hw02

hw02 February 1, 2024 1 Homework 2: Arrays and Tables [1]: # Don't change this cell; just run it. # When you log-in please hit return (not shift + return) after typing in your ␣ , → email import numpy as np from datascience import * Recommended Reading : * Data Types * Sequences * Tables Please complete this notebook by filling in the cells provided. Throughout this homework and all future ones, please be sure to not re-assign variables throughout the notebook! For example, if you use max_temperature in your answer to one question, do not reassign it later on. Before continuing the assignment, select “Save and Checkpoint” in the File menu. 1.1 1. Creating Arrays Question 1. Make an array called weird_numbers containing the following numbers (in the given order): 1. -2 2. the sine of 1.2 3. 3 4. 5 to the power of the cosine of 1.2 Hint: sin and cos are functions in the math module. Note: Python lists are different/behave differently than numpy arrays. In Data 8, we use numpy arrays, so please make an array , not a python list. [2]: # Our solution involved one extra line of code before creating # weird_numbers. import math weird_numbers = make_array( -2 , math . sin( 1.2 ), 3 , 5** math . cos( 1.2 )) weird_numbers [2]: array([-2. , 0.93203909, 3. , 1.79174913]) Question 2. Make an array called book_title_words containing the following three strings: “Eats”, “Shoots”, and “and Leaves”. 1

[3]: book_title_words = make_array( "Eats" , "Shoots" , "and Leaves" ) book_title_words [3]: array(['Eats', 'Shoots', 'and Leaves'], dtype='<U10') Strings have a method called join . join takes one argument, an array of strings. It returns a single string. Specifically, the value of a_string.join(an_array) is a single string that’s the concatenation (“putting together”) of all the strings in an_array , except a_string is inserted in between each string. Question 3. Use the array book_title_words and the method join to make two strings: 1. “Eats, Shoots, and Leaves” (call this one with_commas ) 2. “Eats Shoots and Leaves” (call this one without_commas ) Hint: If you’re not sure what join does, first try just calling, for example, "foo".join(book_title_words) . [4]: with_commas = ", " . join(book_title_words) without_commas = " " . join(book_title_words) # These lines are provided just to print out your answers. print ( 'with_commas:' , with_commas) print ( 'without_commas:' , without_commas) with_commas: Eats, Shoots, and Leaves without_commas: Eats Shoots and Leaves 1.2 2. Indexing Arrays These exercises give you practice accessing individual elements of arrays. In Python (and in many programming languages), elements are accessed by index , so the first element is the element at index 0. Note: Please don’t use bracket notation when indexing (i.e. arr[0] ), as this can yield different data type outputs than what we will be expecting. Question 1. The cell below creates an array of some numbers. Set third_element to the third element of some_numbers . [5]: some_numbers = make_array( -1 , -3 , -6 , -10 , -15 ) third_element = some_numbers . item( 2 ) third_element [5]: -6 Question 2. The next cell creates a table that displays some information about the elements of some_numbers and their order. Run the cell to see the partially-completed table, then fill in the missing information (the cells that say “Ellipsis”) by assigning blank_a , blank_b , blank_c , and blank_d to the correct elements in the table. 2

[9]: president_birth_years = Table . read_table( "president_births.csv" ) . column( 'Birth ␣ , → Year' ) most_recent_birth_year = president_birth_years . item( -1 ) most_recent_birth_year [9]: 1917 Question 5. Finally, assign sum_of_birth_years to the sum of the first, tenth, and last birth year in president_birth_years . [10]: sum_of_birth_years = (president_birth_years . item( 1 ) + president_birth_years . , → item( 10 ) + president_birth_years . item( 37 )) [11]: sum_of_birth_years [11]: 5442 1.3 3. Basic Array Arithmetic Question 1. Multiply the numbers 42, 4224, 42422424, and -250 by 157. Assign each variable below such that first_product is assigned to the result of 42 ∗ 157 , second_product is assigned to the result of 4224 ∗ 157 , and so on. For this question, don’t use arrays. [12]: first_product = 42*157 second_product = 4224*157 third_product = 42422424*157 fourth_product = -250*157 print (first_product, second_product, third_product, fourth_product) 6594 663168 6660320568 -39250 Question 2. Now, do the same calculation, but using an array called numbers and only a single multiplication ( * ) operator. Store the 4 results in an array named products . [13]: numbers = make_array( 42 , 4224 , 42422424 , -250 ) products = numbers * 157 products [13]: array([ 6594, 663168, 6660320568, -39250]) Question 3. Oops, we made a typo! Instead of 157, we wanted to multiply each number by 1577. Compute the correct products in the cell below using array arithmetic. Notice that your job is really easy if you previously defined an array containing the 4 numbers. [14]: correct_products = numbers * 1577 correct_products 4

[14]: array([ 66234, 6661248, 66900162648, -394250]) Question 4. We’ve loaded an array of temperatures in the next cell. Each number is the highest temperature observed on a day at a climate observation station, mostly from the US. Since they’re from the US government agency NOAA , all the temperatures are in Fahrenheit. Convert them all to Celsius by first subtracting 32 from them, then multiplying the results by 5 9 . Make sure to ROUND the final result after converting to Celsius to the nearest integer using the np.round function. [20]: max_temperatures = Table . read_table( "temperatures.csv" ) . column( "Daily Max ␣ , → Temperature" ) celsius_max_temperatures = np . round((max_temperatures -32 ) *5/9 ) celsius_max_temperatures [20]: array([-4., 31., 32., …, 17., 23., 16.]) Question 5. The cell below loads all the lowest temperatures from each day (in Fahrenheit). Compute the size of the daily temperature range for each day. That is, compute the difference between each daily maximum temperature and the corresponding daily minimum temperature. Pay attention to the units, give your answer in Celsius! Make sure NOT to round your answer for this question! [21]: min_temperatures = Table . read_table( "temperatures.csv" ) . column( "Daily Min ␣ , → Temperature" ) celsius_temperature_ranges = (max_temperatures - min_temperatures) *5/9 celsius_temperature_ranges [21]: array([ 6.66666667, 10. , 12.22222222, …, 17.22222222, 11.66666667, 11.11111111]) 1.4 4. World Population The cell below loads a table of estimates of the world population for different years, starting in 1950. The estimates come from the US Census Bureau website . [22]: world = Table . read_table( "world_population.csv" ) . select( 'Year' , 'Population' ) world . show( 4 ) <IPython.core.display.HTML object> The name population is assigned to an array of population estimates. [23]: population = world . column( 1 ) population 5

3993634880, 4072285105, 4151421126, 4230585740, 4308703704, 4386426929, 4464720629, 4543399241, 4621094239, 4698861357]) 1) The total population change between consecutive years, starting at 1951. 2) The total population change between 1950 and each later year, starting at 1951. 3) The total population change between 1950 and each later year, starting inclusively at 1950. [26]: # Assign cumulative_sum_answer to 1, 2, or 3 cumulative_sum_answer = 2 1.5 5. Old Faithful Old Faithful is a geyser in Yellowstone that erupts every 44 to 125 minutes (according to Wikipedia ). People are often told that the geyser erupts every hour , but in fact the waiting time between eruptions is more variable. Let’s take a look. Question 1. The first line below assigns waiting_times to an array of 272 consecutive waiting times between eruptions, taken from a classic 1938 dataset. Assign the names shortest , longest , and average so that the print statement is correct. [29]: waiting_times = Table . read_table( 'old_faithful.csv' ) . column( 'waiting' ) shortest = min (waiting_times) longest = max (waiting_times) average = np . round(np . mean(waiting_times), 1 ) print ( "Old Faithful erupts every" , shortest, "to" , longest, "minutes and ␣ , → every" , average, "minutes on average." ) Old Faithful erupts every 43 to 96 minutes and every 70.9 minutes on average. Question 2. Assign biggest_decrease to the biggest decrease in waiting time between two consecutive eruptions. For example, the third eruption occurred after 74 minutes and the fourth after 62 minutes, so the decrease in waiting time was 74 - 62 = 12 minutes. Hint 1 : You’ll need an array arithmetic function mentioned in the textbook . You have also seen this function earlier in the homework! Hint 2 : We want to return the absolute value of the biggest decrease. [30]: biggest_decrease = abs ( min (np . diff(waiting_times))) biggest_decrease [30]: 45 Question 3. If you expected Old Faithful to erupt every hour, you would expect to wait a total of 60 * k minutes to see k eruptions. Set difference_from_expected to an array with 272 elements, where the element at index i is the absolute difference between the expected and actual total amount of waiting time to see the first i+1 eruptions. 7

Hint : You’ll need to compare a cumulative sum to a range. You’ll go through np.arange more thoroughly in Lab 3, but you can read about it in this textbook section . For example, since the first three waiting times are 79, 54, and 74, the total waiting time for 3 eruptions is 79 + 54 + 74 = 207. The expected waiting time for 3 eruptions is 60 * 3 = 180. Therefore, difference_from_expected.item(2) should be | 207 − 180 | = 27 . [32]: difference_from_expected = np . cumsum(waiting_times) - np . arange( 60 , 273*60 , 60 ) difference_from_expected [32]: array([ 19, 13, 27, 29, 54, 49, 77, 102, 93, 118, 112, 136, 154, 141, 164, 156, 158, 182, 174, 193, 184, 171, 189, 198, 212, 235, 230, 246, 264, 283, 296, 313, 319, 339, 353, 345, 333, 353, 352, 382, 402, 400, 424, 422, 435, 458, 462, 455, 477, 476, 491, 521, 515, 535, 529, 552, 563, 567, 584, 605, 604, 628, 616, 638, 638, 670, 688, 706, 711, 724, 746, 742, 761, 772, 774, 790, 790, 808, 824, 847, 862, 884, 894, 899, 912, 940, 956, 976, 964, 990, 990, 1020, 1010, 1028, 1031, 1043, 1067, 1082, 1073, 1095, 1097, 1125, 1114, 1137, 1158, 1145, 1169, 1161, 1187, 1208, 1223, 1222, 1251, 1270, 1269, 1290, 1280, 1305, 1304, 1331, 1324, 1333, 1350, 1346, 1374, 1395, 1380, 1402, 1397, 1427, 1412, 1435, 1431, 1460, 1446, 1468, 1459, 1485, 1478, 1497, 1518, 1518, 1540, 1557, 1573, 1572, 1592, 1581, 1617, 1610, 1627, 1644, 1649, 1670, 1681, 1691, 1712, 1745, 1738, 1767, 1752, 1778, 1776, 1794, 1800, 1816, 1819, 1847, 1839, 1872, 1861, 1858, 1875, 1883, 1904, 1925, 1938, 1928, 1953, 1967, 1962, 1979, 2002, 2025, 2016, 2034, 2058, 2044, 2067, 2062, 2083, 2080, 2096, 2120, 2137, 2158, 2185, 2202, 2193, 2211, 2211, 2233, 2264, 2257, 2275, 2261, 2278, 2302, 2291, 2314, 2325, 2345, 2334, 2349, 2353, 2369, 2362, 2396, 2391, 2407, 2397, 2419, 2413, 2428, 2446, 2465, 2483, 2501, 2511, 2530, 2540, 2534, 2560, 2550, 2580, 2574, 2568, 2585, 2604, 2608, 2623, 2610, 2636, 2639, 2664, 2686, 2683, 2705, 2712, 2726, 2720, 2743, 2756, 2769, 2797, 2817, 2828, 2851, 2847, 2866, 2884, 2908, 2906, 2929, 2912, 2912, 2927, 2948, 2934, 2964, 2950, 2964]) Question 4. Let’s imagine your guess for the next wait time was always just the length of the previous waiting time. If you always guessed the previous waiting time, how big would your error in guessing the waiting times be, on average? For example, since the first three waiting times are 79, 54, and 74, the average difference between your guess and the actual time for just the second and third eruption would be | 79 − 54 | + | 54 − 74 | 2 = 22 . 5 . [33]: average_error = np . mean( abs (np . diff(waiting_times))) average_error [33]: 20.52029520295203 8

Question 4. The file sales.csv contains the number of fruit sold from each box last Saturday. It has an extra column called “price per fruit ($)” that’s the price per item of fruit for fruit in that box. The rows are in the same order as the inventory table. Load these data into a table called sales . [38]: sales = Table . read_table( 'sales.csv' ) sales [38]: box ID | fruit name | count sold | price per fruit ($) 53686 | kiwi | 3 | 0.5 57181 | strawberry | 101 | 0.2 25274 | apple | 0 | 0.8 48800 | orange | 35 | 0.6 26187 | strawberry | 25 | 0.15 57930 | grape | 355 | 0.06 52357 | strawberry | 102 | 0.25 43566 | peach | 17 | 0.8 Question 5. How many fruits did the store sell in total on that day? [39]: total_fruits_sold = sum (sales . column( "count sold" )) total_fruits_sold [39]: 638 Question 6. What was the store’s total revenue (the total price of all fruits sold) on that day? Hint: If you’re stuck, think first about how you would compute the total revenue from just the grape sales. [45]: total_revenue = sum (sales . column( "count sold" ) * sales . column( "price per fruit ␣ , → ($)" )) total_revenue [45]: 106.85 Question 7. Make a new table called remaining_inventory . It should have the same rows and columns as inventory , except that the amount of fruit sold from each box should be subtracted from that box’s count, so that the “count” is the amount of fruit remaining after Saturday. [46]: remaining_inventory = inventory . with_column( "count" , inventory . column( "count" ) ␣ , → - sales . column( "count sold" )) remaining_inventory [46]: box ID | fruit name | count 53686 | kiwi | 42 57181 | strawberry | 22 25274 | apple | 20 48800 | orange | 0 10

26187 | strawberry | 230 57930 | grape | 162 52357 | strawberry | 0 43566 | peach | 23 11