Lesson 3
What to Do First?
Notes: Check working directory
getwd()## [1] "C:/Dersler/Udacity_R/W3"list.files()## [1] "lesson3_student.html" "lesson3_student.rmd" "pseudo_facebook.tsv"Pseudo-Facebook User Data
Notes:
pf <- read.csv("pseudo_facebook.tsv", sep = '\t')
dim(pf)## [1] 99003 15Histogram of FB Users’ Birthdays
Notes:
#install.packages('ggplot2')
library(ggplot2)
names(pf)## [1] "userid" "age"
## [3] "dob_day" "dob_year"
## [5] "dob_month" "gender"
## [7] "tenure" "friend_count"
## [9] "friendships_initiated" "likes"
## [11] "likes_received" "mobile_likes"
## [13] "mobile_likes_received" "www_likes"
## [15] "www_likes_received"qplot(x = dob_day, data = pf) +
scale_x_discrete(breaks=1:31) 
What are some things that you notice about this histogram?
Response: many people says they were born 1st of the month and less people are born 31st of the month
Moira’s Investigation
Notes: She checked the perception of users’ audience and the real number of audience
Estimating Your Audience Size
Notes: Maybe 25 which is 5%
How many of your friends do you think saw that post?
Response: 25
Think about what percent of your friends on Facebook see any posts or comments that you make in a month. What percent do you think that is?
Response: %10
Perceived Audience Size
Notes: 50
Faceting
Notes: we now expand the graph and plotted dats of the month
qplot(x = dob_day, data = pf) +
scale_x_discrete(breaks=1:31) +
facet_wrap(~dob_month, ncol = 3)
Let’s take another look at our plot. What stands out to you here?
Response: Many of the users were born in 1st of jan! ***
Be Skeptical - Outliers and Anomalies
Notes:
Moira’s Outlier
Notes: She noticed that one user guessed s/he had 10 million audience
Which case do you think applies to Moira’s outlier?
Response: a bad data an extreme outlier
Friend Count
Notes:
What code would you enter to create a histogram of friend counts?
qplot(data =pf, x= friend_count)## stat_bin: binwidth defaulted to range/30. Use 'binwidth = x' to adjust this.
How is this plot similar to Moira’s first plot?
Response:There is a guy with 5000 friends
Limiting the Axes
Notes: we are trimming our graph and omitting the values above 1000
qplot(data = pf, x= friend_count, xlim = c(0,1000))## stat_bin: binwidth defaulted to range/30. Use 'binwidth = x' to adjust this.
# Another way to plot the same data
#qplot(data = pf, x= friend_count) +
# scale_x_continuous(limits = c(0,1000))Exploring with Bin Width
Notes: People tend to say numbers such as 10 20 100 but not 16 47 etc… when bin size is set to 1 this is more clear
Adjusting the Bin Width
Notes:
Faceting Friend Count
## What code would you add to create a facet the histogram by gender?
## Add it to the code below.
qplot(x = friend_count, data = pf, binwidth = 25) +
scale_x_continuous(limits = c(0, 1000),
breaks = seq(0, 1000, 50)) +
facet_wrap(~gender)
# In the alternate solution below, the period or dot in the formula for facet_grid()
# represents all of the other variables in the data set. Essentially, this notation splits
# up the data by gender and produces three histograms, each having their own row.
# qplot(x = friend_count, data = pf) +
# facet_grid(gender ~ .) Omitting NA Values
Notes:
qplot(x = friend_count, data = subset(pf, !is.na(gender)), binwidth = 10) +
scale_x_continuous(limits = c(0, 1000),
breaks = seq(0, 1000, 50)) +
facet_wrap(~gender)
# Other way of ommiting NA values is
# qplot(x = friend_count, data = na.omit(pf), binwidth = 10) +
# scale_x_continuous(limits = c(0, 1000),
# breaks = seq(0, 1000, 50)) +
# facet_wrap(~gender)Statistics ‘by’ Gender
Notes:
table(pf$gender)##
## female male
## 40254 58574by(pf$friend_count,pf$gender,summary)## pf$gender: female
## Min. 1st Qu. Median Mean 3rd Qu. Max.
## 0 37 96 242 244 4923
## --------------------------------------------------------
## pf$gender: male
## Min. 1st Qu. Median Mean 3rd Qu. Max.
## 0 27 74 165 182 4917Who on average has more friends: men or women?
Response:women
What’s the difference between the median friend count for women and men?
Response:22
Why would the median be a better measure than the mean?
Response: distribution is skewed, median would give better representation
Tenure
Notes:
qplot(x = tenure, data = pf, binwidth = 30,
color = I('black'), fill = I('#099DD9'))
How would you create a histogram of tenure by year?
qplot(x = tenure/365, data = pf, binwidth = 0.25,
color = I('black'), fill = I('#099DD9')) +
scale_x_continuous(breaks = seq(1,7,1), limits = c(0,7))
Labeling Plots
Notes:
qplot(x = tenure/365, data = pf, binwidth = 0.25,
xlab = 'Number of years using Facebook',
ylab = 'Number of Users in sample' ,
color = I('black'), fill = I('#099DD9')) +
scale_x_continuous(breaks = seq(1,7,1), limits = c(0,7))
User Ages
Notes: I noticed that there are no users between ages 0-13 due to facebook policy. There are some strange ages over 100 (I don’t think a person over 100 would even fancy a computer)
qplot(x = age, data = pf, binwidth = 2,
xlab = 'Facebook User ages',
ylab = 'Number of Users in sample' ,
color = I('black'), fill = I('#099DD9')) +
scale_x_continuous(breaks = seq(10,90,5), limits = c(10,90))
What do you notice?
Response:
The Spread of Memes
Notes:
Lada’s Money Bag Meme
Notes:
Transforming Data
Notes:
library(gridExtra)## Loading required package: gridsummary(pf$friend_count)## Min. 1st Qu. Median Mean 3rd Qu. Max.
## 0.0 31.0 82.0 196.4 206.0 4923.0summary(log10(pf$friend_count))## Min. 1st Qu. Median Mean 3rd Qu. Max.
## -Inf 1 2 -Inf 2 4summary(log10(pf$friend_count + 1))## Min. 1st Qu. Median Mean 3rd Qu. Max.
## 0.000 1.505 1.919 1.868 2.316 3.692summary(sqrt(pf$friend_count))## Min. 1st Qu. Median Mean 3rd Qu. Max.
## 0.000 5.568 9.055 11.090 14.350 70.160p1 = qplot(x = friend_count, data = pf)
p2 = qplot(x = log10(friend_count+1), data = pf)
p3 = qplot(x = sqrt(friend_count), data = pf)
grid.arrange(p1,p2,p3,ncol=1)## stat_bin: binwidth defaulted to range/30. Use 'binwidth = x' to adjust this.
## stat_bin: binwidth defaulted to range/30. Use 'binwidth = x' to adjust this.
## stat_bin: binwidth defaulted to range/30. Use 'binwidth = x' to adjust this.
Transforming Data Alternate Solution
p1 <- ggplot(aes(x = friend_count), data = pf) + geom_histogram()
p2 <- p1 + scale_x_log10()
p3 <- p1 + scale_x_sqrt()
grid.arrange(p1,p2,p3,ncol=1)## stat_bin: binwidth defaulted to range/30. Use 'binwidth = x' to adjust this.
## stat_bin: binwidth defaulted to range/30. Use 'binwidth = x' to adjust this.
## stat_bin: binwidth defaulted to range/30. Use 'binwidth = x' to adjust this.
Add a Scaling Layer
Notes:
logScale <- qplot(x = log10(friend_count + 1), data = pf)
countScale <- ggplot(aes(x = friend_count), data = pf) +
geom_histogram() +
scale_x_log10()
grid.arrange(logScale,countScale, ncol=2)## stat_bin: binwidth defaulted to range/30. Use 'binwidth = x' to adjust this.
## stat_bin: binwidth defaulted to range/30. Use 'binwidth = x' to adjust this.
qplot(x = friend_count, data =pf) + scale_x_log10()## stat_bin: binwidth defaulted to range/30. Use 'binwidth = x' to adjust this.
Frequency Polygons
qplot(x = friend_count, data = subset(pf, !is.na(gender)), binwidth =10) +
scale_x_continuous(lim = c(0,1000), breaks = seq(0,1000,100)) +
facet_wrap(~gender)
qplot(x = friend_count, y = ..count../sum(..count..),
data = subset(pf, !is.na(gender)),
xlab = 'Friend Count',
ylab = 'Proportion of Users with that friend count',
binwidth =10,
geom ='freqpoly', color = gender) +
scale_x_continuous(lim = c(0,1000), breaks = seq(0,1000,100)) ## Warning: Removed 2 rows containing missing values (geom_path).## Warning: Removed 2 rows containing missing values (geom_path).
Likes on the Web
Notes:
qplot(x = www_likes, data = subset(pf, !is.na(gender)),
geom ='freqpoly', color = gender) +
scale_x_continuous() +
scale_x_log10()## Scale for 'x' is already present. Adding another scale for 'x', which will replace the existing scale.
## stat_bin: binwidth defaulted to range/30. Use 'binwidth = x' to adjust this.
by(pf$www_likes,pf$gender,sum)## pf$gender: female
## [1] 3507665
## --------------------------------------------------------
## pf$gender: male
## [1] 1430175Box Plots
Notes:
qplot(x = friend_count, data = subset(pf, !is.na(gender)),
binwidth = 25) +
scale_x_continuous(limits = c(0,1000), breaks = seq(0,1000,50)) +
facet_wrap(~gender)
qplot(x = gender, y = friend_count,
data = subset(pf, !is.na(gender)),
geom ='boxplot')
Adjust the code to focus on users who have friend counts between 0 and 1000.
qplot(x = gender, y = friend_count,
data = subset(pf, !is.na(gender)),
geom ='boxplot', ylim = c(0,1000))## Warning: Removed 2949 rows containing non-finite values (stat_boxplot).
# Alternatively instead of ylim
qplot(x = gender, y = friend_count,
data = subset(pf, !is.na(gender)),
geom ='boxplot') +
scale_y_continuous(limits = c(0,1000))## Warning: Removed 2949 rows containing non-finite values (stat_boxplot).
# when we use the methods two above we indeed removing some data points!!
# Correct way top do this is with cord_cartesian
qplot(x = gender, y = friend_count,
data = subset(pf, !is.na(gender)),
geom ='boxplot') +
coord_cartesian(ylim = c(0,1000))
Box Plots, Quartiles, and Friendships
Notes:
qplot(x = gender, y = friend_count,
data = subset(pf, !is.na(gender)),
geom ='boxplot') +
coord_cartesian(ylim = c(0,250))
by(pf$friend_count, pf$gender, summary)## pf$gender: female
## Min. 1st Qu. Median Mean 3rd Qu. Max.
## 0 37 96 242 244 4923
## --------------------------------------------------------
## pf$gender: male
## Min. 1st Qu. Median Mean 3rd Qu. Max.
## 0 27 74 165 182 4917On average, who initiated more friendships in our sample: men or women?
Response: Women #### Write about some ways that you can verify your answer. Response: by using by command or plotting a box plot gives us median values
qplot(x = gender, y = friendships_initiated,
data = subset(pf, !is.na(gender)),
geom ='boxplot') +
coord_cartesian(ylim = c(0,150))
by(pf$friendships_initiated,pf$gender,summary)## pf$gender: female
## Min. 1st Qu. Median Mean 3rd Qu. Max.
## 0.0 19.0 49.0 113.9 124.8 3654.0
## --------------------------------------------------------
## pf$gender: male
## Min. 1st Qu. Median Mean 3rd Qu. Max.
## 0.0 15.0 44.0 103.1 111.0 4144.0Response:
Getting Logical
Notes:
summary(pf$mobile_likes)## Min. 1st Qu. Median Mean 3rd Qu. Max.
## 0.0 0.0 4.0 106.1 46.0 25110.0summary(pf$mobile_likes > 0)## Mode FALSE TRUE NA's
## logical 35056 63947 0mobile_check_in <- NA
pf$mobile_checked_in <- ifelse(pf$mobile_likes >0, 1,0)
pf$mobile_checked_in <- factor(pf$mobile_checked_in)
summary(pf$mobile_checked_in)[2]/(summary(pf$mobile_checked_in)[2] +summary(pf$mobile_checked_in)[1])## 1
## 0.6459097Response:
Analyzing One Variable
Reflection:
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