In [1]:
from datascience import *
import numpy as np

%matplotlib inline
import matplotlib.pyplot as plots
plots.style.use('fivethirtyeight')
import warnings
warnings.simplefilter(action='ignore',category=np.VisibleDeprecationWarning)

Lecture 11¶

In this lecture, we will:

  1. Introduce python Lists
  2. Show how to build tables from rows instead of columns
  3. finish our discussion on grouping
  4. Introduce pivots -- a special kind of grouping
  5. Introduce joins

Lists¶

Lists in python are like arrays but they can contain things of different types and you make them by using square brackets:

In [2]:
[1, 5, 'hello', 5.0]
Out[2]:
[1, 5, 'hello', 5.0]
In [3]:
another_list = [1, 5, 'hello', 5.0, make_array(1,2,3)]
another_list
Out[3]:
[1, 5, 'hello', 5.0, array([1, 2, 3])]

You have to use square brakets when accessing elements from a list (you can't use item).

In [4]:
another_list[2]
Out[4]:
'hello'

Recall that for arrays you could do either. (In the real world, most people would probably use square brackets).

In [5]:
my_array = make_array("Apple", "Banana", "Corn")
my_array[1]
Out[5]:
'Banana'

Same thing as:

In [6]:
my_array.item(1)
Out[6]:
'Banana'





Rows from lists¶

We need lists to be able to add a row to our table. Why?

Here we make a table with a column (we have been doing this all semester).

In [7]:
Table().with_columns('Numbers', [1, 2, 3])
Out[7]:
Numbers
1
2
3




Here we make an empty table with just column labels. (This is new!)

In [8]:
drinks = Table(['Drink', 'Cafe', 'Price'])
drinks
Out[8]:
Drink Cafe Price




Here we add rows to our table. Each row is a List of elements -- one for each column.

In [9]:
drinks = drinks.with_rows([
    ['Milk Tea', 'Asha', 5.5],
    ['Espresso', 'Strada',  1.75],
    ['Latte',    'Strada',  3.25],
    ['Espresso', "FSM",   2]
])
drinks
Out[9]:
Drink Cafe Price
Milk Tea Asha 5.5
Espresso Strada 1.75
Latte Strada 3.25
Espresso FSM 2






Return to Slides






Review Grouping Using the Welcome Survey¶

Now let's use the group function to study you:

In [10]:
survey = Table.read_table("welcome_survey_sp23.csv")
survey.show(3)
Year Extraversion Number of textees Hours of sleep Handedness Pant leg Sleep position Pets Piercings
Second Year 2 5 9 Right-handed Right leg in first On your right side Cat, Dog, Fish, Snake, Lizard -3
First Year 2 3 8 Right-handed I don't know On your back None -1
First Year 5 5 8 Right-handed Right leg in first On your left side Bearded dragon 0

... (1498 rows omitted)





Exercise: What happens if we group by "Year" with no additional arguments?

In [11]:
survey.group("Year")
Out[11]:
Year count
Fifth Year or above 9
First Year 606
Fourth Year 109
Graduate Student 7
Second Year 468
Third Year 302






What is the relationship between year and the average amount of sleep?

Exercise: Computer the average "Hours of sleep" for each "Year" and display the result as bar chart with the bars in a meaningful order.

In [12]:
(
    survey
    .select("Year", "Hours of sleep")
    .group("Year", np.average)
    .take(1,4,5,2,0,3)
    .barh("Year")
)
# plots.xlim([6,7.5])


Click for Solution 

(
    survey
    .select("Year", "Hours of sleep")
    .group("Year", np.average)
    .take(1,4,5,2,0,3)
    .barh("Year")
)
plots.xlim([6,7.5])









Understanding Extraversion¶

How extraverted is the class?

In [13]:
survey.hist('Extraversion')







Exercise: Compute the average "Number of textees", "Hours of sleep", "Piercings" for each value of "Extraversion". Save the resulting table as by_extra.

In [14]:
by_extra = (
    survey
    .select("Extraversion", "Number of textees", "Hours of sleep", "Piercings")
    .group("Extraversion", np.mean)
)
by_extra
Out[14]:
Extraversion Number of textees mean Hours of sleep mean Piercings mean
1 3.44828 6.60345 1.72414
2 4.08197 6.98361 1.13115
3 4.75263 7.23158 1.41053
4 5.14953 7.11449 1.49533
5 6.14286 7.07576 1.64502
6 6.74257 6.98267 1.87129
7 10.989 7.09158 3.66301e+06
8 9.7027 7.22297 2.14414
9 11.9608 7.29412 2.41176
10 9.25 6.78571 1.89286


Click for Solution 

by_extra = (
    survey
    .select("Extraversion", "Number of textees", "Hours of sleep", "Piercings")
    .group('Extraversion', np.mean)
)
by_extra



Do you see any trends? Let's plot the data to make it clearer.

In [15]:
by_extra.plot("Extraversion", "Number of textees mean", marker="*")
In [16]:
by_extra.plot("Extraversion", "Hours of sleep mean", marker="*")






Return to Slides












Grouping by Two Columns¶

Do right-handed people tend to sleep on their left side and left-handed people sleep on their right side?



Exercise: Use the group function to count how many people are in each combination of "Handedness" and "Sleep position".

In [17]:
(
    survey
    .group(["Handedness", "Sleep position"])
    .show()
)
Handedness Sleep position count
Ambidextrous On your back 2
Ambidextrous On your left side 4
Ambidextrous On your right side 3
Ambidextrous On your stomach 3
Left-handed On your back 27
Left-handed On your left side 40
Left-handed On your right side 46
Left-handed On your stomach 9
Right-handed On your back 334
Right-handed On your left side 395
Right-handed On your right side 475
Right-handed On your stomach 163


Click for Solution 

(
    survey
    .group(["Handedness", "Sleep position"])
    .show()
)







What combination of "Handedness" and "Sleep position" results in the most sleep on average?

Exercise: Use group to answer the above question.

In [18]:
(
    survey
    .select("Handedness", "Sleep position", "Hours of sleep")
    .group(["Handedness", "Sleep position"], np.average)
    .show()
)
Handedness Sleep position Hours of sleep average
Ambidextrous On your back 7
Ambidextrous On your left side 7.375
Ambidextrous On your right side 7.33333
Ambidextrous On your stomach 9
Left-handed On your back 7
Left-handed On your left side 7.2125
Left-handed On your right side 7.30435
Left-handed On your stomach 6.72222
Right-handed On your back 7.13323
Right-handed On your left side 7.05823
Right-handed On your right side 7.03579
Right-handed On your stomach 7.24847


Click for Solution 

(
    survey
    .select("Handedness", "Sleep position", "Hours of sleep")
    .group(['Handedness', 'Sleep position'], np.average)
    .show()
)



Pivot Tables¶

When grouping by two columns it will sometimes be easier to re-arrange the table so the columns correspond to the values of one of the original grouped columns.

Here is an example:

In [19]:
(
    survey
    .pivot("Handedness", "Sleep position", "Hours of sleep", np.average)
)
Out[19]:
Sleep position Ambidextrous Left-handed Right-handed
On your back 7 7 7.13323
On your left side 7.375 7.2125 7.05823
On your right side 7.33333 7.30435 7.03579
On your stomach 9 6.72222 7.24847

Here I am rewriting the example by explicity naming the function arguments. You wouldn't do this in practice but it helps to highlight the meaning of the four things we passed to the function above.

In [20]:
(
    survey
    .pivot(
        columns = "Handedness", # The column whose unique values become columns
        rows    = "Sleep position", # The column whose unique values become rows
        values  = "Hours of sleep", # The column that will be used to compute the cells
        collect = np.average # How to aggregate the values 
    )
)
Out[20]:
Sleep position Ambidextrous Left-handed Right-handed
On your back 7 7 7.13323
On your left side 7.375 7.2125 7.05823
On your right side 7.33333 7.30435 7.03579
On your stomach 9 6.72222 7.24847

Just as with the regular group function you will often want counts of combinations. In this case, you only need to specify the column to turn into columns and the column to turn into rows.

In [21]:
survey.pivot("Sleep position", "Handedness")
Out[21]:
Handedness On your back On your left side On your right side On your stomach
Ambidextrous 2 4 3 3
Left-handed 27 40 46 9
Right-handed 334 395 475 163

Pivots can be useful for visualizations:

In [22]:
survey.pivot("Sleep position", "Handedness").barh("Handedness")






Return to Slides






Joins¶

We use join to join two tables on a column.

In [23]:
drinks
Out[23]:
Drink Cafe Price
Milk Tea Asha 5.5
Espresso Strada 1.75
Latte Strada 3.25
Espresso FSM 2
In [24]:
discounts = Table().with_columns(
    'Coupon % off', make_array(10, 25, 5),
    'Location', make_array('Asha', 'Strada', 'Asha')
)
discounts
Out[24]:
Coupon % off Location
10 Asha
25 Strada
5 Asha

Exercise: Combine the drinks and discounts table into a new table which has all possible discounts for each drink.

In [25]:
combined = drinks.join('Cafe', discounts, 'Location')
combined
Out[25]:
Cafe Drink Price Coupon % off
Asha Milk Tea 5.5 10
Asha Milk Tea 5.5 5
Strada Espresso 1.75 25
Strada Latte 3.25 25


Click for Solution 

combined = drinks.join('Cafe', discounts, 'Location')
combined






Exercise: Calculate the lowest rate you can get for each drink using the available coupons.

In [26]:
discount_frac = 1-combined.column("Coupon % off")/100.0

(
    combined
    .with_column("Discounted Price", combined.column("Price") * discount_frac)
    .select("Drink", "Discounted Price")
    .group("Drink", min)
)
Out[26]:
Drink Discounted Price min
Espresso 1.3125
Latte 2.4375
Milk Tea 4.95


Click for Solution 

combined = drinks.join('Cafe', discounts, 'Location') # from previous question

discount_frac = 1-combined.column("Coupon % off")/100.0

(
    combined
    .with_column("Discounted Price", combined.column("Price") * discount_frac)
    .select("Drink", "Discounted Price")
    .group("Drink", min)
)






Exercise: What happens if I run the following? How many rows will it produce.

In [27]:
drinks.join('Cafe', drinks, 'Cafe')
Out[27]:
Cafe Drink Price Drink_2 Price_2
Asha Milk Tea 5.5 Milk Tea 5.5
FSM Espresso 2 Espresso 2
Strada Espresso 1.75 Espresso 1.75
Strada Espresso 1.75 Latte 3.25
Strada Latte 3.25 Espresso 1.75
Strada Latte 3.25 Latte 3.25






Return to Slides






Discussion Questions¶

For the following practice questions we will use the sky scraper dataset

In [28]:
# From the CORGIS Dataset Project
# By Austin Cory Bart acbart@vt.edu
# Version 2.0.0, created 3/22/2016
# https://corgis-edu.github.io/corgis/csv/skyscrapers/

this_year = 2023
sky = Table.read_table("skyscrapers.csv")
sky = (
    sky
    .with_column("age", this_year - sky.column("completed"))
    .drop("completed")
)
sky
Out[28]:
name material city height age
One World Trade Center mixed/composite New York City 541.3 9
Willis Tower steel Chicago 442.14 49
432 Park Avenue concrete New York City 425.5 8
Trump International Hotel & Tower concrete Chicago 423.22 14
Empire State Building steel New York City 381 92
Bank of America Tower mixed/composite New York City 365.8 14
Stratosphere Tower concrete Las Vegas 350.22 27
Aon Center steel Chicago 346.26 50
John Hancock Center steel Chicago 343.69 54
Chrysler Building steel New York City 318.9 93

... (1771 rows omitted)




Exercise: For each city, what’s the tallest building for each material?

In [29]:
tall_pivot = sky.pivot("material", "city", "height", max)
tall_pivot
Out[29]:
city concrete mixed/composite steel
Atlanta 264.25 311.8 169.47
Austin 208.15 0 93.6
Baltimore 161.24 0 155.15
Boston 121.92 139 240.79
Charlotte 265.48 239.7 179.23
Chicago 423.22 306.94 442.14
Cincinnati 125 202.69 175
Cleveland 125 288.65 215.8
Columbus 79.25 0 169.3
Dallas 176.48 280.72 270.06

... (25 rows omitted)


Click for Solution 

tall_pivot = sky.pivot("material", "city", "height", max)
tall_pivot



In [30]:
tall_pivot.barh("city")





Exercise: For each city that has at least 50 buildings in the dataset, compute the average height of buildings broken down by material. Make a bar chart with the cities and the average height of buildings for each material.

In [31]:
num_buildings = sky.group("city")
(
    sky
    .join("city", num_buildings)
    .where("count", are.above_or_equal_to(50))
    .pivot("material", "city", "height", np.mean)
    .barh("city")
)


Click for Solution 

num_buildings = sky.group("city")
(
    sky
    .join("city", num_buildings)
    .where("count", are.above_or_equal_to(50))
    .pivot("material", "city", "height", np.mean)
    .barh("city")
)







Exercise: Generate a table of the names and ages of the oldest buildings for each combination of city and materials.

In [32]:
def first(x):
    return x[0]
    
    
(
    sky
    .sort("age", descending=True) 
    .select("city", "material", "name", "age")
    .group(["city", "material"], first)
    #.where("city", "San Francisco") #<- for fun
)
Out[32]:
city material name first age first
Atlanta concrete Westin Peachtree Plaza 47
Atlanta mixed/composite One Atlantic Center 36
Atlanta steel FlatironCity 126
Austin concrete One American Center 39
Austin steel University of Texas Tower 86
Baltimore concrete Charles Towers North Apartments 56
Baltimore steel Emerson Tower 112
Boston concrete Harbor Towers I 52
Boston mixed/composite Ellison Building 31
Boston steel Marriott's Custom House 108

... (81 rows omitted)


Click for Solution 

def first(x):
    return x[0]


(
    sky
    .sort("age", descending=True) 
    .select("city", "material", "name", "age")
    .group(["city", "material"], first)
    #.where("city", "San Francisco") #<- for fun
)



In [ ]: