How to invert the order of elements in a heapq heap with object comparison functions?
First of all, I read this SO question but it actually doesn't include my desired approach. In addition, negating the actual values is not applicable for my use case.
Heapq Docs: https://docs.python.org/3.6/library/heapq.html
Assume I have a list of dataclass objects in my heap. Only the a property determines the order of objects.
import heapq
from dataclasses import dataclass
@dataclass
class C:
a: int
b: int
def __lt__(self, other):
return self.a < other.a
l=[C(2,1),C(9,109),C(2,4),C(9,4)]
print(heapq.heappop(l)) # C(a=2, b=1)
print(heapq.heappop(l)) # C(a=2, b=4)
print(heapq.heappop(l)) # C(a=9, b=109)
print(heapq.heappop(l)) # C(a=9, b=4)
Now I want to have an inverted order. Therefore, I changed the line return self.a < other.a to return self.a > other.a. The result:
import heapq
from dataclasses import dataclass
@dataclass
class C:
a: int
b: int
def __lt__(self, other):
return self.a > other.a
l=[C(2,1),C(9,109),C(2,4),C(9,4)]
print(heapq.heappop(l)) # C(a=2, b=1)
print(heapq.heappop(l)) # C(a=9, b=109)
print(heapq.heappop(l)) # C(a=9, b=4)
print(heapq.heappop(l)) # C(a=2, b=4)
The desired result should be one of the four solutions:
C(a=9, b=109) C(a=9, b=4) C(a=9, b=109) C(a=9, b=4)
C(a=9, b=4) C(a=9, b=109) C(a=9, b=4) C(a=9, b=109)
C(a=2, b=1) C(a=2, b=1) C(a=2, b=4) C(a=2, b=4)
C(a=2, b=4) C(a=2, b=4) C(a=2, b=1) C(a=2, b=1)
Probably, not all pairs of objects are compared by heapq that would explain the strange order. However, is it still possible to get an inverted order?
Do I have to provide more object comparison methods?
object.__lt__(self, other)
object.__le__(self, other)
object.__eq__(self, other)
object.__ne__(self, other)
object.__gt__(self, other)
object.__ge__(self, other)
If you have an completely other approach, do not hesitate!
python python-3.x heap python-3.6
asked Nov 23 '18 at 16:23
d4rtyd4rty
1,36321333
add a comment |
First of all, I read this SO question but it actually doesn't include my desired approach. In addition, negating the actual values is not applicable for my use case.
Heapq Docs: https://docs.python.org/3.6/library/heapq.html
Assume I have a list of dataclass objects in my heap. Only the a property determines the order of objects.
import heapq
from dataclasses import dataclass
@dataclass
class C:
a: int
b: int
def __lt__(self, other):
return self.a < other.a
l=[C(2,1),C(9,109),C(2,4),C(9,4)]
print(heapq.heappop(l)) # C(a=2, b=1)
print(heapq.heappop(l)) # C(a=2, b=4)
print(heapq.heappop(l)) # C(a=9, b=109)
print(heapq.heappop(l)) # C(a=9, b=4)
Now I want to have an inverted order. Therefore, I changed the line return self.a < other.a to return self.a > other.a. The result:
import heapq
from dataclasses import dataclass
@dataclass
class C:
a: int
b: int
def __lt__(self, other):
return self.a > other.a
l=[C(2,1),C(9,109),C(2,4),C(9,4)]
print(heapq.heappop(l)) # C(a=2, b=1)
print(heapq.heappop(l)) # C(a=9, b=109)
print(heapq.heappop(l)) # C(a=9, b=4)
print(heapq.heappop(l)) # C(a=2, b=4)
The desired result should be one of the four solutions:
C(a=9, b=109) C(a=9, b=4) C(a=9, b=109) C(a=9, b=4)
C(a=9, b=4) C(a=9, b=109) C(a=9, b=4) C(a=9, b=109)
C(a=2, b=1) C(a=2, b=1) C(a=2, b=4) C(a=2, b=4)
C(a=2, b=4) C(a=2, b=4) C(a=2, b=1) C(a=2, b=1)
Probably, not all pairs of objects are compared by heapq that would explain the strange order. However, is it still possible to get an inverted order?
Do I have to provide more object comparison methods?
object.__lt__(self, other)
object.__le__(self, other)
object.__eq__(self, other)
object.__ne__(self, other)
object.__gt__(self, other)
object.__ge__(self, other)
If you have an completely other approach, do not hesitate!
python python-3.x heap python-3.6
asked Nov 23 '18 at 16:23
d4rtyd4rty
1,36321333
add a comment |
First of all, I read this SO question but it actually doesn't include my desired approach. In addition, negating the actual values is not applicable for my use case.
Heapq Docs: https://docs.python.org/3.6/library/heapq.html
Assume I have a list of dataclass objects in my heap. Only the a property determines the order of objects.
import heapq
from dataclasses import dataclass
@dataclass
class C:
a: int
b: int
def __lt__(self, other):
return self.a < other.a
l=[C(2,1),C(9,109),C(2,4),C(9,4)]
print(heapq.heappop(l)) # C(a=2, b=1)
print(heapq.heappop(l)) # C(a=2, b=4)
print(heapq.heappop(l)) # C(a=9, b=109)
print(heapq.heappop(l)) # C(a=9, b=4)
Now I want to have an inverted order. Therefore, I changed the line return self.a < other.a to return self.a > other.a. The result:
import heapq
from dataclasses import dataclass
@dataclass
class C:
a: int
b: int
def __lt__(self, other):
return self.a > other.a
l=[C(2,1),C(9,109),C(2,4),C(9,4)]
print(heapq.heappop(l)) # C(a=2, b=1)
print(heapq.heappop(l)) # C(a=9, b=109)
print(heapq.heappop(l)) # C(a=9, b=4)
print(heapq.heappop(l)) # C(a=2, b=4)
The desired result should be one of the four solutions:
C(a=9, b=109) C(a=9, b=4) C(a=9, b=109) C(a=9, b=4)
C(a=9, b=4) C(a=9, b=109) C(a=9, b=4) C(a=9, b=109)
C(a=2, b=1) C(a=2, b=1) C(a=2, b=4) C(a=2, b=4)
C(a=2, b=4) C(a=2, b=4) C(a=2, b=1) C(a=2, b=1)
Probably, not all pairs of objects are compared by heapq that would explain the strange order. However, is it still possible to get an inverted order?
Do I have to provide more object comparison methods?
object.__lt__(self, other)
object.__le__(self, other)
object.__eq__(self, other)
object.__ne__(self, other)
object.__gt__(self, other)
object.__ge__(self, other)
If you have an completely other approach, do not hesitate!
python python-3.x heap python-3.6
asked Nov 23 '18 at 16:23
d4rtyd4rty
1,36321333
First of all, I read this SO question but it actually doesn't include my desired approach. In addition, negating the actual values is not applicable for my use case.
Heapq Docs: https://docs.python.org/3.6/library/heapq.html
Assume I have a list of dataclass objects in my heap. Only the a property determines the order of objects.
import heapq
from dataclasses import dataclass
@dataclass
class C:
a: int
b: int
def __lt__(self, other):
return self.a < other.a
l=[C(2,1),C(9,109),C(2,4),C(9,4)]
print(heapq.heappop(l)) # C(a=2, b=1)
print(heapq.heappop(l)) # C(a=2, b=4)
print(heapq.heappop(l)) # C(a=9, b=109)
print(heapq.heappop(l)) # C(a=9, b=4)
Now I want to have an inverted order. Therefore, I changed the line return self.a < other.a to return self.a > other.a. The result:
import heapq
from dataclasses import dataclass
@dataclass
class C:
a: int
b: int
def __lt__(self, other):
return self.a > other.a
l=[C(2,1),C(9,109),C(2,4),C(9,4)]
print(heapq.heappop(l)) # C(a=2, b=1)
print(heapq.heappop(l)) # C(a=9, b=109)
print(heapq.heappop(l)) # C(a=9, b=4)
print(heapq.heappop(l)) # C(a=2, b=4)
The desired result should be one of the four solutions:
C(a=9, b=109) C(a=9, b=4) C(a=9, b=109) C(a=9, b=4)
C(a=9, b=4) C(a=9, b=109) C(a=9, b=4) C(a=9, b=109)
C(a=2, b=1) C(a=2, b=1) C(a=2, b=4) C(a=2, b=4)
C(a=2, b=4) C(a=2, b=4) C(a=2, b=1) C(a=2, b=1)
Probably, not all pairs of objects are compared by heapq that would explain the strange order. However, is it still possible to get an inverted order?
Do I have to provide more object comparison methods?
object.__lt__(self, other)
object.__le__(self, other)
object.__eq__(self, other)
object.__ne__(self, other)
object.__gt__(self, other)
object.__ge__(self, other)
If you have an completely other approach, do not hesitate!
python python-3.x heap python-3.6
python python-3.x heap python-3.6
asked Nov 23 '18 at 16:23
d4rtyd4rty
1,36321333
asked Nov 23 '18 at 16:23
d4rtyd4rty
1,36321333
edited Nov 23 '18 at 16:29
d4rty
asked Nov 23 '18 at 16:23
d4rtyd4rty
1,36321333
asked Nov 23 '18 at 16:23
d4rtyd4rty
1,36321333
asked Nov 23 '18 at 16:23
d4rtyd4rty
1,36321333
1,36321333
add a comment |
add a comment |
1 Answer
1
active
oldest
votes
You need to make l into a heap using heapify
from heapq import heapify, heappop
from dataclasses import dataclass
@dataclass
class C:
a: int
b: int
def __lt__(self, other):
return self.a > other.a
l=[C(2,1),C(9,109),C(2,4),C(9,4)]
heapify(l)
while l:
print(heappop(l))
prints
C(a=9, b=4)
C(a=9, b=109)
C(a=2, b=1)
C(a=2, b=4)
edited Nov 23 '18 at 16:40
Jon Clements♦
99.5k19174219
answered Nov 23 '18 at 16:30
Patrick HaughPatrick Haugh
29k92747
But why is it working in the first case (in my question)? The list isn't ordered.
– d4rty
Nov 23 '18 at 16:41
1
Because your example list happens to satisfy the heap invariant when<isn't reversed, but isn't a valid heap when<is reversed. The Theory section of theheapqdocumentation would be a good place to start reading, though I also recommend you actually try implementing all of the functions to understand how they work.
– Patrick Haugh
Nov 23 '18 at 16:47
add a comment |
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1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
You need to make l into a heap using heapify
from heapq import heapify, heappop
from dataclasses import dataclass
@dataclass
class C:
a: int
b: int
def __lt__(self, other):
return self.a > other.a
l=[C(2,1),C(9,109),C(2,4),C(9,4)]
heapify(l)
while l:
print(heappop(l))
prints
C(a=9, b=4)
C(a=9, b=109)
C(a=2, b=1)
C(a=2, b=4)
edited Nov 23 '18 at 16:40
Jon Clements♦
99.5k19174219
answered Nov 23 '18 at 16:30
Patrick HaughPatrick Haugh
29k92747
But why is it working in the first case (in my question)? The list isn't ordered.
– d4rty
Nov 23 '18 at 16:41
1
Because your example list happens to satisfy the heap invariant when<isn't reversed, but isn't a valid heap when<is reversed. The Theory section of theheapqdocumentation would be a good place to start reading, though I also recommend you actually try implementing all of the functions to understand how they work.
– Patrick Haugh
Nov 23 '18 at 16:47
add a comment |
You need to make l into a heap using heapify
from heapq import heapify, heappop
from dataclasses import dataclass
@dataclass
class C:
a: int
b: int
def __lt__(self, other):
return self.a > other.a
l=[C(2,1),C(9,109),C(2,4),C(9,4)]
heapify(l)
while l:
print(heappop(l))
prints
C(a=9, b=4)
C(a=9, b=109)
C(a=2, b=1)
C(a=2, b=4)
edited Nov 23 '18 at 16:40
Jon Clements♦
99.5k19174219
answered Nov 23 '18 at 16:30
Patrick HaughPatrick Haugh
29k92747
But why is it working in the first case (in my question)? The list isn't ordered.
– d4rty
Nov 23 '18 at 16:41
1
Because your example list happens to satisfy the heap invariant when<isn't reversed, but isn't a valid heap when<is reversed. The Theory section of theheapqdocumentation would be a good place to start reading, though I also recommend you actually try implementing all of the functions to understand how they work.
– Patrick Haugh
Nov 23 '18 at 16:47
add a comment |
You need to make l into a heap using heapify
from heapq import heapify, heappop
from dataclasses import dataclass
@dataclass
class C:
a: int
b: int
def __lt__(self, other):
return self.a > other.a
l=[C(2,1),C(9,109),C(2,4),C(9,4)]
heapify(l)
while l:
print(heappop(l))
prints
C(a=9, b=4)
C(a=9, b=109)
C(a=2, b=1)
C(a=2, b=4)
edited Nov 23 '18 at 16:40
Jon Clements♦
99.5k19174219
answered Nov 23 '18 at 16:30
Patrick HaughPatrick Haugh
29k92747
You need to make l into a heap using heapify
from heapq import heapify, heappop
from dataclasses import dataclass
@dataclass
class C:
a: int
b: int
def __lt__(self, other):
return self.a > other.a
l=[C(2,1),C(9,109),C(2,4),C(9,4)]
heapify(l)
while l:
print(heappop(l))
prints
C(a=9, b=4)
C(a=9, b=109)
C(a=2, b=1)
C(a=2, b=4)
edited Nov 23 '18 at 16:40
Jon Clements♦
99.5k19174219
answered Nov 23 '18 at 16:30
Patrick HaughPatrick Haugh
29k92747
edited Nov 23 '18 at 16:40
Jon Clements♦
99.5k19174219
edited Nov 23 '18 at 16:40
Jon Clements♦
99.5k19174219
edited Nov 23 '18 at 16:40
Jon Clements♦
99.5k19174219
99.5k19174219
answered Nov 23 '18 at 16:30
Patrick HaughPatrick Haugh
29k92747
answered Nov 23 '18 at 16:30
Patrick HaughPatrick Haugh
29k92747
answered Nov 23 '18 at 16:30
Patrick HaughPatrick Haugh
29k92747
29k92747
But why is it working in the first case (in my question)? The list isn't ordered.
– d4rty
Nov 23 '18 at 16:41
1
Because your example list happens to satisfy the heap invariant when<isn't reversed, but isn't a valid heap when<is reversed. The Theory section of theheapqdocumentation would be a good place to start reading, though I also recommend you actually try implementing all of the functions to understand how they work.
– Patrick Haugh
Nov 23 '18 at 16:47
add a comment |
But why is it working in the first case (in my question)? The list isn't ordered.
– d4rty
Nov 23 '18 at 16:41
1
Because your example list happens to satisfy the heap invariant when<isn't reversed, but isn't a valid heap when<is reversed. The Theory section of theheapqdocumentation would be a good place to start reading, though I also recommend you actually try implementing all of the functions to understand how they work.
– Patrick Haugh
Nov 23 '18 at 16:47
But why is it working in the first case (in my question)? The list isn't ordered.
– d4rty
Nov 23 '18 at 16:41
But why is it working in the first case (in my question)? The list isn't ordered.
– d4rty
Nov 23 '18 at 16:41
1
1
Because your example list happens to satisfy the heap invariant when
< isn't reversed, but isn't a valid heap when < is reversed. The Theory section of the heapq documentation would be a good place to start reading, though I also recommend you actually try implementing all of the functions to understand how they work.– Patrick Haugh
Nov 23 '18 at 16:47
Because your example list happens to satisfy the heap invariant when
< isn't reversed, but isn't a valid heap when < is reversed. The Theory section of the heapq documentation would be a good place to start reading, though I also recommend you actually try implementing all of the functions to understand how they work.– Patrick Haugh
Nov 23 '18 at 16:47
add a comment |
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