Introduction to Pandas, Part 2¶

Advanced Pandas syntax, aggregation, and joining.

In [1]:
import numpy as np
import pandas as pd

Let's start by loading the California baby names again.

In [2]:
import urllib.request
import os.path
import zipfile

data_url = "https://www.ssa.gov/oact/babynames/state/namesbystate.zip"
local_filename = "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 = '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.sample(5)
Out[2]:
State Sex Year Name Count
179400 CA F 2007 Anjelina 8
129290 CA F 1994 Dejanae 13
177087 CA F 2007 Mayra 119
375561 CA M 2014 Jaxxon 11
85174 CA F 1980 Sunnie 7
In [68]:
babynames.query('Sex == "M" and Year == 2020').sort_values("Count", ascending = False)
Out[68]:
State Sex Year Name Count
391409 CA M 2020 Noah 2608
391410 CA M 2020 Liam 2406
391411 CA M 2020 Mateo 2057
391412 CA M 2020 Sebastian 1981
391413 CA M 2020 Julian 1679
... ... ... ... ... ...
393938 CA M 2020 Gavino 5
393937 CA M 2020 Gaspar 5
393936 CA M 2020 Gannon 5
393935 CA M 2020 Galen 5
394178 CA M 2020 Zymir 5

2770 rows × 5 columns

In [69]:
babynames.query('Sex == "M" and Year == 2020').sort_values("Name", ascending = False)
Out[69]:
State Sex Year Name Count
393226 CA M 2020 Zyon 9
394178 CA M 2020 Zymir 5
393352 CA M 2020 Zyan 8
392118 CA M 2020 Zyaire 37
392838 CA M 2020 Zyair 13
... ... ... ... ... ...
393106 CA M 2020 Aamir 9
393822 CA M 2020 Aalam 5
393354 CA M 2020 Aaditya 7
393353 CA M 2020 Aadi 7
392994 CA M 2020 Aaden 10

2770 rows × 5 columns

In [77]:
(
babynames.query('Sex == "M" and Year == 2020')
         .sort_values("Name", key = lambda x: x.str.len(),
                      ascending = False)
)
Out[77]:
State Sex Year Name Count
393478 CA M 2020 Michaelangelo 7
393079 CA M 2020 Michelangelo 10
392964 CA M 2020 Maximilliano 11
394047 CA M 2020 Maxemiliano 5
392610 CA M 2020 Maximillian 16
... ... ... ... ... ...
393110 CA M 2020 Aj 9
392856 CA M 2020 Cy 12
393558 CA M 2020 An 6
393981 CA M 2020 Jj 5
392750 CA M 2020 Om 14

2770 rows × 5 columns

An alternate approach is to create a temporary column corresponding to the length

In [79]:
#create a new series of only the lengths
babyname_lengths = babynames["Name"].str.len()

#add that series to the dataframe as a column
babynames["name_lengths"] = babyname_lengths
babynames.head(5)
Out[79]:
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
In [80]:
#sort by the temporary column
babynames = babynames.sort_values(by = "name_lengths", ascending=False)
babynames.head(25)
Out[80]:
State Sex Year Name Count name_lengths
332035 CA M 1998 Franciscojavier 6 15
329864 CA M 1997 Franciscojavier 5 15
320036 CA M 1993 Ryanchristopher 5 15
314358 CA M 1991 Ryanchristopher 7 15
319923 CA M 1993 Johnchristopher 5 15
337003 CA M 2000 Franciscojavier 6 15
305111 CA M 1987 Franciscojavier 5 15
330040 CA M 1997 Ryanchristopher 5 15
306544 CA M 1988 Franciscojavier 10 15
309451 CA M 1989 Franciscojavier 6 15
326729 CA M 1996 Franciscojavier 8 15
102490 CA F 1986 Mariadelosangel 5 15
314479 CA M 1991 Franciscojavier 6 15
119890 CA F 1991 Mariadelcarmen 5 14
119904 CA F 1991 Meganelizabeth 5 14
130606 CA F 1994 Mayraalejandra 6 14
322064 CA M 1994 Antoniodejesus 6 14
331464 CA M 1998 Michaelanthony 10 14
73527 CA F 1975 Mariadelcarmen 5 14
209392 CA F 2014 Mariaguadalupe 5 14
133016 CA F 1995 Mariaguadalupe 12 14
166772 CA F 2004 Mariaguadalupe 9 14
318793 CA M 1993 Michaelanthony 12 14
101246 CA F 1986 Mariaguadalupe 11 14
102508 CA F 1986 Mayraalejandra 5 14
In [81]:
#drop the temporary column
babynames = babynames.drop("name_lengths", axis = 'columns')
babynames.head(5)
Out[81]:
State Sex Year Name Count
332035 CA M 1998 Franciscojavier 6
329864 CA M 1997 Franciscojavier 5
320036 CA M 1993 Ryanchristopher 5
314358 CA M 1991 Ryanchristopher 7
319923 CA M 1993 Johnchristopher 5

We can also use the Python map function if we want to use an arbitrarily defined function. Suppose we want to sort by the number of occurrences of "dr" plus the number of occurences of "ea".

In [88]:
def dr_ea_count(string):
    return string.count('dr') + string.count('ea')

#create the temporary column
babynames["dr_ea_count"] = babynames["Name"].map(dr_ea_count)

#sort by the temporary column
babynames = babynames.sort_values(by = "dr_ea_count", ascending=False)
babynames.head()
Out[88]:
State Sex Year Name Count dr_ea_count
131013 CA F 1994 Leandrea 5 3
300700 CA M 1985 Deandrea 6 3
108715 CA F 1988 Deandrea 5 3
101962 CA F 1986 Deandrea 6 3
115942 CA F 1990 Deandrea 5 3
In [ ]:
def dr_ea_count(string):
    return string.count('dr') + string.count('ea')

babynames["dr_ea_count"] = babynames["Name"].map(dr_ea_count)
babynames = babynames.sort_values(by = "dr_ea_count", ascending=False)
In [89]:
#drop that column
babynames = babynames.drop("dr_ea_count", axis = 'columns')
babynames.head(5)
Out[89]:
State Sex Year Name Count
131013 CA F 1994 Leandrea 5
300700 CA M 1985 Deandrea 6
108715 CA F 1988 Deandrea 5
101962 CA F 1986 Deandrea 6
115942 CA F 1990 Deandrea 5

Female Name whose popularity has dropped the most.¶

In this exercise, let's find the female name whose popularity has dropped the most since its peak. As an example of a name that has fallen into disfavor, consider "Jennifer", visualized below.

Note: We won't cover plotly in lecture until week 4 after Lisa covers EDA and Regex.

In [6]:
fig = px.line(babynames.query("Name == 'Jennifer' and Sex == 'F'"), x = "Year", y = "Count")
fig.update_layout(font_size = 18)

To answer this question, we'll need a mathematical definition for the change in popularity of a name.

For the purposes of lecture, let’s use the RTP or ratio_to_peak. This is the current count of the name divded by its maximum ever count.

Getting the max Jennifer is easy enough.

In [7]:
max_jennifers = max(babynames.query("Name == 'Jennifer' and Sex == 'F'")["Count"])
max_jennifers
Out[7]:
6064

And we can get the most recent Jennifer count with iloc[-1] (TODO: Maybe make this an exercise)

In [8]:
current_jennifers = babynames.query("Name == 'Jennifer' and Sex == 'F'")["Count"].iloc[-1]
current_jennifers
Out[8]:
141
In [9]:
current_jennifers / max_jennifers
Out[9]:
0.023251978891820582

We can also write a function that produces the ratio_to_peak for a given series.

Here for clarity, let's regenerate the jennifer_counts Series, but let's do so on a DataFrame where the index is the year.

In [10]:
def ratio_to_peak(series):
    return series.iloc[-1] / max(series)
In [11]:
jennifer_counts_series = babynames.query("Name == 'Jennifer' and Sex == 'F'")["Count"]
ratio_to_peak(jennifer_counts_series)
Out[11]:
0.023251978891820582

We can try out various names below:

In [12]:
ratio_to_peak(babynames.query("Name == 'Jessica' and Sex == 'F'")["Count"])
Out[12]:
0.023162134944612285

Since we're only working with female names, let's create a DataFrame with only female names to simplify our later code.

In [13]:
female_babynames = babynames[babynames["Sex"] == "F"] #could have also used query but I want to use various types of syntax to increase your exposure to other ways of writing code

Approach 1: Naive For Loop¶

As a first approach, we can try to use a for loop.

In [14]:
#build dictionary where each entry is the rtp for a given name
#e.g. rtps["jennifer"] should be 0.0231
#rtps = {}
#for name in ??:
#    counts_of_current_name = female_babynames[??]["Count"]
#    rtps[name] = ratio_to_peak(counts_of_current_name)
    
#convert to series
#rtps = pd.Series(rtps) 
#rtps

Answer below. Note that we only used the first 100 names because otherwise the code takes ages to complete running.

In [15]:
#build dictionary where entry i is the ammd for the given name
#e.g. rtps["jennifer"] should be 0.0231
rtps = {}
for name in babynames["Name"].unique()[0:100]:
    counts_of_current_name = female_babynames[female_babynames["Name"] == name]["Count"]
    rtps[name] = ratio_to_peak(counts_of_current_name)
    
#convert to series
rtps = pd.Series(rtps) 
rtps
Out[15]:
Mary        0.030670
Helen       0.130488
Dorothy     0.028133
Margaret    0.093724
Frances     0.134216
              ...   
Lorraine    0.035885
Madeline    0.529617
Maxine      0.570312
Bessie      0.160714
Georgia     0.761628
Length: 100, dtype: float64
In [16]:
rtps.sort_values()
Out[16]:
Patricia     0.005640
Kathleen     0.009516
Barbara      0.012963
Betty        0.014379
Ann          0.021346
               ...   
Genevieve    0.920705
Clara        0.922131
Hazel        0.940171
Eleanor      1.000000
Lena         1.000000
Length: 100, dtype: float64

Approach 2: Use groupby.agg¶

Instead, we can use the very powerful groupby.agg operation, which allows us to simply and efficiently compute what we want.

In [17]:
female_babynames.head()
Out[17]:
State Sex Year Name Count
0 CA F 1910 Mary 295
1 CA F 1910 Helen 239
2 CA F 1910 Dorothy 220
3 CA F 1910 Margaret 163
4 CA F 1910 Frances 134
In [18]:
female_babynames
Out[18]:
State Sex Year Name Count
0 CA F 1910 Mary 295
1 CA F 1910 Helen 239
2 CA F 1910 Dorothy 220
3 CA F 1910 Margaret 163
4 CA F 1910 Frances 134
... ... ... ... ... ...
232097 CA F 2020 Ziana 5
232098 CA F 2020 Zoha 5
232099 CA F 2020 Zuleika 5
232100 CA F 2020 Zuriel 5
232101 CA F 2020 Zyrah 5

232102 rows × 5 columns

In [19]:
rtp_table = female_babynames.groupby("Name").agg(ratio_to_peak)
rtp_table

#Note: If this cell crashes, comment out the code and use the female_babynames.groupby("Name")[["Count"]].agg(ratio_to_peak) instead

/opt/conda/lib/python3.9/site-packages/pandas/core/groupby/generic.py:303: FutureWarning:

Dropping invalid columns in SeriesGroupBy.agg is deprecated. In a future version, a TypeError will be raised. Before calling .agg, select only columns which should be valid for the aggregating function.
Out[19]:
Year Count
Name
Aadhira 1.0 0.600000
Aadhya 1.0 0.720000
Aadya 1.0 0.862069
Aahana 1.0 0.384615
Aahna 1.0 1.000000
... ... ...
Zyanya 1.0 0.571429
Zyla 1.0 1.000000
Zylah 1.0 1.000000
Zyra 1.0 1.000000
Zyrah 1.0 0.833333

13525 rows × 2 columns

This is simply the equivalent of http://data8.org/datascience/_autosummary/datascience.tables.Table.group.html from Data8, e.g. if babynames were using Table, our code would read:

female_babynames.group("Name", ratio_to_peak)

For a visual review of groupby, see this lecture slide.

If you're using an early enough version of pandas, the code above will not crash, and will automatically drop columns for which ratio_to_peak fails, e.g. the Sex column.

However, according to a warning message that is generated as of January 2022, at some point this code will no longer be considered valid Pandas code, and the code will crash on columns for which the aggregation function is undefined.

Whether we're trying to avoid a crash, or just want a clean DataFrame, let's explicitly select only the Count column. The idea here is that we don't really care about the meaningless Year column, so we may as well exclude it when we compute our ratio_to_peak values.

In [20]:
rtp_table = female_babynames.groupby("Name")[["Count"]].agg(ratio_to_peak)
rtp_table
Out[20]:
Count
Name
Aadhira 0.600000
Aadhya 0.720000
Aadya 0.862069
Aahana 0.384615
Aahna 1.000000
... ...
Zyanya 0.571429
Zyla 1.000000
Zylah 1.000000
Zyra 1.000000
Zyrah 0.833333

13525 rows × 1 columns

In [21]:
# this code renames the Count column to RTP. You'll see this syntax in lab
rtp_table = rtp_table.rename(columns = {"Count": "Count RTP"})
rtp_table
Out[21]:
Count RTP
Name
Aadhira 0.600000
Aadhya 0.720000
Aadya 0.862069
Aahana 0.384615
Aahna 1.000000
... ...
Zyanya 0.571429
Zyla 1.000000
Zylah 1.000000
Zyra 1.000000
Zyrah 0.833333

13525 rows × 1 columns

In [22]:
rtp_table.sort_values("Count RTP")
Out[22]:
Count RTP
Name
Debra 0.001260
Debbie 0.002817
Susan 0.004320
Kim 0.004405
Tammy 0.004551
... ...
Natalynn 1.000000
Francella 1.000000
Natalye 1.000000
Natassja 1.000000
Kaylia 1.000000

13525 rows × 1 columns

In [23]:
fig = px.line(babynames.query("Name == 'Debra' and Sex == 'F'"), x = "Year", y = "Count")
fig.update_layout(font_size = 16)
In [24]:
top10 = rtp_table.sort_values("Count RTP").head(10).index
top10
Out[24]:
Index(['Debra', 'Debbie', 'Susan', 'Kim', 'Tammy', 'Terri', 'Shannon',
       'Cheryl', 'Lisa', 'Patricia'],
      dtype='object', name='Name')
In [25]:
fig = px.line(babynames.query("Name in @top10 and Sex == 'F'"), x = "Year", y = "Count", color = "Name")
fig.update_layout(font_size = 13)

Some Additional Groupby Puzzles¶

Groupby puzzle #1: Try to create a groupby.agg call that gives the total babies born with each name.

In [26]:
puzzle1 = female_babynames.groupby("Name")[["Count"]].agg(sum)
puzzle1
Out[26]:
Count
Name
Aadhira 22
Aadhya 368
Aadya 230
Aahana 129
Aahna 7
... ...
Zyanya 174
Zyla 107
Zylah 95
Zyra 71
Zyrah 21

13525 rows × 1 columns

Groupby puzzle #2: Try to create a groupby.agg call that gives total babies born in each year.

In [27]:
puzzle2 = female_babynames.groupby("Year")[["Count"]].agg(sum)
puzzle2
Out[27]:
Count
Year
1910 5950
1911 6602
1912 9804
1913 11860
1914 13815
... ...
2016 203360
2017 195352
2018 188919
2019 183644
2020 172003

111 rows × 1 columns

We can write this using a groupby shorthand aggregation method. Here, sum() is shorthand for groupby.agg(sum).

In [28]:
puzzle2 = female_babynames.groupby("Year")[["Count"]].sum()
In [29]:
fig = px.line(puzzle2, y = "Count")
fig.update_layout(font_size = 15)
In [30]:
#not covered in lecture but we can also compute relative births
relative_births = puzzle2 / max(puzzle2["Count"])
In [31]:
fig = px.line(relative_births, y = "Count")
fig.update_layout(font_size = 15)

groupby Puzzle #4¶

In [32]:
elections = pd.read_csv("elections.csv")
elections.sample(5)
Out[32]:
Year Candidate Party Popular vote Result %
154 2000 Pat Buchanan Reform 448895 loss 0.426819
168 2012 Barack Obama Democratic 65915795 win 51.258484
83 1928 Al Smith Democratic 15015464 loss 40.902853
57 1904 Alton B. Parker Democratic 5083880 loss 37.685116
49 1892 John Bidwell Prohibition 270879 loss 2.249468

We have to be careful when using aggregation functions. For example, the code below might be misinterpreted to say that Woodrow Wilson successfully ran for election in 2020. Why is this happening?

In [33]:
elections.groupby("Party").agg(max).head(10)
Out[33]:
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

Quick Subpuzzle¶

Inspired by above, try to predict the results of the groupby operation shown. The answer is below the image.

groupby_puzzle.png

The top ?? will be "hi", the second ?? will be "tx", and the third ?? will be "sd".

In [34]:
ds = pd.DataFrame(dict(x=[3,1,4,1,5,9,2,5,6], 
                      y=['ak', 'tx', 'fl', 'hi', 'mi', 'ak', 'ca', 'sd', 'nc']), 
                      index=list('ABCABCACB') )
ds
Out[34]:
x y
A 3 ak
B 1 tx
C 4 fl
A 1 hi
B 5 mi
C 9 ak
A 2 ca
C 5 sd
B 6 nc
In [35]:
ds.groupby(ds.index).agg(max)
Out[35]:
x y
A 3 hi
B 6 tx
C 9 sd

Completing groupby puzzle #4¶

Next we'll write code that properly returns the best result by each party. That is, each row should show the Year, Candidate, Popular Vote, Result, and % for the election in which that party saw its best results (rather than mixing them as in the example above), here's what the first rows of the correct output should look like:

parties.png

In [36]:
elections_sorted_by_percent = elections.sort_values("%", ascending=False)
elections_sorted_by_percent.head(5)
Out[36]:
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
In [37]:
elections_sorted_by_percent = elections.sort_values("%", ascending=False)
elections_sorted_by_percent.groupby("Party").agg(lambda x : x.iloc[0]).head(9)
Out[37]:
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
In [38]:
elections_sorted_by_percent = elections.sort_values("%", ascending=False)
elections_sorted_by_percent.groupby("Party").agg(lambda x : x.iloc[0]).head(9)
Out[38]:
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

You'll soon discover that with Pandas' rich tool set, there's typically more than one way to get to the same answer. Each approach has different tradeoffs in terms of readability, performance, memory consumption, complexity and more. It will take some experience for you to develop a sense of which approach is better for each problem, but you should in general try to think if you can at least envision a different solution to a given problem, especially if you find your current solution to be particularly convoluted or hard to read.

Here's a couple of other ways of obtaining the same result (in each case we only show the top part with head()). The first approach uses groupby but finds the location of the maximum value via the idxmax() method (look up its documentation!). We then index and sort by party to match the requested formatting:

In [39]:
elections.head()
Out[39]:
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
In [40]:
elections.groupby('Party')['%'].idxmax()
Out[40]:
Party
American                  22
American Independent     115
Anti-Masonic               6
Anti-Monopoly             38
Citizens                 127
Communist                 89
Constitution             164
Constitutional Union      24
Democratic               114
Democratic-Republican      0
Dixiecrat                103
Farmer–Labor              78
Free Soil                 15
Green                    155
Greenback                 35
Independent              143
Liberal Republican        31
Libertarian              175
National Democratic       50
National Republican        3
National Union            27
Natural Law              148
New Alliance             136
Northern Democratic       26
Populist                  48
Progressive               68
Prohibition               49
Reform                   150
Republican               120
Socialist                 66
Southern Democratic       25
States' Rights           110
Taxpayers                147
Union                     93
Union Labor               42
Whig                      11
Name: %, dtype: int64
In [41]:
best_per_party = elections.loc[elections.groupby('Party')['%'].idxmax()]  # This is the computational part
best_per_party.set_index('Party').sort_index().head()  # This indexes by Party to match the formatting above
Out[41]:
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

And this one doesn't even use groupby! This approach instead uses the drop_duplicates method to keep only the last occurrence of of each party after having sorted by "%", which is the best performance. Again, the 2nd line is purely formatting:

In [42]:
best_per_party2 = elections.sort_values('%').drop_duplicates(['Party'], keep='last')
best_per_party2.set_index('Party').sort_index().head()  # Formatting
Out[42]:
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

Challenge: see if you can find yet another approach that still gives the same answer.

Other DataFrameGroupBy Features¶

The result of groupby is not a DataFrame or a list of DataFrames. It is instead a special type called a DataFrameGroupBy.

In [43]:
grouped_by_year = babynames.groupby("Year")
type(grouped_by_year)
Out[43]:
pandas.core.groupby.generic.DataFrameGroupBy

groupby.size()¶

In [44]:
elections.groupby("Party")
Out[44]:
<pandas.core.groupby.generic.DataFrameGroupBy object at 0x7f44658c9d30>
In [45]:
#size returns a Series giving the size of each group
elections.groupby("Party").size().head(15)
Out[45]:
Party
American                  2
American Independent      3
Anti-Masonic              1
Anti-Monopoly             1
Citizens                  1
Communist                 1
Constitution              3
Constitutional Union      1
Democratic               47
Democratic-Republican     2
Dixiecrat                 1
Farmer–Labor              1
Free Soil                 2
Green                     7
Greenback                 1
dtype: int64

groupby.filter()¶

In [46]:
# filter gives a copy of the original DataFrame where row r is included
# if its group obeys the given condition
#
# Note: Filtering is done per GROUP, not per ROW.
elections.groupby("Year").filter(lambda sf: sf["%"].max() < 45)
Out[46]:
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
115 1968 George Wallace American Independent 9901118 loss 13.571218
116 1968 Hubert Humphrey Democratic 31271839 loss 42.863537
117 1968 Richard Nixon Republican 31783783 win 43.565246
139 1992 Andre Marrou Libertarian 290087 loss 0.278516
140 1992 Bill Clinton Democratic 44909806 win 43.118485
141 1992 Bo Gritz Populist 106152 loss 0.101918
142 1992 George H. W. Bush Republican 39104550 loss 37.544784
143 1992 Ross Perot Independent 19743821 loss 18.956298
In [49]:
# the code below lets us peek into the groups and see why they were rejected or not
for i, (n, g) in enumerate(elections.groupby("Party")):
    print(n)
    display(g)
    if i>3: break

American
Year Candidate Party Popular vote Result %
22 1856 Millard Fillmore American 873053 loss 21.554001
126 1976 Thomas J. Anderson American 158271 loss 0.194862 
American Independent
Year Candidate Party Popular vote Result %
115 1968 George Wallace American Independent 9901118 loss 13.571218
119 1972 John G. Schmitz American Independent 1100868 loss 1.421524
124 1976 Lester Maddox American Independent 170274 loss 0.209640 
Anti-Masonic
Year Candidate Party Popular vote Result %
6 1832 William Wirt Anti-Masonic 100715 loss 7.821583 
Anti-Monopoly
Year Candidate Party Popular vote Result %
38 1884 Benjamin Butler Anti-Monopoly 134294 loss 1.335838 
Citizens
Year Candidate Party Popular vote Result %
127 1980 Barry Commoner Citizens 233052 loss 0.270182

Puzzle 5: Finding the number of babies born in each year of each sex.¶

Earlier we saw how to add up the total number of babies born in each year, but what if we want the number born in each and year for each sex separately?

In [57]:
babynames.groupby("Year").sum()
Out[57]:
Count
Year
1910 9163
1911 9983
1912 17946
1913 22094
1914 26926
... ...
2016 426658
2017 410614
2018 394826
2019 385777
2020 359229

111 rows × 1 columns

It is possible to group a DataFrame by multiple features. For example, if we group by Year and Sex we get back a DataFrame with the total number of babies of each sex born in each year.

In [54]:
babynames.head(10)
Out[54]:
State Sex Year Name Count
0 CA F 1910 Mary 295
1 CA F 1910 Helen 239
2 CA F 1910 Dorothy 220
3 CA F 1910 Margaret 163
4 CA F 1910 Frances 134
5 CA F 1910 Ruth 128
6 CA F 1910 Evelyn 126
7 CA F 1910 Alice 118
8 CA F 1910 Virginia 101
9 CA F 1910 Elizabeth 93
In [50]:
babynames.groupby(["Year", "Sex"]).agg(sum).head(6)
Out[50]:
Count
Year Sex
1910 F 5950
M 3213
1911 F 6602
M 3381
1912 F 9804
M 8142
In [51]:
babynames.groupby(["Sex", "Year"]).agg(sum).head(6)
Out[51]:
Count
Sex Year
F 1910 5950
1911 6602
1912 9804
1913 11860
1914 13815
1915 18643

The DataFrame resulting from an aggregation operation on a table grouped by a list of columns is multi-indexed. That is, it has more than one dimension to its index. We will explore this in a future lecture.

A more natural approach is to use a pivot table (like we saw in data 8).

In [58]:
babynames.head(5)
Out[58]:
State Sex Year Name Count
0 CA F 1910 Mary 295
1 CA F 1910 Helen 239
2 CA F 1910 Dorothy 220
3 CA F 1910 Margaret 163
4 CA F 1910 Frances 134
In [65]:
babynames_pivot = babynames.pivot_table(
    index='Year',     # the rows (turned into index)
    columns='Sex',    # the column values
    values=['Count'], # the field(s) to processed in each group
    aggfunc=np.sum,   # group operation
)
babynames_pivot.head(6)
Out[65]:
Count
Sex F M
Year
1910 5950 3213
1911 6602 3381
1912 9804 8142
1913 11860 10234
1914 13815 13111
1915 18643 17192
In [62]:
babynames_pivot = babynames.pivot_table(
    index='Year',     # the rows (turned into index)
    columns='Sex',    # the column values
    values=['Count', 'Name'], 
    aggfunc=np.max,   # group operation
)
babynames_pivot.head(6)
Out[62]:
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

The basic idea behind pivot tables is shown in the image below.

pivot_picture.png

Merging Tables¶

In [91]:
elections
Out[91]:
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
... ... ... ... ... ... ...
177 2016 Jill Stein Green 1457226 loss 1.073699
178 2020 Joseph Biden Democratic 81268924 win 51.311515
179 2020 Donald Trump Republican 74216154 loss 46.858542
180 2020 Jo Jorgensen Libertarian 1865724 loss 1.177979
181 2020 Howard Hawkins Green 405035 loss 0.255731

182 rows × 6 columns

In [107]:
male_2020_babynames = babynames.query('Sex == "M" and Year == 2020')
male_2020_babynames
Out[107]:
State Sex Year Name Count
392447 CA M 2020 Deandre 19
394024 CA M 2020 Leandre 5
392438 CA M 2020 Andreas 19
391863 CA M 2020 Leandro 72
392562 CA M 2020 Rudra 17
... ... ... ... ... ...
393418 CA M 2020 Demetri 7
393881 CA M 2020 Casimir 5
393393 CA M 2020 Bridger 7
393391 CA M 2020 Brayson 7
393392 CA M 2020 Brecken 7

2770 rows × 5 columns

In [96]:
elections["First Name"] = elections["Candidate"].str.split().str[0]
In [97]:
elections
Out[97]:
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
... ... ... ... ... ... ... ...
177 2016 Jill Stein Green 1457226 loss 1.073699 Jill
178 2020 Joseph Biden Democratic 81268924 win 51.311515 Joseph
179 2020 Donald Trump Republican 74216154 loss 46.858542 Donald
180 2020 Jo Jorgensen Libertarian 1865724 loss 1.177979 Jo
181 2020 Howard Hawkins Green 405035 loss 0.255731 Howard

182 rows × 7 columns

In [105]:
merged = pd.merge(left = elections, right = male_2020_babynames, 
                  left_on = "First Name", right_on = "Name")
merged
Out[105]:
Year_x Candidate Party Popular vote Result % First Name State Sex Year_y Name Count
0 1824 Andrew Jackson Democratic-Republican 151271 loss 57.210122 Andrew CA M 2020 Andrew 867
1 1828 Andrew Jackson Democratic 642806 win 56.203927 Andrew CA M 2020 Andrew 867
2 1832 Andrew Jackson Democratic 702735 win 54.574789 Andrew CA M 2020 Andrew 867
3 1824 John Quincy Adams Democratic-Republican 113142 win 42.789878 John CA M 2020 John 617
4 1828 John Quincy Adams National Republican 500897 loss 43.796073 John CA M 2020 John 617
... ... ... ... ... ... ... ... ... ... ... ... ...
141 2016 Darrell Castle Constitution 203091 loss 0.149640 Darrell CA M 2020 Darrell 10
142 2016 Donald Trump Republican 62984828 win 46.407862 Donald CA M 2020 Donald 39
143 2020 Donald Trump Republican 74216154 loss 46.858542 Donald CA M 2020 Donald 39
144 2016 Evan McMullin Independent 732273 loss 0.539546 Evan CA M 2020 Evan 595
145 2020 Joseph Biden Democratic 81268924 win 51.311515 Joseph CA M 2020 Joseph 989

146 rows × 12 columns

In [106]:
merged.sort_values("Count")
Out[106]:
Year_x Candidate Party Popular vote Result % First Name State Sex Year_y Name Count
129 1996 Bill Clinton Democratic 47400125 win 49.296938 Bill CA M 2020 Bill 5
106 1956 Dwight Eisenhower Republican 35579180 win 57.650654 Dwight CA M 2020 Dwight 5
105 1952 Dwight Eisenhower Republican 34075529 win 55.325173 Dwight CA M 2020 Dwight 5
131 1992 Ross Perot Independent 19743821 loss 18.956298 Ross CA M 2020 Ross 5
132 1996 Ross Perot Reform 8085294 loss 8.408844 Ross CA M 2020 Ross 5
... ... ... ... ... ... ... ... ... ... ... ... ...
121 2004 David Cobb Green 119859 loss 0.098088 David CA M 2020 David 1155
120 1984 David Bergland Libertarian 228111 loss 0.247245 David CA M 2020 David 1155
66 1884 Benjamin Butler Anti-Monopoly 134294 loss 1.335838 Benjamin CA M 2020 Benjamin 1634
67 1888 Benjamin Harrison Republican 5443633 win 47.858041 Benjamin CA M 2020 Benjamin 1634
68 1892 Benjamin Harrison Republican 5176108 loss 42.984101 Benjamin CA M 2020 Benjamin 1634

146 rows × 12 columns

In [ ]: