pd.merge function, and we'll see few examples of how this can work in practice.display() functionality from the previous section:
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)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.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.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.DataFrames which contain information on several employees in a company:
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')DataFrame, we can use the pd.merge() function:
df3 = pd.merge(df1, df2)
df3pd.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).DataFrame will preserve those duplicate entries as appropriate.
Consider the following example of a many-to-one join:
df4 = pd.DataFrame({'group': ['Accounting', 'Engineering', 'HR'],
'supervisor': ['Carly', 'Guido', 'Steve']})
display('df3', 'df4', 'pd.merge(df3, df4)')DataFrame has an aditional column with the "supervisor" information, where the information is repeated in one or more locations as required by the inputs.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:
df5 = pd.DataFrame({'group': ['Accounting', 'Accounting',
'Engineering', 'Engineering', 'HR', 'HR'],
'skills': ['math', 'spreadsheets', 'coding', 'linux',
'spreadsheets', 'organization']})
display('df1', 'df5', "pd.merge(df1, df5)")pd.merge() that enable you to tune how the join operations work.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.on keywordon keyword, which takes a column name or a list of column names:
display('df1', 'df2', "pd.merge(df1, df2, on='employee')")DataFrames have the specified column name.left_on and right_on keywordsleft_on and right_on keywords to specify the two column names:
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")')drop() method of DataFrames:
pd.merge(df1, df3, left_on="employee", right_on="name").drop('name', axis=1)left_index and right_index keywords
df1a = df1.set_index('employee')
df2a = df2.set_index('employee')
display('df1a', 'df2a')left_index and/or right_index flags in pd.merge():
display('df1a', 'df2a',
"pd.merge(df1a, df2a, left_index=True, right_index=True)")DataFrames implement the join() method, which performs a merge that defaults to joining on indices:
display('df1a', 'df2a', 'df1a.join(df2a)')left_index with right_on or left_on with right_index to get the desired behavior:
display('df1a', 'df3', "pd.merge(df1a, df3, left_index=True, right_on='name')")
df6 = pd.DataFrame({'name': ['Peter', 'Paul', 'Mary'],
'food': ['fish', 'beans', 'bread']},
columns=['name', 'food'])
df7 = pd.DataFrame({'name': ['Mary', 'Joseph'],
'drink': ['wine', 'beer']},
columns=['name', 'drink'])
display('df6', 'df7', 'pd.merge(df6, df7)')how keyword, which defaults to "inner":
pd.merge(df6, df7, how='inner')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:
display('df6', 'df7', "pd.merge(df6, df7, how='outer')")
display('df6', 'df7', "pd.merge(df6, df7, how='left')")how='right' works in a similar manner.suffixes KeywordDataFrames have conflicting column names.
Consider this example:
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")')_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:
display('df8', 'df9', 'pd.merge(df8, df9, on="name", suffixes=["_L", "_R"])')
# 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.csvread_csv() function:
pop = pd.read_csv('data/state-population.csv')
areas = pd.read_csv('data/state-areas.csv')
abbrevs = pd.read_csv('data/state-abbrevs.csv')
display('pop.head()', 'areas.head()', 'abbrevs.head()')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.
merged = pd.merge(pop, abbrevs, how='outer',
left_on='state/region', right_on='abbreviation')
merged = merged.drop('abbreviation', 1) # drop duplicate info
merged.head()
merged.isnull().any()population info is null; let's figure out which these are!
merged[merged['population'].isnull()].head()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:
merged.loc[merged['state'].isnull(), 'state/region'].unique()
merged.loc[merged['state/region'] == 'PR', 'state'] = 'Puerto Rico'
merged.loc[merged['state/region'] == 'USA', 'state'] = 'United States'
merged.isnull().any()state column: we're all set!state column in both:
final = pd.merge(merged, areas, on='state', how='left')
final.head()
final.isnull().any()area column; we can take a look to see which regions were ignored here:
final['state'][final['area (sq. mi)'].isnull()].unique()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:
final.dropna(inplace=True)
final.head()query() function to do this quickly (this requires the numexpr package to be installed; see High-Performance Pandas: eval() and query()):
data2010 = final.query("year == 2010 & ages == 'total'")
data2010.head()
data2010.set_index('state', inplace=True)
density = data2010['population'] / data2010['area (sq. mi)']
density.sort_values(ascending=False, inplace=True)
density.head()
density.tail()