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# Combining Datasets: Merge and Join¶

One essential feature offered by Pandas is its high-performance, in-memory join and merge operations. If you have ever worked with databases, you should be familiar with this type of data interaction. The main interface for this is the pd.merge function, and we'll see few examples of how this can work in practice.

For convenience, we will start by redefining the display() functionality from the previous section:

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

class display(object):
"""Display HTML representation of multiple objects"""
template = """<div style="float: left; padding: 10px;">
<p style='font-family:"Courier New", Courier, monospace'>{0}</p>{1}
</div>"""
def __init__(self, *args):
self.args = args

def _repr_html_(self):
return '\n'.join(self.template.format(a, eval(a)._repr_html_())
for a in self.args)

def __repr__(self):
return '\n\n'.join(a + '\n' + repr(eval(a))
for a in self.args)


## Relational Algebra¶

The behavior implemented in pd.merge() is a subset of what is known as relational algebra, which is a formal set of rules for manipulating relational data, and forms the conceptual foundation of operations available in most databases. The strength of the relational algebra approach is that it proposes several primitive operations, which become the building blocks of more complicated operations on any dataset. With this lexicon of fundamental operations implemented efficiently in a database or other program, a wide range of fairly complicated composite operations can be performed.

Pandas implements several of these fundamental building-blocks in the pd.merge() function and the related join() method of Series and Dataframes. As we will see, these let you efficiently link data from different sources.

## Categories of Joins¶

The pd.merge() function implements a number of types of joins: the one-to-one, many-to-one, and many-to-many joins. All three types of joins are accessed via an identical call to the pd.merge() interface; the type of join performed depends on the form of the input data. Here we will show simple examples of the three types of merges, and discuss detailed options further below.

### One-to-one joins¶

Perhaps the simplest type of merge expresion is the one-to-one join, which is in many ways very similar to the column-wise concatenation seen in Combining Datasets: Concat & Append. As a concrete example, consider the following two DataFrames which contain information on several employees in a company:

In [2]:
df1 = pd.DataFrame({'employee': ['Bob', 'Jake', 'Lisa', 'Sue'],
'group': ['Accounting', 'Engineering', 'Engineering', 'HR']})
df2 = pd.DataFrame({'employee': ['Lisa', 'Bob', 'Jake', 'Sue'],
'hire_date': [2004, 2008, 2012, 2014]})
display('df1', 'df2')

Out[2]:

df1

employee group
0 Bob Accounting
1 Jake Engineering
2 Lisa Engineering
3 Sue HR

df2

employee hire_date
0 Lisa 2004
1 Bob 2008
2 Jake 2012
3 Sue 2014

To combine this information into a single DataFrame, we can use the pd.merge() function:

In [3]:
df3 = pd.merge(df1, df2)
df3

Out[3]:
employee group hire_date
0 Bob Accounting 2008
1 Jake Engineering 2012
2 Lisa Engineering 2004
3 Sue HR 2014

The pd.merge() function recognizes that each DataFrame has an "employee" column, and automatically joins using this column as a key. The result of the merge is a new DataFrame that combines the information from the two inputs. Notice that the order of entries in each column is not necessarily maintained: in this case, the order of the "employee" column differs between df1 and df2, and the pd.merge() function correctly accounts for this. Additionally, keep in mind that the merge in general discards the index, except in the special case of merges by index (see the left_index and right_index keywords, discussed momentarily).

### Many-to-one joins¶

Many-to-one joins are joins in which one of the two key columns contains duplicate entries. For the many-to-one case, the resulting DataFrame will preserve those duplicate entries as appropriate. Consider the following example of a many-to-one join:

In [4]:
df4 = pd.DataFrame({'group': ['Accounting', 'Engineering', 'HR'],
'supervisor': ['Carly', 'Guido', 'Steve']})
display('df3', 'df4', 'pd.merge(df3, df4)')

Out[4]:

df3

employee group hire_date
0 Bob Accounting 2008
1 Jake Engineering 2012
2 Lisa Engineering 2004
3 Sue HR 2014

df4

group supervisor
0 Accounting Carly
1 Engineering Guido
2 HR Steve

pd.merge(df3, df4)

employee group hire_date supervisor
0 Bob Accounting 2008 Carly
1 Jake Engineering 2012 Guido
2 Lisa Engineering 2004 Guido
3 Sue HR 2014 Steve

The resulting DataFrame has an aditional column with the "supervisor" information, where the information is repeated in one or more locations as required by the inputs.

### Many-to-many joins¶

Many-to-many joins are a bit confusing conceptually, but are nevertheless well defined. If the key column in both the left and right array contains duplicates, then the result is a many-to-many merge. This will be perhaps most clear with a concrete example. Consider the following, where we have a DataFrame showing one or more skills associated with a particular group. By performing a many-to-many join, we can recover the skills associated with any individual person:

In [5]:
df5 = pd.DataFrame({'group': ['Accounting', 'Accounting',
'Engineering', 'Engineering', 'HR', 'HR'],
display('df1', 'df5', "pd.merge(df1, df5)")

Out[5]:

df1

employee group
0 Bob Accounting
1 Jake Engineering
2 Lisa Engineering
3 Sue HR

df5

group skills
0 Accounting math
2 Engineering coding
3 Engineering linux
5 HR organization

pd.merge(df1, df5)

employee group skills
0 Bob Accounting math
2 Jake Engineering coding
3 Jake Engineering linux
4 Lisa Engineering coding
5 Lisa Engineering linux
7 Sue HR organization

These three types of joins can be used with other Pandas tools to implement a wide array of functionality. But in practice, datasets are rarely as clean as the one we're working with here. In the following section we'll consider some of the options provided by pd.merge() that enable you to tune how the join operations work.

## Specification of the Merge Key¶

We've already seen the default behavior of pd.merge(): it looks for one or more matching column names between the two inputs, and uses this as the key. However, often the column names will not match so nicely, and pd.merge() provides a variety of options for handling this.

### The on keyword¶

Most simply, you can explicitly specify the name of the key column using the on keyword, which takes a column name or a list of column names:

In [6]:
display('df1', 'df2', "pd.merge(df1, df2, on='employee')")

Out[6]:

df1

employee group
0 Bob Accounting
1 Jake Engineering
2 Lisa Engineering
3 Sue HR

df2

employee hire_date
0 Lisa 2004
1 Bob 2008
2 Jake 2012
3 Sue 2014

pd.merge(df1, df2, on='employee')

employee group hire_date
0 Bob Accounting 2008
1 Jake Engineering 2012
2 Lisa Engineering 2004
3 Sue HR 2014

This option works only if both the left and right DataFrames have the specified column name.

### The left_on and right_on keywords¶

At times you may wish to merge two datasets with different column names; for example, we may have a dataset in which the employee name is labeled as "name" rather than "employee". In this case, we can use the left_on and right_on keywords to specify the two column names:

In [7]:
df3 = pd.DataFrame({'name': ['Bob', 'Jake', 'Lisa', 'Sue'],
'salary': [70000, 80000, 120000, 90000]})
display('df1', 'df3', 'pd.merge(df1, df3, left_on="employee", right_on="name")')

Out[7]:

df1

employee group
0 Bob Accounting
1 Jake Engineering
2 Lisa Engineering
3 Sue HR

df3

name salary
0 Bob 70000
1 Jake 80000
2 Lisa 120000
3 Sue 90000

pd.merge(df1, df3, left_on="employee", right_on="name")

employee group name salary
0 Bob Accounting Bob 70000
1 Jake Engineering Jake 80000
2 Lisa Engineering Lisa 120000
3 Sue HR Sue 90000

The result has a redundant column that we can drop if desired–for example, by using the drop() method of DataFrames:

In [8]:
pd.merge(df1, df3, left_on="employee", right_on="name").drop('name', axis=1)

Out[8]:
employee group salary
0 Bob Accounting 70000
1 Jake Engineering 80000
2 Lisa Engineering 120000
3 Sue HR 90000

### The left_index and right_index keywords¶

Sometimes, rather than merging on a column, you would instead like to merge on an index. For example, your data might look like this:

In [9]:
df1a = df1.set_index('employee')
df2a = df2.set_index('employee')
display('df1a', 'df2a')

Out[9]:

df1a

group
employee
Bob Accounting
Jake Engineering
Lisa Engineering
Sue HR

df2a

hire_date
employee
Lisa 2004
Bob 2008
Jake 2012
Sue 2014

You can use the index as the key for merging by specifying the left_index and/or right_index flags in pd.merge():

In [10]:
display('df1a', 'df2a',
"pd.merge(df1a, df2a, left_index=True, right_index=True)")

Out[10]:

df1a

group
employee
Bob Accounting
Jake Engineering
Lisa Engineering
Sue HR

df2a

hire_date
employee
Lisa 2004
Bob 2008
Jake 2012
Sue 2014

pd.merge(df1a, df2a, left_index=True, right_index=True)

group hire_date
employee
Lisa Engineering 2004
Bob Accounting 2008
Jake Engineering 2012
Sue HR 2014

For convenience, DataFrames implement the join() method, which performs a merge that defaults to joining on indices:

In [11]:
display('df1a', 'df2a', 'df1a.join(df2a)')

Out[11]:

df1a

group
employee
Bob Accounting
Jake Engineering
Lisa Engineering
Sue HR

df2a

hire_date
employee
Lisa 2004
Bob 2008
Jake 2012
Sue 2014

df1a.join(df2a)

group hire_date
employee
Bob Accounting 2008
Jake Engineering 2012
Lisa Engineering 2004
Sue HR 2014

If you'd like to mix indices and columns, you can combine left_index with right_on or left_on with right_index to get the desired behavior:

In [12]:
display('df1a', 'df3', "pd.merge(df1a, df3, left_index=True, right_on='name')")

Out[12]:

df1a

group
employee
Bob Accounting
Jake Engineering
Lisa Engineering
Sue HR

df3

name salary
0 Bob 70000
1 Jake 80000
2 Lisa 120000
3 Sue 90000

pd.merge(df1a, df3, left_index=True, right_on='name')

group name salary
0 Accounting Bob 70000
1 Engineering Jake 80000
2 Engineering Lisa 120000
3 HR Sue 90000

All of these options also work with multiple indices and/or multiple columns; the interface for this behavior is very intuitive. For more information on this, see the "Merge, Join, and Concatenate" section of the Pandas documentation.

## Specifying Set Arithmetic for Joins¶

In all the preceding examples we have glossed over one important consideration in performing a join: the type of set arithmetic used in the join. This comes up when a value appears in one key column but not the other. Consider this example:

In [13]:
df6 = pd.DataFrame({'name': ['Peter', 'Paul', 'Mary'],
columns=['name', 'food'])
df7 = pd.DataFrame({'name': ['Mary', 'Joseph'],
'drink': ['wine', 'beer']},
columns=['name', 'drink'])
display('df6', 'df7', 'pd.merge(df6, df7)')

Out[13]:

df6

name food
0 Peter fish
1 Paul beans

df7

name drink
0 Mary wine
1 Joseph beer

pd.merge(df6, df7)

name food drink

Here we have merged two datasets that have only a single "name" entry in common: Mary. By default, the result contains the intersection of the two sets of inputs; this is what is known as an inner join. We can specify this explicitly using the how keyword, which defaults to "inner":

In [14]:
pd.merge(df6, df7, how='inner')

Out[14]:
name food drink

Other options for the how keyword are 'outer', 'left', and 'right'. An outer join returns a join over the union of the input columns, and fills in all missing values with NAs:

In [15]:
display('df6', 'df7', "pd.merge(df6, df7, how='outer')")

Out[15]:

df6

name food
0 Peter fish
1 Paul beans

df7

name drink
0 Mary wine
1 Joseph beer

pd.merge(df6, df7, how='outer')

name food drink
0 Peter fish NaN
1 Paul beans NaN
3 Joseph NaN beer

The left join and right join return joins over the left entries and right entries, respectively. For example:

In [16]:
display('df6', 'df7', "pd.merge(df6, df7, how='left')")

Out[16]:

df6

name food
0 Peter fish
1 Paul beans

df7

name drink
0 Mary wine
1 Joseph beer

pd.merge(df6, df7, how='left')

name food drink
0 Peter fish NaN
1 Paul beans NaN

The output rows now correspond to the entries in the left input. Using how='right' works in a similar manner.

All of these options can be applied straightforwardly to any of the preceding join types.

## Overlapping Column Names: The suffixes Keyword¶

Finally, you may end up in a case where your two input DataFrames have conflicting column names. Consider this example:

In [17]:
df8 = pd.DataFrame({'name': ['Bob', 'Jake', 'Lisa', 'Sue'],
'rank': [1, 2, 3, 4]})
df9 = pd.DataFrame({'name': ['Bob', 'Jake', 'Lisa', 'Sue'],
'rank': [3, 1, 4, 2]})
display('df8', 'df9', 'pd.merge(df8, df9, on="name")')

Out[17]:

df8

name rank
0 Bob 1
1 Jake 2
2 Lisa 3
3 Sue 4

df9

name rank
0 Bob 3
1 Jake 1
2 Lisa 4
3 Sue 2

pd.merge(df8, df9, on="name")

name rank_x rank_y
0 Bob 1 3
1 Jake 2 1
2 Lisa 3 4
3 Sue 4 2

Because the output would have two conflicting column names, the merge function automatically appends a suffix _x or _y to make the output columns unique. If these defaults are inappropriate, it is possible to specify a custom suffix using the suffixes keyword:

In [18]:
display('df8', 'df9', 'pd.merge(df8, df9, on="name", suffixes=["_L", "_R"])')

Out[18]:

df8

name rank
0 Bob 1
1 Jake 2
2 Lisa 3
3 Sue 4

df9

name rank
0 Bob 3
1 Jake 1
2 Lisa 4
3 Sue 2

pd.merge(df8, df9, on="name", suffixes=["_L", "_R"])

name rank_L rank_R
0 Bob 1 3
1 Jake 2 1
2 Lisa 3 4
3 Sue 4 2

These suffixes work in any of the possible join patterns, and work also if there are multiple overlapping columns.

For more information on these patterns, see Aggregation and Grouping where we dive a bit deeper into relational algebra. Also see the Pandas "Merge, Join and Concatenate" documentation for further discussion of these topics.

## Example: US States Data¶

Merge and join operations come up most often when combining data from different sources. Here we will consider an example of some data about US states and their populations. The data files can be found at http://github.com/jakevdp/data-USstates/:

In [19]:
# Following are shell commands to download the data
# !curl -O https://raw.githubusercontent.com/jakevdp/data-USstates/master/state-population.csv
# !curl -O https://raw.githubusercontent.com/jakevdp/data-USstates/master/state-areas.csv
# !curl -O https://raw.githubusercontent.com/jakevdp/data-USstates/master/state-abbrevs.csv


Let's take a look at the three datasets, using the Pandas read_csv() function:

In [20]:
pop = pd.read_csv('data/state-population.csv')


Out[20]:

state/region ages year population
0 AL under18 2012 1117489.0
1 AL total 2012 4817528.0
2 AL under18 2010 1130966.0
3 AL total 2010 4785570.0
4 AL under18 2011 1125763.0

state area (sq. mi)
0 Alabama 52423
2 Arizona 114006
3 Arkansas 53182
4 California 163707

state abbreviation
0 Alabama AL
2 Arizona AZ
3 Arkansas AR
4 California CA

Given this information, say we want to compute a relatively straightforward result: rank US states and territories by their 2010 population density. We clearly have the data here to find this result, but we'll have to combine the datasets to find the result.

We'll start with a many-to-one merge that will give us the full state name within the population DataFrame. We want to merge based on the state/region column of pop, and the abbreviation column of abbrevs. We'll use how='outer' to make sure no data is thrown away due to mismatched labels.

In [21]:
merged = pd.merge(pop, abbrevs, how='outer',
left_on='state/region', right_on='abbreviation')
merged = merged.drop('abbreviation', 1) # drop duplicate info

Out[21]:
state/region ages year population state
0 AL under18 2012 1117489.0 Alabama
1 AL total 2012 4817528.0 Alabama
2 AL under18 2010 1130966.0 Alabama
3 AL total 2010 4785570.0 Alabama
4 AL under18 2011 1125763.0 Alabama

Let's double-check whether there were any mismatches here, which we can do by looking for rows with nulls:

In [22]:
merged.isnull().any()

Out[22]:
state/region    False
ages            False
year            False
population       True
state            True
dtype: bool

Some of the population info is null; let's figure out which these are!

In [23]:
merged[merged['population'].isnull()].head()

Out[23]:
state/region ages year population state
2448 PR under18 1990 NaN NaN
2449 PR total 1990 NaN NaN
2450 PR total 1991 NaN NaN
2451 PR under18 1991 NaN NaN
2452 PR total 1993 NaN NaN

It appears that all the null population values are from Puerto Rico prior to the year 2000; this is likely due to this data not being available from the original source.

More importantly, we see also that some of the new state entries are also null, which means that there was no corresponding entry in the abbrevs key! Let's figure out which regions lack this match:

In [24]:
merged.loc[merged['state'].isnull(), 'state/region'].unique()

Out[24]:
array(['PR', 'USA'], dtype=object)

We can quickly infer the issue: our population data includes entries for Puerto Rico (PR) and the United States as a whole (USA), while these entries do not appear in the state abbreviation key. We can fix these quickly by filling in appropriate entries:

In [25]:
merged.loc[merged['state/region'] == 'PR', 'state'] = 'Puerto Rico'
merged.loc[merged['state/region'] == 'USA', 'state'] = 'United States'
merged.isnull().any()

Out[25]:
state/region    False
ages            False
year            False
population       True
state           False
dtype: bool

No more nulls in the state column: we're all set!

Now we can merge the result with the area data using a similar procedure. Examining our results, we will want to join on the state column in both:

In [26]:
final = pd.merge(merged, areas, on='state', how='left')

Out[26]:
state/region ages year population state area (sq. mi)
0 AL under18 2012 1117489.0 Alabama 52423.0
1 AL total 2012 4817528.0 Alabama 52423.0
2 AL under18 2010 1130966.0 Alabama 52423.0
3 AL total 2010 4785570.0 Alabama 52423.0
4 AL under18 2011 1125763.0 Alabama 52423.0

Again, let's check for nulls to see if there were any mismatches:

In [27]:
final.isnull().any()

Out[27]:
state/region     False
ages             False
year             False
population        True
state            False
area (sq. mi)     True
dtype: bool

There are nulls in the area column; we can take a look to see which regions were ignored here:

In [28]:
final['state'][final['area (sq. mi)'].isnull()].unique()

Out[28]:
array(['United States'], dtype=object)

We see that our areas DataFrame does not contain the area of the United States as a whole. We could insert the appropriate value (using the sum of all state areas, for instance), but in this case we'll just drop the null values because the population density of the entire United States is not relevant to our current discussion:

In [29]:
final.dropna(inplace=True)

Out[29]:
state/region ages year population state area (sq. mi)
0 AL under18 2012 1117489.0 Alabama 52423.0
1 AL total 2012 4817528.0 Alabama 52423.0
2 AL under18 2010 1130966.0 Alabama 52423.0
3 AL total 2010 4785570.0 Alabama 52423.0
4 AL under18 2011 1125763.0 Alabama 52423.0

Now we have all the data we need. To answer the question of interest, let's first select the portion of the data corresponding with the year 2000, and the total population. We'll use the query() function to do this quickly (this requires the numexpr package to be installed; see High-Performance Pandas: eval() and query()):

In [30]:
data2010 = final.query("year == 2010 & ages == 'total'")

Out[30]:
state/region ages year population state area (sq. mi)
3 AL total 2010 4785570.0 Alabama 52423.0
91 AK total 2010 713868.0 Alaska 656425.0
101 AZ total 2010 6408790.0 Arizona 114006.0
189 AR total 2010 2922280.0 Arkansas 53182.0
197 CA total 2010 37333601.0 California 163707.0

Now let's compute the population density and display it in order. We'll start by re-indexing our data on the state, and then compute the result:

In [31]:
data2010.set_index('state', inplace=True)
density = data2010['population'] / data2010['area (sq. mi)']

In [32]:
density.sort_values(ascending=False, inplace=True)

Out[32]:
state
District of Columbia    8898.897059
Puerto Rico             1058.665149
New Jersey              1009.253268
Rhode Island             681.339159
Connecticut              645.600649
dtype: float64

The result is a ranking of US states plus Washington, DC, and Puerto Rico in order of their 2010 population density, in residents per square mile. We can see that by far the densest region in this dataset is Washington, DC (i.e., the District of Columbia); among states, the densest is New Jersey.

We can also check the end of the list:

In [33]:
density.tail()

Out[33]:
state
South Dakota    10.583512
North Dakota     9.537565
Montana          6.736171
Wyoming          5.768079
dtype: float64