{ "cells": [ { "cell_type": "code", "execution_count": 1, "metadata": { "tags": [ "remove_input" ] }, "outputs": [], "source": [ "path_data = '../../data/'\n", "\n", "import numpy as np\n", "import pandas as pd\n", "\n", "%matplotlib inline\n", "import matplotlib.pyplot as plt\n", "plt.style.use('fivethirtyeight')\n", "\n", "import warnings\n", "warnings.filterwarnings('ignore')" ] }, { "cell_type": "markdown", "metadata": {}, "source": [ "# Classifying by One Variable\n", "\n", "Data scientists often need to classify individuals into groups according to shared features, and then identify some characteristics of the groups. For example, in the example using Galton's data on heights, we saw that it was useful to classify families according to the parents' midparent heights, and then find the average height of the children in each group.\n", "\n", "This section is about classifying individuals into categories that are not numerical. We begin by recalling the basic use of `group`. " ] }, { "cell_type": "markdown", "metadata": {}, "source": [ "## Counting the Number in Each Category\n", "The `group` method with a single argument counts the number of rows for each category in a column. The result contains one row per unique value in the grouped column.\n", "\n", "Here is a small table of data on ice cream cones. The `group` method can be used to list the distinct flavors and provide the counts of each flavor." ] }, { "cell_type": "code", "execution_count": 19, "metadata": {}, "outputs": [ { "data": { "text/html": [ "
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FlavorPrice
0strawberry3.55
1chocolate4.75
2chocolate6.55
3strawberry5.25
4chocolate5.25
\n", "
" ], "text/plain": [ " Flavor Price\n", "0 strawberry 3.55\n", "1 chocolate 4.75\n", "2 chocolate 6.55\n", "3 strawberry 5.25\n", "4 chocolate 5.25" ] }, "execution_count": 19, "metadata": {}, "output_type": "execute_result" } ], "source": [ "cones = pd.DataFrame({\n", " 'Flavor':np.array(['strawberry', 'chocolate', 'chocolate', 'strawberry', 'chocolate']),\n", " 'Price':np.array([3.55, 4.75, 6.55, 5.25, 5.25])}\n", ")\n", "cones" ] }, { "cell_type": "code", "execution_count": 20, "metadata": {}, "outputs": [ { "data": { "text/html": [ "
\n", "\n", "\n", " \n", " \n", " \n", " \n", " \n", " \n", " \n", " \n", " \n", " \n", " \n", " \n", " \n", " \n", " \n", " \n", " \n", " \n", " \n", " \n", "
count
Flavor
chocolate3
strawberry2
\n", "
" ], "text/plain": [ " count\n", "Flavor \n", "chocolate 3\n", "strawberry 2" ] }, "execution_count": 20, "metadata": {}, "output_type": "execute_result" } ], "source": [ "df_grouped = cones.groupby([\"Flavor\"]).agg(\n", " count=pd.NamedAgg(column=\"Flavor\", aggfunc=\"count\")\n", ")\n", "\n", "df_grouped" ] }, { "cell_type": "markdown", "metadata": {}, "source": [ "There are two distinct categories, chocolate and strawberry. When we call `groupby` we must state what we want to do with the group data e.g. `count()`. Applying the `count()` method will create a column of counts whcih takes the names of the first column in the df by default, and contains the number of rows in each category. To make this easier to read we could change the count column to 'count'.\n", "\n", "Notice that this can all be worked out from just the `Flavor` column. Only the `Price` column name has been used, the data has not been used.\n", "\n", "But what if we wanted the total price of the cones of each different flavor? In this case we can apply a different method e.g. `sum()`, to `groupby`." ] }, { "cell_type": "markdown", "metadata": {}, "source": [ "### Finding a Characteristic of Each Category\n", "The optional second argument of `group` names the function that will be used to aggregate values in other columns for all of those rows. For instance, `sum` will sum up the prices in all rows that match each category. This result also contains one row per unique value in the grouped column, but it has the same number of columns as the original table.\n", "\n", "To find the total price of each flavor, we call `group` again, with `Flavor` as its first argument as before. But this time there is a second argument: the function name `sum`." ] }, { "cell_type": "code", "execution_count": 21, "metadata": {}, "outputs": [ { "data": { "text/html": [ "
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Price_sum
Flavor
chocolate16.55
strawberry8.80
\n", "
" ], "text/plain": [ " Price_sum\n", "Flavor \n", "chocolate 16.55\n", "strawberry 8.80" ] }, "execution_count": 21, "metadata": {}, "output_type": "execute_result" } ], "source": [ "df_grouped_by_price = cones.groupby([\"Flavor\"]).agg(\n", " Price_sum=pd.NamedAgg(column=\"Price\", aggfunc=\"sum\")\n", ")\n", "\n", "df_grouped_by_price" ] }, { "cell_type": "markdown", "metadata": {}, "source": [ "To create this new table, `groupby` has calculated the **sum** of the `Price` entries in all the rows corresponding to each distinct flavor. The prices in the three `chocolate` rows add up to 16.55 (in whatever currency). The prices in the two `strawberry` rows have a total of 8.80.\n", "\n", "Using pandas `groupby aggregation` we can compute a summary statistic (or statistics). The label of the newly created column is `Price sum`, which is created by taking the label of the column being summed, and appending the word `sum` in the *aggregation pipeline*. \n", "\n", "In this insatnce there are only two columns so when `group` finds the `sum` of all columns other than the one with the categories, there is no need to specify that it has to `sum` the prices. Using the [`Pandas NamedAgg`](https://pandas.pydata.org/pandas-docs/stable/user_guide/groupby.html#aggregation) function we can name the column contining the results of the aggregation.\n", "\n", "To see in more detail what `group` is doing, notice that you could have figured out the total prices yourself, not only by mental arithmetic but also using code. For example, to find the total price of all the chocolate cones, you could start by creating a new table consisting of only the chocolate cones, and then accessing the column of prices:" ] }, { "cell_type": "code", "execution_count": 22, "metadata": {}, "outputs": [ { "data": { "text/plain": [ "1 4.75\n", "2 6.55\n", "4 5.25\n", "Name: Price, dtype: float64" ] }, "execution_count": 22, "metadata": {}, "output_type": "execute_result" } ], "source": [ "cones[cones['Flavor'] == 'chocolate']['Price']" ] }, { "cell_type": "code", "execution_count": 23, "metadata": {}, "outputs": [ { "data": { "text/html": [ "
\n", "\n", "\n", " \n", " \n", " \n", " \n", " \n", " \n", " \n", " \n", " \n", " \n", " \n", " \n", " \n", " \n", " \n", " \n", " \n", " \n", " \n", " \n", "
Price
14.75
26.55
45.25
\n", "
" ], "text/plain": [ " Price\n", "1 4.75\n", "2 6.55\n", "4 5.25" ] }, "execution_count": 23, "metadata": {}, "output_type": "execute_result" } ], "source": [ "cones[cones['Flavor'] == 'chocolate'][['Price']]" ] }, { "cell_type": "code", "execution_count": 24, "metadata": {}, "outputs": [ { "data": { "text/plain": [ "Price 16.55\n", "dtype: float64" ] }, "execution_count": 24, "metadata": {}, "output_type": "execute_result" } ], "source": [ "np.sum(cones[cones['Flavor'] == 'chocolate'][['Price']])" ] }, { "cell_type": "code", "execution_count": 25, "metadata": {}, "outputs": [ { "data": { "text/plain": [ "16.55" ] }, "execution_count": 25, "metadata": {}, "output_type": "execute_result" } ], "source": [ "np.sum([cones[cones['Flavor'] == 'chocolate'][['Price']]])" ] }, { "cell_type": "markdown", "metadata": {}, "source": [ "This is what `groupby` is doing for each distinct value in `Flavor`." ] }, { "cell_type": "code", "execution_count": 26, "metadata": {}, "outputs": [ { "data": { "text/html": [ "
\n", "\n", "\n", " \n", " \n", " \n", " \n", " \n", " \n", " \n", " \n", " \n", " \n", " \n", " \n", " \n", " \n", " \n", " \n", " \n", " \n", " \n", " \n", " \n", " \n", "
FlavorArray of All the PricesSum of the Array
0chocolate[4.75, 6.55, 5.25]16.55
1strawberry[3.55, 5.25]8.80
\n", "
" ], "text/plain": [ " Flavor Array of All the Prices Sum of the Array\n", "0 chocolate [4.75, 6.55, 5.25] 16.55\n", "1 strawberry [3.55, 5.25] 8.80" ] }, "execution_count": 26, "metadata": {}, "output_type": "execute_result" } ], "source": [ "# For each distinct value in `Flavor, access all the rows\n", "# and create an array of `Price`\n", "\n", "cones_choc = cones[cones['Flavor'] == 'chocolate']['Price']\n", "\n", "cones_strawb = cones[cones['Flavor'] =='strawberry']['Price']\n", "\n", "# Display the arrays in a table\n", "\n", "cones_choc = np.array(cones_choc)\n", "cones_strawb = np.array(cones_strawb)\n", "\n", "grouped_cones = pd.DataFrame({\n", " 'Flavor':np.array(['chocolate', 'strawberry']),\n", " 'Array of All the Prices':[cones_choc, cones_strawb]}\n", ")\n", "\n", "#priceTotals\n", "\n", "# Append a column with the sum of the `Price` values in each array\n", "\n", "price_totals = grouped_cones\n", "\n", "price_totals['Sum of the Array'] = np.array([sum(cones_choc), sum(cones_strawb)])\n", "\n", "price_totals" ] }, { "cell_type": "markdown", "metadata": {}, "source": [ "You can replace `sum` by any other functions that work on arrays. For example, you could use `max` to find the largest price in each category:" ] }, { "cell_type": "code", "execution_count": 27, "metadata": {}, "outputs": [ { "data": { "text/html": [ "
\n", "\n", "\n", " \n", " \n", " \n", " \n", " \n", " \n", " \n", " \n", " \n", " \n", " \n", " \n", " \n", " \n", " \n", " \n", " \n", " \n", " \n", " \n", "
Price
Flavor
chocolate6.55
strawberry5.25
\n", "
" ], "text/plain": [ " Price\n", "Flavor \n", "chocolate 6.55\n", "strawberry 5.25" ] }, "execution_count": 27, "metadata": {}, "output_type": "execute_result" } ], "source": [ "cones.groupby('Flavor').max()" ] }, { "cell_type": "markdown", "metadata": {}, "source": [ "### *Or*" ] }, { "cell_type": "code", "execution_count": 28, "metadata": {}, "outputs": [ { "data": { "text/html": [ "
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Price_Max
Flavor
chocolate6.55
strawberry5.25
\n", "
" ], "text/plain": [ " Price_Max\n", "Flavor \n", "chocolate 6.55\n", "strawberry 5.25" ] }, "execution_count": 28, "metadata": {}, "output_type": "execute_result" } ], "source": [ "price_max = cones.groupby([\"Flavor\"]).agg(\n", " Price_Max=pd.NamedAgg(column=\"Price\", aggfunc=\"max\")\n", ")\n", "\n", "price_max" ] }, { "cell_type": "markdown", "metadata": {}, "source": [ "Once again, `groupby` creates arrays of the prices in each `Flavor` category. But now it finds the `max` of each array:" ] }, { "cell_type": "code", "execution_count": 29, "metadata": {}, "outputs": [ { "data": { "text/html": [ "
\n", "\n", "\n", " \n", " \n", " \n", " \n", " \n", " \n", " \n", " \n", " \n", " \n", " \n", " \n", " \n", " \n", " \n", " \n", " \n", " \n", " \n", " \n", " \n", " \n", " \n", " \n", " \n", "
FlavorArray of All the PricesSum of the ArrayMax of the Array
0chocolate[4.75, 6.55, 5.25]16.556.55
1strawberry[3.55, 5.25]8.805.25
\n", "
" ], "text/plain": [ " Flavor Array of All the Prices Sum of the Array Max of the Array\n", "0 chocolate [4.75, 6.55, 5.25] 16.55 6.55\n", "1 strawberry [3.55, 5.25] 8.80 5.25" ] }, "execution_count": 29, "metadata": {}, "output_type": "execute_result" } ], "source": [ "price_max = grouped_cones.copy()\n", "\n", "price_max['Max of the Array'] = np.array([max(cones_choc), max(cones_strawb)])\n", "\n", "price_max" ] }, { "cell_type": "markdown", "metadata": {}, "source": [ "Indeed, the original call to `group` with just one argument has the same effect as using `len` as the function and then cleaning up the table." ] }, { "cell_type": "code", "execution_count": 30, "metadata": {}, "outputs": [ { "data": { "text/html": [ "
\n", "\n", "\n", " \n", " \n", " \n", " \n", " \n", " \n", " \n", " \n", " \n", " \n", " \n", " \n", " \n", " \n", " \n", " \n", " \n", " \n", " \n", " \n", " \n", " \n", " \n", " \n", " \n", "
FlavorArray of All the PricesSum of the ArrayLength of the Array
0chocolate[4.75, 6.55, 5.25]16.553
1strawberry[3.55, 5.25]8.802
\n", "
" ], "text/plain": [ " Flavor Array of All the Prices Sum of the Array Length of the Array\n", "0 chocolate [4.75, 6.55, 5.25] 16.55 3\n", "1 strawberry [3.55, 5.25] 8.80 2" ] }, "execution_count": 30, "metadata": {}, "output_type": "execute_result" } ], "source": [ "array_length = grouped_cones.copy()\n", "\n", "array_length['Length of the Array'] = np.array([len(cones_choc), len(cones_strawb)])\n", "\n", "array_length" ] }, { "cell_type": "markdown", "metadata": {}, "source": [ "### Example: NBA Salaries\n", "The table `nba` contains data on the 2015-2016 players in the National Basketball Association. We have examined these data earlier. Recall that salaries are measured in millions of dollars." ] }, { "cell_type": "code", "execution_count": 31, "metadata": {}, "outputs": [ { "data": { "text/html": [ "
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PLAYERPOSITIONTEAMSALARY
0Paul MillsapPFAtlanta Hawks18.671659
1Al HorfordCAtlanta Hawks12.000000
2Tiago SplitterCAtlanta Hawks9.756250
3Jeff TeaguePGAtlanta Hawks8.000000
4Kyle KorverSGAtlanta Hawks5.746479
...............
412Gary NealPGWashington Wizards2.139000
413DeJuan BlairCWashington Wizards2.000000
414Kelly Oubre Jr.SFWashington Wizards1.920240
415Garrett TempleSGWashington Wizards1.100602
416Jarell EddieSGWashington Wizards0.561716
\n", "

417 rows × 4 columns

\n", "
" ], "text/plain": [ " PLAYER POSITION TEAM SALARY\n", "0 Paul Millsap PF Atlanta Hawks 18.671659\n", "1 Al Horford C Atlanta Hawks 12.000000\n", "2 Tiago Splitter C Atlanta Hawks 9.756250\n", "3 Jeff Teague PG Atlanta Hawks 8.000000\n", "4 Kyle Korver SG Atlanta Hawks 5.746479\n", ".. ... ... ... ...\n", "412 Gary Neal PG Washington Wizards 2.139000\n", "413 DeJuan Blair C Washington Wizards 2.000000\n", "414 Kelly Oubre Jr. SF Washington Wizards 1.920240\n", "415 Garrett Temple SG Washington Wizards 1.100602\n", "416 Jarell Eddie SG Washington Wizards 0.561716\n", "\n", "[417 rows x 4 columns]" ] }, "execution_count": 31, "metadata": {}, "output_type": "execute_result" } ], "source": [ "nba1 = pd.read_csv(path_data + 'nba_salaries.csv')\n", "\n", "nba = nba1.rename(columns={\"'15-'16 SALARY\": 'SALARY'})\n", "\n", "nba" ] }, { "cell_type": "markdown", "metadata": {}, "source": [ "**1.** How much money did each team pay for its players' salaries?\n", "\n", "The only columns involved are `TEAM` and `SALARY`. We have to `group` the rows by `TEAM` and then `sum` the salaries of the groups. " ] }, { "cell_type": "code", "execution_count": 32, "metadata": {}, "outputs": [ { "data": { "text/html": [ "
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SALARY
TEAM
Atlanta Hawks69.573103
Boston Celtics50.285499
Brooklyn Nets57.306976
Charlotte Hornets84.102397
Chicago Bulls78.820890
Cleveland Cavaliers102.312412
Dallas Mavericks65.762559
Denver Nuggets62.429404
Detroit Pistons42.211760
Golden State Warriors94.085137
Houston Rockets85.285837
Indiana Pacers62.695023
Los Angeles Clippers66.074113
Los Angeles Lakers68.607944
Memphis Grizzlies93.796439
Miami Heat81.528667
Milwaukee Bucks52.258355
Minnesota Timberwolves65.847421
New Orleans Pelicans80.514606
New York Knicks69.404994
Oklahoma City Thunder96.832165
Orlando Magic77.623940
Philadelphia 76ers42.481345
Phoenix Suns50.520815
Portland Trail Blazers45.446878
Sacramento Kings68.384890
San Antonio Spurs84.652074
Toronto Raptors74.672620
Utah Jazz52.631878
Washington Wizards90.047498
\n", "
" ], "text/plain": [ " SALARY\n", "TEAM \n", "Atlanta Hawks 69.573103\n", "Boston Celtics 50.285499\n", "Brooklyn Nets 57.306976\n", "Charlotte Hornets 84.102397\n", "Chicago Bulls 78.820890\n", "Cleveland Cavaliers 102.312412\n", "Dallas Mavericks 65.762559\n", "Denver Nuggets 62.429404\n", "Detroit Pistons 42.211760\n", "Golden State Warriors 94.085137\n", "Houston Rockets 85.285837\n", "Indiana Pacers 62.695023\n", "Los Angeles Clippers 66.074113\n", "Los Angeles Lakers 68.607944\n", "Memphis Grizzlies 93.796439\n", "Miami Heat 81.528667\n", "Milwaukee Bucks 52.258355\n", "Minnesota Timberwolves 65.847421\n", "New Orleans Pelicans 80.514606\n", "New York Knicks 69.404994\n", "Oklahoma City Thunder 96.832165\n", "Orlando Magic 77.623940\n", "Philadelphia 76ers 42.481345\n", "Phoenix Suns 50.520815\n", "Portland Trail Blazers 45.446878\n", "Sacramento Kings 68.384890\n", "San Antonio Spurs 84.652074\n", "Toronto Raptors 74.672620\n", "Utah Jazz 52.631878\n", "Washington Wizards 90.047498" ] }, "execution_count": 32, "metadata": {}, "output_type": "execute_result" } ], "source": [ "teams_and_money = nba[['TEAM', 'SALARY']]\n", "\n", "teams_and_money.groupby('TEAM').sum()" ] }, { "cell_type": "markdown", "metadata": {}, "source": [ "**2.** How many NBA players were there in each of the five positions?\n", "\n", "We have to classify by `POSITION`, and count. This can be achieved by applying the `count()` method to a `groupby` or and the aggregation method with `aggfunc=\"count\"`." ] }, { "cell_type": "code", "execution_count": 33, "metadata": {}, "outputs": [ { "data": { "text/html": [ "
\n", "\n", "\n", " \n", " \n", " \n", " \n", " \n", " \n", " \n", " \n", " \n", " \n", " \n", " \n", " \n", " \n", " \n", " \n", " \n", " \n", " \n", " \n", " \n", " \n", " \n", " \n", " \n", " \n", " \n", " \n", " \n", " \n", " \n", " \n", "
count
POSITION
C69
PF85
PG85
SF82
SG96
\n", "
" ], "text/plain": [ " count\n", "POSITION \n", "C 69\n", "PF 85\n", "PG 85\n", "SF 82\n", "SG 96" ] }, "execution_count": 33, "metadata": {}, "output_type": "execute_result" } ], "source": [ "#nba.groupby('POSITION').count()\n", "\n", "# -- or\n", "\n", "position_count = nba.groupby([\"POSITION\"]).agg(\n", " count=pd.NamedAgg(column=\"PLAYER\", aggfunc=\"count\")\n", ")\n", "\n", "position_count" ] }, { "cell_type": "markdown", "metadata": {}, "source": [ "**3.** What was the average salary of the players at each of the five positions?\n", "\n", "This time, we have to group by `POSITION` and take the mean of the salaries. For clarity, we will work with a table of just the positions and the salaries." ] }, { "cell_type": "code", "execution_count": 34, "metadata": {}, "outputs": [ { "data": { "text/html": [ "
\n", "\n", "\n", " \n", " \n", " \n", " \n", " \n", " \n", " \n", " \n", " \n", " \n", " \n", " \n", " \n", " \n", " \n", " \n", " \n", " \n", " \n", " \n", " \n", " \n", " \n", " \n", " \n", " \n", " \n", " \n", " \n", " \n", " \n", " \n", "
SALARY
POSITION
C6.082913
PF4.951344
PG5.165487
SF5.532675
SG3.988195
\n", "
" ], "text/plain": [ " SALARY\n", "POSITION \n", "C 6.082913\n", "PF 4.951344\n", "PG 5.165487\n", "SF 5.532675\n", "SG 3.988195" ] }, "execution_count": 34, "metadata": {}, "output_type": "execute_result" } ], "source": [ "positions_and_money = nba[['POSITION', 'SALARY']]\n", "\n", "positions_and_money.groupby('POSITION').mean()" ] }, { "cell_type": "markdown", "metadata": {}, "source": [ "Center was the most highly paid position, at an average of over 6 million dollars.\n", "\n", "If we had not selected the two columns as our first step, `group` would not attempt to \"average\" the categorical columns in `nba`. (It is impossible to average two strings like \"Atlanta Hawks\" and \"Boston Celtics\".) It performs arithmetic only on numerical columns and leaves the rest blank." ] }, { "cell_type": "code", "execution_count": 35, "metadata": {}, "outputs": [ { "data": { "text/html": [ "
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SALARY_mean
POSITION
C6.082913
PF4.951344
PG5.165487
SF5.532675
SG3.988195
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" ], "text/plain": [ " SALARY_mean\n", "POSITION \n", "C 6.082913\n", "PF 4.951344\n", "PG 5.165487\n", "SF 5.532675\n", "SG 3.988195" ] }, "execution_count": 35, "metadata": {}, "output_type": "execute_result" } ], "source": [ "nba_mean = nba.groupby('POSITION').mean()\n", "\n", "nba_mean\n", "\n", "nba_mean = nba.groupby([\"POSITION\"]).agg(\n", " SALARY_mean=pd.NamedAgg(column=\"SALARY\", aggfunc=\"mean\")\n", ")\n", "\n", "nba_mean" ] } ], "metadata": { "kernelspec": { "display_name": "Python 3", "language": "python", "name": "python3" }, "language_info": { "codemirror_mode": { "name": "ipython", "version": 3 }, "file_extension": ".py", "mimetype": "text/x-python", "name": "python", "nbconvert_exporter": "python", "pygments_lexer": "ipython3", "version": "3.8.5" } }, "nbformat": 4, "nbformat_minor": 1 }