HarvardX PH125.1x: Chapter 2 Exercises
Dimple K. Patel
2023-12-31
Section 2.3
1. What is the sum of the first 100 positive integers? The formula for the sum of integers 1 through n is \(n(n+1)/2\). Define n=100 then use R to compute the sum of 11 through 100 using the formula. What is the sum?
n<-100
num100<-(100)*(100+1)/2
num100## [1] 5050- Now use the same formula to compute the sum of the integers from 1 through 1,000.
n<-1000
num1000<-(1000)*(100+1)/2
num1000## [1] 50500- Look at the result of typing the following code into R:
n <- 1000
x <- seq(1, n)
sum(x)## [1] 500500x## [1] 1 2 3 4 5 6 7 8 9 10 11 12 13 14
## [15] 15 16 17 18 19 20 21 22 23 24 25 26 27 28
## [29] 29 30 31 32 33 34 35 36 37 38 39 40 41 42
## [43] 43 44 45 46 47 48 49 50 51 52 53 54 55 56
## [57] 57 58 59 60 61 62 63 64 65 66 67 68 69 70
## [71] 71 72 73 74 75 76 77 78 79 80 81 82 83 84
## [85] 85 86 87 88 89 90 91 92 93 94 95 96 97 98
## [99] 99 100 101 102 103 104 105 106 107 108 109 110 111 112
## [113] 113 114 115 116 117 118 119 120 121 122 123 124 125 126
## [127] 127 128 129 130 131 132 133 134 135 136 137 138 139 140
## [141] 141 142 143 144 145 146 147 148 149 150 151 152 153 154
## [155] 155 156 157 158 159 160 161 162 163 164 165 166 167 168
## [169] 169 170 171 172 173 174 175 176 177 178 179 180 181 182
## [183] 183 184 185 186 187 188 189 190 191 192 193 194 195 196
## [197] 197 198 199 200 201 202 203 204 205 206 207 208 209 210
## [211] 211 212 213 214 215 216 217 218 219 220 221 222 223 224
## [225] 225 226 227 228 229 230 231 232 233 234 235 236 237 238
## [239] 239 240 241 242 243 244 245 246 247 248 249 250 251 252
## [253] 253 254 255 256 257 258 259 260 261 262 263 264 265 266
## [267] 267 268 269 270 271 272 273 274 275 276 277 278 279 280
## [281] 281 282 283 284 285 286 287 288 289 290 291 292 293 294
## [295] 295 296 297 298 299 300 301 302 303 304 305 306 307 308
## [309] 309 310 311 312 313 314 315 316 317 318 319 320 321 322
## [323] 323 324 325 326 327 328 329 330 331 332 333 334 335 336
## [337] 337 338 339 340 341 342 343 344 345 346 347 348 349 350
## [351] 351 352 353 354 355 356 357 358 359 360 361 362 363 364
## [365] 365 366 367 368 369 370 371 372 373 374 375 376 377 378
## [379] 379 380 381 382 383 384 385 386 387 388 389 390 391 392
## [393] 393 394 395 396 397 398 399 400 401 402 403 404 405 406
## [407] 407 408 409 410 411 412 413 414 415 416 417 418 419 420
## [421] 421 422 423 424 425 426 427 428 429 430 431 432 433 434
## [435] 435 436 437 438 439 440 441 442 443 444 445 446 447 448
## [449] 449 450 451 452 453 454 455 456 457 458 459 460 461 462
## [463] 463 464 465 466 467 468 469 470 471 472 473 474 475 476
## [477] 477 478 479 480 481 482 483 484 485 486 487 488 489 490
## [491] 491 492 493 494 495 496 497 498 499 500 501 502 503 504
## [505] 505 506 507 508 509 510 511 512 513 514 515 516 517 518
## [519] 519 520 521 522 523 524 525 526 527 528 529 530 531 532
## [533] 533 534 535 536 537 538 539 540 541 542 543 544 545 546
## [547] 547 548 549 550 551 552 553 554 555 556 557 558 559 560
## [561] 561 562 563 564 565 566 567 568 569 570 571 572 573 574
## [575] 575 576 577 578 579 580 581 582 583 584 585 586 587 588
## [589] 589 590 591 592 593 594 595 596 597 598 599 600 601 602
## [603] 603 604 605 606 607 608 609 610 611 612 613 614 615 616
## [617] 617 618 619 620 621 622 623 624 625 626 627 628 629 630
## [631] 631 632 633 634 635 636 637 638 639 640 641 642 643 644
## [645] 645 646 647 648 649 650 651 652 653 654 655 656 657 658
## [659] 659 660 661 662 663 664 665 666 667 668 669 670 671 672
## [673] 673 674 675 676 677 678 679 680 681 682 683 684 685 686
## [687] 687 688 689 690 691 692 693 694 695 696 697 698 699 700
## [701] 701 702 703 704 705 706 707 708 709 710 711 712 713 714
## [715] 715 716 717 718 719 720 721 722 723 724 725 726 727 728
## [729] 729 730 731 732 733 734 735 736 737 738 739 740 741 742
## [743] 743 744 745 746 747 748 749 750 751 752 753 754 755 756
## [757] 757 758 759 760 761 762 763 764 765 766 767 768 769 770
## [771] 771 772 773 774 775 776 777 778 779 780 781 782 783 784
## [785] 785 786 787 788 789 790 791 792 793 794 795 796 797 798
## [799] 799 800 801 802 803 804 805 806 807 808 809 810 811 812
## [813] 813 814 815 816 817 818 819 820 821 822 823 824 825 826
## [827] 827 828 829 830 831 832 833 834 835 836 837 838 839 840
## [841] 841 842 843 844 845 846 847 848 849 850 851 852 853 854
## [855] 855 856 857 858 859 860 861 862 863 864 865 866 867 868
## [869] 869 870 871 872 873 874 875 876 877 878 879 880 881 882
## [883] 883 884 885 886 887 888 889 890 891 892 893 894 895 896
## [897] 897 898 899 900 901 902 903 904 905 906 907 908 909 910
## [911] 911 912 913 914 915 916 917 918 919 920 921 922 923 924
## [925] 925 926 927 928 929 930 931 932 933 934 935 936 937 938
## [939] 939 940 941 942 943 944 945 946 947 948 949 950 951 952
## [953] 953 954 955 956 957 958 959 960 961 962 963 964 965 966
## [967] 967 968 969 970 971 972 973 974 975 976 977 978 979 980
## [981] 981 982 983 984 985 986 987 988 989 990 991 992 993 994
## [995] 995 996 997 998 999 1000Based on the result, what do you think the functions seq
and sum do? You can use help.
seq creates a list of numbers and sum
adds them up.
- In math and programming, we say that we evaluate a function when we
replace the argument with a given number. So if we
typeÂ
sqrt(4), we evaluate theÂsqrt function. In R, you can evaluate a function inside another function. The evaluations happen from the inside out. Use one line of code to compute the log, in base 10, of the square root of 100.
log(sqrt(100), base=10)## [1] 15. Which of the following will always return the numeric value stored
in x? You can try out examples and use the help system if
you want.
log(exp(2))## [1] 2Section 2.5
1. Load the US murders dataset.
Use the function str to examine the structure of
the murders object. Which of the following best describes
the variables represented in this data frame?
library(dslabs)
data(murders)
str(murders)## 'data.frame': 51 obs. of 5 variables:
## $ state : chr "Alabama" "Alaska" "Arizona" "Arkansas" ...
## $ abb : chr "AL" "AK" "AZ" "AR" ...
## $ region : Factor w/ 4 levels "Northeast","South",..: 2 4 4 2 4 4 1 2 2 2 ...
## $ population: num 4779736 710231 6392017 2915918 37253956 ...
## $ total : num 135 19 232 93 1257 ...The state name, the abbreviation of the state name, the state’s region, and the state’s population and total number of murders for 2010.
- What are the column names used by the data frame for these five variables?
names(murders)## [1] "state" "abb" "region" "population" "total"3. Use the accessor $ to extract the state abbreviations
and assign them to the object a. What is the class of this
object?
a<-murders$abb
class(a)## [1] "character"4. Now use the square brackets to extract the state abbreviations and
assign them to the object b. Use
the identical function to determine
if a and b are the same.
b<-murders[,2]
b## [1] "AL" "AK" "AZ" "AR" "CA" "CO" "CT" "DE" "DC" "FL" "GA" "HI" "ID" "IL" "IN"
## [16] "IA" "KS" "KY" "LA" "ME" "MD" "MA" "MI" "MN" "MS" "MO" "MT" "NE" "NV" "NH"
## [31] "NJ" "NM" "NY" "NC" "ND" "OH" "OK" "OR" "PA" "RI" "SC" "SD" "TN" "TX" "UT"
## [46] "VT" "VA" "WA" "WV" "WI" "WY"identical(a, b)## [1] TRUE5. We saw that the region column stores a factor. You
can corroborate this by typing:
With one line of code, use the function levels and
length to determine the number of regions defined by this
dataset.
class(murders$region)## [1] "factor"length(levels(murders$region))## [1] 46. The function table takes a vector and returns the
frequency of each element. You can quickly see how many states are in
each region by applying this function. Use this function in one line of
code to create a table of states per region.
table(murders$region)##
## Northeast South North Central West
## 9 17 12 13Section 2.8
1. Use the function c to create a vector with the
average high temperatures in January for Beijing, Lagos, Paris, Rio de
Janeiro, San Juan, and Toronto, which are 35, 88, 42, 84, 81, and 30
degrees Fahrenheit. Call the object temp.
temp<-c(35, 88, 42, 84, 81, 30)2. Now create a vector with the city names and call the object
city.
city<-c("Beijing", "Lagos", "Paris", "Rio de Janeiro", "San Juan", "Toronto")3. Use the names function and the objects defined in the
previous exercises to associate the temperature data with its
corresponding city.
names(temp)<-city
temp## Beijing Lagos Paris Rio de Janeiro San Juan
## 35 88 42 84 81
## Toronto
## 304. Use the [ and : operators to access the
temperature of the first three cities on the list.
temp[1:3]## Beijing Lagos Paris
## 35 88 425. Use the [ operator to access the temperature of Paris
and San Juan.
temp[c(3,5)]## Paris San Juan
## 42 816. Use the : operator to create a sequence of numbers
12,13,14,…, 73
c(12:73)## [1] 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36
## [26] 37 38 39 40 41 42 43 44 45 46 47 48 49 50 51 52 53 54 55 56 57 58 59 60 61
## [51] 62 63 64 65 66 67 68 69 70 71 72 737. Create a vector containing all the positive odd numbers smaller than 100.
seq(1, 100, by=2)## [1] 1 3 5 7 9 11 13 15 17 19 21 23 25 27 29 31 33 35 37 39 41 43 45 47 49
## [26] 51 53 55 57 59 61 63 65 67 69 71 73 75 77 79 81 83 85 87 89 91 93 95 97 998. Create a vector of numbers that starts at 6, does not pass 55, and
adds numbers in increments of 4/7: 6, 6 + 4/7, 6 + 8/7, and so on. How
many numbers does the list have? Hint: use seq and
length.
length(seq(6, 55, 4/7))## [1] 869. What is the class of the following object
a <- seq(1, 10, 0.5)?
a<-seq(1, 10, .5)
class(a)## [1] "numeric"10. What is the class of the following object
a <- seq(1, 10)?
a<-seq(1, 10)
class(a)## [1] "integer"11. The class of class(a<-1) is numeric, not integer.
R defaults to numeric and to force an integer, you need to add the
letter L. Confirm that the class of 1L is
integer.
class(a<-1)## [1] "numeric"class(a<-1L)## [1] "integer"12. Define the following vector:
x <- c("1", "3", "5")and coerce it to get integers.
x <-c(1L, 3L, 5L)
class(x)## [1] "integer"Section 2.10
For these exercises we will use the US murders dataset. Make sure you load it prior to starting.
library(dslabs)
data("murders")1. Use the $ operator to access the population size data
and store it as the object pop. Then use the
sort function to redefine pop so that it is
sorted. Finally, use the [ operator to report the smallest
population size.
library(dslabs)
data("murders")
pop<-sort(murders$population)
pop[1]## [1] 5636262. Now instead of the smallest population size, find the index of the
entry with the smallest population size. Hint: use order
instead of sort.
popo<-order(murders$population)
popo[1]## [1] 513. We can actually perform the same operation as in the previous
exercise using the function which.min. Write one line of
code that does this.
which.min(murders$population)## [1] 514. Now we know how small the smallest state is and we know which row
represents it. Which state is it? Define a variable states
to be the state names from the murders data frame. Report
the name of the state with the smallest population.
states<-murders$state
states[51]## [1] "Wyoming"5. You can create a data frame using the data.frame
function. Here is a quick example:
temp <- c(35, 88, 42, 84, 81, 30)
city <- c("Beijing", "Lagos", "Paris", "Rio de Janeiro", "San Juan", "Toronto")
city_temps <- data.frame(name = city, temperature = temp)Use the rank function to determine the population rank
of each state from smallest population size to biggest. Save these ranks
in an object called ranks, then create a data frame with
the state name and its rank. Call the data frame my_df.
ranks<-rank(murders$population)
my_df<-data.frame(states, ranks)
my_df## states ranks
## 1 Alabama 29
## 2 Alaska 5
## 3 Arizona 36
## 4 Arkansas 20
## 5 California 51
## 6 Colorado 30
## 7 Connecticut 23
## 8 Delaware 7
## 9 District of Columbia 2
## 10 Florida 49
## 11 Georgia 44
## 12 Hawaii 12
## 13 Idaho 13
## 14 Illinois 47
## 15 Indiana 37
## 16 Iowa 22
## 17 Kansas 19
## 18 Kentucky 26
## 19 Louisiana 27
## 20 Maine 11
## 21 Maryland 33
## 22 Massachusetts 38
## 23 Michigan 43
## 24 Minnesota 31
## 25 Mississippi 21
## 26 Missouri 34
## 27 Montana 8
## 28 Nebraska 14
## 29 Nevada 17
## 30 New Hampshire 10
## 31 New Jersey 41
## 32 New Mexico 16
## 33 New York 48
## 34 North Carolina 42
## 35 North Dakota 4
## 36 Ohio 45
## 37 Oklahoma 24
## 38 Oregon 25
## 39 Pennsylvania 46
## 40 Rhode Island 9
## 41 South Carolina 28
## 42 South Dakota 6
## 43 Tennessee 35
## 44 Texas 50
## 45 Utah 18
## 46 Vermont 3
## 47 Virginia 40
## 48 Washington 39
## 49 West Virginia 15
## 50 Wisconsin 32
## 51 Wyoming 16. Repeat the previous exercise, but this time order
my_df so that the states are ordered from least populous to
most populous. Hint: create an object ind that stores the
indexes needed to order the population values. Then use the bracket
operator [ to re-order each column in the data frame.
ind<-order(murders$population)
my_df2<-data.frame(states[ind], ranks[ind])
my_df2## states.ind. ranks.ind.
## 1 Wyoming 1
## 2 District of Columbia 2
## 3 Vermont 3
## 4 North Dakota 4
## 5 Alaska 5
## 6 South Dakota 6
## 7 Delaware 7
## 8 Montana 8
## 9 Rhode Island 9
## 10 New Hampshire 10
## 11 Maine 11
## 12 Hawaii 12
## 13 Idaho 13
## 14 Nebraska 14
## 15 West Virginia 15
## 16 New Mexico 16
## 17 Nevada 17
## 18 Utah 18
## 19 Kansas 19
## 20 Arkansas 20
## 21 Mississippi 21
## 22 Iowa 22
## 23 Connecticut 23
## 24 Oklahoma 24
## 25 Oregon 25
## 26 Kentucky 26
## 27 Louisiana 27
## 28 South Carolina 28
## 29 Alabama 29
## 30 Colorado 30
## 31 Minnesota 31
## 32 Wisconsin 32
## 33 Maryland 33
## 34 Missouri 34
## 35 Tennessee 35
## 36 Arizona 36
## 37 Indiana 37
## 38 Massachusetts 38
## 39 Washington 39
## 40 Virginia 40
## 41 New Jersey 41
## 42 North Carolina 42
## 43 Michigan 43
## 44 Georgia 44
## 45 Ohio 45
## 46 Pennsylvania 46
## 47 Illinois 47
## 48 New York 48
## 49 Florida 49
## 50 Texas 50
## 51 California 517. The na_example vector represents a series of counts.
You can quickly examine the object using:
data("na_example")
str(na_example)However, when we compute the average with the function
mean, we obtain an NA:
mean(na_example) The is.na function returns a logical vector that tells
us which entries are NA. Assign this logical vector to an
object called ind and determine how many NAs
does na_example have.
ind<-is.na(na_example)
sum(ind)## [1] 1458. Now compute the average again, but only for the entries that are
not NA. Hint: remember the ! operator.
mean(na_example[!is.na(na_example)])## [1] 2.301754Section 2.12
1. Previously we created this data frame:
temp <- c(35, 88, 42, 84, 81, 30)
city <- c("Beijing", "Lagos", "Paris", "Rio de Janeiro", "San Juan", "Toronto")
city_temps <- data.frame(name = city, temperature = temp)Remake the data frame using the code above, but add a line that converts the temperature from Fahrenheit to Celsius. The conversion is \(C=5/9*(F-32)\)
temp <- c(35, 88, 42, 84, 81, 30)
tempc<-(temp-32)*5/9
city <- c("Beijing", "Lagos", "Paris", "Rio de Janeiro", "San Juan", "Toronto")
city_temps <- data.frame(name=city, temperature=tempc)
city_temps## name temperature
## 1 Beijing 1.666667
## 2 Lagos 31.111111
## 3 Paris 5.555556
## 4 Rio de Janeiro 28.888889
## 5 San Juan 27.222222
## 6 Toronto -1.1111112. What is the following sum \(1/1^2 + 1/2^2 + 1/3^2 +...+ 1/100^2\)? Hint: thanks to Euler, we know it should be close to \[(pi^2)/6\]
j<-seq(1,100)
y<-seq(1,1000)
ysum<-sum(1/(y*y))
eul<-pi*pi/6
ydiff<-ysum-eul
jsum<-sum(1/(j*j))
jdiff<-jsum-eul
jsum## [1] 1.6349843. Compute the per 100,000 murder rate for each state and store it in
the object murder_rate. Then compute the average murder
rate for the US using the function mean. What is the
average?
murder_rate<-(murders$total)/((murders$population)*1/100000)
mean(murder_rate)## [1] 2.779125Section 2.14
Start by loading the library and data.
library(dslabs)
data(murders)1. Compute the per 100,000 murder rate for each state and store it in
an object called murder_rate. Then use logical operators to
create a logical vector named low that tells us which
entries of murder_rate are lower than 1.
murder_rate<-(murders$total)/((murders$population)*1/100000)
low<-murder_rate<1
sum(low)## [1] 122. Now use the results from the previous exercise and the function
which to determine the indices of murder_rate
associated with values lower than 1.
which(murder_rate<1)## [1] 12 13 16 20 24 30 35 38 42 45 46 513. Use the results from the previous exercise to report the names of the states with murder rates lower than 1.
lowname<-murders$state[which(murder_rate<1)]
lowname## [1] "Hawaii" "Idaho" "Iowa" "Maine"
## [5] "Minnesota" "New Hampshire" "North Dakota" "Oregon"
## [9] "South Dakota" "Utah" "Vermont" "Wyoming"4. Now extend the code from exercises 2 and 3 to report the states in
the Northeast with murder rates lower than 1. Hint: use the previously
defined logical vector low and the logical operator
&.
levels(murders$region)## [1] "Northeast" "South" "North Central" "West"lowne<-murders$state[which(murder_rate<1 & murders$region=="Northeast")]
lowne## [1] "Maine" "New Hampshire" "Vermont"5. In a previous exercise we computed the murder rate for each state and the average of these numbers. How many states are below the average?
murder_rate<-(murders$total)/((murders$population)*1/100000)
mean(murder_rate)## [1] 2.779125belowavg<-murders$state[which(murder_rate<mean(murder_rate))]
nobelowavg<-sum(murder_rate<mean(murder_rate))
length(belowavg)## [1] 276. Use the match function to identify the states with abbreviations
AK, MI, and IA. Hint: start by defining an index of the entries of
murders$abb that match the three abbreviations, then use
the [ operator to extract the states.
ind<-match(c("AK", "MI", "IA"), murders$abb)
murders$state[ind]## [1] "Alaska" "Michigan" "Iowa"7. Use the %in% operator to create a logical vector that
answers the question: which of the following are actual abbreviations:
MA, ME, MI, MO, MU?
y<-c("MA", "ME", "MI", "MO", "MU") %in% murders$abb
y## [1] TRUE TRUE TRUE TRUE FALSE8. Extend the code you used in exercise 7 to report the one entry
that is not an actual abbreviation. Hint: use the
! operator, which turns FALSE into
TRUE and vice versa, then which to obtain an
index.
which(!c("MA", "ME", "MI", "MO", "MU")%in%murders$abb)## [1] 5Section 2.16
1. We made a plot of total murders versus population and noted a strong relationship. Not surprisingly, states with larger populations had more murders.
library(dslabs)
data(murders)
population_in_millions <- murders$population/10^6
total_gun_murders <- murders$total
plot(population_in_millions, total_gun_murders)Keep in mind that many states have populations below 5 million and
are bunched up. We may gain further insights from making this plot in
the log scale. Transform the variables using the log10
transformation and then plot them.
library("dslabs")
data(murders)
population_in_millions <- murders$population/10^6
total_gun_murders <- murders$total
plot(population_in_millions, total_gun_murders)
plot(log(population_in_millions, log(total_gun_murders)))
2. Create a histogram of the state populations.
hist(murders$population)
3. Generate boxplots of the state populations by region.
boxplot(murders$population~region, data=murders)