In [ ]:
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 [7]:
# Pandas Tutor setup
%reload_ext pandas_tutor
%set_pandas_tutor_options {"maxDisplayCols": 8, "nohover": True, "projectorMode": True}
import warnings
warnings.simplefilter(action='ignore', category=FutureWarning)

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

DSC 80, Fall 2024¶

Announcements 📣¶

  • Project 1 checkpoint due tonight. No extensions allowed!
  • Lab 2 is due on Fri, Oct 11th.
  • Project 1 is due on Tue, Oct 15th.

Agenda¶

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

Other DataFrameGroupBy methods¶

Split-apply-combine, revisited¶

When we introduced the split-apply-combine pattern, the "apply" step involved aggregation – our final DataFrame had one row for each group.

No description has been provided for this image

Instead of aggregating during the apply step, we could instead perform a:

  • Transformation, in which we perform operations to every value within each group.

  • Filtration, in which we keep only the groups that satisfy some condition.

Transformations¶

Suppose we want to convert the 'body_mass_g' column to to z-scores (i.e. standard units):

$$z(x_i) = \frac{x_i - \text{mean of } x}{\text{SD of } x}$$

In [8]:
import seaborn as sns
penguins = sns.load_dataset('penguins').dropna()
In [9]:
def z_score(x):
    return (x - x.mean()) / x.std(ddof=0)
In [10]:
z_score(penguins['body_mass_g'])
Out[10]:
0     -0.57
1     -0.51
2     -1.19
       ... 
341    1.92
342    1.23
343    1.48
Name: body_mass_g, Length: 333, dtype: float64

Transformations within groups¶

  • Now, what if we wanted the z-score within each group?

  • To do so, we can use the transform method on a DataFrameGroupBy object. The transform method takes in a function, which itself takes in a Series and returns a new Series.

  • A transformation produces a DataFrame or Series of the same size – it is not an aggregation!

In [11]:
z_mass = (penguins
          .groupby('species')
          ['body_mass_g']
          .transform(z_score))
z_mass
Out[11]:
0      0.10
1      0.21
2     -1.00
       ... 
341    1.32
342    0.22
343    0.62
Name: body_mass_g, Length: 333, dtype: float64
In [12]:
penguins.assign(z_mass=z_mass)
Out[12]:
species island bill_length_mm bill_depth_mm flipper_length_mm body_mass_g sex z_mass
0 Adelie Torgersen 39.1 18.7 181.0 3750.0 Male 0.10
1 Adelie Torgersen 39.5 17.4 186.0 3800.0 Female 0.21
2 Adelie Torgersen 40.3 18.0 195.0 3250.0 Female -1.00
... ... ... ... ... ... ... ... ...
341 Gentoo Biscoe 50.4 15.7 222.0 5750.0 Male 1.32
342 Gentoo Biscoe 45.2 14.8 212.0 5200.0 Female 0.22
343 Gentoo Biscoe 49.9 16.1 213.0 5400.0 Male 0.62

333 rows × 8 columns

In [13]:
display_df(penguins.assign(z_mass=z_mass), rows=8)
species island bill_length_mm bill_depth_mm flipper_length_mm body_mass_g sex z_mass
0 Adelie Torgersen 39.1 18.7 181.0 3750.0 Male 0.10
1 Adelie Torgersen 39.5 17.4 186.0 3800.0 Female 0.21
2 Adelie Torgersen 40.3 18.0 195.0 3250.0 Female -1.00
4 Adelie Torgersen 36.7 19.3 193.0 3450.0 Female -0.56
... ... ... ... ... ... ... ... ...
340 Gentoo Biscoe 46.8 14.3 215.0 4850.0 Female -0.49
341 Gentoo Biscoe 50.4 15.7 222.0 5750.0 Male 1.32
342 Gentoo Biscoe 45.2 14.8 212.0 5200.0 Female 0.22
343 Gentoo Biscoe 49.9 16.1 213.0 5400.0 Male 0.62

333 rows × 8 columns

Note that above, penguin 340 has a larger 'body_mass_g' than penguin 0, but a lower 'z_mass'.

  • Penguin 0 has an above average 'body_mass_g' among 'Adelie' penguins.
  • Penguin 340 has a below average 'body_mass_g' among 'Gentoo' penguins. Remember from earlier that the average 'body_mass_g' of 'Gentoo' penguins is much higher than for other species.
In [14]:
penguins.groupby('species')['body_mass_g'].mean()
Out[14]:
species
Adelie       3706.16
Chinstrap    3733.09
Gentoo       5092.44
Name: body_mass_g, dtype: float64

Filtering groups¶

  • To keep only the groups that satisfy a particular condition, use the filter method on a DataFrameGroupBy object.

  • The filter method takes in a function, which itself takes in a DataFrame/Series and return a single Boolean. The result is a new DataFrame/Series with only the groups for which the filter function returned True.

For example, suppose we want only the 'species' whose average 'bill_length_mm' is above 39.

In [15]:
(penguins
 .groupby('species')
 .filter(lambda df: df['bill_length_mm'].mean() > 39)
)
Out[15]:
species island bill_length_mm bill_depth_mm flipper_length_mm body_mass_g sex
152 Chinstrap Dream 46.5 17.9 192.0 3500.0 Female
153 Chinstrap Dream 50.0 19.5 196.0 3900.0 Male
154 Chinstrap Dream 51.3 19.2 193.0 3650.0 Male
... ... ... ... ... ... ... ...
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

187 rows × 7 columns

No more 'Adelie's!

Or, as another example, suppose we only want 'species' with at least 100 penguins:

In [16]:
(penguins
 .groupby('species')
 .filter(lambda df: df.shape[0] > 100)
)
Out[16]:
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

265 rows × 7 columns

No more 'Chinstrap's!

Question 🤔 (Answer at dsc80.com/q)

Code: aggs

Answer the following questions about grouping:

  • In .agg(fn), what is the input to fn? What is the output of fn?
  • In .transform(fn), what is the input to fn? What is the output of fn?
  • In .filter(fn), what is the input to fn? What is the output of fn?

Grouping with multiple columns¶

When we group with multiple columns, one group is created for every unique combination of elements in the specified columns.

In [17]:
penguins
Out[17]:
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

In [18]:
species_and_island = (
    penguins
    .groupby(['species', 'island'])
    [['bill_length_mm', 'body_mass_g']]
    .mean()
)
species_and_island
Out[18]:
bill_length_mm body_mass_g
species island
Adelie Biscoe 38.98 3709.66
Dream 38.52 3701.36
Torgersen 39.04 3708.51
Chinstrap Dream 48.83 3733.09
Gentoo Biscoe 47.57 5092.44

Grouping and indexes¶

  • The groupby method creates an index based on the specified columns.
  • When grouping by multiple columns, the resulting DataFrame has a MultiIndex.
  • Advice: When working with a MultiIndex, use reset_index or set as_index=False in groupby.
In [19]:
species_and_island
Out[19]:
bill_length_mm body_mass_g
species island
Adelie Biscoe 38.98 3709.66
Dream 38.52 3701.36
Torgersen 39.04 3708.51
Chinstrap Dream 48.83 3733.09
Gentoo Biscoe 47.57 5092.44
In [20]:
species_and_island['body_mass_g']
Out[20]:
species    island   
Adelie     Biscoe       3709.66
           Dream        3701.36
           Torgersen    3708.51
Chinstrap  Dream        3733.09
Gentoo     Biscoe       5092.44
Name: body_mass_g, dtype: float64
In [21]:
species_and_island.loc['Adelie']
Out[21]:
bill_length_mm body_mass_g
island
Biscoe 38.98 3709.66
Dream 38.52 3701.36
Torgersen 39.04 3708.51
In [22]:
species_and_island.loc[('Adelie', 'Torgersen')]
Out[22]:
bill_length_mm      39.04
body_mass_g       3708.51
Name: (Adelie, Torgersen), dtype: float64
In [23]:
species_and_island.reset_index()
Out[23]:
species island bill_length_mm body_mass_g
0 Adelie Biscoe 38.98 3709.66
1 Adelie Dream 38.52 3701.36
2 Adelie Torgersen 39.04 3708.51
3 Chinstrap Dream 48.83 3733.09
4 Gentoo Biscoe 47.57 5092.44
In [24]:
(penguins
 .groupby(['species', 'island'], as_index=False)
 [['bill_length_mm', 'body_mass_g']]
 .mean()
)
Out[24]:
species island bill_length_mm body_mass_g
0 Adelie Biscoe 38.98 3709.66
1 Adelie Dream 38.52 3701.36
2 Adelie Torgersen 39.04 3708.51
3 Chinstrap Dream 48.83 3733.09
4 Gentoo Biscoe 47.57 5092.44

Question 🤔 (Answer at dsc80.com/q)

Code: names

Find the most popular Male and Female baby Name for each Year in baby. Exclude Years where there were fewer than 1 million births recorded.

In [25]:
baby_path = Path('data') / 'baby.csv'
baby = pd.read_csv(baby_path)
baby
Out[25]:
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 [26]:
# Your code goes here.

Pivot tables using the pivot_table method¶

Pivot tables: an extension of grouping¶

Pivot tables are a compact way to display tables for humans to read:

Sex F M
Year
2018 1698373 1813377
2019 1675139 1790682
2020 1612393 1721588
2021 1635800 1743913
2022 1628730 1733166
  • Notice that each value in the table is a sum over the counts, split by year and sex.
  • You can think of pivot tables as grouping using two columns, then "pivoting" one of the group labels into columns.

pivot_table¶

The pivot_table DataFrame method aggregates a DataFrame using two columns. To use it:

df.pivot_table(index=index_col,
               columns=columns_col,
               values=values_col,
               aggfunc=func)

The resulting DataFrame will have:

  • One row for every unique value in index_col.
  • One column for every unique value in columns_col.
  • Values determined by applying func on values in values_col.
In [27]:
last_5_years = baby.query('Year >= 2018')
last_5_years
Out[27]:
Name Sex Count Year
0 Liam M 20456 2022
1 Noah M 18621 2022
2 Olivia F 16573 2022
... ... ... ... ...
159444 Zyrie M 5 2018
159445 Zyron M 5 2018
159446 Zzyzx M 5 2018

159447 rows × 4 columns

In [28]:
last_5_years.pivot_table(
    index='Year',
    columns='Sex',
    values='Count',
    aggfunc='sum',
)
Out[28]:
Sex F M
Year
2018 1698373 1813377
2019 1675139 1790682
2020 1612393 1721588
2021 1635800 1743913
2022 1628730 1733166
In [29]:
# Look at the similarity to the snippet above!
(last_5_years
 .groupby(['Year', 'Sex'])
 [['Count']]
 .sum()
)
Out[29]:
Count
Year Sex
2018 F 1698373
M 1813377
2019 F 1675139
... ... ...
2021 M 1743913
2022 F 1628730
M 1733166

10 rows × 1 columns

Example¶

Find the number of penguins per 'island' and 'species'.

In [30]:
penguins
Out[30]:
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

In [31]:
penguins.pivot_table(
    index='species',
    columns='island',
    values='bill_length_mm', # Choice of column here doesn't actually matter!
    aggfunc='count',
)
Out[31]:
island Biscoe Dream Torgersen
species
Adelie 44.0 55.0 47.0
Chinstrap NaN 68.0 NaN
Gentoo 119.0 NaN NaN

Note that there is a NaN at the intersection of 'Biscoe' and 'Chinstrap', because there were no Chinstrap penguins on Biscoe Island.

We can either use the fillna method afterwards or the fill_value argument to fill in NaNs.

In [32]:
penguins.pivot_table(
    index='species',
    columns='island',
    values='bill_length_mm',
    aggfunc='count',
    fill_value=0,
)
Out[32]:
island Biscoe Dream Torgersen
species
Adelie 44 55 47
Chinstrap 0 68 0
Gentoo 119 0 0

Granularity, revisited¶

Take another look at the pivot table from the previous slide. Each row of the original penguins DataFrame represented a single penguin, and each column represented features of the penguins.

What is the granularity of the DataFrame below?

In [33]:
penguins.pivot_table(
    index='species',
    columns='island',
    values='bill_length_mm',
    aggfunc='count',
    fill_value=0,
)
Out[33]:
island Biscoe Dream Torgersen
species
Adelie 44 55 47
Chinstrap 0 68 0
Gentoo 119 0 0

Reshaping¶

  • pivot_table reshapes DataFrames from "long" to "wide".
  • Other DataFrame reshaping methods:
    • melt: Un-pivots a DataFrame. Very useful in data cleaning.
    • pivot: Like pivot_table, but doesn't do aggregation.
    • stack: Pivots multi-level columns to multi-indices.
    • unstack: Pivots multi-indices to columns.
    • Google and the documentation are your friends!

Distributions¶

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 [34]:
counts = penguins.pivot_table(
    index='species',
    columns='sex',
    values='body_mass_g',
    aggfunc='count',
    fill_value=0,
)
counts
Out[34]:
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 [35]:
joint = counts / counts.sum().sum()
joint
Out[35]:
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 [36]:
joint
Out[36]:
sex Female Male
species
Adelie 0.22 0.22
Chinstrap 0.10 0.10
Gentoo 0.17 0.18
In [37]:
# Recall, joint.sum(axis=0) sums across the rows,
# which computes the sum of the **columns**.
joint.sum(axis=0)
Out[37]:
sex
Female    0.5
Male      0.5
dtype: float64
In [38]:
joint.sum(axis=1)
Out[38]:
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 [39]:
counts
Out[39]:
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 [40]:
counts
Out[40]:
sex Female Male
species
Adelie 73 73
Chinstrap 34 34
Gentoo 58 61
In [41]:
counts.sum(axis=0)
Out[41]:
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 [42]:
counts / counts.sum(axis=0)
Out[42]:
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 dsc80.com/q)

Code: cond

Find the conditional distribution of 'sex' given 'species'.

Hint: Use .T.

In [43]:
# 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 [44]:
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 [45]:
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 [46]:
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[46]:
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 dsc80.com/q)

Code: gpa

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

In [80]:
# Helper function to show lisa and bart side-by-side to save screen space
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 [48]:
# Your code goes here.

What happened?¶

In [49]:
(quarterly_gpas
 .assign(Lisa_Units=lisa['Units'],
         Bart_Units=bart['Units'])
 .iloc[:, [0, 2, 1, 3]]
)
Out[49]:
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 [50]:
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.

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 [51]:
IFrame('https://www.youtube-nocookie.com/embed/zeuW1Z2EtLs?si=l2Dl7P-5RCq3ODpo',
       width=800, height=450)
Out[51]:

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.

Questions? 🤔

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 [52]:
baby_path = Path('data') / 'baby.csv'
baby = pd.read_csv(baby_path)
baby
Out[52]:
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 [53]:
nyt_path = Path('data') / 'nyt_names.csv'
nyt = pd.read_csv(nyt_path)
nyt
Out[53]:
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 [54]:
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 [55]:
%%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 [56]:
%%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 [57]:
%%pt
# Note the NaNs!
baby_small.merge(nyt_small, left_on='Name', right_on='nyt_name', how='left')
In [58]:
%%pt
baby_small.merge(nyt_small, left_on='Name', right_on='nyt_name', how='right')
In [59]:
%%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 [60]:
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 [61]:
df1['a'] + df2['b']
Out[61]:
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 [62]:
# 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 [63]:
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 [64]:
%%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 [65]:
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 [66]:
%%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 dsc80.com/q)

Code: merge

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 [67]:
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 [68]:
# 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 [69]:
cate_counts = (
    baby
    .merge(nyt, left_on='Name', right_on='nyt_name')
    .groupby(['category', 'Year'])
    ['Count']
    .sum()
    .reset_index()
)
cate_counts
Out[69]:
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 [70]:
# 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)

Questions? 🤔

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 [71]:
baby
Out[71]:
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 [72]:
def number_of_vowels(string):
    return sum(c in 'aeiou' for c in string.lower())

baby['Name'].apply(number_of_vowels)
Out[72]:
0          2
1          2
2          4
          ..
2085155    1
2085156    1
2085157    4
Name: Name, Length: 2085158, dtype: int64
In [73]:
# Built-in functions work with apply, too.
baby['Name'].apply(len)
Out[73]:
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 [74]:
%%timeit
baby['Name'].apply(number_of_vowels)

1.09 s ± 6.86 ms per loop (mean ± std. dev. of 7 runs, 1 loop each)
In [75]:
%%timeit
res = []
for name in baby['Name']:
    res.append(number_of_vowels(name))

946 ms ± 30.9 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 [76]:
%%timeit
baby['Year'] // 10 * 10 # Rounds down to the nearest multiple of 10.

3.12 ms ± 12.4 μs per loop (mean ± std. dev. of 7 runs, 100 loops each)
In [77]:
%%timeit
baby['Year'].apply(lambda y: y // 10 * 10)

394 ms ± 2.18 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 [78]:
%%timeit
baby['Name'].str.len()

243 ms ± 353 μs per loop (mean ± std. dev. of 7 runs, 1 loop each)
In [79]:
%%timeit
baby['Name'].apply(len)

265 ms ± 2.06 ms per loop (mean ± std. dev. of 7 runs, 1 loop each)

It's very convenient and runs about the same speed as apply!

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!