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CS 8803 Social Computing Data Mini-Project
                      Harish Kanakaraju Prashanth Palanthandalam




Problem I


Method:

To analyze the prominence of people who are following a particular celebrity. Three
celebrities who were analyzed are

      Britney Spears
      Mariah Carey
      Ashley Tisdale

These celebrities are all singers and among the top 11 influential celebrities in twitter.
Britney spears has close to 7.7 million followers with Ashley Tisdale and Mariah Carey
having approximately 4.3 millions each.

The samples of followers of these celebrities were analyzed to find out how many of
them were prominent. The prominence of each followers were found out using
The formula No of followers/No of following, higher the value, higher the prominence.

We used the sample sizes of 1500, 2000 and 3000. The confidence interval is 1.8 and
confidence level is 95% for the sample size of 3000, considering the total population of
the celebritys followers.

The initial analysis with a sample size of 1500 was done to find the effect of sample size
on the prominence ratio.

Results:

SS = 1500                                     Prominence Ratio
                  Mean         Median          SD            Chi square        P-value
Britney Spears    0.288        0.056           2.047
Mariah Carey      0.265        0.132           1.383
Ashley Tisdale    0.239        0.115           0.880

SS = 2000                                     Prominence Ratio
                  Mean         Median          SD            Chi square        P-value
Britney Spears    0.546       0.111          3.067
Mariah Carey      0.289       0.163          1.230
Ashley Tisdale    0.406       0.130          7.007

SS = 3000                                   Prominence Ratio
                  Mean        Median         SD            Chi square      P-value
Britney Spears    0.493       0.081          3.403
Mariah Carey      0.258       0.154          1.014
Ashley Tisdale    0.348       0.133          5.734


Basic Analysis:


The mean and the standard deviation may swing either ways based on the sample due
to the outliers. If the sample contains one very prominent person, it would boost the
mean and SD values. But the median trend always remains the same.

Using Median: Mariah Carey has prominent followers than Ashley Tisdale. And Ashley
Tisdale has more prominent followers than Britney spears.

From Fig 1, we can see that Britney spears has relatively high number of low prominent
followers (ratio close to zero), while Ashley and Mariah have large number of followers
with a decent prominence value, while number of followers for Britney in this region is
low. Thats why her median is the lowest among the three.

From Fig 2, we can find that Britney Spears has relatively more number of very
prominent followers compared to Ashley and Mariah. But the very prominent followers
are very very less in number compared to the whole population set.
R Commands used:

The below sequence was executed for the three celebrities,

at4 <- getUser("ashleytisdale")
at4Fl <- at4$getFollowers(n=3000)
at4FFl <- sapply(at4Fl,followersCount)
at4FFd <- sapply(at4Fl,friendsCount)
at4Ratio <- mapply("/", at4FFl, at4FFd)
med <- median(sort(at4Ratio))
stad<- sd(at4Ratio)
meanRatio <- mean(at4Ratio)
at4sum <- sum(at4Ratio)
Chi-square test

Chisq.test(c(at4sum,bs4sum,mc4sum))

Plotting graph (executed only once)

xyz <- cbind(bs4Ratio, at4Ratio, mc4Ratio, deparse.level = 1)
data = melt(xyz, id=c("bs4Ratio"))
lowProminence <- qplot(value, data = data, geom = "histogram", color = X2, binwidth =
50)
highP <- ggplot(data, aes(x=X2, y=value))
highP + geom_point(position = "jitter")




                         Fig 1: Low prominent followers




                             Fig 2: High prominent followers
Problem II

Method:

To extract tweets from two different geographic locations in the world, and select the
tweets which contain the phrase I want. A comparison of preferences of the twitter
users from the two locations has been done, with respect to the terms I want a pizza
and I want to sleep. Also, the mood of the users on Monday and Friday has been
studied, by extracting the tweets with the terms Monday and I hate; and Friday
and Thank God.

The searchTwitter() functionality of the twitteR package for R Studio has been used.
The two cities chosen were Seattle, Washington and Southampton, UK.
1000 tweets with the phrase I want were extracted within a 20 mile radius of the two
cities.

southamTweets = searchTwitter("I
want",1000,NULL,NULL,NULL,NULL,'50.903,-1.40625,20mi',NULL)

The list of 1000 tweets is then converted into text form by using the lapply() command.

southamTweets.text = lapply(southamTweets, function(southampton)
southampton$getText())



The grep() command is used to extract incidences of the term pizza in the tweet list.

southamTweets.spec = grep("pizza",southamTweets.text,TRUE)

The procedure is repeated for Seattle:

seattleTweets = searchTwitter("I
want",1000,NULL,NULL,NULL,NULL,'47.606,-122.299,20mi',NULL)
> seattleTweets.text = lapply(seattleTweets,function(seattle)
seattle$getText())
> seattle.spec = grep("pizza",seattleTweets.text,TRUE)

Variations of the I want a pizza phrase have also been tried.

seattleSpecific.spec = grep("I want pizza",seattleTweets.text,TRUE)



Instead of pizza, the tweets containing the phrase sleep or I want to sleep were
used.

southamTweetsSleep.spec = grep("sleep",southamTweets.text,TRUE)

southamTweetsSleepSpecific.spec = grep("I want to
sleep",southamTweets.text,TRUE)

seattleSleep.spec = grep("sleep",seattleTweets.text,TRUE)

seattleSleepSpecific.spec = grep("I want to
sleep",seattleTweets.text,TRUE)

seattleSleepSpecific.spec = grep("I want
sleep",seattleTweets.text,TRUE)

Another variant of the above experiment was done, with the terms Monday and
Friday and respectively, the phrases I hate and Thank God
seattleMonday =
searchTwitter("Monday",1000,NULL,NULL,NULL,NULL,'47.606,-
122.299,20mi',NULL)
> seattleFriday =
searchTwitter("Friday",1000,NULL,NULL,NULL,NULL,'47.606,-
122.299,20mi',NULL)
> southamMonday = searchTwitter("I
want",1000,NULL,NULL,NULL,NULL,'50.903,-1.40625,20mi',NULL)
> southamMonday =
searchTwitter("Monday",1000,NULL,NULL,NULL,NULL,'50.903,-
1.40625,20mi',NULL)
> southamFriday =
searchTwitter("Friday",1000,NULL,NULL,NULL,NULL,'50.903,-
1.40625,20mi',NULL)
> southamMonday.text = lapply(southamMonday, function(southampton)
southampton$getText())
> southamFriday.text = lapply(southamFriday, function(southampton)
southampton$getText())
>
> seattleFriday.text = lapply(seattleFriday, function(seattle)
seattle$getText())
>
> seattleMonday.text = lapply(seattleMonday, function(seattle)
seattle$getText())
>
> seattleMonday.spec = grep("I hate",seattleMonday.text,TRUE)
> seattleFriday.spec = grep("Thank God",seattleFriday.text,TRUE)
> southamFriday.spec = grep("Thank God",southamFriday.text,TRUE)
> southamMonday.spec = grep("I hate",southamMonday.text,TRUE)

The Chi-Square Statistical test was then done on the data obtained using the chisq.test()
command.

The results obtained were plotted using the following commands:

x   <- rchisq(southamFriday.spec,southamMonday.spec)
>   hist(x,prob = TRUE)
>   curve( dchisq(x, df=5), col='green', add=TRUE)
>   curve( dchisq(x, df=10), col='red', add=TRUE )
>   lines( density(x), col='orange')



Both histogram and density line plots have been used to depict the results.

Result:

Broadly, it was found that the terms I want and pizza featured together in only six
out of 1000 tweets in Seattle, and the single phrase I want pizza returned three
tweets.

The issue with searchTwitter() is that I want is not considered as a continuous term,
and the command also returned tweets such as I really think I want or I dont think
he wants..
Seattle threw up 10 tweets out of 1000 with the term sleep. However, I want to
sleep did not return any values, and I want sleep returned just one result.




In Southampton, only one tweet out of 1000 expressed the desire to have pizza, indeed,
there was only one tweet with comprised of I want and pizza in the same tweet,
while I want a pizza returned no results. It appears that pizza is more popular in
cosmopolitan Seattle than the relatively more conservative Southampton.

23 tweets were returned by the query for the term sleep in Southampton, and two for
I want to sleep, which is marginally higher than the results for Seattle.
In the experiment with tweets posted on Mondays and Fridays, it appears that citizens
of both cities rant more on Mondays, in comparison to feeling thankful on Fridays. The
search for I hate and Monday returned 54 tweets in Seattle, while Thank God and
Friday returned just one, which is surprising. Southampton returned 8 tweets for the
former query (Monday), and two for the latter.
Thus, it is seen that Southampton returns an almost symmetric plot as compared to
Seattle, where the difference between Monday and Friday is more substantial.

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Lily Ray

Soc prashanth harish

  • 1. CS 8803 Social Computing Data Mini-Project Harish Kanakaraju Prashanth Palanthandalam Problem I Method: To analyze the prominence of people who are following a particular celebrity. Three celebrities who were analyzed are Britney Spears Mariah Carey Ashley Tisdale These celebrities are all singers and among the top 11 influential celebrities in twitter. Britney spears has close to 7.7 million followers with Ashley Tisdale and Mariah Carey having approximately 4.3 millions each. The samples of followers of these celebrities were analyzed to find out how many of them were prominent. The prominence of each followers were found out using The formula No of followers/No of following, higher the value, higher the prominence. We used the sample sizes of 1500, 2000 and 3000. The confidence interval is 1.8 and confidence level is 95% for the sample size of 3000, considering the total population of the celebritys followers. The initial analysis with a sample size of 1500 was done to find the effect of sample size on the prominence ratio. Results: SS = 1500 Prominence Ratio Mean Median SD Chi square P-value Britney Spears 0.288 0.056 2.047 Mariah Carey 0.265 0.132 1.383 Ashley Tisdale 0.239 0.115 0.880 SS = 2000 Prominence Ratio Mean Median SD Chi square P-value
  • 2. Britney Spears 0.546 0.111 3.067 Mariah Carey 0.289 0.163 1.230 Ashley Tisdale 0.406 0.130 7.007 SS = 3000 Prominence Ratio Mean Median SD Chi square P-value Britney Spears 0.493 0.081 3.403 Mariah Carey 0.258 0.154 1.014 Ashley Tisdale 0.348 0.133 5.734 Basic Analysis: The mean and the standard deviation may swing either ways based on the sample due to the outliers. If the sample contains one very prominent person, it would boost the mean and SD values. But the median trend always remains the same. Using Median: Mariah Carey has prominent followers than Ashley Tisdale. And Ashley Tisdale has more prominent followers than Britney spears. From Fig 1, we can see that Britney spears has relatively high number of low prominent followers (ratio close to zero), while Ashley and Mariah have large number of followers with a decent prominence value, while number of followers for Britney in this region is low. Thats why her median is the lowest among the three. From Fig 2, we can find that Britney Spears has relatively more number of very prominent followers compared to Ashley and Mariah. But the very prominent followers are very very less in number compared to the whole population set. R Commands used: The below sequence was executed for the three celebrities, at4 <- getUser("ashleytisdale") at4Fl <- at4$getFollowers(n=3000) at4FFl <- sapply(at4Fl,followersCount) at4FFd <- sapply(at4Fl,friendsCount) at4Ratio <- mapply("/", at4FFl, at4FFd) med <- median(sort(at4Ratio)) stad<- sd(at4Ratio) meanRatio <- mean(at4Ratio) at4sum <- sum(at4Ratio)
  • 3. Chi-square test Chisq.test(c(at4sum,bs4sum,mc4sum)) Plotting graph (executed only once) xyz <- cbind(bs4Ratio, at4Ratio, mc4Ratio, deparse.level = 1) data = melt(xyz, id=c("bs4Ratio")) lowProminence <- qplot(value, data = data, geom = "histogram", color = X2, binwidth = 50) highP <- ggplot(data, aes(x=X2, y=value)) highP + geom_point(position = "jitter") Fig 1: Low prominent followers Fig 2: High prominent followers
  • 4. Problem II Method: To extract tweets from two different geographic locations in the world, and select the tweets which contain the phrase I want. A comparison of preferences of the twitter users from the two locations has been done, with respect to the terms I want a pizza and I want to sleep. Also, the mood of the users on Monday and Friday has been studied, by extracting the tweets with the terms Monday and I hate; and Friday and Thank God. The searchTwitter() functionality of the twitteR package for R Studio has been used. The two cities chosen were Seattle, Washington and Southampton, UK.
  • 5. 1000 tweets with the phrase I want were extracted within a 20 mile radius of the two cities. southamTweets = searchTwitter("I want",1000,NULL,NULL,NULL,NULL,'50.903,-1.40625,20mi',NULL) The list of 1000 tweets is then converted into text form by using the lapply() command. southamTweets.text = lapply(southamTweets, function(southampton) southampton$getText()) The grep() command is used to extract incidences of the term pizza in the tweet list. southamTweets.spec = grep("pizza",southamTweets.text,TRUE) The procedure is repeated for Seattle: seattleTweets = searchTwitter("I want",1000,NULL,NULL,NULL,NULL,'47.606,-122.299,20mi',NULL) > seattleTweets.text = lapply(seattleTweets,function(seattle) seattle$getText()) > seattle.spec = grep("pizza",seattleTweets.text,TRUE) Variations of the I want a pizza phrase have also been tried. seattleSpecific.spec = grep("I want pizza",seattleTweets.text,TRUE) Instead of pizza, the tweets containing the phrase sleep or I want to sleep were used. southamTweetsSleep.spec = grep("sleep",southamTweets.text,TRUE) southamTweetsSleepSpecific.spec = grep("I want to sleep",southamTweets.text,TRUE) seattleSleep.spec = grep("sleep",seattleTweets.text,TRUE) seattleSleepSpecific.spec = grep("I want to sleep",seattleTweets.text,TRUE) seattleSleepSpecific.spec = grep("I want sleep",seattleTweets.text,TRUE) Another variant of the above experiment was done, with the terms Monday and Friday and respectively, the phrases I hate and Thank God
  • 6. seattleMonday = searchTwitter("Monday",1000,NULL,NULL,NULL,NULL,'47.606,- 122.299,20mi',NULL) > seattleFriday = searchTwitter("Friday",1000,NULL,NULL,NULL,NULL,'47.606,- 122.299,20mi',NULL) > southamMonday = searchTwitter("I want",1000,NULL,NULL,NULL,NULL,'50.903,-1.40625,20mi',NULL) > southamMonday = searchTwitter("Monday",1000,NULL,NULL,NULL,NULL,'50.903,- 1.40625,20mi',NULL) > southamFriday = searchTwitter("Friday",1000,NULL,NULL,NULL,NULL,'50.903,- 1.40625,20mi',NULL) > southamMonday.text = lapply(southamMonday, function(southampton) southampton$getText()) > southamFriday.text = lapply(southamFriday, function(southampton) southampton$getText()) > > seattleFriday.text = lapply(seattleFriday, function(seattle) seattle$getText()) > > seattleMonday.text = lapply(seattleMonday, function(seattle) seattle$getText()) > > seattleMonday.spec = grep("I hate",seattleMonday.text,TRUE) > seattleFriday.spec = grep("Thank God",seattleFriday.text,TRUE) > southamFriday.spec = grep("Thank God",southamFriday.text,TRUE) > southamMonday.spec = grep("I hate",southamMonday.text,TRUE) The Chi-Square Statistical test was then done on the data obtained using the chisq.test() command. The results obtained were plotted using the following commands: x <- rchisq(southamFriday.spec,southamMonday.spec) > hist(x,prob = TRUE) > curve( dchisq(x, df=5), col='green', add=TRUE) > curve( dchisq(x, df=10), col='red', add=TRUE ) > lines( density(x), col='orange') Both histogram and density line plots have been used to depict the results. Result: Broadly, it was found that the terms I want and pizza featured together in only six out of 1000 tweets in Seattle, and the single phrase I want pizza returned three tweets. The issue with searchTwitter() is that I want is not considered as a continuous term, and the command also returned tweets such as I really think I want or I dont think he wants..
  • 7. Seattle threw up 10 tweets out of 1000 with the term sleep. However, I want to sleep did not return any values, and I want sleep returned just one result. In Southampton, only one tweet out of 1000 expressed the desire to have pizza, indeed, there was only one tweet with comprised of I want and pizza in the same tweet, while I want a pizza returned no results. It appears that pizza is more popular in cosmopolitan Seattle than the relatively more conservative Southampton. 23 tweets were returned by the query for the term sleep in Southampton, and two for I want to sleep, which is marginally higher than the results for Seattle.
  • 8. In the experiment with tweets posted on Mondays and Fridays, it appears that citizens of both cities rant more on Mondays, in comparison to feeling thankful on Fridays. The search for I hate and Monday returned 54 tweets in Seattle, while Thank God and Friday returned just one, which is surprising. Southampton returned 8 tweets for the former query (Monday), and two for the latter.
  • 9. Thus, it is seen that Southampton returns an almost symmetric plot as compared to Seattle, where the difference between Monday and Friday is more substantial.