submit assignment 4

Assignment 4.Rmd

@@ -8,13 +8,14 @@ https://www.cs.uic.edu/~wilkinson/Applets/cluster.html
 
 
 ```{r}
-library()
+library(tidyr)
+library(dplyr)
 ```
 
 Now, upload the file "Class_Motivation.csv" from the Assignment 4 Repository as a data frame called "K1""
 ```{r}
 
-K1 <- read.csv(...)
+K1 <- read.csv("Class_Motivation.csv")
 
 ```
 
@@ -26,13 +27,14 @@ The algorithm will treat each row as a value belonging to a person, so we need t
 
 ```{r}
 
-K2 <- 
+K2 <- select(K1, motivation1, motivation2, motivation3, motivation4, motivation5)
 
 ```
 
 It is important to think about the meaning of missing values when clustering. We could treat them as having meaning or we could remove those people who have them. Neither option is ideal. What problems do you foresee if we recode or remove these values? Write your answers below:
 
-
+If we remove these values, then the result patterns will be biased, since values are missing.
+If we recode these values, people might have talked about the scores with classmates after the first time of self report, which will influence the authenticity of recode values.  
 
 We will remove people with missing values for this assignment, but keep in mind the issues that you have identified.
 
@@ -41,13 +43,16 @@ We will remove people with missing values for this assignment, but keep in mind
 
 K3 <- na.omit(K2) #This command create a data frame with only those people with no missing values. It "omits" all rows with missing values, also known as a "listwise deletion". EG - It runs down the list deleting rows as it goes.
 
+K3 <- K2
+
+K3[is.na(K3)] <- 0
 ```
 
 Another pre-processing step used in K-means is to standardize the values so that they have the same range. We do this because we want to treat each week as equally important - if we do not standardise then the week with the largest range will have the greatest impact on which clusters are formed. We standardise the values by using the "scale()" command.
 
 ```{r}
 
-K3 <- 
+K3 <- scale(K3)
 
 ```
 
@@ -66,19 +71,35 @@ Also, we need to choose the number of clusters we think are in the data. We will
 
 ```{r}
 
-fit <- 
+fit1a <- kmeans(K3, 2)
+fit1b <- kmeans(K3, 2)
+fit1c <- kmeans(K3, 2)
 
 #We have created an object called "fit" that contains all the details of our clustering including which observations belong to each cluster.
 
 #We can access the list of clusters by typing "fit$cluster", the top row corresponds to the original order the rows were in. Notice we have deleted some rows.
 
-
+fit1a$cluster
 
 #We can also attach these clusters to the original dataframe by using the "data.frame" command to create a new data frame called K4.
 
-K4
+K4 <- data.frame(K3, fit1a$cluster, fit1b$cluster, fit1c$cluster)
+
+fit1a$withinss
+fit1b$withinss
+fit1c$withinss
+
+fit1a$tot.withinss
+fit1b$tot.withinss
+fit1c$tot.withinss
 
+fit1a$betweenss
+fit1b$betweenss
+fit1c$betweenss
+
+K4 <- data.frame(K3, fit1c$cluster)
 #Have a look at the K4 dataframe. Lets change the names of the variables to make it more convenient with the names() command.
+names(K4) <- c("1", "2", "3", "4", "5", "cluster")
 
 
 ```
@@ -88,14 +109,14 @@ Now we need to visualize the clusters we have created. To do so we want to play
 First lets use tidyr to convert from wide to long format.
 ```{r}
 
-K5 <- gather(K4, "week", "motivation", 1:5)
+K5 <- tidyr::gather(K4, "week", "motivation", 1:5)
 ```
 
 Now lets use dplyr to average our motivation values by week and by cluster.
 
 ```{r}
 
-K6 <- K5 %>% group_by(week, cluster) %>% summarise(K6, avg = mean(motivation))
+K6 <- K5 %>% group_by(week, cluster) %>% summarise(avg = mean(motivation))
 
 ```
 
@@ -113,9 +134,9 @@ Likewise, since "cluster" is not numeric but rather a categorical label we want
 
 ```{r}
 
-K6$week <- 
+K6$week <- as.numeric(K6$week)
 
-K6$cluster <- 
+K6$cluster <- as.factor(K6$cluster)
 
 ```
 
@@ -127,7 +148,7 @@ Now we can plot our line plot using the ggplot command, "ggplot()".
 - Finally we are going to clean up our axes labels: xlab("Week") & ylab("Average Motivation")
 
 ```{r}
-
+library(ggplot2)
 ggplot(K6, aes(week, avg, colour = cluster)) + geom_line() + xlab("Week") + ylab("Average Motivation")
 
 ```
@@ -147,13 +168,56 @@ Look at the number of people in each cluster, now repeat this process for 3 rath
 ##Part II
 
 Using the data collected in the HUDK4050 entrance survey (HUDK4050-cluster.csv) use K-means to cluster the students first according location (lat/long) and then according to their answers to the questions, each student should belong to two clusters.
+```{r}
+library(tidyverse) 
+library(dplyr)
+M1 <- read.csv("HUDK405020-cluster.csv", header = TRUE)
+M2 <- select(M1, 4:9)
+
+fit2a <- kmeans(M2, 1)
+fit2b <- kmeans(M2, 2)
+fit2c <- kmeans(M2, 3)
+fit2d <- kmeans(M2, 4)
+fit2e <- kmeans(M2, 5)
+fit2f <- kmeans(M2, 6)
+fit2g <- kmeans(M2, 7)
+
+mss <- c(fit2a$tot.withinss, fit2b$tot.withinss, fit2c$tot.withinss, fit2d$tot.withinss, fit2e$tot.withinss, fit2f$tot.withinss, fit2g$tot.withinss, fit2a$betweenss, fit2b$betweenss, fit2c$betweenss, fit2d$betweenss, fit2e$betweenss, fit2f$betweenss, fit2g$betweenss)
+
+clusters <- c(seq(1,7,1), seq(1,7,1))
+col <- c(rep("blue",7), rep("red",7))
+plot(clusters, mss, col = col)
+
+L1 <- select(M1, 2:3)
+plot(L1$long,L1$lat)
+
+fit3a <- kmeans(L1, 2)
+fit3b <- kmeans(L1, 2)
+fit3c <- kmeans(L1, 2)
+
+fit3a$tot.withinss
+fit3b$tot.withinss
+fit3c$tot.withinss
+
+ML <- data.frame(M1$compare.features,M1$math.accuracy,M1$planner.use,M1$enjoy.discuss,M1$meet.deadline, fit2c$cluster,M1$lat,M1$long, fit3a$cluster)
+
+pairs(ML)
+```
 
 ##Part III
 
 Create a visualization that shows the overlap between the two clusters each student belongs to in Part II. IE - Are there geographical patterns that correspond to the answers? 
 
 ```{r}
+table(ML$fit2c.cluster,ML$fit3a.cluster)
+
+ML2 <- ML %>% group_by(fit2c.cluster,fit3a.cluster) %>% summarize(count = n())
+ggplot(ML2, aes(x = fit2c.cluster, y = fit3a.cluster, size = count)) + geom_point()
+
 
+library(vcd)
+P1 <- structable(fit2c$cluster ~ fit3a$cluster)
+mosaic(P1, shade=TRUE, legent=TRUE)
 ```