Data Science Intro

Question 1

cylinders = set(d[‘cyl’] for d in mpg)

Answer 1

Use set to return the unique values for the number of cylinders the cars in our dataset have.

Question 2

sum(float(d[‘hwy’]) for d in mpg) / len(mpg)

Answer 2

This is how to find the average hwy fuel economy across all cars.

Question 3

len(mpg) - mpg is the title of a list that includes dictionary keys.

Answer 3

csv.Dictreader has read in each row of our csv file as a dictionary. len shows that our list is comprised of 234 dictionaries.

Question 4

import csv

%precision 2

with open('mpg.csv') as csvfile:
    mpg = list(csv.DictReader(csvfile))

mpg[:3] # The first three dictionaries in our list.

Answer 4

Reads csv file and make a list named mpg

Question 5

What will the output be?

sales_record = {
‘price’: 3.24,
‘num_items’: 4,
‘person’: ‘Chris’}

sales_statement = ‘{} bought {} item(s) at a price of {} each for a total of {}’

print(sales_statement.format(sales_record[‘person’],
sales_record[‘num_items’],
sales_record[‘price’],
sales_record[‘num_items’]*sales_record[‘price’]))

Answer 5

Chris bought 4 item(s) at a price of 3.24 each for a total of 12.96

Question 6

x = (‘Christopher’, ‘Brooks’, ‘brooksch@umich.edu’)
fname, lname, email = x

print(fname)

Answer 6

Christopher

Question 7

Tuple format?

Answer 7

list = (“Hi”, “Dave”, 4)

Question 8

List format?

Answer 8

list = [“hi”, 4, 2, “Dave”]

Question 9

x = {‘Christopher Brooks’: ‘brooksch@umich.edu’, ‘Bill Gates’: ‘billg@microsoft.com’}

x[‘Christopher Brooks’]

Answer 9

Retrieve a value by using the indexing operator

‘brooksch@umich.edu’

Question 10

CtyMpgByCyl = []

for c in cylinders: # iterate over all the cylinder levels
summpg = 0
cyltypecount = 0
for d in mpg: # iterate over all dictionaries
if d[‘cyl’] == c: # if the cylinder level type matches,
summpg += float(d[‘cty’]) # add the cty mpg
cyltypecount += 1 # increment the count
CtyMpgByCyl.append((c, summpg / cyltypecount)) # append the tuple (‘cylinder’, ‘avg mpg’)

CtyMpgByCyl.sort(key=lambda x: x[0])
CtyMpgByCyl

Answer 10

Prints the average mpg for each cylinder size

Lambda sorts CityMpgByCyl by first key index.

[(‘4’, 21.01), (‘5’, 20.50), (‘6’, 16.22), (‘8’, 12.57)]

Question 11

vehicleclass = set(d[‘class’] for d in mpg)

vehicleclass

Answer 11

What are the class types? Only show me one of each

{‘2seater’, ‘compact’, ‘midsize’, ‘minivan’, ‘pickup’, ‘subcompact’, ‘suv’}

Question 12

HwyMpgByClass = []

for t in vehicleclass: # iterate over all the vehicle classes
summpg = 0
vclasscount = 0
for d in mpg: # iterate over all dictionaries
if d[‘class’] == t: # if the cylinder amount type matches,
summpg += float(d[‘hwy’]) # add the hwy mpg
vclasscount += 1 # increment the count
HwyMpgByClass.append((t, summpg / vclasscount)) # append the tuple (‘class’, ‘avg mpg’)

HwyMpgByClass.sort(key=lambda x: x[1])
HwyMpgByClass

Answer 12

example of how to find the average hwy mpg for each class of vehicle in our dataset.

[('pickup', 16.88),
 ('suv', 18.13),
 ('minivan', 22.36),
 ('2seater', 24.80),
 ('midsize', 27.29),
 ('subcompact', 28.14),
 ('compact', 28.30)]

Question 13

import datetime as dt
import time as tm

tm.time()

Answer 13

time returns the current time in seconds

Question 14

import datetime as dt
import time as tm

dtnow = dt.datetime.fromtimestamp(tm.time())
dtnow

Answer 14

Convert the timestamp to datetime.

Question 15

dtnow.year, dtnow.month, dtnow.day, dtnow.hour, dtnow.minute, dtnow.second

Answer 15

get year, month, day, etc. from a datetime

Question 16

delta = dt.timedelta(days = 100)
# create a timedelta of 100 days

delta

Answer 16

timedelta is a duration expressing the difference between two dates.

Question 17

delta = dt.timedelta(days = 100)
today = dt.date.today()

today - delta

Answer 17

Returns date 100 days ago.

datetime.date(2016, 8, 13)

Question 18

today > today-delta

Answer 18

compare dates

returns True

Question 19

store1 = [10.00, 11.00, 12.34, 2.34]
store2 = [9.00, 11.10, 12.34, 2.01]
cheapest = map(min, store1, store2)
cheapest

Answer 19

stores the lowest values as a list in cheapest

Question 20

my_function = lambda a, b, c : a + b

my_function(1, 2, 3)

Answer 20

Here’s an example of lambda that takes in three parameters and adds the first two.

Question 21

my_list = []
for number in range(0, 1000):
    if number % 2 == 0:
        my_list.append(number)
my_list

Answer 21

appends even numbers in range

Question 22

my_list = [number for number in range(0,1000) if number % 2 == 0]
my_list

Answer 22

shorthand version of :

my_list = []
for number in range(0, 1000):
    if number % 2 == 0:
        my_list.append(number)
my_list

Question 23

m = np.array([[7, 8, 9], [10, 11, 12]]) # create array w/ numpy

m.shape

Answer 23

Use the shape method to find the dimensions of the array. (rows, columns)

(2, 3)

Question 24

n = np.arange(0, 30, 2)

n

Answer 24

arange returns evenly spaced values within a given interval.

start at 0 count up by 2, stop before 30

array([ 0, 2, 4, 6, 8, 10, 12, 14, 16, 18, 20, 22, 24, 26, 28])

Question 25

n = [ 0, 2, 4, 6, 8, 10, 12, 14, 16, 18, 20, 22, 24, 26, 28]

n = n.reshape(3, 5)

Answer 25

reshape returns an array with the same data with a new shape.

reshape array to be 3x5

array([[ 0, 2, 4, 6, 8],
[10, 12, 14, 16, 18],
[20, 22, 24, 26, 28]])

Question 26

o = np.linspace(0, 4, 9)

o

Answer 26

linspace returns evenly spaced numbers over a specified interval.

return 9 evenly spaced values from 0 to 4

array([ 0. , 0.5, 1. , 1.5, 2. , 2.5, 3. , 3.5, 4. ])

Question 27

o = [ 0. , 0.5, 1. , 1.5, 2. , 2.5, 3. , 3.5, 4. ]

o.resize(3, 3)
o

Answer 27

resize changes the shape and size of array in-place.

array([[ 0. , 0.5, 1. ],
[ 1.5, 2. , 2.5],
[ 3. , 3.5, 4. ]])

Question 28

np.ones((3, 2))

Answer 28

ones returns a new array of given shape and type, filled with ones.

array([[ 1., 1.],
[ 1., 1.],
[ 1., 1.]])

Question 29

np.zeros((2, 3))

Answer 29

zeros returns a new array of given shape and type, filled with zeros.

array([[ 0., 0., 0.],
[ 0., 0., 0.]])

Question 30

np.eye(3)

Answer 30

eye returns a 2-D array with ones on the diagonal and zeros elsewhere.

array([[ 1., 0., 0.],
[ 0., 1., 0.],
[ 0., 0., 1.]])

Question 31

np.diag(y)

Answer 31

diag extracts a diagonal or constructs a diagonal array.

array([[4, 0, 0],
[0, 5, 0],
[0, 0, 6]])

Question 32

np.array([1, 2, 3] * 3)

Answer 32

Create an array using repeating list (or see np.tile)

array([1, 2, 3, 1, 2, 3, 1, 2, 3])

Question 33

np.repeat([1, 2, 3], 3)

Answer 33

Repeat elements of an array using repeat.

array([1, 1, 1, 2, 2, 2, 3, 3, 3])

Question 34

p = ([[1, 1, 1],
[1, 1, 1]])

np.vstack([p, 2*p])

Answer 34

Use vstack to stack arrays in sequence vertically (row wise).

array([[1, 1, 1],
[1, 1, 1],
[2, 2, 2],
[2, 2, 2]])

Question 35

p = ([[1, 1, 1],
[1, 1, 1]])

np.hstack([p, 2*p])

Answer 35

Use hstack to stack arrays in sequence horizontally (column wise).

array([[1, 1, 1, 2, 2, 2],
[1, 1, 1, 2, 2, 2]])

Question 36

x = [1 2 3]
y = [4 5 6]

print(x + y)
print(x - y)

Answer 36

[5 7 9]

[-3 -3 -3]

Question 37

x = [1 2 3]
y = [4 5 6]

print(x * y)
print(x / y)

Answer 37

[ 4 10 18]

[ 0.25 0.4 0.5 ]

Question 38

x = [1 2 3]

print(x**2)

Answer 38

raises all elements to power of 2

[1 4 9]

Question 39

x.dot(y) # dot product 14 + 25 + 3*6

Answer 39

dot product 14 + 25 + 3*6

x[0]y[0] + x[1]y[1] + x[2]y[2] = 32

Question 40

z = np.array([y, y**2])

print(len(z))

Answer 40

prints number of rows in 2d array

Question 41

z = ([[ 4, 5, 6],
[16, 25, 36]])

z. shape is (2,3)
z. T

Answer 41

Transposing changes shape of array

array([[ 4, 16],
[ 5, 25],
[ 6, 36]])

Question 42

z.dtype

Answer 42

Use .dtype to see the data type of the elements in the array.

dtype(‘int64’)

Question 43

z starts off as int

z = z.astype(‘f’)
z.dtype

Answer 43

Use .astype to cast to a specific type.

dtype(‘float32’)

Question 44

a. argmax()

a. argmin()

Answer 44

argmax and argmin return the index of the maximum and minimum values in the array.

Question 45

s = np.arange(13)**2

s

Answer 45

fills array w/ squares of first 13 index spots

array([ 0, 1, 4, 9, 16, 25, 36, 49, 64, 81, 100, 121, 144])

Question 46

s = ([ 0, 1, 4, 9, 16, 25, 36, 49, 64, 81, 100, 121, 144])

s[0], s[4], s[-1]

Answer 46

Use bracket notation to get the value at a specific index. Remember that indexing starts at 0.

(0, 16, 144)

Question 47

s = ([ 0, 1, 4, 9, 16, 25, 36, 49, 64, 81, 100, 121, 144])

s[1:5]

Answer 47

Use : to indicate a range. array[start:stop]
Leaving start or stop empty will default to the beginning/end of the array.

array([ 1, 4, 9, 16])

Question 48

s = ([ 0, 1, 4, 9, 16, 25, 36, 49, 64, 81, 100, 121, 144])

s[1:5]

Answer 48

Use negatives to count from the back.

array([ 81, 100, 121, 144])

Question 49

s = ([ 0, 1, 4, 9, 16, 25, 36, 49, 64, 81, 100, 121, 144])

s[-5::-2]

Answer 49

A second : can be used to indicate step-size. array[start:stop:stepsize]

Here we are starting 5th element from the end, and counting backwards by 2 until the beginning of the array is reached.

array([64, 36, 16, 4, 0])

Question 50

r = ([[ 0,  1,  2,  3,  4,  5],
       [ 6,  7,  8,  9, 10, 11],
       [12, 13, 14, 15, 16, 17],
       [18, 19, 20, 21, 22, 23],
       [24, 25, 26, 27, 28, 29],
       [30, 31, 32, 33, 34, 35]])

r[3, 3:6]

Answer 50

use : to select a range of rows or columns

array([21, 22, 23])

Question 51

r = ([[ 0,  1,  2,  3,  4,  5],
       [ 6,  7,  8,  9, 10, 11],
       [12, 13, 14, 15, 16, 17],
       [18, 19, 20, 21, 22, 23],
       [24, 25, 26, 27, 28, 29],
       [30, 31, 32, 33, 34, 35]])

r[:2, :-1]

Answer 51

Here we are selecting all the rows up to (and not including) row 2, and all the columns up to (and not including) the last column.

array([[ 0, 1, 2, 3, 4],
[ 6, 7, 8, 9, 10]])

Question 52

r = ([[ 0,  1,  2,  3,  4,  5],
       [ 6,  7,  8,  9, 10, 11],
       [12, 13, 14, 15, 16, 17],
       [18, 19, 20, 21, 22, 23],
       [24, 25, 26, 27, 28, 29],
       [30, 31, 32, 33, 34, 35]])

r[-1, ::2]

Answer 52

This is a slice of the last row, and only every other element.

array([30, 32, 34])

Question 53

r = ([[ 0,  1,  2,  3,  4,  5],
       [ 6,  7,  8,  9, 10, 11],
       [12, 13, 14, 15, 16, 17],
       [18, 19, 20, 21, 22, 23],
       [24, 25, 26, 27, 28, 29],
       [30, 31, 32, 33, 34, 35]])

r[r > 30]

Answer 53

We can also perform conditional indexing. Here we are selecting values from the array that are greater than 30. (Also see np.where)

array([31, 32, 33, 34, 35])

Question 54

r = ([[ 0,  1,  2,  3,  4,  5],
       [ 6,  7,  8,  9, 10, 11],
       [12, 13, 14, 15, 16, 17],
       [18, 19, 20, 21, 22, 23],
       [24, 25, 26, 27, 28, 29],
       [30, 31, 32, 33, 34, 35]])

r[r > 30] = 30
r

Answer 54

Here we are assigning all values in the array that are greater than 30 to the value of 30.

array([[ 0,  1,  2,  3,  4,  5],
       [ 6,  7,  8,  9, 10, 11],
       [12, 13, 14, 15, 16, 17],
       [18, 19, 20, 21, 22, 23],
       [24, 25, 26, 27, 28, 29],
       [30, 30, 30, 30, 30, 30]])

Question 55

r2[:] = 0

r2

Answer 55

Set this slice’s values to zero ([:] selects the entire array)

if r2 is the result of a slice of r’s values, r will also change in the sliced positions. Need to copy.

Question 56

test = np.random.randint(0, 10, (4,3))

test

Answer 56

Create a new 4 by 3 array of random numbers 0-9.

array([[0, 8, 0],
[0, 5, 7],
[3, 7, 4],
[3, 4, 9]])

Question 57

test = ([[0, 8, 0],
[0, 5, 7],
[3, 7, 4],
[3, 4, 9]])

for row in test:
print(row)

Answer 57

Iterate by row:

Each row is an index because this is multidimensional array.

[0 8 0]
[0 5 7]
[3 7 4]
[3 4 9]

Question 58

test = ([[0, 8, 0],
[0, 5, 7],
[3, 7, 4],
[3, 4, 9]])

for i in range(len(test)):
print(test[i])

Answer 58

Iterate by index:

[6 9 4]
[8 1 9]
[4 8 1]
[7 2 2]

Question 59

test = ([[0, 8, 0],
[0, 5, 7],
[3, 7, 4],
[3, 4, 9]])

for i, row in enumerate(test):
print(‘row’, i, ‘is’, row)

Answer 59

Iterate by row and index:

row 0 is [6 9 4]
row 1 is [8 1 9]
row 2 is [4 8 1]
row 3 is [7 2 2]

Question 60

test = ([[0, 8, 0],
[0, 5, 7],
[3, 7, 4],
[3, 4, 9]])

test2 = test**2
test2 prints as:
array([[36, 81, 16],
       [64,  1, 81],
       [16, 64,  1],
       [49,  4,  4]])

for i, j in zip(test, test2):
print(i, ‘+’, j, ‘=’, i+j)

Answer 60

Use zip to iterate over multiple iterables.

for i, j in zip(test, test2):
print(i, ‘+’, j, ‘=’, i+j)

[6 9 4] + [36 81 16] = [42 90 20]
[8 1 9] + [64 1 81] = [72 2 90]
[4 8 1] + [16 64 1] = [20 72 2]
[7 2 2] + [49 4 4] = [56 6 6]