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Problem 17: Spectral graph partitioning
Version 1.3
Changelog:
1.3: Added another more informative error message. [Dec 5, 2019]
1.2: Added more hints and more informative error messages. [Dec 3, 2019]
1.1: Added more examples; no changes to code or test cells. [Dec 2, 2019]
1.0: Initial version
In this problem, you'll consider the data mining task of "clustering" a graph. That is, given a graph or network of
relationships, can you identify distinct communities within it? This problem assesses your Pandas and Numpy
skills.
Exercises.
There are six exercises, numbered 0-5, worth a total of ten (10) points. However, only Exercises 0-4
require you to write code. If you have done them correctly, then Exercise 5, which consists only of a hidden test
cell, should pass when submitted to the autograder.
Regarding dependencies and partial credit:
Exercise 1 (1 points) depends on a correct Exercise 0 (2 points).
Exercises 2 (2 points) and 3 (2 points) are independent. Neither depends on Exercises 0 or 1.
Exercise 4 (2 points) relies on all earlier exercises.
Exercise 5 (1 point) depends on Exercise 4.
As always
, it is possible that you will pass Exercises 0-4 but not pass Exercise 5 if there is a subtle bug that the
test cells happen not to catch, so be prepared for that possibility!
Setup
The main modules you'll need are pandas
, numpy
, and scipy
. The rest are auxiliary functions for loading data,
plotting results, and test code support. Start by running the following cell.
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In [1]:
# Main modules you'll need:
import
numpy
as
np
import
scipy
as
sp
import
pandas
as
pd
from
pandas
import
DataFrame
# Support code for data loading, code testing, and visualization:
import
sys
sys.path.insert(0, 'resource/asnlib/public')
from
cse6040utils
import
tibbles_are_equivalent, pandas_df_to_markdown
_table, hidden_cell_template_msg, cspy
from
matplotlib.pyplot
import
figure, subplots
%
matplotlib
inline
from
networkx.convert_matrix
import
from_pandas_edgelist
from
networkx.drawing.nx_pylab
import
draw
from
networkx
import
DiGraph
from
networkx.drawing
import
circular_layout
# Location of input data:
def
dataset_path(base_filename):
return
f"resource/asnlib/publicdata/
{base_filename}
"
Background: Relationship networks and partitioning
Suppose we have data on the following five people, stored in the nodes
data frame (run the next cell):
In [2]:
nodes = DataFrame({'name': ['alice', 'bob', 'carol', 'dave', 'edith'],
'age': [35, 18, 27, 57, 41]})
nodes
Also suppose we have some information on their relationships, as might be captured in a social network or
database of person-to-person transactions. In particular, if some person follows another person , we say is
the source
and is the target
. For the people listed above, suppose these relationships are stored in a data
frame named edges
(run this code cell):
Out[2]:
name age
0
alice
35
1
bob
18
2
carol
27
3
dave
57
4
edith
41
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In [3]:
edges = DataFrame({'source': ['alice', 'alice', 'dave', 'dave', 'dav
e', 'bob'],
'target': ['dave', 'edith', 'alice', 'edith', 'car
ol', 'carol']})
edges
We can visualize these relationships as a directed graph
or directed network
, where the people are shown as
nodes
(circles) and the follows-relationships are shown as edges from source to target. Run the next code cell
to see the relationships in our example.
In [4]:
G = from_pandas_edgelist(edges, source='source', target='target', crea
te_using=DiGraph())
figure(figsize=(4, 4))
draw(G, arrows=
True
, with_labels=
True
, pos=circular_layout(G),
node_size=1200, node_color=
None
, font_color='w',
width=2, arrowsize=20)
Out[3]:
source target
0
alice
dave
1
alice
edith
2
dave
alice
3
dave
edith
4
dave
carol
5
bob
carol
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Observation 0:
From the edges
data frame, recall that alice
follows edith
; therefore, there is an arrow (or
edge) pointing from alice
to edith
. Since alice
and dave
both follow one another, there is a double-headed
arrow between them.
Observation 1:
One can arguably say there are two distinct "groups" in this picture. One group consists of
alice
, dave
, and edith
, who have at least 1 one-way follow-relationships among all pairs of them. Similarly,
bob
and carol
have a one-way relationship between them. However, there is only one relationship between
someone from the first group and someone from the second group.
In this problem, we might ask a data mining question, namely, whether we can automatically identify these
clusters or groups, given only the known relationships. The method you will implement is known as spectral
graph partitioning
or spectral graph clustering
, which is formulated as a linear algebra problem.
Exercises
Exercise 0
(2 points -- 0.5 exposed, 1.5 hidden). For our analysis, we won't care whether follows or follows , only that there is some interaction between them. To do so, let's write some code to "symmetrize" the
edges. That is, if there is a directed edge from node to node , then symmetrize will ensure there is also a
directed edge to , unless one already exists. (
Recall Notebook 10!
)
For example, a symmetrized version of the edges
data frame from above would look like the following:
source target
alice
dave
alice
edith
bob
carol
carol
bob
carol
dave
dave
alice
dave
carol
dave
edith
edith
alice
edith
dave
Complete the function, symmetrize_df(edges)
, below, so that it symmetrizes edges
. Assume that edges
is
a pandas DataFrame
with source
and target
columns. Your function should return a new pandas
DataFrame
with the edges symmetrized. Your function should also reset the index, so that the output is a
proper "tibble."
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Note 0:
The order of the edges in the output does not
matter.
Note 1:
You may assume the input data frame has columns named 'source'
and 'target'
.
Note 2:
You should drop any duplicate edges. In the example, the edges 'dave'
'alice'
and 'alice'
'dave'
already exist in the input. Therefore, observe that they appear in the
output, but only once each.
Note 3:
Your function should work even if there is a "self-edge," i.e., an edge . The
example above does not contain such a case, but the hidden test might check it.
In [5]:
def
symmetrize_df(edges):
assert
'source' in
edges.columns
assert
'target' in
edges.columns
### BEGIN SOLUTION
from
pandas
import
concat
edges_transpose = edges.rename(columns={'source': 'target', 'targe
t': 'source'})
edges_all = concat([edges, edges_transpose], sort=
False
) \
.drop_duplicates() \
.reset_index(drop=
True
)
return
edges_all
### END SOLUTION
# Demo of your function:
symmetrize_df(edges)
Out[5]:
source target
0
alice
dave
1
alice
edith
2
dave
alice
3
dave
edith
4
dave
carol
5
bob
carol
6
edith
alice
7
edith
dave
8
carol
dave
9
carol
bob
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6/25
In [6]:
# Test cell: `ex0_symmetrize_df__visible` (1 point)
edges_input = DataFrame({'source': ['alice', 'alice', 'dave', 'dave', 'dave', 'bob'],
'target': ['dave', 'edith', 'alice', 'edit
h', 'carol', 'carol']})
edges_output = symmetrize_df(edges_input)
# The following comment block suggests there is hidden content in this cell, but there really isn't.
### BEGIN HIDDEN TESTS
from
os.path
import
isfile
if
not
isfile(dataset_path('symmetrize_soln.csv')):
symmetrize_df_soln0__ = edges_output.sample(frac=1) \
.sort_values(by=['source', 'ta
rget'])
print(pandas_df_to_markdown_table(symmetrize_df_soln0__))
symmetrize_df_soln0__.to_csv(dataset_path('symmetrize_soln.csv'), index=
False
)
### END HIDDEN TESTS
edges_output_soln = pd.read_csv(dataset_path('symmetrize_soln.csv'))
assert
tibbles_are_equivalent(edges_output, edges_output_soln), \
"Your solution does not produce the expected output."
print("
\n
(Passed.)")
(Passed.)
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Which of the following is NOT true about [[a,b,c][a,d],d]?a. It can be unified with LIlIb. It can be unified with (la,b, c], X/Y]c. It can be unified with LI]d. It can be unified with [(a, b,c], X,Y]
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Recommended textbooks for you
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