In [1]:
from dsc80_utils import *

def show_paradox_slides():
    src = 'https://docs.google.com/presentation/d/e/2PACX-1vSbFSaxaYZ0NcgrgqZLvjhkjX-5MQzAITWAsEFZHnix3j1c0qN8Vd1rogTAQP7F7Nf5r-JWExnGey7h/embed?start=false&rm=minimal'
    width = 960
    height = 569
    display(IFrame(src, width, height))
In [2]:
# Pandas Tutor setup
%reload_ext pandas_tutor
%set_pandas_tutor_options {"maxDisplayCols": 8, "nohover": True, "projectorMode": True}

Lecture 4 – Simpson's Paradox, Joining, and Transforming¶

DSC 80, Spring 2024¶

Announcements 📣¶

  • Project 1 checkpoint due tomorrow. No extensions allowed!
  • Discussion tomorrow and Monday are Zoom-only since Praveen is traveling.
  • Lab 2 is due on Wed, April 17th.
  • Project 1 is due on Fri, April 19th.

Agenda¶

  • Distributions.
  • Simpson's paradox.
  • Merging.
    • Many-to-one & many-to-many joins.
  • Transforming.
    • The price of apply.
  • Other data representations.

Distributions¶

Example: Palmer Penguins¶

In [3]:
penguins = sns.load_dataset('penguins').dropna()
penguins
Out[3]:
species island bill_length_mm bill_depth_mm flipper_length_mm body_mass_g sex
0 Adelie Torgersen 39.1 18.7 181.0 3750.0 Male
1 Adelie Torgersen 39.5 17.4 186.0 3800.0 Female
2 Adelie Torgersen 40.3 18.0 195.0 3250.0 Female
... ... ... ... ... ... ... ...
341 Gentoo Biscoe 50.4 15.7 222.0 5750.0 Male
342 Gentoo Biscoe 45.2 14.8 212.0 5200.0 Female
343 Gentoo Biscoe 49.9 16.1 213.0 5400.0 Male

333 rows × 7 columns

Question 🤔 (Answer at q.dsc80.com)

Suppose I draw one penguin from penguins at random. What is the exact (not empirical) probability that:

  • The penguin is a Gentoo penguin?
  • The penguin is a male Gentoo penguin?
In [4]:
# Your code goes here.

Let's compute probabilities using an easier way.

We'll start by using the pivot_table method to recreate the DataFrame shown below.

sex Female Male
species
Adelie 73 73
Chinstrap 34 34
Gentoo 58 61

Joint distribution¶

When using aggfunc='count', a pivot table describes the joint distribution of two categorical variables. This is also called a contingency table.

In [5]:
counts = penguins.pivot_table(
    index='species', 
    columns='sex', 
    values='body_mass_g', 
    aggfunc='count', 
    fill_value=0,
)
counts
Out[5]:
sex Female Male
species
Adelie 73 73
Chinstrap 34 34
Gentoo 58 61

We can normalize the DataFrame by dividing by the total number of penguins. The resulting numbers can be interpreted as probabilities that a randomly selected penguin from the dataset belongs to a given combination of species and sex.

In [6]:
joint = counts / counts.sum().sum()
joint
Out[6]:
sex Female Male
species
Adelie 0.22 0.22
Chinstrap 0.10 0.10
Gentoo 0.17 0.18

Marginal probabilities¶

If we sum over one of the axes, we can compute marginal probabilities, i.e. unconditional probabilities.

In [7]:
joint
Out[7]:
sex Female Male
species
Adelie 0.22 0.22
Chinstrap 0.10 0.10
Gentoo 0.17 0.18
In [8]:
# Recall, joint.sum(axis=0) sums across the rows, 
# which computes the sum of the **columns**.
joint.sum(axis=0)
Out[8]:
sex
Female    0.5
Male      0.5
dtype: float64
In [9]:
joint.sum(axis=1)
Out[9]:
species
Adelie       0.44
Chinstrap    0.20
Gentoo       0.36
dtype: float64

For instance, the second Series tells us that a randomly selected penguin has a 0.36 chance of being of species 'Gentoo'.

Conditional probabilities¶

Using counts, how might we compute conditional probabilities like $$P(\text{species } = \text{"Adelie"} \mid \text{sex } = \text{"Female"})?$$

In [10]:
counts
Out[10]:
sex Female Male
species
Adelie 73 73
Chinstrap 34 34
Gentoo 58 61
$$\begin{align*} P(\text{species} = c \mid \text{sex} = x) &= \frac{\# \: (\text{species} = c \text{ and } \text{sex} = x)}{\# \: (\text{sex} = x)} \end{align*}$$
➡️ Click here to see more of a derivation. $$\begin{align*} P(\text{species} = c \mid \text{sex} = x) &= \frac{P(\text{species} = c \text{ and } \text{sex} = x)}{P(\text{sex = }x)} \\ &= \frac{\frac{\# \: (\text{species } = \: c \text{ and } \text{sex } = \: x)}{N}}{\frac{\# \: (\text{sex } = \: x)}{N}} \\ &= \frac{\# \: (\text{species} = c \text{ and } \text{sex} = x)}{\# \: (\text{sex} = x)} \end{align*}$$

Answer: To find conditional probabilities of 'species' given 'sex', divide by column sums. To find conditional probabilities of 'sex' given 'species', divide by row sums.

Conditional probabilities¶

To find conditional probabilities of 'species' given 'sex', divide by column sums. To find conditional probabilities of 'sex' given 'species', divide by row sums.

In [11]:
counts
Out[11]:
sex Female Male
species
Adelie 73 73
Chinstrap 34 34
Gentoo 58 61
In [12]:
counts.sum(axis=0)
Out[12]:
sex
Female    165
Male      168
dtype: int64

The conditional distribution of 'species' given 'sex' is below. Note that in this new DataFrame, the 'Female' and 'Male' columns each sum to 1.

In [13]:
counts / counts.sum(axis=0)
Out[13]:
sex Female Male
species
Adelie 0.44 0.43
Chinstrap 0.21 0.20
Gentoo 0.35 0.36

For instance, the above DataFrame tells us that the probability that a randomly selected penguin is of 'species' 'Adelie' given that they are of 'sex' 'Female' is 0.442424.

Question 🤔 (Answer at q.dsc80.com)

Find the conditional distribution of 'sex' given 'species'.
Hint: Use .T.

In [14]:
# Your code goes here.

Simpson's paradox¶


No description has been provided for this image

Example: Grades¶

  • Two students, Lisa and Bart, just finished their first year at UCSD. They both took a different number of classes in Fall, Winter, and Spring.

  • Each quarter, Lisa had a higher GPA than Bart.

  • But Bart has a higher overall GPA.

  • How is this possible? 🤔

Run this cell to create DataFrames that contain each students' grades.

In [15]:
lisa = pd.DataFrame([[20, 46], [18, 54], [5, 20]],
    columns=['Units', 'Grade Points Earned'], 
    index=['Fall', 'Winter', 'Spring'],
)
lisa.columns.name = 'Lisa' # This allows us to see the name "Lisa" in the top left of the DataFrame.

bart = pd.DataFrame([[5, 10], [5, 13.5], [22, 81.4]],
    columns=['Units', 'Grade Points Earned'], 
    index=['Fall', 'Winter', 'Spring'],
)
bart.columns.name = 'Bart'

Quarter-specific vs. overall GPAs¶

Note: The number of "grade points" earned for a course is

$$\text{number of units} \cdot \text{grade (out of 4)}$$

For instance, an A- in a 4 unit course earns $3.7 \cdot 4 = 14.8$ grade points.

In [16]:
dfs_side_by_side(lisa, bart)
Lisa Units Grade Points Earned
Fall 20 46
Winter 18 54
Spring 5 20
Bart Units Grade Points Earned
Fall 5 10.0
Winter 5 13.5
Spring 22 81.4

Lisa had a higher GPA in all three quarters.

In [17]:
quarterly_gpas = pd.DataFrame({
    "Lisa's Quarter GPA": lisa['Grade Points Earned'] / lisa['Units'],
    "Bart's Quarter GPA": bart['Grade Points Earned'] / bart['Units'],
})

quarterly_gpas
Out[17]:
Lisa's Quarter GPA Bart's Quarter GPA
Fall 2.3 2.0
Winter 3.0 2.7
Spring 4.0 3.7

Question 🤔 (Answer at q.dsc80.com)

Use the DataFrame lisa to compute Lisa's overall GPA, and use the DataFrame bartto compute Bart's overall GPA.

In [18]:
dfs_side_by_side(lisa, bart)
Lisa Units Grade Points Earned
Fall 20 46
Winter 18 54
Spring 5 20
Bart Units Grade Points Earned
Fall 5 10.0
Winter 5 13.5
Spring 22 81.4
In [19]:
# Your code goes here.

What happened?¶

In [20]:
(quarterly_gpas
 .assign(Lisa_Units=lisa['Units'],
         Bart_Units=bart['Units']) 
 .iloc[:, [0, 2, 1, 3]]
)
Out[20]:
Lisa's Quarter GPA Lisa_Units Bart's Quarter GPA Bart_Units
Fall 2.3 20 2.0 5
Winter 3.0 18 2.7 5
Spring 4.0 5 3.7 22

  • When Lisa and Bart both performed poorly, Lisa took more units than Bart. This brought down 📉 Lisa's overall average.

  • When Lisa and Bart both performed well, Bart took more units than Lisa. This brought up 📈 Bart's overall average.

Simpson's paradox¶

  • Simpson's paradox occurs when grouped data and ungrouped data show opposing trends.

    • It is named after Edward H. Simpson, not Lisa or Bart Simpson.

  • It often happens because there is a hidden factor (i.e. a confounder) within the data that influences results.

  • Question: What is the "correct" way to summarize your data? What if you had to act on these results?

Example: How Berkeley was almost sued for gender discrimination (1973)¶

What do you notice?

No description has been provided for this image
In [21]:
show_paradox_slides()

What happened?¶

  • The overall acceptance rate for women (30%) was lower than it was for men (45%).

  • However, most departments (A, B, D, F) had a higher acceptance rate for women.

  • Department A had a 62% acceptance rate for men and an 82% acceptance rate for women!

    • 31% of men applied to Department A.
    • 6% of women applied to Department A.

  • Department F had a 6% acceptance rate for men and a 7% acceptance rate for women!

    • 14% of men applied to Department F.
    • 19% of women applied to Department F.

  • Conclusion: Women tended to apply to departments with a lower acceptance rate; the data don't support the hypothesis that there was major gender discrimination against women.

Caution!¶

This doesn't mean that admissions are free from gender discrimination!

From Moss-Racusin et al., 2012, PNAS (cited 2600+ times):

In a randomized double-blind study (n = 127), science faculty from research-intensive universities rated the application materials of a student—who was randomly assigned either a male or female name—for a laboratory manager position. Faculty participants rated the male applicant as significantly more competent and hireable than the (identical) female applicant. These participants also selected a higher starting salary and offered more career mentoring to the male applicant. The gender of the faculty participants did not affect responses, such that female and male faculty were equally likely to exhibit bias against the female student.

But then...¶

From Williams and Ceci, 2015, PNAS:

Here we report five hiring experiments in which faculty evaluated hypothetical female and male applicants, using systematically varied profiles disguising identical scholarship, for assistant professorships in biology, engineering, economics, and psychology. Contrary to prevailing assumptions, men and women faculty members from all four fields preferred female applicants 2:1 over identically qualified males with matching lifestyles (single, married, divorced), with the exception of male economists, who showed no gender preference.

Do these conflict?¶

Not necessarily. One explanation, from William and Ceci:

Instead, past studies have used ratings of students’ hirability for a range of posts that do not include tenure-track jobs, such as managing laboratories or performing math assignments for a company. However, hiring tenure-track faculty differs from hiring lower-level staff: it entails selecting among highly accomplished candidates, all of whom have completed Ph.D.s and amassed publications and strong letters of support. Hiring bias may occur when applicants’ records are ambiguous, as was true in studies of hiring bias for lower-level staff posts, but such bias may not occur when records are clearly strong, as is the case with tenure-track hiring.

Example: Restaurant reviews and phone types¶

  • You are deciding whether to eat at Dirty Birds or The Loft.

  • Suppose Yelp shows ratings aggregated by phone type (Android vs. iPhone).

Phone Type Stars for Dirty Birds Stars for The Loft
Android 4.24 4.0
iPhone 2.99 2.79
All 3.32 3.37

  • Question: Should you choose Dirty Birds or The Loft?

  • Answer: The type of phone you use likely has nothing to do with your taste in food – pick the restaurant that is rated higher overall.

Rule of thumb 👍¶

  • Let $(X, Y)$ be a pair of variables of interest. Simpson's paradox occurs when the association between $X$ and $Y$ reverses when we condition on $Z$, a third variable.

  • If $Z$ has a causal connection to both $X$ and $Y$, we should condition on $Z$ and use deaggregated data.

  • If not, we shouldn't condition on $Z$ and use the aggregated data instead.

  • Berkeley gender discrimination: $X$ is gender, $Y$ is acceptance rate. $Z$ is the department.

    • $Z$ has a plausible causal effect on both $X$ and $Y$, so we should condition on $Z$.

  • Yelp ratings: $X$ is the restaurant, $Y$ is the average stars. $Z$ is the phone type.

    • $Z$ doesn't plausibly cause $X$ to change, so we should not condition on $Z$.

Takeaways¶

Be skeptical of...

  • Aggregate statistics.
  • People misusing statistics to "prove" that discrimination doesn't exist.
  • Drawing conclusions from individual publications ($p$-hacking, publication bias, narrow focus, etc.).
  • Everything!

We need to apply domain knowledge and human judgement calls to decide what to do when Simpson's paradox is present.

Really?¶

To handle Simpson's paradox with rigor, we need some ideas from causal inference which we don't have time to cover in DSC 80. This video has a good example of how to approach Simpson's paradox using a minimal amount of causal inference, if you're curious (not required for DSC 80).

In [22]:
IFrame('https://www.youtube-nocookie.com/embed/zeuW1Z2EtLs?si=l2Dl7P-5RCq3ODpo',
       width=800, height=450)
Out[22]:

Further reading¶

  • Gender Bias in Admission Statistics?
    • Contains a great visualization, but seems to be paywalled now.
  • What is Simpson's Paradox?
  • Understanding Simpson's Paradox
    • Requires more statistics background, but gives a rigorous understanding of when to use aggregated vs. unaggregated data.

Question 🤔 (Answer at q.dsc80.com)

What questions do you have?

Merging¶

Example: Name categories¶

The New York Times article from Lecture 1 claims that certain categories of names are becoming more popular. For example:

  • Forbidden names like Lucifer, Lilith, Kali, and Danger.

  • Evangelical names like Amen, Savior, Canaan, and Creed.

  • Mythological names.

  • It also claims that baby boomer names are becoming less popular.

Let's see if we can verify these claims using data!

Loading in the data¶

Our first DataFrame, baby, is the same as we saw in Lecture 1. It has one row for every combination of 'Name', 'Sex', and 'Year'.

In [23]:
baby_path = Path('data') / 'baby.csv'
baby = pd.read_csv(baby_path)
baby
Out[23]:
Name Sex Count Year
0 Liam M 20456 2022
1 Noah M 18621 2022
2 Olivia F 16573 2022
... ... ... ... ...
2085155 Wright M 5 1880
2085156 York M 5 1880
2085157 Zachariah M 5 1880

2085158 rows × 4 columns

Our second DataFrame, nyt, contains the New York Times' categorization of each of several names, based on the aforementioned article.

In [24]:
nyt_path = Path('data') / 'nyt_names.csv'
nyt = pd.read_csv(nyt_path)
nyt
Out[24]:
nyt_name category
0 Lucifer forbidden
1 Lilith forbidden
2 Danger forbidden
... ... ...
20 Venus celestial
21 Celestia celestial
22 Skye celestial

23 rows × 2 columns

Issue: To find the number of babies born with (for example) forbidden names each year, we need to combine information from both baby and nyt.

Merging¶

  • We want to link rows from baby and nyt together whenever the names match up.
  • This is a merge (pandas term), i.e. a join (SQL term).
  • A merge is appropriate when we have two sources of information about the same individuals that is linked by a common column(s).
  • The common column(s) are called the join key.

Example merge¶

Let's demonstrate on a small subset of baby and nyt.

In [25]:
nyt_small = nyt.iloc[[11, 12, 14]].reset_index(drop=True)

names_to_keep = ['Julius', 'Karen', 'Noah']
baby_small = (baby
 .query("Year == 2020 and Name in @names_to_keep")
 .reset_index(drop=True)
)

dfs_side_by_side(baby_small, nyt_small)
Name Sex Count Year
0 Noah M 18407 2020
1 Julius M 966 2020
2 Karen F 330 2020
3 Noah F 306 2020
4 Karen M 6 2020
nyt_name category
0 Karen boomer
1 Julius mythology
2 Freya mythology
In [26]:
%%pt
baby_small.merge(nyt_small, left_on='Name', right_on='nyt_name')

The merge method¶

  • The merge DataFrame method joins two DataFrames by columns or indexes.

    • As mentioned before, "merge" is just the pandas word for "join."

  • When using the merge method, the DataFrame before merge is the "left" DataFrame, and the DataFrame passed into merge is the "right" DataFrame.

    • In baby_small.merge(nyt_small), baby_small is considered the "left" DataFrame and nyt_small is the "right" DataFrame; the columns from the left DataFrame appear to the left of the columns from right DataFrame.

  • By default:

    • If join keys are not specified, all shared columns between the two DataFrames are used.
    • The "type" of join performed is an inner join. This is the only type of join you saw in DSC 10, but there are more, as we'll now see!

Join types: inner joins¶

In [27]:
%%pt
baby_small.merge(nyt_small, left_on='Name', right_on='nyt_name')
  • Note that 'Noah' and 'Freya' do not appear in the merged DataFrame.
  • This is because there is:
    • no 'Noah' in the right DataFrame (nyt_small), and
    • no 'Freya' in the left DataFrame (baby_small).
  • The default type of join that merge performs is an inner join, which keeps the intersection of the join keys.
No description has been provided for this image

Different join types¶

We can change the type of join performed by changing the how argument in merge. Let's experiment!

In [28]:
%%pt
# Note the NaNs!
baby_small.merge(nyt_small, left_on='Name', right_on='nyt_name', how='left')
In [29]:
%%pt
baby_small.merge(nyt_small, left_on='Name', right_on='nyt_name', how='right')
In [30]:
%%pt
baby_small.merge(nyt_small, left_on='Name', right_on='nyt_name', how='outer')

Different join types handle mismatches differently¶

There are four types of joins.

  • Inner: keep only matching keys (intersection).
  • Outer: keep all keys in both DataFrames (union).
  • Left: keep all keys in the left DataFrame, whether or not they are in the right DataFrame.
  • Right: keep all keys in the right DataFrame, whether or not they are in the left DataFrame.
    • Note that a.merge(b, how='left') contains the same information as b.merge(a, how='right'), just in a different order.
No description has been provided for this image

Notes on the merge method¶

  • merge is flexible – you can merge using a combination of columns, or the index of the DataFrame.
  • If the two DataFrames have the same column names, pandas will add _x and _y to the duplicated column names to avoid having columns with the same name (change these the suffixes argument).
  • There is, in fact, a join method, but it's actually a wrapper around merge with fewer options.
  • As always, the documentation is your friend!

Lots of pandas operations do an implicit outer join!¶

  • pandas will almost always try to match up index values using an outer join.
  • It won't tell you that it's doing an outer join, it'll just throw NaNs in your result!
In [31]:
df1 = pd.DataFrame({'a': [1, 2, 3]}, index=['hello', 'dsc80', 'students'])
df2 = pd.DataFrame({'b': [10, 20, 30]}, index=['dsc80', 'is', 'awesome'])
dfs_side_by_side(df1, df2)
a
hello 1
dsc80 2
students 3
b
dsc80 10
is 20
awesome 30
In [32]:
df1['a'] + df2['b']
Out[32]:
awesome      NaN
dsc80       12.0
hello        NaN
is           NaN
students     NaN
dtype: float64

Many-to-one & many-to-many joins¶

One-to-one joins¶

  • So far in this lecture, the joins we have worked with are called one-to-one joins.
  • Neither the left DataFrame (baby_small) nor the right DataFrame (nyt_small) contained any duplicates in the join key.
  • What if there are duplicated join keys, in one or both of the DataFrames we are merging?
In [33]:
# Run this cell to set up the next example.
profs = pd.DataFrame(
[['Sam', 'UCB', 5],
 ['Sam', 'UCSD', 5],
 ['Janine', 'UCSD', 8],
 ['Marina', 'UIC', 7],
 ['Justin', 'OSU', 5],
 ['Soohyun', 'UCSD', 2],
 ['Suraj', 'UCB', 2]],
    columns=['Name', 'School', 'Years']
)

schools = pd.DataFrame({
    'Abr': ['UCSD', 'UCLA', 'UCB', 'UIC'],
    'Full': ['University of California San Diego', 'University of California, Los Angeles', 'University of California, Berkeley', 'University of Illinois Chicago']
})

programs = pd.DataFrame({
    'uni': ['UCSD', 'UCSD', 'UCSD', 'UCB', 'OSU', 'OSU'],
    'dept': ['Math', 'HDSI', 'COGS', 'CS', 'Math', 'CS'],
    'grad_students': [205, 54, 281, 439, 304, 193]
})

Many-to-one joins¶

  • Many-to-one joins are joins where one of the DataFrames contains duplicate values in the join key.
  • The resulting DataFrame will preserve those duplicate entries as appropriate.
In [34]:
dfs_side_by_side(profs, schools)
Name School Years
0 Sam UCB 5
1 Sam UCSD 5
2 Janine UCSD 8
3 Marina UIC 7
4 Justin OSU 5
5 Soohyun UCSD 2
6 Suraj UCB 2
Abr Full
0 UCSD University of California San Diego
1 UCLA University of California, Los Angeles
2 UCB University of California, Berkeley
3 UIC University of Illinois Chicago

Note that when merging profs and schools, the information from schools is duplicated.

  • 'University of California, San Diego' appears three times.
  • 'University of California, Berkeley' appears twice.
In [35]:
%%pt
profs.merge(schools, left_on='School', right_on='Abr', how='left')

Many-to-many joins¶

Many-to-many joins are joins where both DataFrames have duplicate values in the join key.

In [36]:
dfs_side_by_side(profs, programs)
Name School Years
0 Sam UCB 5
1 Sam UCSD 5
2 Janine UCSD 8
3 Marina UIC 7
4 Justin OSU 5
5 Soohyun UCSD 2
6 Suraj UCB 2
uni dept grad_students
0 UCSD Math 205
1 UCSD HDSI 54
2 UCSD COGS 281
3 UCB CS 439
4 OSU Math 304
5 OSU CS 193

Before running the following cell, try predicting the number of rows in the output.

In [37]:
%%pt
profs.merge(programs, left_on='School', right_on='uni')
  • merge stitched together every UCSD row in profs with every UCSD row in programs.
  • Since there were 3 UCSD rows in profs and 3 in programs, there are $3 \cdot 3 = 9$ UCSD rows in the output. The same applies for all other schools.

Question 🤔 (Answer at q.dsc80.com)

Fill in the blank so that the last statement evaluates to True.

df = profs.merge(programs, left_on='School', right_on='uni')
df.shape[0] == (____).sum()

Don't use merge (or join) in your solution!

In [38]:
dfs_side_by_side(profs, programs)
Name School Years
0 Sam UCB 5
1 Sam UCSD 5
2 Janine UCSD 8
3 Marina UIC 7
4 Justin OSU 5
5 Soohyun UCSD 2
6 Suraj UCB 2
uni dept grad_students
0 UCSD Math 205
1 UCSD HDSI 54
2 UCSD COGS 281
3 UCB CS 439
4 OSU Math 304
5 OSU CS 193
In [39]:
# Your code goes here.

Returning back to our original question¶

Let's find the popularity of baby name categories over time. To start, we'll define a DataFrame that has one row for every combination of 'category' and 'Year'.

In [40]:
cate_counts = (
    baby
    .merge(nyt, left_on='Name', right_on='nyt_name')
    .groupby(['category', 'Year'])
    ['Count']
    .sum()
    .reset_index()
)
cate_counts
Out[40]:
category Year Count
0 boomer 1880 292
1 boomer 1881 298
2 boomer 1882 326
... ... ... ...
659 mythology 2020 3516
660 mythology 2021 3895
661 mythology 2022 4049

662 rows × 3 columns

In [41]:
# We'll talk about plotting code soon!
import plotly.express as px
fig = px.line(cate_counts, x='Year', y='Count',
              facet_col='category', facet_col_wrap=3,
              facet_row_spacing=0.15,
              width=600, height=400)
fig.update_yaxes(matches=None, showticklabels=False)

Question 🤔 (Answer at q.dsc80.com)

What questions do you have?

Transforming¶

Transforming values¶

  • A transformation results from performing some operation on every element in a sequence, e.g. a Series.

  • While we haven't discussed it yet in DSC 80, you learned how to transform Series in DSC 10, using the apply method. apply is very flexible – it takes in a function, which itself takes in a single value as input and returns a single value.

In [42]:
baby
Out[42]:
Name Sex Count Year
0 Liam M 20456 2022
1 Noah M 18621 2022
2 Olivia F 16573 2022
... ... ... ... ...
2085155 Wright M 5 1880
2085156 York M 5 1880
2085157 Zachariah M 5 1880

2085158 rows × 4 columns

In [43]:
def number_of_vowels(string):
    return sum(c in 'aeiou' for c in string.lower())

baby['Name'].apply(number_of_vowels)
Out[43]:
0          2
1          2
2          4
          ..
2085155    1
2085156    1
2085157    4
Name: Name, Length: 2085158, dtype: int64
In [44]:
# Built-in functions work with apply, too.
baby['Name'].apply(len)
Out[44]:
0          4
1          4
2          6
          ..
2085155    6
2085156    4
2085157    9
Name: Name, Length: 2085158, dtype: int64

The price of apply¶

Unfortunately, apply runs really slowly!

In [45]:
%%timeit
baby['Name'].apply(number_of_vowels)

1.33 s ± 16 ms per loop (mean ± std. dev. of 7 runs, 1 loop each)
In [46]:
%%timeit
res = []
for name in baby['Name']:
    res.append(number_of_vowels(name))

1.21 s ± 8.01 ms per loop (mean ± std. dev. of 7 runs, 1 loop each)

Internally, apply actually just runs a for-loop!

So, when possible – say, when applying arithmetic operations – we should work on Series objects directly and avoid apply!

The price of apply¶

In [47]:
%%timeit
baby['Year'] // 10 * 10 # Rounds down to the nearest multiple of 10.

4.81 ms ± 25.2 µs per loop (mean ± std. dev. of 7 runs, 100 loops each)
In [48]:
%%timeit
baby['Year'].apply(lambda y: y // 10 * 10)

437 ms ± 9.74 ms per loop (mean ± std. dev. of 7 runs, 1 loop each)

100x slower!

The .str accessor¶

For string operations, pandas provides a convenient .str accessor.

In [49]:
%%timeit
baby['Name'].str.len()

366 ms ± 2.26 ms per loop (mean ± std. dev. of 7 runs, 1 loop each)
In [50]:
%%timeit
baby['Name'].apply(len)

284 ms ± 1.51 ms per loop (mean ± std. dev. of 7 runs, 1 loop each)

It's very convenient, but runs even slower than apply!

Even though it's slow, we use .str anyway because it makes life easier.

Other data representations¶

Representations of tabular data¶

  • In DSC 80, we work with DataFrames in pandas.

    • When we say pandas DataFrame, we're talking about the pandas API for its DataFrame objects.
      • API stands for "application programming interface." We'll learn about these more soon.
    • When we say "DataFrame", we're referring to a general way to represent data (rows and columns, with labels for both rows and columns).

  • There many other ways to work with data tables!

    • Examples: R data frames, SQL databases, spreadsheets, or even matrices from linear algebra.
    • When you learn SQL in DSC 100, you'll find many similaries (e.g. slicing columns, filtering rows, grouping, joining, etc.).
    • Relational algebra captures common data operations between many data table systems.

  • Why use DataFrames over something else?

DataFrames vs. spreadsheets¶

  • DataFrames give us a data lineage: the code records down data changes. Not so in spreadsheets!
  • Using a general-purpose programming language gives us the ability to handle much larger datasets, and we can use distributed computing systems to handle massive datasets.

DataFrames vs. matrices¶

\begin{split} \begin{aligned} \mathbf{X} = \begin{bmatrix} 1 & 0 \\ 0 & 4 \\ 0 & 0 \\ \end{bmatrix} \end{aligned} \end{split}
  • Matrices are mathematical objects. They only hold numbers, but have many useful properties (which you've learned about in your linear algebra class, Math 18).
  • Often, we process data from a DataFrame into matrix format for machine learning models. You saw this a bit in DSC 40A, and we'll see this more in DSC 80 in a few weeks.

DataFrames vs. relations¶

  • Relations are the data representation for relational database systems (e.g. MySQL, PostgreSQL, etc.).
  • You'll learn all about these in DSC 100.
  • Database systems are much better than DataFrames at storing many large data tables and handling concurrency (many people reading and writing data at the same time).
  • Common workflow: load a subset of data in from a database system into pandas, then make a plot.
  • Or: load and clean data in pandas, then store it in a database system for others to use.

Summary¶

  • There is no "formula" to automatically resolve Simpson's paradox! Domain knowledge is important.
  • We've covered most of the primary DataFrame operations: subsetting, aggregating, joining, and transforming.

Next time¶

Data cleaning: applying what we've already learned to real-world, messy data!