In [1]:
# Run this cell to set up packages for lecture.
from lec08_imports import *

Lecture 8 – Functions and Applying¶

DSC 10, Summer 2024¶

Agenda¶

  • Functions.
  • Applying functions to DataFrames.
    • Example: Student names.

Reminder: Use the DSC 10 Reference Sheet.

Functions¶

Defining functions¶

  • We've learned how to do quite a bit in Python:
    • Manipulate arrays, Series, and DataFrames.
    • Perform operations on strings.
    • Create visualizations.
  • But so far, we've been restricted to using existing functions (e.g. max, np.sqrt, len) and methods (e.g. .groupby, .assign, .plot).

Motivation¶

Suppose you drive to a restaurant 🥘 in LA, located exactly 100 miles away.

  • For the first 50 miles, you drive at 80 miles per hour.
  • For the last 50 miles, you drive at 60 miles per hour.
  • Question: What is your average speed throughout the journey?
  • 🚨 The answer is not 70 miles per hour! Remember, from Homework 1, you need to use the fact that $\text{speed} = \frac{\text{distance}}{\text{time}}$.

$$\text{average speed} = \frac{\text{total distance}}{\text{total time}} = \frac{50 + 50}{\text{time}_1 + \text{time}_2} \text{ miles per hour}$$

In segment 1, when you drove 50 miles at 80 miles per hour, you drove for $\frac{50}{80}$ hours:

$$\text{speed}_1 = \frac{\text{distance}_1}{\text{time}_1}$$

$$80 \text{ miles per hour} = \frac{50 \text{ miles}}{\text{time}_1} \implies \text{time}_1 = \frac{50}{80} \text{ hours}$$

Similarly, in segment 2, when you drove 50 miles at 60 miles per hour, you drove for $\text{time}_2 = \frac{50}{60} \text{ hours}$.

Then,

$$\text{average speed} = \frac{50 + 50}{\frac{50}{80} + \frac{50}{60}} \text{ miles per hour} $$

$$\begin{align*}\text{average speed} &= \frac{50}{50} \cdot \frac{1 + 1}{\frac{1}{80} + \frac{1}{60}} \text{ miles per hour} \\ &= \frac{2}{\frac{1}{80} + \frac{1}{60}} \text{ miles per hour} \end{align*}$$

Example: Harmonic mean¶

The harmonic mean ($\text{HM}$) of two positive numbers, $a$ and $b$, is defined as

$$\text{HM} = \frac{2}{\frac{1}{a} + \frac{1}{b}}$$

It is often used to find the average of multiple rates.

Finding the harmonic mean of 80 and 60 is not hard:

In [2]:
2 / (1 / 80 + 1 / 60)
Out[2]:
68.57142857142857

But what if we want to find the harmonic mean of 80 and 70? 80 and 90? 20 and 40? This would require a lot of copy-pasting, which is prone to error.

It turns out that we can define our own "harmonic mean" function just once, and re-use it multiple times.

In [3]:
def harmonic_mean(a, b):
    return 2 / (1 / a + 1 / b)
In [4]:
harmonic_mean(80, 60)
Out[4]:
68.57142857142857
In [5]:
harmonic_mean(20, 40)
Out[5]:
26.666666666666664

Note that we only had to specify how to calculate the harmonic mean once!

Functions¶

Functions are a way to divide our code into small subparts to prevent us from writing repetitive code. Each time we define our own function in Python, we will use the following pattern.

In [6]:
show_def()

Functions are "recipes"¶

  • Functions take in inputs, known as arguments, do something, and produce some outputs.
  • The beauty of functions is that you don't need to know how they are implemented in order to use them!
    • For instance, you've been using the function bpd.read_csv without knowing how it works.
    • This is the premise of the idea of abstraction in computer science – you'll hear a lot about this if you take DSC 20.
In [7]:
harmonic_mean(20, 40)
Out[7]:
26.666666666666664
In [8]:
harmonic_mean(79, 894)
Out[8]:
145.17163412127442
In [9]:
harmonic_mean(-2, 4)
Out[9]:
-8.0

Parameters and arguments¶

triple has one parameter, x.

In [10]:
def triple(x):
    return x * 3

When we call triple with the argument 5, within the body of triple, x means 5.

In [11]:
triple(5)
Out[11]:
15

We can change the argument we call triple with – we can even call it with strings!

In [12]:
triple(7 + 8)
Out[12]:
45
In [13]:
triple('triton')
Out[13]:
'tritontritontriton'

Scope 🩺¶

The names you choose for a function’s parameters are only known to that function (known as local scope). The rest of your notebook is unaffected by parameter names.

In [14]:
def triple(x):
    return x * 3
In [15]:
triple(7)
Out[15]:
21

Since we haven't defined an x outside of the body of triple, our notebook doesn't know what x means.

In [16]:
x

---------------------------------------------------------------------------
NameError                                 Traceback (most recent call last)
Cell In[16], line 1
----> 1 x

NameError: name 'x' is not defined

We can define an x outside of the body of triple, but that doesn't change how triple works.

In [17]:
x = 15
In [18]:
# When triple(12) is called, you can pretend
# there's an invisible line inside the body of x
# that says x = 12.
# The x = 15 above is ignored.
triple(12)
Out[18]:
36

Functions can take 0 or more arguments¶

Functions can take any number of arguments. So far, we've created a function with two arguments, harmonic_mean, and a function with one argument, triple.

greeting takes no arguments!

In [19]:
def greeting():
    return 'Hi! 👋'
In [20]:
greeting()
Out[20]:
'Hi! 👋'

Functions don't run until you call them!¶

The body of a function is not run until you use (call) the function.

Here, we can define where_is_the_error without seeing an error message.

In [21]:
def where_is_the_error(something):
    '''You can describe your function within triple quotes. For example, this function 
    illustrates that errors don't occur until functions are executed (called).'''
    return (1 / 0) + something

It is only when we call where_is_the_error that Python gives us an error message.

In [22]:
where_is_the_error(5)

---------------------------------------------------------------------------
ZeroDivisionError                         Traceback (most recent call last)
Cell In[22], line 1
----> 1 where_is_the_error(5)

Cell In[21], line 4, in where_is_the_error(something)
      1 def where_is_the_error(something):
      2     '''You can describe your function within triple quotes. For example, this function 
      3     illustrates that errors don't occur until functions are executed (called).'''
----> 4     return (1 / 0) + something

ZeroDivisionError: division by zero

Example: first_name¶

Let's create a function called first_name that takes in someone's full name and returns their first name. Example behavior is shown below.

>>> first_name('Pradeep Khosla')
'Pradeep'

Hint: Use the string method .split.

General strategy for writing functions:

  1. First, try and get the behavior to work on a single example.
  2. Then, encapsulate that behavior inside a function.
In [23]:
'Pradeep Khosla'.split(' ')[0]
Out[23]:
'Pradeep'
In [24]:
def first_name(full_name):
    '''Returns the first name given a full name.'''
    return full_name.split(' ')[0]
In [25]:
first_name('Pradeep Khosla')
Out[25]:
'Pradeep'
In [26]:
# What if there are three names?
first_name('Chancellor Pradeep Khosla')
Out[26]:
'Chancellor'

Returning¶

  • The return keyword specifies what the output of your function should be, i.e. what a call to your function will evaluate to.
  • Most functions we write will use return, but using return is not strictly required.
    • If you want to be able to save the output of your function to a variable, you must use return!
  • Be careful: print and return work differently!
In [27]:
def pythagorean(a, b):
    '''Computes the hypotenuse length of a right triangle with legs a and b.'''
    c = (a ** 2 + b ** 2) ** 0.5
    print(c)
In [28]:
x = pythagorean(3, 4)

5.0
In [29]:
# No output – why?
x
In [30]:
# Errors – why?
x + 10

---------------------------------------------------------------------------
TypeError                                 Traceback (most recent call last)
Cell In[30], line 2
      1 # Errors – why?
----> 2 x + 10

TypeError: unsupported operand type(s) for +: 'NoneType' and 'int'
In [31]:
def better_pythagorean(a, b):
    '''Computes the hypotenuse length of a right triangle with legs a and b, 
       and actually returns the result.
    '''
    c = (a ** 2 + b ** 2) ** 0.5
    return c
In [32]:
x = better_pythagorean(3, 4)
x
Out[32]:
5.0
In [33]:
x + 10
Out[33]:
15.0

Returning¶

Once a function executes a return statement, it stops running.

In [34]:
def motivational(quote):
    return 0
    print("Here's a motivational quote:", quote)
In [35]:
motivational('Fall seven times and stand up eight.')
Out[35]:
0

Applying functions to DataFrames¶

DSC 10 student data¶

The DataFrame roster contains the names and lecture sections of all students enrolled in DSC 10 in a previous quarter. The first names are real, while the last names have been anonymized for privacy.

In [36]:
roster = bpd.read_csv('data/roster-anon.csv')
roster
Out[36]:
name section
0 Allie Sazhma 11AM
1 Amina Igxazd 10AM
2 Jazmine Enesxr 9AM
... ... ...
219 Ismayl Gwuiij 10AM
220 Neil Dkaqgm 10AM
221 Maggie Ldfgau 9AM

222 rows × 2 columns

Example: Common first names¶

What is the most common first name among DSC 10 students? (Any guesses?)

In [37]:
roster
Out[37]:
name section
0 Allie Sazhma 11AM
1 Amina Igxazd 10AM
2 Jazmine Enesxr 9AM
... ... ...
219 Ismayl Gwuiij 10AM
220 Neil Dkaqgm 10AM
221 Maggie Ldfgau 9AM

222 rows × 2 columns

  • Problem: We can't answer that right now, since we don't have a column with first names. If we did, we could group by it.
  • Solution: Use our function that extracts first names on every element of the 'name' column.

Using our first_name function¶

Somehow, we need to call first_name on every student's 'name'.

In [38]:
roster
Out[38]:
name section
0 Allie Sazhma 11AM
1 Amina Igxazd 10AM
2 Jazmine Enesxr 9AM
... ... ...
219 Ismayl Gwuiij 10AM
220 Neil Dkaqgm 10AM
221 Maggie Ldfgau 9AM

222 rows × 2 columns

In [39]:
roster.get('name').iloc[0]
Out[39]:
'Allie Sazhma'
In [40]:
first_name(roster.get('name').iloc[0])
Out[40]:
'Allie'
In [41]:
first_name(roster.get('name').iloc[1])
Out[41]:
'Amina'

Ideally, there's a better solution than doing this hundreds of times...

.apply¶

  • To apply the function func_name to every element of column 'col' in DataFrame df, use
df.get('col').apply(func_name)
  • The .apply method is a Series method.
    • Important: We use .apply on Series, not DataFrames.
    • The output of .apply is also a Series.
  • Pass just the name of the function – don't call it!
    • Good ✅: .apply(first_name).
    • Bad ❌: .apply(first_name()).
In [42]:
roster.get('name')
Out[42]:
0        Allie Sazhma
1        Amina Igxazd
2      Jazmine Enesxr
            ...      
219     Ismayl Gwuiij
220       Neil Dkaqgm
221     Maggie Ldfgau
Name: name, Length: 222, dtype: object
In [43]:
roster.get('name').apply(first_name)
Out[43]:
0        Allie
1        Amina
2      Jazmine
        ...   
219     Ismayl
220       Neil
221     Maggie
Name: name, Length: 222, dtype: object

Example: Common first names¶

In [44]:
roster = roster.assign(
    first=roster.get('name').apply(first_name)
)
roster
Out[44]:
name section first
0 Allie Sazhma 11AM Allie
1 Amina Igxazd 10AM Amina
2 Jazmine Enesxr 9AM Jazmine
... ... ... ...
219 Ismayl Gwuiij 10AM Ismayl
220 Neil Dkaqgm 10AM Neil
221 Maggie Ldfgau 9AM Maggie

222 rows × 3 columns

Now that we have a column containing first names, we can find the distribution of first names.

In [45]:
name_counts = (
    roster
    .groupby('first')
    .count()
    .sort_values('name', ascending=False)
    .get(['name'])
)
name_counts
Out[45]:
name
first
Kevin 4
Ryan 4
Noah 3
... ...
Hongyu 1
Hriday 1
Zixuan 1

200 rows × 1 columns

Activity¶

Below:

  • Create a bar chart showing the number of students with each first name, but only include first names shared by at least two students.
  • Determine the proportion of students in DSC 10 who have a first name that is shared by at least two students.

Hint: Start by defining a DataFrame with only the names in name_counts that appeared at least twice. You can use this DataFrame to answer both questions.


✅ Click here to see the solutions after you've tried it yourself. 

shared_names = name_counts[name_counts.get('name') >= 2]

# Bar chart.
shared_names.sort_values('name').plot(kind='barh', y='name');

# Proportion = # students with a shared name / total # of students.
shared_names.get('name').sum() / roster.shape[0]
In [46]:
...
Out[46]:
Ellipsis
In [47]:
...
Out[47]:
Ellipsis

.apply works with built-in functions, too!¶

In [48]:
name_counts.get('name')
Out[48]:
first
Kevin     4
Ryan      4
Noah      3
         ..
Hongyu    1
Hriday    1
Zixuan    1
Name: name, Length: 200, dtype: int64
In [49]:
# Not necessarily meaningful, but doable.
name_counts.get('name').apply(np.sqrt)
Out[49]:
first
Kevin     2.00
Ryan      2.00
Noah      1.73
          ... 
Hongyu    1.00
Hriday    1.00
Zixuan    1.00
Name: name, Length: 200, dtype: float64

Aside: Resetting the index¶

In name_counts, first names are stored in the index, which is not a Series. This means we can't use .apply on it.

In [50]:
name_counts.index
Out[50]:
Index(['Kevin', 'Ryan', 'Noah', 'Kristen', 'Jimmy', 'Felix', 'Edward', 'David',
       'Olivia', 'Brandon',
       ...
       'Hailey', 'Hannah', 'Haotian', 'Harrison', 'Helen', 'Henry', 'Hongan',
       'Hongyu', 'Hriday', 'Zixuan'],
      dtype='object', name='first', length=200)
In [51]:
name_counts.index.apply(max)

---------------------------------------------------------------------------
AttributeError                            Traceback (most recent call last)
Cell In[51], line 1
----> 1 name_counts.index.apply(max)

AttributeError: 'Index' object has no attribute 'apply'

To help, we can use .reset_index() to turn the index of a DataFrame into a column, and to reset the index back to the default of 0, 1, 2, 3, and so on.

In [52]:
# What is the max of an individual string?
name_counts.reset_index().get('first').apply(max)
Out[52]:
0      v
1      y
2      o
      ..
197    y
198    y
199    x
Name: first, Length: 200, dtype: object
In [53]:
name_counts.reset_index()
Out[53]:
first name
0 Kevin 4
1 Ryan 4
2 Noah 3
... ... ...
197 Hongyu 1
198 Hriday 1
199 Zixuan 1

200 rows × 2 columns

In [54]:
name_counts
Out[54]:
name
first
Kevin 4
Ryan 4
Noah 3
... ...
Hongyu 1
Hriday 1
Zixuan 1

200 rows × 1 columns

Example: Shared first names and sections¶

  • Suppose you were one of the $\approx$17% of students in DSC 10 who had a first name that is shared with at least one other student.
  • Let's try and determine whether someone in your lecture section shares the same first name as you.
    • For example, maybe 'Olivia Kcjqla' wants to see if there's another 'Olivia' in their section.

Strategy:

  1. Which section is 'Olivia Kcjqla' in?
  2. How many people in that section have a first name of 'Olivia'?
In [55]:
roster
Out[55]:
name section first
0 Allie Sazhma 11AM Allie
1 Amina Igxazd 10AM Amina
2 Jazmine Enesxr 9AM Jazmine
... ... ... ...
219 Ismayl Gwuiij 10AM Ismayl
220 Neil Dkaqgm 10AM Neil
221 Maggie Ldfgau 9AM Maggie

222 rows × 3 columns

In [56]:
which_section = roster[roster.get('name') == 'Olivia Kcjqla'].get('section').iloc[0]
which_section
Out[56]:
'10AM'
In [57]:
first_cond = roster.get('first') == 'Olivia' # A Series of Booleans!
section_cond = roster.get('section') == which_section # A Series of Booleans!
how_many = roster[first_cond & section_cond].shape[0]
how_many
Out[57]:
1

Another function: shared_first_and_section¶

Let's create a function named shared_first_and_section. It will take in the full name of a student and return the number of students in their section with the same first name and section (including them).

Note: This is the first function we're writing that involves using a DataFrame within the function – this is fine!

In [58]:
def shared_first_and_section(name):
    # First, find the row corresponding to that full name in roster.
    # We're assuming that full names are unique.
    row = roster[roster.get('name') == name]
    
    # Then, get that student's first name and section.
    first = row.get('first').iloc[0]
    section = row.get('section').iloc[0]
    
    # Now, find all the students with the same first name and section.
    shared_info = roster[(roster.get('first') == first) & (roster.get('section') == section)]
    
    # Return the number of such students.
    return shared_info.shape[0]
In [59]:
shared_first_and_section('Olivia Kcjqla')
Out[59]:
1
In [60]:
# This means that there is another Jimmy in the same section as Jimmy Xvngxm.
shared_first_and_section('Jimmy Xvngxm')
Out[60]:
2

Now, let's add a column to roster that contains the values returned by shared_first_and_section.

In [61]:
roster = roster.assign(shared=roster.get('name').apply(shared_first_and_section))
roster
Out[61]:
name section first shared
0 Allie Sazhma 11AM Allie 1
1 Amina Igxazd 10AM Amina 1
2 Jazmine Enesxr 9AM Jazmine 1
... ... ... ... ...
219 Ismayl Gwuiij 10AM Ismayl 1
220 Neil Dkaqgm 10AM Neil 1
221 Maggie Ldfgau 9AM Maggie 1

222 rows × 4 columns

Let's find all of the students who are in a section with someone that has the same first name as them.

In [62]:
roster[(roster.get('shared') >= 2)].sort_values('shared', ascending=False)
Out[62]:
name section first shared
192 Ryan Oogwno 11AM Ryan 4
36 Ryan Mgetat 11AM Ryan 4
41 Ryan Nwdowi 11AM Ryan 4
... ... ... ... ...
203 Felix Fnrqck 10AM Felix 2
43 Jimmy Xvngxm 11AM Jimmy 2
107 Noah Llpjpu 10AM Noah 2

21 rows × 4 columns

We can narrow this down to a particular lecture section if we'd like.

In [63]:
one_section_only = (
    roster[(roster.get('shared') >= 2) & 
           (roster.get('section') == '10AM')]
    .sort_values('shared', ascending=False)
)
one_section_only
Out[63]:
name section first shared
11 Noah Qodtvo 10AM Noah 2
70 John Ubarsl 10AM John 2
74 John Paqmwc 10AM John 2
85 Felix Pgqrnv 10AM Felix 2
107 Noah Llpjpu 10AM Noah 2
203 Felix Fnrqck 10AM Felix 2

For instance, the above DataFrame preview is telling us that there are 2 Noahs, 2 Johns, and 2 Felixes in the 10AM section.

In [64]:
# All of the names shared by multiple students in the 10AM section.
one_section_only.get('first').unique()
Out[64]:
array(['Noah', 'John', 'Felix'], dtype=object)

Sneak peek¶

While the DataFrames on the previous slide contain the info we were looking for, they're not organized very conveniently. For instance, there are two rows containing the fact that there are 2 Noahs in the 10AM lecture section.

Wouldn't it be great if we could create a DataFrame like the one below? We'll see how next time!

section first name shared
0 10AM Noah 2 2
1 11AM Jimmy 2 2
2 10AM Ismayl 1 1

Activity¶

Find the longest first name in the class that is shared by at least two students in the same section.

Hint: You'll have to use both .assign and .apply.


✅ Click here to see the answer after you've tried it yourself. 

with_len = roster.assign(name_len=roster.get('first').apply(len))
with_len[with_len.get('shared') >= 2].sort_values('name_len', ascending=False).get('first').iloc[0]
In [65]:
...
Out[65]:
Ellipsis

Summary, next time¶

Summary¶

  • Functions are a way to divide our code into small subparts to prevent us from writing repetitive code.
  • The .apply method allows us to call a function on every single element of a Series, which usually comes from .getting a column of a DataFrame.

Next time¶

More advanced DataFrame manipulations!