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Boxiang Sun
Pyston
Commits
6dab00bb
Commit
6dab00bb
authored
Feb 24, 2015
by
Kevin Modzelewski
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Add the schulze minibenchmark
parent
82a63d19
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+1389
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minibenchmarks/schulze.py
minibenchmarks/schulze.py
+89
-0
minibenchmarks/schulze.votes
minibenchmarks/schulze.votes
+1300
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minibenchmarks/schulze.py
0 → 100644
View file @
6dab00bb
# Read the votes
import
os
fp
=
open
(
os
.
path
.
join
(
os
.
path
.
dirname
(
__file__
),
'schulze.votes'
),
'r'
)
lines
=
fp
.
readlines
()
fp
.
close
()
# Strip superfluous information
lines
=
[
line
[
14
:].
rstrip
()
for
line
in
lines
]
votes
=
[
line
.
split
(
' '
)
for
line
in
lines
]
# Make a canonical set of all the candidates
candidates
=
{}
for
line
in
votes
:
for
memberId
in
line
:
candidates
[
memberId
]
=
1
# Go from member number to an index
for
i
,
k
in
enumerate
(
candidates
.
keys
()):
candidates
[
k
]
=
i
# And vice versa
reverseCandidates
=
{}
for
k
,
v
in
candidates
.
items
():
reverseCandidates
[
v
]
=
k
# Turn the votes in to an index number
numbers
=
[[
candidates
[
memberId
]
for
memberId
in
line
]
for
line
in
votes
]
size
=
len
(
candidates
)
# Initialize the d and p matrixes
row
=
[]
for
i
in
range
(
size
):
row
.
append
(
0
)
d
=
[]
p
=
[]
for
i
in
range
(
size
):
d
.
append
(
row
[:])
p
.
append
(
row
[:])
# Fill in the preferences in the d matrix
for
i
in
range
(
size
):
for
line
in
numbers
:
for
entry
in
line
:
if
entry
==
i
:
break
d
[
entry
][
i
]
+=
1
# Calculate the p matrix. Algorithm copied straight from wikipedia
# article http://en.wikipedia.org/wiki/Schulze_method
for
i
in
range
(
size
):
for
j
in
range
(
size
):
if
i
!=
j
:
if
d
[
i
][
j
]
>
d
[
j
][
i
]:
p
[
i
][
j
]
=
d
[
i
][
j
]
else
:
p
[
i
][
j
]
=
0
for
i
in
range
(
size
):
for
j
in
range
(
size
):
if
i
!=
j
:
for
k
in
range
(
size
):
if
i
!=
k
:
if
j
!=
k
:
p
[
j
][
k
]
=
max
(
p
[
j
][
k
],
min
(
p
[
j
][
i
],
p
[
i
][
k
]))
# Find the best candidate (p[candidate, X] >= p[X, candidate])
# Put the candidate on the final list, remove the candidate from p and
# repeat
order
=
[]
cl
=
range
(
size
)
while
cl
:
for
c
in
cl
:
for
n
in
cl
:
if
c
!=
n
:
if
p
[
c
][
n
]
<
p
[
n
][
c
]:
break
# Found a better candidate
else
:
order
.
append
(
c
)
cl
.
remove
(
c
)
# Display the results
j
=
0
for
i
in
order
:
j
+=
1
print
'%3d %s'
%
(
j
,
reverseCandidates
[
i
])
minibenchmarks/schulze.votes
0 → 100644
View file @
6dab00bb
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