Learning Outcomes
  • Perform advanced aggregation using .groupby()
  • Use the pd.pivot_table method to construct a pivot table
  • Perform simple merges between DataFrames using pd.merge()

We will introduce the concept of aggregating data – we will familiarize ourselves with GroupBy objects and used them as tools to consolidate and summarize aDataFrame. In this lecture, we will explore working with the different aggregation functions and dive into some advanced .groupby methods to show just how powerful of a resource they can be for understanding our data. We will also introduce other techniques for data aggregation to provide flexibility in how we manipulate our tables.

4.1 Custom Sorts

First, let’s finish our discussion about sorting. Let’s try to solve a sorting problem using different approaches. Assume we want to find the longest baby names and sort our data accordingly.

We’ll start by loading the babynames dataset. Note that this dataset is filtered to only contain data from California.

Code
# This code pulls census data and loads it into a DataFrame
# We won't cover it explicitly in this class, but you are welcome to explore it on your own
import pandas as pd
import numpy as np
import urllib.request
import os.path
import zipfile

data_url = "https://www.ssa.gov/oact/babynames/state/namesbystate.zip"
local_filename = "data/babynamesbystate.zip"
if not os.path.exists(local_filename): # If the data exists don't download again
    with urllib.request.urlopen(data_url) as resp, open(local_filename, 'wb') as f:
        f.write(resp.read())

zf = zipfile.ZipFile(local_filename, 'r')

ca_name = 'STATE.CA.TXT'
field_names = ['State', 'Sex', 'Year', 'Name', 'Count']
with zf.open(ca_name) as fh:
    babynames = pd.read_csv(fh, header=None, names=field_names)

babynames.tail(10)
State Sex Year Name Count
407418 CA M 2022 Zach 5
407419 CA M 2022 Zadkiel 5
407420 CA M 2022 Zae 5
407421 CA M 2022 Zai 5
407422 CA M 2022 Zay 5
407423 CA M 2022 Zayvier 5
407424 CA M 2022 Zia 5
407425 CA M 2022 Zora 5
407426 CA M 2022 Zuriel 5
407427 CA M 2022 Zylo 5

4.1.1 Approach 1: Create a Temporary Column

One method to do this is to first start by creating a column that contains the lengths of the names.

# Create a Series of the length of each name
babyname_lengths = babynames["Name"].str.len()

# Add a column named "name_lengths" that includes the length of each name
babynames["name_lengths"] = babyname_lengths
babynames.head(5)
State Sex Year Name Count name_lengths
0 CA F 1910 Mary 295 4
1 CA F 1910 Helen 239 5
2 CA F 1910 Dorothy 220 7
3 CA F 1910 Margaret 163 8
4 CA F 1910 Frances 134 7

We can then sort the DataFrame by that column using .sort_values():

# Sort by the temporary column
babynames = babynames.sort_values(by="name_lengths", ascending=False)
babynames.head(5)
State Sex Year Name Count name_lengths
334166 CA M 1996 Franciscojavier 8 15
337301 CA M 1997 Franciscojavier 5 15
339472 CA M 1998 Franciscojavier 6 15
321792 CA M 1991 Ryanchristopher 7 15
327358 CA M 1993 Johnchristopher 5 15

Finally, we can drop the name_length column from babynames to prevent our table from getting cluttered.

# Drop the 'name_length' column
babynames = babynames.drop("name_lengths", axis='columns')
babynames.head(5)
State Sex Year Name Count
334166 CA M 1996 Franciscojavier 8
337301 CA M 1997 Franciscojavier 5
339472 CA M 1998 Franciscojavier 6
321792 CA M 1991 Ryanchristopher 7
327358 CA M 1993 Johnchristopher 5

4.1.2 Approach 2: Sorting using the key Argument

Another way to approach this is to use the key argument of .sort_values(). Here we can specify that we want to sort "Name" values by their length.

babynames.sort_values("Name", key=lambda x: x.str.len(), ascending=False).head()
State Sex Year Name Count
334166 CA M 1996 Franciscojavier 8
327472 CA M 1993 Ryanchristopher 5
337301 CA M 1997 Franciscojavier 5
337477 CA M 1997 Ryanchristopher 5
312543 CA M 1987 Franciscojavier 5

4.1.3 Approach 3: Sorting using the map Function

We can also use the map function on a Series to solve this. Say we want to sort the babynames table by the number of "dr"’s and "ea"’s in each "Name". We’ll define the function dr_ea_count to help us out.

# First, define a function to count the number of times "dr" or "ea" appear in each name
def dr_ea_count(string):
    return string.count('dr') + string.count('ea')

# Then, use `map` to apply `dr_ea_count` to each name in the "Name" column
babynames["dr_ea_count"] = babynames["Name"].map(dr_ea_count)

# Sort the DataFrame by the new "dr_ea_count" column so we can see our handiwork
babynames = babynames.sort_values(by="dr_ea_count", ascending=False)
babynames.head()
State Sex Year Name Count dr_ea_count
115957 CA F 1990 Deandrea 5 3
101976 CA F 1986 Deandrea 6 3
131029 CA F 1994 Leandrea 5 3
108731 CA F 1988 Deandrea 5 3
308131 CA M 1985 Deandrea 6 3

We can drop the dr_ea_count once we’re done using it to maintain a neat table.

# Drop the `dr_ea_count` column
babynames = babynames.drop("dr_ea_count", axis = 'columns')
babynames.head(5)
State Sex Year Name Count
115957 CA F 1990 Deandrea 5
101976 CA F 1986 Deandrea 6
131029 CA F 1994 Leandrea 5
108731 CA F 1988 Deandrea 5
308131 CA M 1985 Deandrea 6

4.2 Aggregating Data with .groupby

Up until this point, we have been working with individual rows of DataFrames. As data scientists, we often wish to investigate trends across a larger subset of our data. For example, we may want to compute some summary statistic (the mean, median, sum, etc.) for a group of rows in our DataFrame. To do this, we’ll use pandas GroupBy objects. Our goal is to group together rows that fall under the same category and perform an operation that aggregates across all rows in the category.

Let’s say we wanted to aggregate all rows in babynames for a given year.

babynames.groupby("Year")
<pandas.core.groupby.generic.DataFrameGroupBy object at 0x105e8db20>

What does this strange output mean? Calling .groupby (documentation) has generated a GroupBy object. You can imagine this as a set of “mini” sub-DataFrames, where each subframe contains all of the rows from babynames that correspond to a particular year.

The diagram below shows a simplified view of babynames to help illustrate this idea.

We can’t work with a GroupBy object directly – that is why you saw that strange output earlier rather than a standard view of a DataFrame. To actually manipulate values within these “mini” DataFrames, we’ll need to call an aggregation method. This is a method that tells pandas how to aggregate the values within the GroupBy object. Once the aggregation is applied, pandas will return a normal (now grouped) DataFrame.

The first aggregation method we’ll consider is .agg. The .agg method takes in a function as its argument; this function is then applied to each column of a “mini” grouped DataFrame. We end up with a new DataFrame with one aggregated row per subframe. Let’s see this in action by finding the sum of all counts for each year in babynames – this is equivalent to finding the number of babies born in each year.

babynames[["Year", "Count"]].groupby("Year").agg(sum).head(5)
Count
Year
1910 9163
1911 9983
1912 17946
1913 22094
1914 26926

We can relate this back to the diagram we used above. Remember that the diagram uses a simplified version of babynames, which is why we see smaller values for the summed counts.

Performing an aggregation

Calling .agg has condensed each subframe back into a single row. This gives us our final output: a DataFrame that is now indexed by "Year", with a single row for each unique year in the original babynames DataFrame.

There are many different aggregation functions we can use, all of which are useful in different applications.

babynames[["Year", "Count"]].groupby("Year").agg(min).head(5)
Count
Year
1910 5
1911 5
1912 5
1913 5
1914 5
babynames[["Year", "Count"]].groupby("Year").agg(max).head(5)
Count
Year
1910 295
1911 390
1912 534
1913 614
1914 773
# Same result, but now we explicitly tell pandas to only consider the "Count" column when summing
babynames.groupby("Year")[["Count"]].agg(sum).head(5)
Count
Year
1910 9163
1911 9983
1912 17946
1913 22094
1914 26926

There are many different aggregations that can be applied to the grouped data. The primary requirement is that an aggregation function must:

  • Take in a Series of data (a single column of the grouped subframe).
  • Return a single value that aggregates this Series.

4.2.1 Aggregation Functions

Because of this fairly broad requirement, pandas offers many ways of computing an aggregation.

In-built Python operations – such as sum, max, and min – are automatically recognized by pandas.

# What is the minimum count for each name in any year?
babynames.groupby("Name")[["Count"]].agg(min).head()
Count
Name
Aadan 5
Aadarsh 6
Aaden 10
Aadhav 6
Aadhini 6
# What is the largest single-year count of each name?
babynames.groupby("Name")[["Count"]].agg(max).head()
Count
Name
Aadan 7
Aadarsh 6
Aaden 158
Aadhav 8
Aadhini 6

As mentioned previously, functions from the NumPy library, such as np.mean, np.max, np.min, and np.sum, are also fair game in pandas.

# What is the average count for each name across all years?
babynames.groupby("Name")[["Count"]].agg(np.mean).head()
Count
Name
Aadan 6.000000
Aadarsh 6.000000
Aaden 46.214286
Aadhav 6.750000
Aadhini 6.000000

pandas also offers a number of in-built functions. Functions that are native to pandas can be referenced using their string name within a call to .agg. Some examples include:

  • .agg("sum")
  • .agg("max")
  • .agg("min")
  • .agg("mean")
  • .agg("first")
  • .agg("last")

The latter two entries in this list – "first" and "last" – are unique to pandas. They return the first or last entry in a subframe column. Why might this be useful? Consider a case where multiple columns in a group share identical information. To represent this information in the grouped output, we can simply grab the first or last entry, which we know will be identical to all other entries.

Let’s illustrate this with an example. Say we add a new column to babynames that contains the first letter of each name.

# Imagine we had an additional column, "First Letter". We'll explain this code next week
babynames["First Letter"] = babynames["Name"].str[0]

# We construct a simplified DataFrame containing just a subset of columns
babynames_new = babynames[["Name", "First Letter", "Year"]]
babynames_new.head()
Name First Letter Year
115957 Deandrea D 1990
101976 Deandrea D 1986
131029 Leandrea L 1994
108731 Deandrea D 1988
308131 Deandrea D 1985

If we form groups for each name in the dataset, "First Letter" will be the same for all members of the group. This means that if we simply select the first entry for "First Letter" in the group, we’ll represent all data in that group.

We can use a dictionary to apply different aggregation functions to each column during grouping.

Aggregating using “first”
babynames_new.groupby("Name").agg({"First Letter":"first", "Year":"max"}).head()
First Letter Year
Name
Aadan A 2014
Aadarsh A 2019
Aaden A 2020
Aadhav A 2019
Aadhini A 2022

4.2.2 Plotting Birth Counts

Let’s use .agg to find the total number of babies born in each year. Recall that using .agg with .groupby() follows the format: df.groupby(column_name).agg(aggregation_function). The line of code below gives us the total number of babies born in each year.

Code
babynames.groupby("Year")[["Count"]].agg(sum).head(5)
# Alternative 1
# babynames.groupby("Year")[["Count"]].sum()
# Alternative 2
# babynames.groupby("Year").sum(numeric_only=True)
Count
Year
1910 9163
1911 9983
1912 17946
1913 22094
1914 26926

Here’s an illustration of the process:

aggregation

Plotting the Dataframe we obtain tells an interesting story.

Code
import plotly.express as px
puzzle2 = babynames.groupby("Year")[["Count"]].agg(sum)
px.line(puzzle2, y = "Count")

A word of warning: we made an enormous assumption when we decided to use this dataset to estimate birth rate. According to this article from the Legistlative Analyst Office, the true number of babies born in California in 2020 was 421,275. However, our plot shows 362,882 babies —— what happened?

4.2.3 Summary of the .groupby() Function

A groupby operation involves some combination of splitting a DataFrame into grouped subframes, applying a function, and combining the results.

For some arbitrary DataFrame df below, the code df.groupby("year").agg(sum) does the following:

  • Splits the DataFrame into sub-DataFrames with rows belonging to the same year.
  • Applies the sum function to each column of each sub-DataFrame.
  • Combines the results of sum into a single DataFrame, indexed by year.

groupby_demo

4.2.4 Revisiting the .agg() Function

.agg() can take in any function that aggregates several values into one summary value. Some commonly-used aggregation functions can even be called directly, without explicit use of .agg(). For example, we can call .mean() on .groupby():

babynames.groupby("Year").mean().head()

We can now put this all into practice. Say we want to find the baby name with sex “F” that has fallen in popularity the most in California. To calculate this, we can first create a metric: “Ratio to Peak” (RTP). The RTP is the ratio of babies born with a given name in 2022 to the maximum number of babies born with the name in any year.

Let’s start with calculating this for one baby, “Jennifer”.

# We filter by babies with sex "F" and sort by "Year"
f_babynames = babynames[babynames["Sex"] == "F"]
f_babynames = f_babynames.sort_values(["Year"])

# Determine how many Jennifers were born in CA per year
jenn_counts_series = f_babynames[f_babynames["Name"] == "Jennifer"]["Count"]

# Determine the max number of Jennifers born in a year and the number born in 2022 
# to calculate RTP
max_jenn = max(f_babynames[f_babynames["Name"] == "Jennifer"]["Count"])
curr_jenn = f_babynames[f_babynames["Name"] == "Jennifer"]["Count"].iloc[-1]
rtp = curr_jenn / max_jenn
rtp
0.018796372629843364

By creating a function to calculate RTP and applying it to our DataFrame by using .groupby(), we can easily compute the RTP for all names at once!

def ratio_to_peak(series):
    return series.iloc[-1] / max(series)

#Using .groupby() to apply the function
rtp_table = f_babynames.groupby("Name")[["Year", "Count"]].agg(ratio_to_peak)
rtp_table.head()
Year Count
Name
Aadhini 1.0 1.000000
Aadhira 1.0 0.500000
Aadhya 1.0 0.660000
Aadya 1.0 0.586207
Aahana 1.0 0.269231

In the rows shown above, we can see that every row shown has a Year value of 1.0.

This is the “pandas-ification” of logic you saw in Data 8. Much of the logic you’ve learned in Data 8 will serve you well in Data 100.

4.2.5 Nuisance Columns

Note that you must be careful with which columns you apply the .agg() function to. If we were to apply our function to the table as a whole by doing f_babynames.groupby("Name").agg(ratio_to_peak), executing our .agg() call would result in a TypeError.

error

We can avoid this issue (and prevent unintentional loss of data) by explicitly selecting column(s) we want to apply our aggregation function to BEFORE calling .agg(),

4.2.6 Renaming Columns After Grouping

By default, .groupby will not rename any aggregated columns. As we can see in the table above, the aggregated column is still named Count even though it now represents the RTP. For better readability, we can rename Count to Count RTP

rtp_table = rtp_table.rename(columns = {"Count": "Count RTP"})
rtp_table
Year Count RTP
Name
Aadhini 1.0 1.000000
Aadhira 1.0 0.500000
Aadhya 1.0 0.660000
Aadya 1.0 0.586207
Aahana 1.0 0.269231
... ... ...
Zyanya 1.0 0.466667
Zyla 1.0 1.000000
Zylah 1.0 1.000000
Zyra 1.0 1.000000
Zyrah 1.0 0.833333

13782 rows × 2 columns

4.2.7 Some Data Science Payoff

By sorting rtp_table, we can see the names whose popularity has decreased the most.

rtp_table = rtp_table.rename(columns = {"Count": "Count RTP"})
rtp_table.sort_values("Count RTP").head()
Year Count RTP
Name
Debra 1.0 0.001260
Debbie 1.0 0.002815
Carol 1.0 0.003180
Tammy 1.0 0.003249
Susan 1.0 0.003305

To visualize the above DataFrame, let’s look at the line plot below:

Code
import plotly.express as px
px.line(f_babynames[f_babynames["Name"] == "Debra"], x = "Year", y = "Count")

We can get the list of the top 10 names and then plot popularity with the following code:

top10 = rtp_table.sort_values("Count RTP").head(10).index
px.line(
    f_babynames[f_babynames["Name"].isin(top10)], 
    x = "Year", 
    y = "Count", 
    color = "Name"
)

As a quick exercise, consider what code would compute the total number of babies with each name.

Code
babynames.groupby("Name")[["Count"]].agg(sum).head()
# alternative solution: 
# babynames.groupby("Name")[["Count"]].sum()
Count
Name
Aadan 18
Aadarsh 6
Aaden 647
Aadhav 27
Aadhini 6

4.3 .groupby(), Continued

We’ll work with the elections DataFrame again.

Code
import pandas as pd
import numpy as np

elections = pd.read_csv("data/elections.csv")
elections.head(5)
Year Candidate Party Popular vote Result %
0 1824 Andrew Jackson Democratic-Republican 151271 loss 57.210122
1 1824 John Quincy Adams Democratic-Republican 113142 win 42.789878
2 1828 Andrew Jackson Democratic 642806 win 56.203927
3 1828 John Quincy Adams National Republican 500897 loss 43.796073
4 1832 Andrew Jackson Democratic 702735 win 54.574789

4.3.1 Raw GroupBy Objects

The result of groupby applied to a DataFrame is a DataFrameGroupBy object, not a DataFrame.

grouped_by_year = elections.groupby("Year")
type(grouped_by_year)
pandas.core.groupby.generic.DataFrameGroupBy

There are several ways to look into DataFrameGroupBy objects:

grouped_by_party = elections.groupby("Party")
grouped_by_party.groups
{'American': [22, 126], 'American Independent': [115, 119, 124], 'Anti-Masonic': [6], 'Anti-Monopoly': [38], 'Citizens': [127], 'Communist': [89], 'Constitution': [160, 164, 172], 'Constitutional Union': [24], 'Democratic': [2, 4, 8, 10, 13, 14, 17, 20, 28, 29, 34, 37, 39, 45, 47, 52, 55, 57, 64, 70, 74, 77, 81, 83, 86, 91, 94, 97, 100, 105, 108, 111, 114, 116, 118, 123, 129, 134, 137, 140, 144, 151, 158, 162, 168, 176, 178], 'Democratic-Republican': [0, 1], 'Dixiecrat': [103], 'Farmer–Labor': [78], 'Free Soil': [15, 18], 'Green': [149, 155, 156, 165, 170, 177, 181], 'Greenback': [35], 'Independent': [121, 130, 143, 161, 167, 174], 'Liberal Republican': [31], 'Libertarian': [125, 128, 132, 138, 139, 146, 153, 159, 163, 169, 175, 180], 'National Democratic': [50], 'National Republican': [3, 5], 'National Union': [27], 'Natural Law': [148], 'New Alliance': [136], 'Northern Democratic': [26], 'Populist': [48, 61, 141], 'Progressive': [68, 82, 101, 107], 'Prohibition': [41, 44, 49, 51, 54, 59, 63, 67, 73, 75, 99], 'Reform': [150, 154], 'Republican': [21, 23, 30, 32, 33, 36, 40, 43, 46, 53, 56, 60, 65, 69, 72, 79, 80, 84, 87, 90, 96, 98, 104, 106, 109, 112, 113, 117, 120, 122, 131, 133, 135, 142, 145, 152, 157, 166, 171, 173, 179], 'Socialist': [58, 62, 66, 71, 76, 85, 88, 92, 95, 102], 'Southern Democratic': [25], 'States' Rights': [110], 'Taxpayers': [147], 'Union': [93], 'Union Labor': [42], 'Whig': [7, 9, 11, 12, 16, 19]}
grouped_by_party.get_group("Socialist")
Year Candidate Party Popular vote Result %
58 1904 Eugene V. Debs Socialist 402810 loss 2.985897
62 1908 Eugene V. Debs Socialist 420852 loss 2.850866
66 1912 Eugene V. Debs Socialist 901551 loss 6.004354
71 1916 Allan L. Benson Socialist 590524 loss 3.194193
76 1920 Eugene V. Debs Socialist 913693 loss 3.428282
85 1928 Norman Thomas Socialist 267478 loss 0.728623
88 1932 Norman Thomas Socialist 884885 loss 2.236211
92 1936 Norman Thomas Socialist 187910 loss 0.412876
95 1940 Norman Thomas Socialist 116599 loss 0.234237
102 1948 Norman Thomas Socialist 139569 loss 0.286312

4.3.2 Other GroupBy Methods

There are many aggregation methods we can use with .agg. Some useful options are:

  • .mean: creates a new DataFrame with the mean value of each group
  • .sum: creates a new DataFrame with the sum of each group
  • .max and .min: creates a new DataFrame with the maximum/minimum value of each group
  • .first and .last: creates a new DataFrame with the first/last row in each group
  • .size: creates a new Series with the number of entries in each group
  • .count: creates a new DataFrame with the number of entries, excluding missing values.

Let’s illustrate some examples by creating a DataFrame called df.

df = pd.DataFrame({'letter':['A','A','B','C','C','C'], 
                   'num':[1,2,3,4,np.NaN,4], 
                   'state':[np.NaN, 'tx', 'fl', 'hi', np.NaN, 'ak']})
df
letter num state
0 A 1.0 NaN
1 A 2.0 tx
2 B 3.0 fl
3 C 4.0 hi
4 C NaN NaN
5 C 4.0 ak

Note the slight difference between .size() and .count(): while .size() returns a Series and counts the number of entries including the missing values, .count() returns a DataFrame and counts the number of entries in each column excluding missing values.

df.groupby("letter").size()
letter
A    2
B    1
C    3
dtype: int64
df.groupby("letter").count()
num state
letter
A 2 1
B 1 1
C 2 2

You might recall that the value_counts() function in the previous note does something similar. It turns out value_counts() and groupby.size() are the same, except value_counts() sorts the resulting Series in descending order automatically.

df["letter"].value_counts()
C    3
A    2
B    1
Name: letter, dtype: int64

These (and other) aggregation functions are so common that pandas allows for writing shorthand. Instead of explicitly stating the use of .agg, we can call the function directly on the GroupBy object.

For example, the following are equivalent:

  • elections.groupby("Candidate").agg(mean)
  • elections.groupby("Candidate").mean()

There are many other methods that pandas supports. You can check them out on the pandas documentation.

4.3.3 Filtering by Group

Another common use for GroupBy objects is to filter data by group.

groupby.filter takes an argument func, where func is a function that:

  • Takes a DataFrame object as input
  • Returns a single True or False.

groupby.filter applies func to each group/sub-DataFrame:

  • If func returns True for a group, then all rows belonging to the group are preserved.
  • If func returns False for a group, then all rows belonging to that group are filtered out.

In other words, sub-DataFrames that correspond to True are returned in the final result, whereas those with a False value are not. Importantly, groupby.filter is different from groupby.agg in that an entire sub-DataFrame is returned in the final DataFrame, not just a single row. As a result, groupby.filter preserves the original indices and the column we grouped on does NOT become the index!

groupby_demo

To illustrate how this happens, let’s go back to the elections dataset. Say we want to identify “tight” election years – that is, we want to find all rows that correspond to election years where all candidates in that year won a similar portion of the total vote. Specifically, let’s find all rows corresponding to a year where no candidate won more than 45% of the total vote.

In other words, we want to:

  • Find the years where the maximum % in that year is less than 45%
  • Return all DataFrame rows that correspond to these years

For each year, we need to find the maximum % among all rows for that year. If this maximum % is lower than 45%, we will tell pandas to keep all rows corresponding to that year.

elections.groupby("Year").filter(lambda sf: sf["%"].max() < 45).head(9)
Year Candidate Party Popular vote Result %
23 1860 Abraham Lincoln Republican 1855993 win 39.699408
24 1860 John Bell Constitutional Union 590901 loss 12.639283
25 1860 John C. Breckinridge Southern Democratic 848019 loss 18.138998
26 1860 Stephen A. Douglas Northern Democratic 1380202 loss 29.522311
66 1912 Eugene V. Debs Socialist 901551 loss 6.004354
67 1912 Eugene W. Chafin Prohibition 208156 loss 1.386325
68 1912 Theodore Roosevelt Progressive 4122721 loss 27.457433
69 1912 William Taft Republican 3486242 loss 23.218466
70 1912 Woodrow Wilson Democratic 6296284 win 41.933422

What’s going on here? In this example, we’ve defined our filtering function, func, to be lambda sf: sf["%"].max() < 45. This filtering function will find the maximum "%" value among all entries in the grouped sub-DataFrame, which we call sf. If the maximum value is less than 45, then the filter function will return True and all rows in that grouped sub-DataFrame will appear in the final output DataFrame.

Examine the DataFrame above. Notice how, in this preview of the first 9 rows, all entries from the years 1860 and 1912 appear. This means that in 1860 and 1912, no candidate in that year won more than 45% of the total vote.

You may ask: how is the groupby.filter procedure different to the boolean filtering we’ve seen previously? Boolean filtering considers individual rows when applying a boolean condition. For example, the code elections[elections["%"] < 45] will check the "%" value of every single row in elections; if it is less than 45, then that row will be kept in the output. groupby.filter, in contrast, applies a boolean condition across all rows in a group. If not all rows in that group satisfy the condition specified by the filter, the entire group will be discarded in the output.

4.3.4 Aggregation with lambda Functions

What if we wish to aggregate our DataFrame using a non-standard function – for example, a function of our own design? We can do so by combining .agg with lambda expressions.

Let’s first consider a puzzle to jog our memory. We will attempt to find the Candidate from each Party with the highest % of votes.

A naive approach may be to group by the Party column and aggregate by the maximum.

elections.groupby("Party").agg(max).head(10)
Year Candidate Popular vote Result %
Party
American 1976 Thomas J. Anderson 873053 loss 21.554001
American Independent 1976 Lester Maddox 9901118 loss 13.571218
Anti-Masonic 1832 William Wirt 100715 loss 7.821583
Anti-Monopoly 1884 Benjamin Butler 134294 loss 1.335838
Citizens 1980 Barry Commoner 233052 loss 0.270182
Communist 1932 William Z. Foster 103307 loss 0.261069
Constitution 2016 Michael Peroutka 203091 loss 0.152398
Constitutional Union 1860 John Bell 590901 loss 12.639283
Democratic 2020 Woodrow Wilson 81268924 win 61.344703
Democratic-Republican 1824 John Quincy Adams 151271 win 57.210122

This approach is clearly wrong – the DataFrame claims that Woodrow Wilson won the presidency in 2020.

Why is this happening? Here, the max aggregation function is taken over every column independently. Among Democrats, max is computing:

  • The most recent Year a Democratic candidate ran for president (2020)
  • The Candidate with the alphabetically “largest” name (“Woodrow Wilson”)
  • The Result with the alphabetically “largest” outcome (“win”)

Instead, let’s try a different approach. We will:

  1. Sort the DataFrame so that rows are in descending order of %
  2. Group by Party and select the first row of each sub-DataFrame

While it may seem unintuitive, sorting elections by descending order of % is extremely helpful. If we then group by Party, the first row of each GroupBy object will contain information about the Candidate with the highest voter %.

elections_sorted_by_percent = elections.sort_values("%", ascending=False)
elections_sorted_by_percent.head(5)
Year Candidate Party Popular vote Result %
114 1964 Lyndon Johnson Democratic 43127041 win 61.344703
91 1936 Franklin Roosevelt Democratic 27752648 win 60.978107
120 1972 Richard Nixon Republican 47168710 win 60.907806
79 1920 Warren Harding Republican 16144093 win 60.574501
133 1984 Ronald Reagan Republican 54455472 win 59.023326
elections_sorted_by_percent.groupby("Party").agg(lambda x : x.iloc[0]).head(10)

# Equivalent to the below code
# elections_sorted_by_percent.groupby("Party").agg('first').head(10)
Year Candidate Popular vote Result %
Party
American 1856 Millard Fillmore 873053 loss 21.554001
American Independent 1968 George Wallace 9901118 loss 13.571218
Anti-Masonic 1832 William Wirt 100715 loss 7.821583
Anti-Monopoly 1884 Benjamin Butler 134294 loss 1.335838
Citizens 1980 Barry Commoner 233052 loss 0.270182
Communist 1932 William Z. Foster 103307 loss 0.261069
Constitution 2008 Chuck Baldwin 199750 loss 0.152398
Constitutional Union 1860 John Bell 590901 loss 12.639283
Democratic 1964 Lyndon Johnson 43127041 win 61.344703
Democratic-Republican 1824 Andrew Jackson 151271 loss 57.210122

Here’s an illustration of the process:

groupby_demo

Notice how our code correctly determines that Lyndon Johnson from the Democratic Party has the highest voter %.

More generally, lambda functions are used to design custom aggregation functions that aren’t pre-defined by Python. The input parameter x to the lambda function is a GroupBy object. Therefore, it should make sense why lambda x : x.iloc[0] selects the first row in each groupby object.

In fact, there’s a few different ways to approach this problem. Each approach has different tradeoffs in terms of readability, performance, memory consumption, complexity, etc. We’ve given a few examples below.

Note: Understanding these alternative solutions is not required. They are given to demonstrate the vast number of problem-solving approaches in pandas.

# Using the idxmax function
best_per_party = elections.loc[elections.groupby('Party')['%'].idxmax()]
best_per_party.head(5)
Year Candidate Party Popular vote Result %
22 1856 Millard Fillmore American 873053 loss 21.554001
115 1968 George Wallace American Independent 9901118 loss 13.571218
6 1832 William Wirt Anti-Masonic 100715 loss 7.821583
38 1884 Benjamin Butler Anti-Monopoly 134294 loss 1.335838
127 1980 Barry Commoner Citizens 233052 loss 0.270182
# Using the .drop_duplicates function
best_per_party2 = elections.sort_values('%').drop_duplicates(['Party'], keep='last')
best_per_party2.head(5)
Year Candidate Party Popular vote Result %
148 1996 John Hagelin Natural Law 113670 loss 0.118219
164 2008 Chuck Baldwin Constitution 199750 loss 0.152398
110 1956 T. Coleman Andrews States' Rights 107929 loss 0.174883
147 1996 Howard Phillips Taxpayers 184656 loss 0.192045
136 1988 Lenora Fulani New Alliance 217221 loss 0.237804

4.4 Aggregating Data with Pivot Tables

We know now that .groupby gives us the ability to group and aggregate data across our DataFrame. The examples above formed groups using just one column in the DataFrame. It’s possible to group by multiple columns at once by passing in a list of column names to .groupby.

Let’s consider the babynames dataset again. In this problem, we will find the total number of baby names associated with each sex for each year. To do this, we’ll group by both the "Year" and "Sex" columns.

babynames.head()
State Sex Year Name Count First Letter
115957 CA F 1990 Deandrea 5 D
101976 CA F 1986 Deandrea 6 D
131029 CA F 1994 Leandrea 5 L
108731 CA F 1988 Deandrea 5 D
308131 CA M 1985 Deandrea 6 D
# Find the total number of baby names associated with each sex for each 
# year in the data
babynames.groupby(["Year", "Sex"])[["Count"]].agg(sum).head(6)
Count
Year Sex
1910 F 5950
M 3213
1911 F 6602
M 3381
1912 F 9804
M 8142

Notice that both "Year" and "Sex" serve as the index of the DataFrame (they are both rendered in bold). We’ve created a multi-index DataFrame where two different index values, the year and sex, are used to uniquely identify each row.

This isn’t the most intuitive way of representing this data – and, because multi-indexed DataFrames have multiple dimensions in their index, they can often be difficult to use.

Another strategy to aggregate across two columns is to create a pivot table. You saw these back in Data 8. One set of values is used to create the index of the pivot table; another set is used to define the column names. The values contained in each cell of the table correspond to the aggregated data for each index-column pair.

Here’s an illustration of the process:

groupby_demo

The best way to understand pivot tables is to see one in action. Let’s return to our original goal of summing the total number of names associated with each combination of year and sex. We’ll call the pandas .pivot_table method to create a new table.

# The `pivot_table` method is used to generate a Pandas pivot table
import numpy as np
babynames.pivot_table(
    index = "Year",
    columns = "Sex",    
    values = "Count", 
    aggfunc = np.sum, 
).head(5)
Sex F M
Year
1910 5950 3213
1911 6602 3381
1912 9804 8142
1913 11860 10234
1914 13815 13111

Looks a lot better! Now, our DataFrame is structured with clear index-column combinations. Each entry in the pivot table represents the summed count of names for a given combination of "Year" and "Sex".

Let’s take a closer look at the code implemented above.

  • index = "Year" specifies the column name in the original DataFrame that should be used as the index of the pivot table
  • columns = "Sex" specifies the column name in the original DataFrame that should be used to generate the columns of the pivot table
  • values = "Count" indicates what values from the original DataFrame should be used to populate the entry for each index-column combination
  • aggfunc = np.sum tells pandas what function to use when aggregating the data specified by values. Here, we are summing the name counts for each pair of "Year" and "Sex"

We can even include multiple values in the index or columns of our pivot tables.

babynames_pivot = babynames.pivot_table(
    index="Year",     # the rows (turned into index)
    columns="Sex",    # the column values
    values=["Count", "Name"], 
    aggfunc=max,      # group operation
)
babynames_pivot.head(6)
Count Name
Sex F M F M
Year
1910 295 237 Yvonne William
1911 390 214 Zelma Willis
1912 534 501 Yvonne Woodrow
1913 584 614 Zelma Yoshio
1914 773 769 Zelma Yoshio
1915 998 1033 Zita Yukio

Note that each row provides the number of girls and number of boys having that year’s most common name, and also lists the alphabetically largest girl name and boy name. The counts for number of girls/boys in the resulting DataFrame do not correspond to the names listed. For example, in 1910, the most popular girl name is given to 295 girls, but that name was likely not Yvonne.

4.5 Joining Tables

When working on data science projects, we’re unlikely to have absolutely all the data we want contained in a single DataFrame – a real-world data scientist needs to grapple with data coming from multiple sources. If we have access to multiple datasets with related information, we can join two or more tables into a single DataFrame.

To put this into practice, we’ll revisit the elections dataset.

elections.head(5)
Year Candidate Party Popular vote Result %
0 1824 Andrew Jackson Democratic-Republican 151271 loss 57.210122
1 1824 John Quincy Adams Democratic-Republican 113142 win 42.789878
2 1828 Andrew Jackson Democratic 642806 win 56.203927
3 1828 John Quincy Adams National Republican 500897 loss 43.796073
4 1832 Andrew Jackson Democratic 702735 win 54.574789

Say we want to understand the popularity of the names of each presidential candidate in 2022. To do this, we’ll need the combined data of babynames and elections.

We’ll start by creating a new column containing the first name of each presidential candidate. This will help us join each name in elections to the corresponding name data in babynames.

# This `str` operation splits each candidate's full name at each 
# blank space, then takes just the candidate's first name
elections["First Name"] = elections["Candidate"].str.split().str[0]
elections.head(5)
Year Candidate Party Popular vote Result % First Name
0 1824 Andrew Jackson Democratic-Republican 151271 loss 57.210122 Andrew
1 1824 John Quincy Adams Democratic-Republican 113142 win 42.789878 John
2 1828 Andrew Jackson Democratic 642806 win 56.203927 Andrew
3 1828 John Quincy Adams National Republican 500897 loss 43.796073 John
4 1832 Andrew Jackson Democratic 702735 win 54.574789 Andrew
# Here, we'll only consider `babynames` data from 2022
babynames_2022 = babynames[babynames["Year"]==2022]
babynames_2022.head()
State Sex Year Name Count First Letter
237964 CA F 2022 Leandra 10 L
404916 CA M 2022 Leandro 99 L
405892 CA M 2022 Andreas 14 A
235927 CA F 2022 Andrea 322 A
405695 CA M 2022 Deandre 18 D

Now, we’re ready to join the two tables. pd.merge is the pandas method used to join DataFrames together.

merged = pd.merge(left = elections, right = babynames_2022, \
                  left_on = "First Name", right_on = "Name")
merged.head()
# Notice that pandas automatically specifies `Year_x` and `Year_y` 
# when both merged DataFrames have the same column name to avoid confusion

# Second option
# merged = elections.merge(right = babynames_2022, \
    # left_on = "First Name", right_on = "Name")
Year_x Candidate Party Popular vote Result % First Name State Sex Year_y Name Count First Letter
0 1824 Andrew Jackson Democratic-Republican 151271 loss 57.210122 Andrew CA M 2022 Andrew 741 A
1 1828 Andrew Jackson Democratic 642806 win 56.203927 Andrew CA M 2022 Andrew 741 A
2 1832 Andrew Jackson Democratic 702735 win 54.574789 Andrew CA M 2022 Andrew 741 A
3 1824 John Quincy Adams Democratic-Republican 113142 win 42.789878 John CA M 2022 John 490 J
4 1828 John Quincy Adams National Republican 500897 loss 43.796073 John CA M 2022 John 490 J

Let’s take a closer look at the parameters:

  • left and right parameters are used to specify the DataFrames to be joined.
  • left_on and right_on parameters are assigned to the string names of the columns to be used when performing the join. These two on parameters tell pandas what values should act as pairing keys to determine which rows to merge across the DataFrames. We’ll talk more about this idea of a pairing key next lecture.

4.6 Parting Note

Congratulations! We finally tackled pandas. Don’t worry if you are still not feeling very comfortable with it—you will have plenty of chances to practice over the next few weeks.

Next, we will get our hands dirty with some real-world datasets and use our pandas knowledge to conduct some exploratory data analysis.