PA1_template.Rmd

---
title: "Reproducible Research: Peer Assessment 1"
output: 
  html_document:
    keep_md: true
---

Exam time again!!!


Here's a useful link to help Rmarkdown writing, by Rstudio : [R Markdown cheat sheet](https://www.rstudio.com/wp-content/uploads/2015/02/rmarkdown-cheatsheet.pdf)


# Loading libraries
```{r echo = TRUE,  comment=F,warning=T}
library(lubridate,warn.conflicts = F)
library(dplyr,warn.conflicts = F)
library(ggplot2)

#To make AWESOME tables
library(xtable)
```
# Loading and preprocessing the data

``` {r tidy=T}

if(!file.exists('activity.csv')) unzip('activity.zip')

d<-read.csv('activity.csv', stringsAsFactors = F)

#data<-data[complete.cases(data),]

data <- mutate(d, date=ymd(date))
```
# What is mean total number of steps taken per day?

Aggregating datas

``` {r tidy=T}
#dplyr version
#sumSteps<-d %>% group_by(date) %>% summarise( sum(steps))

#Basic version
sumSteps<-aggregate(steps~date, data=d, FUN = sum)

#renaming columns for convenience
colnames(sumSteps) = c('Date', 'Steps')

#Computing median and mean

Mean1 <-as.integer(round(mean(sumSteps$Steps, na.rm=T),0))
Median1 <-as.integer(round(median(sumSteps$Steps, na.rm=T),0))
```

Now plotting time

``` {r tidy=T}
ggplot(sumSteps, aes(Steps))+
  geom_histogram( bins=9,col='black', fill='green')+
  ggtitle('Histogram of Steps')
```

The _mean_ total number of steps taken per day is **`r  Mean1`** and the _median_ is
**`r Median1`**


The choice of 9 beans instead the default of 30 bins by ggplot has been made to have kind of a 'Gaussian' look, with no empty bean.


From this graphic, we can note 
* **Data looks like normally distributed**
* Th mean and the median being very close, reinforcing this impression

## Let's go play and superpose a bell-shaped curve to the graph

We'll draw the corresponding Gaussian distribution:

- Mean= `r mean(sumSteps$Steps)`
- Standard deviation = `r sd(sumSteps$Steps)`


``` {r }
ggplot(sumSteps, aes(Steps))+
  geom_histogram( aes(y=..density..), bins=9,col='black', fill='green')+
  stat_function(fun=dnorm, colour='red', lwd=2, args=list(mean=mean(sumSteps$Steps), sd=sd(sumSteps$Steps)))+
  ggtitle('Comparing histogram and normal distribution')
```



The graphic belows shows:

- Histogram : in green
- Density plot: in blue
- Gaussian distribution : in red


``` {r}

ggplot(sumSteps, aes(Steps))+
  geom_histogram( aes(y=..density..), bins=30,col='black', fill='green')+
  geom_density(colour='blue', lwd=1, fill='blue', alpha=0.3)+
  stat_function(fun=dnorm, colour='red', lwd=2, args=list(mean=mean(sumSteps$Steps), sd=sd(sumSteps$Steps)))+
  ggtitle('Comparing histogram and normal distribution with 30 bins histograms')

```


Comparison not so obvious.... for the histogram, as it depends on number of bins

However, we get a better feel with the density plotting (in blue)

# What is the average daily activity pattern?

First, we do aggregate data by interval

I enclosed 2 versions of the aggregation code: 

- One using the **dplyr** package, as My personal goal is to mater it, as it is more flexible and advanced than the 'base' package
- Another one using Ocam's razor **aggregate()** function

```{r tidy=T}
# dplyr version
#daily<-d %>% group_by(interval) %>% summarise(steps=mean(steps, na.rm=T))
daily<-aggregate(steps~interval, data=d, FUN=function(x) {mean(x,na.rm=T)})
```


Now about the plotting, the assigment required a line plot, that you can find below.
However, this plot does'nt show the missing intervals, so I built a 'barplot', with the line plot ovrlay clearly showing the data 'holes'

## Line plot of  activity pattern
```{r}
ggplot(daily, aes(x=interval,y=steps) ) + 
  geom_line(col='red')
```

## Barplot with Line plot overlay
This one shows clearly how ggplot operates
- barplot in blue
- lineplot in red

Both using the same dataset

```{r}
ggplot(daily, aes(x=interval,y=steps) ) + 
  geom_bar(stat = 'identity',col='blue')+ 
  geom_line(col='red')
```


The interval having the maximal mean step value being

**

```{r}
daily[[which.max(daily$steps),1]]
```

**

# Imputing missing values

## Let's count missing values

```{r results='asis'}

t<-table(is.na(d), dnn = 'Missing Values')

#now making it look AWESOME with xtable
print(xtable(t), type = 'html')
```

So there are `r t['TRUE']` missing values out of `r sum(t)` (about `r round(t['TRUE']/sum(t)*100 , 2)`%)

# Now filling missing values

The filling strategy is as following:
We'll replace the missing intervals for a specific day with the average of this interval

```{r}

#First: Build a table of means by interval
meanByIntervalAndDate <- d %>%group_by(interval,date) %>% summarize(steps=mean(steps, rm.na=T))
meanByInterval  <-aggregate(steps~interval, data=meanByIntervalAndDate, FUN=function(x) {mean(x,na.rm=TRUE)})

# Then we split data in two parts: the NA part and the non-NA part
naIndexes <-is.na(d$steps)

# We keep the NA lines and drop the 'steps' (containig only NAs) column
naPart= d[naIndexes,]
naPart$steps=NULL

#We drop the NA lines from the data
nonNaPart=d[!naIndexes,]

#Then we do a left join with the 'Mean by interval' table 
naPart<-left_join(naPart, meanByInterval, by='interval')

# We check that all lines have been filled
length(naPart[is.na(naPart$steps),'steps'])


#Then we go for row_bind betwen the naPartand the nonNaPart

d <- bind_rows(naPart, nonNaPart)


```

Now that we have a new dataframe,it is time to bilsdan histogram to check if the fact of filling data had an impact


``` {r tidy=T}
sumSteps2<-aggregate(steps~date, data=d, FUN = sum)

#renaming columns for convenience
colnames(sumSteps2) = c('Date', 'Steps')

#computing median and mean
Mean2 <-as.integer(round(mean(sumSteps2$Steps, na.rm=T),0))
Median2 <-as.integer(round(median(sumSteps2$Steps, na.rm=T),0))


ggplot(sumSteps2, aes(Steps))+
  geom_histogram( bins=9,col='black', fill='green')+
  ggtitle('Histogram of Steps')
```

** For this new dataset **

The _mean_ total number of steps taken per day is **`r  Mean2`** and the _median_ is **`r Median2`**

So we notice that the mean didn't change after having filled data.
However, the median increased by `r Median2-Median1` unit (`r Median1` to `r Median2`), a **`r round((Median2-Median1)/Median1*100,2)`% change** what is non significative


# Are there differences in activity patterns between weekdays and weekends?

This part is two fold:
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.

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). See the README file in the GitHub repository to see an example of what this plot should look like using simulated data.

## Step one: adding the 'daytype' factor to the dataset

``` {r}
# adding the weekday factor

#To do so, we use the wday() function what return 1 for sunday, 7for saturday, and 2..6 for the other weekday
d<-mutate(d,dayType= ifelse(wday(date) %in% c(1,7), 'weekend', 'weekday'))

#Makes it a factor, as requested
d$dayType<-as.factor(d$dayType)
```

## Step two: Creating the plot

```{r tidy=T}

dailyByDayType<-aggregate(steps~interval+dayType, data=d, FUN=function(x) {mean(x,na.rm=T)})

ggplot(dailyByDayType, aes(interval, steps )) + geom_area(col='red', fill='red') + facet_grid(dayType ~ .)

```

So we can see there is a difference in activity level between the weekdays and the weekends!

##Bonus track

A variant of the plot, what shows better the difference in the activity levels

```{r tidy=T}

dailyByDayType<-aggregate(steps~interval+dayType, data=d, FUN=function(x) {mean(x,na.rm=T)})

ggplot(dailyByDayType, aes(interval, steps, coloour=dayType, fill=dayType)) + 
  geom_area(col='black',alpha=0.5) 

```