nrows <- dim(activity_full)[1]
activity_fill_na <- data.frame(activity_full)
for( i in 1:nrows){
if ( is.na(activity_fill_na$steps[i])){
index_mean_intrvl <- which(interval_avrg$interval==activity_fill_na$interval[i])
activity_fill_na$steps[i] <- interval_avrg$steps[index_mean_intrvl]
}
}
kable( head(activity_fill_na, 10), align='c')steps date interval
1.7169811 2012-10-01 0
0.3396226 2012-10-01 5
0.1320755 2012-10-01 10
0.1509434 2012-10-01 15
0.0754717 2012-10-01 20
2.0943396 2012-10-01 25
0.5283019 2012-10-01 30
0.8679245 2012-10-01 35
0.0000000 2012-10-01 40
1.4716981 2012-10-01 45
Now let's plot a histogram of the total number of steps taken each day:
activity_sum2 <- aggregate(steps~date, data=activity_fill_na, sum)
qplot(steps, data = activity_sum2, bins=50)
averages2 <- c( round( mean(activity_sum2$steps), 2), round( median(activity_sum2$steps), 2))
averages2## [1] 10766.19 10766.19
The mean of total steps is 10766.19 and their median is 10766.19.
Comparison: The value of the mean is the same before and after imputing missing values, while the median changed and became equal to the mean.
For this part I use the dataset with the filled-in missing values and create a new factor variable days in the dataset with two levels – “weekday” and “weekend” indicating whether a given date is a weekday or weekend day.
Note that in my r version, the language is french, and so I had to use the french translation for the weekdays.
weekdays_FR <- c("lundi", "mardi", "mercredi", "jeudi", "vendredi")
activity_fill_na$days <- factor(ifelse(weekdays(activity_fill_na$date) %in% weekdays_FR, "weekday", "weekend")) Now I create two ggplots each one corresponding to weekdays or weekends, and then plotting them together using the ggarrange function from the ggpubr package:
activity_wkdays <- subset(activity_fill_na, days=='weekday')
interval_avrg_wkdays <- aggregate(steps~interval, data=activity_wkdays, mean)
pwd <- ggplot(data=interval_avrg_wkdays, aes(x=interval, y=steps)) +
geom_line()+
ggtitle("The average daily activity pattern\n through weekdays") +
xlab(" 5-minute intervals") + ylab("average steps")+
theme_minimal()
activity_wkends <- subset(activity_fill_na, days=='weekend')
interval_avrg_wkends <- aggregate(steps~interval, data=activity_wkends, mean)
pwe <- ggplot(data=interval_avrg_wkends, aes(x=interval, y=steps)) +
geom_line()+
ggtitle("The average daily activity pattern\n through weekends") +
xlab(" 5-minute intervals") + ylab("average steps")+
theme_minimal()
library(ggpubr)
ggarrange(pwe, pwd,nrow = 2, ncol = 1)