diff --git a/01 Intro to R.Rmd b/01 Intro to R.Rmd index 198e48c..1db046d 100644 --- a/01 Intro to R.Rmd +++ b/01 Intro to R.Rmd @@ -31,6 +31,9 @@ output: + + + ```{r echo=FALSE} .counterExercise <- 0 @@ -157,9 +160,9 @@ paste0(my.states) What does the `nchar()` function do? The `paste()` function? Does it make a difference to use `sep=""` or `collapse=","`? What about `paste0()`? ## Exercises -`r exNum("Print all even numbers less than 100")` -`r exNum("What is the mean of even numbers less than 100")` -`r exNum('Have R put in alphabetical order \x60c("WA","DC","CA","PA","MD","VA","OH")\x60')` +`r .exNum("Print all even numbers less than 100")` +`r .exNum("What is the mean of even numbers less than 100")` +`r .exNum('Have R put in alphabetical order \x60c("WA","DC","CA","PA","MD","VA","OH")\x60')` # Assignment of values to variables @@ -215,11 +218,11 @@ state.names[i[1:3]] # show me those three states Note that in the last example we used square brackets within square brackets. First, we asked R to give us the indices of the first three states in alphabetical order and that was `r i[1:3]`. Then R took those three values and plugged them into the second set of square brackets to show you the state names in those positions in the collection. ## Exercises -`r exNum("What's the last state in the \x60state.names\x60?")` -`r exNum('Pick out states that begin with "M" using their indices')` -`r exNum("Pick out states where you have lived")` -`r exNum("What's the last state in alphabetical order?")` -`r exNum("What are the last three states in alphabetical order?")` +`r .exNum("What's the last state in the \x60state.names\x60?")` +`r .exNum('Pick out states that begin with "M" using their indices')` +`r .exNum("Pick out states where you have lived")` +`r .exNum("What's the last state in alphabetical order?")` +`r .exNum("What are the last three states in alphabetical order?")` # Logical values and operations @@ -305,9 +308,9 @@ sum(my.states %in% c("CA","OR","WA","AK","HI")) Note in the last line we used `sum()` to count for how many of the elements in `my.states` did `%in%` evaluate to be `TRUE`. ## Exercises -`r exNum("Report \x60TRUE\x60 or \x60FALSE\x60 for each state depending on if you have lived there")` -`r exNum("With \x60a <- 1:100\x60, pick out odd numbers between 50 and 75")` -`r exNum("Use greater than less than signs to get all state names that begin with M")` +`r .exNum("Report \x60TRUE\x60 or \x60FALSE\x60 for each state depending on if you have lived there")` +`r .exNum("With \x60a <- 1:100\x60, pick out odd numbers between 50 and 75")` +`r .exNum("Use greater than less than signs to get all state names that begin with M")` # Sampling The function `sample()` randomly shuffles a collection of values. @@ -328,10 +331,10 @@ table(a) max(table(a)) # find out which value appears most frequently ``` ## Exercises -`r exNum("Use \x60sample()\x60 to estimate the probability of rolling a 6")` -`r exNum("Use \x60sample()\x60 to estimate the probability that the sum of two die equal 7")` -`r exNum("Use \x60sample()\x60 to select randomly five states without replacement")` -`r exNum("Use \x60sample()\x60 to select randomly 1000 states with replacement")` +`r .exNum("Use \x60sample()\x60 to estimate the probability of rolling a 6")` +`r .exNum("Use \x60sample()\x60 to estimate the probability that the sum of two die equal 7")` +`r .exNum("Use \x60sample()\x60 to select randomly five states without replacement")` +`r .exNum("Use \x60sample()\x60 to select randomly 1000 states with replacement")` + Tabulate how often each state was selected + Which state was selected the least? Make R do this for you @@ -439,12 +442,12 @@ ls() Assuming you are using R Studio, you can also see the objects stored in memory by clicking on the Environment tab. ## Exercises -`r exNum('Fix \x60state.list\x60 so that "DC" is in "other" rather than "east"')`. Here are a few hints +`r .exNum('Fix \x60state.list\x60 so that "DC" is in "other" rather than "east"')`. Here are a few hints + access "other" using `$` + combine things using `c()` + assign values using `<-` + remove values using `[]` with a negative index or using a logical statement -`r exNum("Print out east and central states together sorted")` +`r .exNum("Print out east and central states together sorted")` # Functions @@ -480,9 +483,9 @@ IQR You can see that it computes the 0.25 quantile and the 0.75 quantile and uses `diff()` to compute their difference. ## Exercises -`r exNum('Make a function \x60is.island(x)\x60 returns \x60TRUE\x60 if \x60x\x60 is an island')`. Islands are "HI", "FM", "MH", "PW", "AS", "GU", "MP", "PR", "VI", "UM". Borrow the template I used for `give.first.and.last()`. Then try using the `%in%` operator -`r exNum("Count how many islands are within each region. Use an \x60sapply()\x60 (or two) and your new \x60is.island()\x60 function")` -`r exNum("Which components of \x60b\x60 having missing values? Use \x60is.na()\x60")`. `b` was defined earlier +`r .exNum('Make a function \x60is.island(x)\x60 returns \x60TRUE\x60 if \x60x\x60 is an island')`. Islands are "HI", "FM", "MH", "PW", "AS", "GU", "MP", "PR", "VI", "UM". Borrow the template I used for `give.first.and.last()`. Then try using the `%in%` operator +`r .exNum("Count how many islands are within each region. Use an \x60sapply()\x60 (or two) and your new \x60is.island()\x60 function")` +`r .exNum("Which components of \x60b\x60 having missing values? Use \x60is.na()\x60")`. `b` was defined earlier # Matrices and apply() @@ -600,9 +603,9 @@ with(chicagoCrime, sort(table(Primary.Type[District==10]))) Much easier to read and understand! ## Exercises -`r exNum("Display three randomly selected rows")` -`r exNum("Count \x60NA\x60s in each column")` -`r exNum("Look up \x60Location.Description\x60, \x60Block\x60, \x60Beat\x60, and \x60Ward\x60 for those missing \x60Latitude\x60")` +`r .exNum("Display three randomly selected rows")` +`r .exNum("Count \x60NA\x60s in each column")` +`r .exNum("Look up \x60Location.Description\x60, \x60Block\x60, \x60Beat\x60, and \x60Ward\x60 for those missing \x60Latitude\x60")` # For loops Sometimes we need to have R repeat certain tasks multiple times, such as marching through each row of a dataset and modifying values. For loops accomplish this. Later in this course we will be using Google Maps to extract information about addresses. So we might need to iterate through every row in the dataset, check whether the latitude and longitude are missing, and if missing try to retrieve the latitude and longitude from Google Maps. The last crime in the dataset missing coordinates is in row 9954. @@ -697,11 +700,11 @@ chicagoCrime$google.maps.url <- paste("https://www.google.com/maps/place/", This took `r timeWithoutForLoop[3]` seconds. That's `r round(time4ForLoop[3]/timeWithoutForLoop[3],1)` times faster than the for loop. ## Exercises -`r exNum('Use a for loop to create a variable \x60Coordinates\x60 that looks like "(X.Coordinate,Y.Coordinate)"')` +`r .exNum('Use a for loop to create a variable \x60Coordinates\x60 that looks like "(X.Coordinate,Y.Coordinate)"')` + Use `paste()` with the `X.Coordinate` and `Y.Coordinate` variables + Remember the `sep=` option in `paste()` + You might find using the `with()` function to simplify your code and avoid having a lot of `chicagoCrime$`s -`r exNum("Redo the previous exercise without using a for loop and compare computation time")` +`r .exNum("Redo the previous exercise without using a for loop and compare computation time")` # More tabulating, aggregating, and breaking statistics down by group The variable `Arrest` indicates whether someone was arrested for the crime. Here are the first 10 values. @@ -750,8 +753,8 @@ barplot(a$`(Arrest == "true")`, ``` ## Exercises -`r exNum('How many assaults occurred in the street? (\x60Location.Description=="STREET"\x60)')`. Try using `subset()` even though there are other ways -`r exNum("What percentage of assaults occurred in the street by Ward?")` +`r .exNum('How many assaults occurred in the street? (\x60Location.Description=="STREET"\x60)')`. Try using `subset()` even though there are other ways +`r .exNum("What percentage of assaults occurred in the street by Ward?")` # Plotting Data @@ -809,14 +812,14 @@ text(ifelse(tab<80, 180, tab-5), # x-coord of text, adj=1) # right justify text ``` -# Exercises -`r exNum("Make a barplot indicating how many states are in each region. Use \x60state.list\x60")` -`r exNum("Identify the beat with the most crimes")` -`r exNum("Identify the beat with the most domestic violence incidents")` -`r exNum("Part 1 crimes are homicide, robbery, assault, arson, burglary, theft, sex offense, motor vehicle theft. Calculate the number of Part 1 crimes in Chicago")` +## Exercises +`r .exNum("Make a barplot indicating how many states are in each region. Use \x60state.list\x60")` +`r .exNum("Identify the beat with the most crimes")` +`r .exNum("Identify the beat with the most domestic violence incidents")` +`r .exNum("Part 1 crimes are homicide, robbery, assault, arson, burglary, theft, sex offense, motor vehicle theft. Calculate the number of Part 1 crimes in Chicago")` # Solutions to the exercises -1. `r exerciseQuestions[1]` +1. `r .exerciseQuestions[1]` ```{r comment=""} (1:49)*2 ``` @@ -825,22 +828,22 @@ or seq(2,98,by=2) ``` -2. `r exerciseQuestions[2]` +2. `r .exerciseQuestions[2]` ```{r comment=""} mean((1:49)*2) ``` -3. `r exerciseQuestions[3]` +3. `r .exerciseQuestions[3]` ```{r comment=""} sort(c("WA","DC","CA","PA","MD","VA","OH")) ``` -4. `r exerciseQuestions[4]` +4. `r .exerciseQuestions[4]` ```{r comment=""} state.names[51] ``` -5. `r exerciseQuestions[5]` +5. `r .exerciseQuestions[5]` ```{r comment=""} state.names[c(7,8,21,24,28,32,35,46)] ``` @@ -853,13 +856,13 @@ Here's another possible answer that uses `substring` (which we haven't covered y state.names[substring(state.names, 1, 1)=="M"] ``` -6. `r exerciseQuestions[6]` +6. `r .exerciseQuestions[6]` Of course, these may vary depending on where you have lived. ```{r comment=""} state.names[c(1, 4, 10, 26)] ``` -7. `r exerciseQuestions[7]` +7. `r .exerciseQuestions[7]` ```{r comment=""} sort(state.names)[51] ``` @@ -868,29 +871,29 @@ or rev(sort(state.names))[1] ``` -8. `r exerciseQuestions[8]` +8. `r .exerciseQuestions[8]` ```{r comment=""} rev(sort(state.names))[1:3] ``` -9. `r exerciseQuestions[9]` +9. `r .exerciseQuestions[9]` ```{r comment=""} my.states <- c("PA", "NJ", "NY", "MD", "DE", "MA", "RI", "CT", "ME", "LA", "IN") state.names %in% my.states ``` -10. `r exerciseQuestions[10]` +10. `r .exerciseQuestions[10]` ```{r comment=""} a <- 1:100 a[a %% 2==1 & a>50 & a<75] ``` -11. `r exerciseQuestions[11]` +11. `r .exerciseQuestions[11]` ```{r comment=""} state.names[state.names>"LZ" & state.names<"N"] ``` -12. `r exerciseQuestions[12]` +12. `r .exerciseQuestions[12]` ```{r comment=""} a <- sample(1:6, size=100000, replace=TRUE) table(a)[6]/length(a) @@ -904,7 +907,7 @@ Or mean(a==6) ``` -13. `r exerciseQuestions[13]` +13. `r .exerciseQuestions[13]` ```{r comment=""} dice1 <- sample(1:6, size=1000, replace=TRUE) dice2 <- sample(1:6, size=1000, replace=TRUE) @@ -912,12 +915,12 @@ doubleroll <- dice1 + dice2 mean(doubleroll==7) # should be close to 1/6 or 0.1666... ``` -14. `r exerciseQuestions[14]` (Answers will vary) +14. `r .exerciseQuestions[14]` (Answers will vary) ```{r comment=""} sample(state.names, size=5, replace=FALSE) ``` -15. `r exerciseQuestions[15]` +15. `r .exerciseQuestions[15]` + Tabulate how often each state was selected (Answers will vary) ```{r comment=""} a <- sample(state.names, size=1000, replace=TRUE) @@ -929,7 +932,7 @@ table(a) sort(table(a))[1] ``` -16. `r exerciseQuestions[16]` +16. `r .exerciseQuestions[16]` ```{r comment=""} state.list$east <- state.list$east[state.list$east!="DC"] state.list$other <- c(state.list$other, "DC") @@ -942,7 +945,7 @@ state.list$other <- c(state.list$other, "DC") state.list ``` -17. `r exerciseQuestions[17]` +17. `r .exerciseQuestions[17]` ```{r comment=""} sort(c(state.list$east, state.list$central)) ``` @@ -951,7 +954,7 @@ Or with(state.list, sort(c(east, central))) ``` -18. `r exerciseQuestions[18]` +18. `r .exerciseQuestions[18]` ```{r comment=""} is.island <- function(x) { @@ -959,7 +962,7 @@ is.island <- function(x) } ``` -19. `r exerciseQuestions[19]` +19. `r .exerciseQuestions[19]` First, this `lapply()` asks each state if they are an island. ```{r comment=""} @@ -970,7 +973,7 @@ Now we want to count up how many `TRUE`s there are in each component, so wrap th sapply(lapply(state.list, is.island), sum) ``` -20. `r exerciseQuestions[20]` +20. `r .exerciseQuestions[20]` ```{r comment=""} sapply(lapply(b, is.na), any) ``` @@ -980,12 +983,12 @@ b <- list(0:9, c("A","B","C"), c(TRUE,FALSE,NA)) sapply(b, function(x) any(is.na(x))) ``` -21. `r exerciseQuestions[21]` +21. `r .exerciseQuestions[21]` ```{r comment=""} chicagoCrime[sample(1:nrow(chicagoCrime), size=3),] ``` -22. `r exerciseQuestions[22]` +22. `r .exerciseQuestions[22]` ```{r comment=""} sapply(lapply(chicagoCrime, is.na), sum) ``` @@ -994,7 +997,7 @@ Or sapply(chicagoCrime, function(x) sum(is.na(x))) ``` -23. `r exerciseQuestions[23]` +23. `r .exerciseQuestions[23]` ```{r comment=""} i <- is.na(chicagoCrime$Latitude) # Let's just show the first 5 rows @@ -1007,7 +1010,7 @@ subset(chicagoCrime, is.na(chicagoCrime$Latitude), select=c("Location.Description","Block","Beat","Ward"))[1:5,] ``` -24. `r exerciseQuestions[24]` +24. `r .exerciseQuestions[24]` ```{r comment=""} system.time( for (i in 1:nrow(chicagoCrime)) @@ -1028,7 +1031,7 @@ for (i in 1:nrow(chicagoCrime)) } ) ``` -25. `r exerciseQuestions[25]` +25. `r .exerciseQuestions[25]` ```{r comment=""} system.time( chicagoCrime$coords3 <- with(chicagoCrime, @@ -1036,25 +1039,25 @@ chicagoCrime$coords3 <- with(chicagoCrime, ) ``` -26. `r exerciseQuestions[26]` +26. `r .exerciseQuestions[26]` ```{r comment=""} with(subset(chicagoCrime, Primary.Type=="ASSAULT"), sum(chicagoCrime$Location.Description=="STREET")) ``` -27. `r exerciseQuestions[27]` +27. `r .exerciseQuestions[27]` ```{r comment=""} aggregate((Location.Description=="STREET")~Ward, data=subset(chicagoCrime, Primary.Type=="ASSAULT"), mean) ``` -28. `r exerciseQuestions[28]` +28. `r .exerciseQuestions[28]` ```{r comment=""} barplot(sapply(state.list, length)) ``` -29. `r exerciseQuestions[29]` +29. `r .exerciseQuestions[29]` ```{r comment=""} names(rev(sort(table(chicagoCrime$Beat)))[1]) ``` @@ -1063,13 +1066,13 @@ Or names(which.max(table(chicagoCrime$Beat))) ``` -30. `r exerciseQuestions[30]` +30. `r .exerciseQuestions[30]` ```{r comment=""} with(subset(chicagoCrime, Description=="DOMESTIC BATTERY SIMPLE"), names(which.max(table(Beat)))) ``` -31. `r exerciseQuestions[31]` +31. `r .exerciseQuestions[31]` ```{r comment=""} sum(chicagoCrime$Primary.Type %in% c("HOMICIDE", "ROBBERY", "ASSAULT", "ARSON", "BURGLARY", "THEFT", "SEX OFFENSE", diff --git a/01_Intro_to_R.html b/01_Intro_to_R.html index cfa7daa..5056d1b 100644 --- a/01_Intro_to_R.html +++ b/01_Intro_to_R.html @@ -10,7 +10,7 @@ - + Introduction to R @@ -92,6 +92,9 @@ .tabbed-pane { padding-top: 12px; } +.html-widget { + margin-bottom: 20px; +} button.code-folding-btn:focus { outline: none; } @@ -128,7 +131,7 @@

Ruth Moyer

University of Pennsylvania
moyruth@upenn.edu
-

July 26, 2018

+

September 04, 2018

@@ -139,7 +142,7 @@

July 26, 2018

- + @@ -148,7 +151,7 @@

July 26, 2018

- +

Introduction

This is the first set of notes for an introduction to R programming from criminology and criminal justice. These notes assume that you have the latest version of R and R Studio installed. We are also assuming that you know how to start a new script file and submit code to the R console. From that basic knowledge about using R, we are going to start with 2+2 and by the end of this set of notes you will load in a small Chicago crime dataset, create a few plots, count some crimes, and be able to subset the data. Our aim is to build a firm foundation on which we will build throughout this set of notes.

@@ -454,9 +457,9 @@

Sampling

sample(1:10) a <- sample(1:1000,size=10) # pick 10 numbers between 1-1000 a <- sample(1:6,size=1000,replace=TRUE) # roll a die 1000 times -
 [1]  6  9  5  4  2  8  3  1 10  7
- [1]  5  2  7  1  4 10  6  3  9  8
- [1]  4  1  9 10  7  5  3  2  6  8
+
 [1]  6  4  2  9  7  1  5 10  3  8
+ [1]  3  7  9 10  1  6  4  5  8  2
+ [1]  4  6  1  2 10  9  3  5  8  7

Notice that sample() has several options including size= to indicate how many to select and replace= to indicate whether to sample with or without replacement. You can access the help on the sample() function by typing ?sample at the R prompt.

@@ -466,8 +469,8 @@

Tabulating

max(table(a)) # find out which value appears most frequently 
a
   1   2   3   4   5   6 
-182 165 168 157 178 150 
-[1] 182
+155 185 168 147 181 164 +[1] 185

Exercises

    @@ -570,13 +573,13 @@

    Lists

    [15] "WI" "WY" 
    lapply(state.list,sample,size=3,replace=FALSE)
    $west
-[1] "UT" "ID" "AZ"
+[1] "ID" "AZ" "CA"
 
 $east
-[1] "MS" "DC" "OH"
+[1] "MD" "NH" "OH"
 
 $central
-[1] "NE" "WI" "MI"
    +[1] "OK" "WI" "SD" 
    sapply(state.list,length)
       west    east central 
      11      24      16
    @@ -600,9 +603,8 @@

    Lists

    This creates a separate object called other, unconnected to our state.list. By using the $ we add our new collection of states (other) to state.list.

    We have now created a lot of objects. At any time you can run ls() to list all the objects that R has in memory.

    ls()
    -
    [1] "a"                 "b"                 "counterExercise"  
-[4] "exerciseQuestions" "exNum"             "i"                
-[7] "my.states"         "state.list"        "state.names"
    +
    [1] "a"           "b"           "i"           "my.states"   "state.list" 
+[6] "state.names"

    Assuming you are using R Studio, you can also see the objects stored in memory by clicking on the Environment tab.

    Exercises

    @@ -657,7 +659,7 @@

    Functions

    function (x, na.rm = FALSE, type = 7) 
 diff(quantile(as.numeric(x), c(0.25, 0.75), na.rm = na.rm, names = FALSE, 
     type = type))
-<bytecode: 0x000000001d36aad8>
+<bytecode: 0x0000000016585f08>
 <environment: namespace:stats>

    You can see that it computes the 0.25 quantile and the 0.75 quantile and uses diff() to compute their difference.

    @@ -675,10 +677,10 @@

    Matrices and apply()

    a <- matrix(sample(1:5,size=12,replace=TRUE),nrow=4)
 a
         [,1] [,2] [,3]
-[1,]    1    4    1
-[2,]    2    4    2
-[3,]    4    2    3
-[4,]    2    1    3
    +[1,] 3 4 1 +[2,] 4 3 2 +[3,] 2 3 5 +[4,] 3 1 1

    This matrix has two dimensions, 4 rows and 3 columns. You can use square brackets to select elements from the matrix.

    a[1,2]     # element in first row, second column
 a[1,]      # the entire first row
@@ -686,36 +688,36 @@ 
    Matrices and apply()
    
 a[-1,-1]   # dropping the first row and first column
 a[3:4,2:3] # rows 3 & 4, columns 2 & 3
    [1] 4
-[1] 1 4 1
-[1] 4 4 2 1
+[1] 3 4 1
+[1] 4 3 3 1
      [,1] [,2]
-[1,]    4    2
-[2,]    2    3
-[3,]    1    3
+[1,]    3    2
+[2,]    3    5
+[3,]    1    1
      [,1] [,2]
-[1,]    2    3
-[2,]    1    3
    +[1,] 3 5 +[2,] 1 1

    The numbers to the left of the comma index rows and the numbers to the right of the comma index columns. The apply() function, like the lapply() and sapply() functions, allows you to apply a function to all the rows or all the columns of a matrix. apply() needs the name of the matrix, whether you want to apply the function to the first dimension (rows) or the second dimension (columns), and the name of the function to apply.

    apply(a, 1, sum)     # compute sum of each row
 apply(a, 2, sum)     # compute sum of each column
 apply(a, 1, mean)    # compute mean of each row
 apply(a, 1, summary) # summarize each row
    -
    [1] 6 8 9 6
-[1]  9 11  9
-[1] 2.000000 2.666667 3.000000 2.000000
-        [,1]     [,2] [,3] [,4]
-Min.     1.0 2.000000  2.0  1.0
-1st Qu.  1.0 2.000000  2.5  1.5
-Median   1.0 2.000000  3.0  2.0
-Mean     2.0 2.666667  3.0  2.0
-3rd Qu.  2.5 3.000000  3.5  2.5
-Max.     4.0 4.000000  4.0  3.0
    +
    [1]  8  9 10  5
+[1] 12 11  9
+[1] 2.666667 3.000000 3.333333 1.666667
+            [,1] [,2]     [,3]     [,4]
+Min.    1.000000  2.0 2.000000 1.000000
+1st Qu. 2.000000  2.5 2.500000 1.000000
+Median  3.000000  3.0 3.000000 1.000000
+Mean    2.666667  3.0 3.333333 1.666667
+3rd Qu. 3.500000  3.5 4.000000 2.000000
+Max.    4.000000  4.0 5.000000 3.000000

    We can also create a new function right on the spot to compute something on each row or column. Let’s find the minimum and maximum values in each row and find out if all the values are greater than 1.

    apply(a, 1, function(x) {c(min(x),max(x))}) # there is also a function range()
 apply(a, 1, function(x) {all(x>1)})
         [,1] [,2] [,3] [,4]
 [1,]    1    2    2    1
-[2,]    4    4    4    3
+[2,]    4    4    5    3
 [1] FALSE  TRUE  TRUE FALSE
    @@ -737,10 +739,9 @@

    Data frames

    load("chicago crime 20141124-20141209.RData")

    List the objects R now has in memory and you will see that there is a new object, chicagoCrime.

    ls()
    -
     [1] "a"                   "b"                   "chicagoCrime"       
- [4] "counterExercise"     "exerciseQuestions"   "exNum"              
- [7] "give.first.and.last" "i"                   "my.states"          
-[10] "state.list"          "state.names"
    +
    [1] "a"                   "b"                   "chicagoCrime"       
+[4] "give.first.and.last" "i"                   "my.states"          
+[7] "state.list"          "state.names"

    If you did not spell the name of the .RData file exactly correctly, then R will give you an error. A common occurrence when downloading the same file from the web multiple times is for your web browser to add numbers to the multiple versions you’ve downloaded. So check the file name carefully. Here’s what happens when I request a file that doesn’t exist.

    load("chicago crime.RData")
    Warning in readChar(con, 5L, useBytes = TRUE): cannot open compressed file
@@ -1629,7 +1630,7 @@ 
    For loops
    
                                             ",+Chicago,+IL",sep="")
 }
 )
    -

    Note that we’ve wrapped the for loop with a call to system.time(). This will keep the time on how long this for loop takes. When creating these notes on a laptop it took 0.56 seconds. Not bad. Much faster than having to type out these 10,000 URLs. However, if we had one million addresses, then this code is going to take much more time.

    +

    Note that we’ve wrapped the for loop with a call to system.time(). This will keep the time on how long this for loop takes. When creating these notes on a laptop it took 0.58 seconds. Not bad. Much faster than having to type out these 10,000 URLs. However, if we had one million addresses, then this code is going to take much more time.

    In fact, in R for loops are very slow. They are so slow that R programmers attempt to avoid them whenever possible. We can actually accomplish the same task without using a for loop. gsub() will accept a whole collection of addresses and modify them all at once. paste() also will accept a collection of text values and paste them together with the other parts.

    timeWithoutForLoop <- system.time(
 {
@@ -1639,7 +1640,7 @@ 
    For loops
    
                                       ",+Chicago,+IL",sep="")
 }
 )
    -

    This took 0.02 seconds. That’s 28 times faster than the for loop.

    +

    This took 0.02 seconds. That’s 29 times faster than the for loop.

    Exercises

      @@ -1805,9 +1806,8 @@

      Plotting Data

      cex=0.7, # shrink text (cex=character expansion) adj=1) # right justify text

      -
    -
    -

    Exercises

    +
    +

    Exercises

    1. Make a barplot indicating how many states are in each region. Use state.list
    2. Identify the beat with the most crimes
      3. @@ -1815,6 +1815,7 @@

      Exercises

    3. Part 1 crimes are homicide, robbery, assault, arson, burglary, theft, sex offense, motor vehicle theft. Calculate the number of Part 1 crimes in Chicago
    +

    Solutions to the exercises

      @@ -1899,14 +1900,14 @@

      Solutions to the exercises

    a <- sample(1:6, size=100000, replace=TRUE)
 table(a)[6]/length(a)
    -
          6 
-0.16624
    +
         6 
+0.1672

    Or

    sum(a==6)/length(a)
    -
    [1] 0.16624
    +
    [1] 0.1672

    Or

    mean(a==6)
    -
    [1] 0.16624
    +
    [1] 0.1672
    1. Use sample() to estimate the probability that the sum of two die equal 7
    @@ -1914,12 +1915,12 @@

    Solutions to the exercises

    dice2 <- sample(1:6, size=1000, replace=TRUE) doubleroll <- dice1 + dice2 mean(doubleroll==7) # should be close to 1/6 or 0.1666... -
    [1] 0.159
    +
    [1] 0.156
    1. Use sample() to select randomly five states without replacement (Answers will vary)
    sample(state.names, size=5, replace=FALSE)
    -
    [1] "MS" "NJ" "MD" "DE" "IL"
    +
    [1] "OR" "TX" "KY" "HI" "AR"
    1. Use sample() to select randomly 1000 states with replacement
    @@ -1930,17 +1931,17 @@

    Solutions to the exercises

    table(a) 
    a
 AK AL AR AZ CA CO CT DC DE FL GA HI IA ID IL IN KS KY LA MA MD ME MI MN MO 
-19 23 14 15 20 28 20 22 15 20 12 18 22 17 20 16 23 28 21 17 18 27 21 19 21 
+14 10 18 18 28 13 21 21 26 19 19 17 26 13 21 19 17 23 13 25 26 19 26 18 17 
 MS MT NC ND NE NH NJ NM NV NY OH OK OR PA RI SC SD TN TX UT VA VT WA WI WV 
-18 26 21 17 20 17 18 18 28 21 16 11 23 19 22 15 17 15 29 20 20 22 12 20 15 
+25 19 20 16 25 19 23 23 25 23 18 18 19 10 16 28 13 12 21 19 18 25 18 17 23 
 WY 
-24
    +20
    • Which state was selected the least? (Answers will vary)
    sort(table(a))[1]
    -
    OK 
-11
    +
    AL 
+10
    1. Fix state.list so that “DC” is in “other” rather than “east”
    @@ -2032,30 +2033,30 @@

    Solutions to the exercises

  1. Display three randomly selected rows
chicagoCrime[sample(1:nrow(chicagoCrime), size=3),]
-
          ID Case.Number                   Date              Block IUCR
-4323 9881744    HX532834 12/03/2014 11:30:00 AM  033XX W CERMAK RD 1310
-3326 9879694    HX530445 12/04/2014 07:30:00 PM 002XX S CICERO AVE 0530
-9524 9868923    HX519453 11/24/2014 09:15:00 PM  048XX W GEORGE ST 1310
-        Primary.Type                   Description Location.Description
-4323 CRIMINAL DAMAGE                   TO PROPERTY            RESIDENCE
-3326         ASSAULT AGGRAVATED: OTHER DANG WEAPON               STREET
-9524 CRIMINAL DAMAGE                   TO PROPERTY            APARTMENT
-     Arrest Domestic Beat District Ward Community.Area FBI.Code
-4323  false    false 1024       10   22             30       14
-3326  false    false 1113       11   28             25      04A
-9524  false    false 2521       25   31             19       14
-     X.Coordinate Y.Coordinate Year             Updated.On Latitude
-4323      1154560      1889141 2014 12/10/2014 12:42:02 PM 41.85162
-3326      1144474      1898281 2014 12/11/2014 12:47:57 PM 41.87690
-9524      1143684      1918886 2014 12/01/2014 12:41:33 PM 41.93346
-     Longitude                      Location
-4323 -87.70821 (41.851621827, -87.708213779)
-3326 -87.74500 (41.876898678, -87.745002411)
-9524 -87.74739 (41.933455891, -87.747386305)
-                                                       google.maps.url
-4323  https://www.google.com/maps/place/03350 W CERMAK RD,+Chicago,+IL
-3326 https://www.google.com/maps/place/00250 S CICERO AVE,+Chicago,+IL
-9524  https://www.google.com/maps/place/04850 W GEORGE ST,+Chicago,+IL
+
          ID Case.Number                   Date               Block IUCR
+9791 9868419    HX518714 11/24/2014 12:30:00 PM 065XX S WESTERN AVE 0460
+8039 9871729    HX522206 11/27/2014 11:30:00 AM  096XX S HALSTED ST 0486
+3951 9878332    HX529282 12/03/2014 08:55:00 PM   055XX W CORTEZ ST 2024
+     Primary.Type             Description Location.Description Arrest
+9791      BATTERY                  SIMPLE           RESTAURANT  false
+8039      BATTERY DOMESTIC BATTERY SIMPLE            RESIDENCE  false
+3951    NARCOTICS     POSS: HEROIN(WHITE)            RESIDENCE   true
+     Domestic Beat District Ward Community.Area FBI.Code X.Coordinate
+9791    false  832        8   15             66      08B      1161532
+8039     true 2223       22   21             73      08B      1172717
+3951    false 1524       15   37             25       18      1139051
+     Y.Coordinate Year             Updated.On Latitude Longitude
+9791      1861424 2014 12/01/2014 12:41:33 PM 41.77542 -87.68339
+8039      1840878 2014 12/04/2014 12:50:41 PM 41.71880 -87.64300
+3951      1906410 2014 12/10/2014 12:42:02 PM 41.89931 -87.76472
+                          Location
+9791 (41.775420908, -87.683394137)
+8039 (41.718800734, -87.642995516)
+3951 (41.899305964, -87.764716528)
+                                                        google.maps.url
+9791 https://www.google.com/maps/place/06550 S WESTERN AVE,+Chicago,+IL
+8039  https://www.google.com/maps/place/09650 S HALSTED ST,+Chicago,+IL
+3951   https://www.google.com/maps/place/05550 W CORTEZ ST,+Chicago,+IL
  1. Count NAs in each column
@@ -2127,7 +2128,7 @@

Solutions to the exercises

} ) 
   user  system elapsed 
-   0.62    0.00    0.63
+ 0.58 0.02 0.60

Or

system.time(
 for (i in 1:nrow(chicagoCrime)) 
@@ -2138,7 +2139,7 @@ 
Solutions to the exercises

 }
 )
   user  system elapsed 
-   0.63    0.00    0.63
+ 0.53 0.00 0.53
  1. Redo the previous exercise without using a for loop and compare computation time
diff --git a/01_Intro_to_R.pdf b/01_Intro_to_R.pdf index 9e74a76..cbc0bcd 100644 Binary files a/01_Intro_to_R.pdf and b/01_Intro_to_R.pdf differ