Why does my code take different values when i switch the order in a set (knowing that order doesn't matter...
I'm trying to implement a method that returns the edges of a graph, represented by an adjacency list/dictionary.
So to iterate through the dictionary, first I iterated through the keys, then through every value stored in the corresponding key. Inside the nested for-loop, I had a condition where, if a particular edge, say (a,b) is not in the set of edges, then add it to the set -- pass otherwise. On my first run, the method took in edges that are the same -- that is, in the set of edges, there are (a,b) and (b,a).
class Graph():
def __init__(self, grph={}):
self.graph = grph
def get_vertices(self):
for keys in self.graph:
yield keys
def get_edges(self):
edges = set()
for key in self.graph:
for adj_node in self.graph[key]:
if (key, adj_node) not in edges:
edge = (key, adj_node)
edges.add(edge)
else:
pass
return edges
def main():
graph1 = {
'A': ['B','C','D'],
'B': ['A','E'],
'C': ['A', 'D'],
'D': ['A', 'C'],
'E': ['B'],
}
graph_one = Graph(graph1)
print(list(graph_one.get_vertices()))
print(graph_one.get_edges())
if __name__ =='__main__':
main()
the output is:
{('A','B'),('D','A'),('B','A'),('B','E'),('A','D'),('D','C'),('E','B'),('C','D'),('A','C'),('C','A')}
So what I did was that, I just changed the if statement:
"if (adj_node, key) not in edges:"
def get_edges(self):
edges = set()
for key in self.graph:
for adj_node in self.graph[key]:
if (adj_node, key) not in edges:
edge = (key, adj_node)
edges.add(edge)
else:
pass
return edges
Now the output was:
{('C','D'),('A','B'),('E','B'),('A','C'),('A','D')}
Im very curious as to why it is so, and I'd be so thankful if you guys could explain it to me. Thanks in advance!
python python-3.x data-structures
New contributor
add a comment |
I'm trying to implement a method that returns the edges of a graph, represented by an adjacency list/dictionary.
So to iterate through the dictionary, first I iterated through the keys, then through every value stored in the corresponding key. Inside the nested for-loop, I had a condition where, if a particular edge, say (a,b) is not in the set of edges, then add it to the set -- pass otherwise. On my first run, the method took in edges that are the same -- that is, in the set of edges, there are (a,b) and (b,a).
class Graph():
def __init__(self, grph={}):
self.graph = grph
def get_vertices(self):
for keys in self.graph:
yield keys
def get_edges(self):
edges = set()
for key in self.graph:
for adj_node in self.graph[key]:
if (key, adj_node) not in edges:
edge = (key, adj_node)
edges.add(edge)
else:
pass
return edges
def main():
graph1 = {
'A': ['B','C','D'],
'B': ['A','E'],
'C': ['A', 'D'],
'D': ['A', 'C'],
'E': ['B'],
}
graph_one = Graph(graph1)
print(list(graph_one.get_vertices()))
print(graph_one.get_edges())
if __name__ =='__main__':
main()
the output is:
{('A','B'),('D','A'),('B','A'),('B','E'),('A','D'),('D','C'),('E','B'),('C','D'),('A','C'),('C','A')}
So what I did was that, I just changed the if statement:
"if (adj_node, key) not in edges:"
def get_edges(self):
edges = set()
for key in self.graph:
for adj_node in self.graph[key]:
if (adj_node, key) not in edges:
edge = (key, adj_node)
edges.add(edge)
else:
pass
return edges
Now the output was:
{('C','D'),('A','B'),('E','B'),('A','C'),('A','D')}
Im very curious as to why it is so, and I'd be so thankful if you guys could explain it to me. Thanks in advance!
python python-3.x data-structures
New contributor
you mean tosort
the data or use an helper set with sorted data to avoid insertion. What's the type of key and adj_node?
– Jean-François Fabre
2 days ago
The check is irrelevant anyway because the set can only contain one of each value, so you can add the same value multiple times and you'll still only have one instance at the end in your set.
– roganjosh
2 days ago
sorry, i forgot to say (or emphasize, I suppose) that I'm not asking how to remove the duplicates! I was just wondering why it ignored the duplicates when I changed the condition on the if-statement
– Francis Magnusson
2 days ago
That's fine, my comment was somewhat tangential to what you're asking, but I still think it's valid. It's very possible I'm not seeing the full picture but I can't see the purpose of the check at all.
– roganjosh
2 days ago
So simple explaination is thatset
will only holds one instance of each value, and you usedif
to check opposite direction's tuple, so in effect both directions' duplicates are avoided.
– Tiw
2 days ago
add a comment |
I'm trying to implement a method that returns the edges of a graph, represented by an adjacency list/dictionary.
So to iterate through the dictionary, first I iterated through the keys, then through every value stored in the corresponding key. Inside the nested for-loop, I had a condition where, if a particular edge, say (a,b) is not in the set of edges, then add it to the set -- pass otherwise. On my first run, the method took in edges that are the same -- that is, in the set of edges, there are (a,b) and (b,a).
class Graph():
def __init__(self, grph={}):
self.graph = grph
def get_vertices(self):
for keys in self.graph:
yield keys
def get_edges(self):
edges = set()
for key in self.graph:
for adj_node in self.graph[key]:
if (key, adj_node) not in edges:
edge = (key, adj_node)
edges.add(edge)
else:
pass
return edges
def main():
graph1 = {
'A': ['B','C','D'],
'B': ['A','E'],
'C': ['A', 'D'],
'D': ['A', 'C'],
'E': ['B'],
}
graph_one = Graph(graph1)
print(list(graph_one.get_vertices()))
print(graph_one.get_edges())
if __name__ =='__main__':
main()
the output is:
{('A','B'),('D','A'),('B','A'),('B','E'),('A','D'),('D','C'),('E','B'),('C','D'),('A','C'),('C','A')}
So what I did was that, I just changed the if statement:
"if (adj_node, key) not in edges:"
def get_edges(self):
edges = set()
for key in self.graph:
for adj_node in self.graph[key]:
if (adj_node, key) not in edges:
edge = (key, adj_node)
edges.add(edge)
else:
pass
return edges
Now the output was:
{('C','D'),('A','B'),('E','B'),('A','C'),('A','D')}
Im very curious as to why it is so, and I'd be so thankful if you guys could explain it to me. Thanks in advance!
python python-3.x data-structures
New contributor
I'm trying to implement a method that returns the edges of a graph, represented by an adjacency list/dictionary.
So to iterate through the dictionary, first I iterated through the keys, then through every value stored in the corresponding key. Inside the nested for-loop, I had a condition where, if a particular edge, say (a,b) is not in the set of edges, then add it to the set -- pass otherwise. On my first run, the method took in edges that are the same -- that is, in the set of edges, there are (a,b) and (b,a).
class Graph():
def __init__(self, grph={}):
self.graph = grph
def get_vertices(self):
for keys in self.graph:
yield keys
def get_edges(self):
edges = set()
for key in self.graph:
for adj_node in self.graph[key]:
if (key, adj_node) not in edges:
edge = (key, adj_node)
edges.add(edge)
else:
pass
return edges
def main():
graph1 = {
'A': ['B','C','D'],
'B': ['A','E'],
'C': ['A', 'D'],
'D': ['A', 'C'],
'E': ['B'],
}
graph_one = Graph(graph1)
print(list(graph_one.get_vertices()))
print(graph_one.get_edges())
if __name__ =='__main__':
main()
the output is:
{('A','B'),('D','A'),('B','A'),('B','E'),('A','D'),('D','C'),('E','B'),('C','D'),('A','C'),('C','A')}
So what I did was that, I just changed the if statement:
"if (adj_node, key) not in edges:"
def get_edges(self):
edges = set()
for key in self.graph:
for adj_node in self.graph[key]:
if (adj_node, key) not in edges:
edge = (key, adj_node)
edges.add(edge)
else:
pass
return edges
Now the output was:
{('C','D'),('A','B'),('E','B'),('A','C'),('A','D')}
Im very curious as to why it is so, and I'd be so thankful if you guys could explain it to me. Thanks in advance!
python python-3.x data-structures
python python-3.x data-structures
New contributor
New contributor
edited 2 days ago
roganjosh
6,31731228
6,31731228
New contributor
asked 2 days ago
Francis MagnussonFrancis Magnusson
464
464
New contributor
New contributor
you mean tosort
the data or use an helper set with sorted data to avoid insertion. What's the type of key and adj_node?
– Jean-François Fabre
2 days ago
The check is irrelevant anyway because the set can only contain one of each value, so you can add the same value multiple times and you'll still only have one instance at the end in your set.
– roganjosh
2 days ago
sorry, i forgot to say (or emphasize, I suppose) that I'm not asking how to remove the duplicates! I was just wondering why it ignored the duplicates when I changed the condition on the if-statement
– Francis Magnusson
2 days ago
That's fine, my comment was somewhat tangential to what you're asking, but I still think it's valid. It's very possible I'm not seeing the full picture but I can't see the purpose of the check at all.
– roganjosh
2 days ago
So simple explaination is thatset
will only holds one instance of each value, and you usedif
to check opposite direction's tuple, so in effect both directions' duplicates are avoided.
– Tiw
2 days ago
add a comment |
you mean tosort
the data or use an helper set with sorted data to avoid insertion. What's the type of key and adj_node?
– Jean-François Fabre
2 days ago
The check is irrelevant anyway because the set can only contain one of each value, so you can add the same value multiple times and you'll still only have one instance at the end in your set.
– roganjosh
2 days ago
sorry, i forgot to say (or emphasize, I suppose) that I'm not asking how to remove the duplicates! I was just wondering why it ignored the duplicates when I changed the condition on the if-statement
– Francis Magnusson
2 days ago
That's fine, my comment was somewhat tangential to what you're asking, but I still think it's valid. It's very possible I'm not seeing the full picture but I can't see the purpose of the check at all.
– roganjosh
2 days ago
So simple explaination is thatset
will only holds one instance of each value, and you usedif
to check opposite direction's tuple, so in effect both directions' duplicates are avoided.
– Tiw
2 days ago
you mean to
sort
the data or use an helper set with sorted data to avoid insertion. What's the type of key and adj_node?– Jean-François Fabre
2 days ago
you mean to
sort
the data or use an helper set with sorted data to avoid insertion. What's the type of key and adj_node?– Jean-François Fabre
2 days ago
The check is irrelevant anyway because the set can only contain one of each value, so you can add the same value multiple times and you'll still only have one instance at the end in your set.
– roganjosh
2 days ago
The check is irrelevant anyway because the set can only contain one of each value, so you can add the same value multiple times and you'll still only have one instance at the end in your set.
– roganjosh
2 days ago
sorry, i forgot to say (or emphasize, I suppose) that I'm not asking how to remove the duplicates! I was just wondering why it ignored the duplicates when I changed the condition on the if-statement
– Francis Magnusson
2 days ago
sorry, i forgot to say (or emphasize, I suppose) that I'm not asking how to remove the duplicates! I was just wondering why it ignored the duplicates when I changed the condition on the if-statement
– Francis Magnusson
2 days ago
That's fine, my comment was somewhat tangential to what you're asking, but I still think it's valid. It's very possible I'm not seeing the full picture but I can't see the purpose of the check at all.
– roganjosh
2 days ago
That's fine, my comment was somewhat tangential to what you're asking, but I still think it's valid. It's very possible I'm not seeing the full picture but I can't see the purpose of the check at all.
– roganjosh
2 days ago
So simple explaination is that
set
will only holds one instance of each value, and you used if
to check opposite direction's tuple, so in effect both directions' duplicates are avoided.– Tiw
2 days ago
So simple explaination is that
set
will only holds one instance of each value, and you used if
to check opposite direction's tuple, so in effect both directions' duplicates are avoided.– Tiw
2 days ago
add a comment |
2 Answers
2
active
oldest
votes
When we say that sets have no order or that order doesn't matter, it means that {x, y} == {y, x}
. But (a, b)
and (b, a)
are tuples, order matters for them, so (a, b) != (b, a)
and therefore {(a, b), (b, a)}
is a set with two distinct elements, although it's equal to {(b, a), (a, b)}
.
When your code looks like this:
if (adj_node, key) not in edges:
edge = (key, adj_node)
edges.add(edge)
then when the edge a <-> b
is first encountered, it's as (key, adj_node) == (a, b)
and is added to the set. When it's encountered the second (and only other) time, it's as (key, adj_node) == (b, a)
, meaning (adj_node, key) == (a, b)
which is already in the set so (adj_node, key) not in edges
is false and (b, a)
doesn't get added to the set.
1
Looks like i got tuples and sets mixed up! thanks for pointing that out! It took me quite a while to understand your explanation on the if statement, but makes sense to me now; when it checks at "if (adj_node, key) not in edges" it still takes into account the tuple stored previously by the "edge = (key, adj_node)" statement, which is basically the same. Thanks!
– Francis Magnusson
2 days ago
add a comment |
I think it just needs a little change, try this:
def get_edges(self):
edges = set()
for key in self.graph:
for adj_node in self.graph[key]:
if ((key, adj_node) not in edges) and ((adj_node, key) not in edges):
edge = (key, adj_node)
edges.add(edge)
else:
pass
return edges
Update:
So it is an Undigraph.
And it's me overcomplicated this.
And your way is actually better than my checking both ways.
The reason your code succeed, is that set
will only contain one instance of any value.
So each time do the add
, if there's already same tuple exists, it simply won't change the set.
And you already used the if
to check the tuple of opposite direction, so it won't create duplicate edges.
For example, when (a, b)
hits the if
checking, it will check (b,a)
exists in the set or not, if exists, then pass. If not, add (a, b) in the set, if (a, b) exists, the set won't change since only one instace will be in the set.
And later when looped to (b, a), since (a, b) already in the set, the if
will be false and passed.
So by this way, the set is safe, free of duplicate edges.
1
I still don't get the point of the check because the set will only contain one instance of any value
– roganjosh
2 days ago
So just add them to the set. What is theif
check preventing? "If it's not in the set, add it" is no different, from what I can see, to "add it to the set and let the set mechanics mean I only have one instance"
– roganjosh
2 days ago
Or, just add both? The question has 3 upvotes so I'm guessing it's me that's missing something here, but I genuinely don't see it :)
– roganjosh
2 days ago
1
@FrancisMagnusson that's not something to be sorry about. A big, and interesting, part of SO is helping people align their thinking :)
– roganjosh
2 days ago
1
oh is that so haha... thanks, then! :D
– Francis Magnusson
2 days ago
|
show 7 more comments
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2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
When we say that sets have no order or that order doesn't matter, it means that {x, y} == {y, x}
. But (a, b)
and (b, a)
are tuples, order matters for them, so (a, b) != (b, a)
and therefore {(a, b), (b, a)}
is a set with two distinct elements, although it's equal to {(b, a), (a, b)}
.
When your code looks like this:
if (adj_node, key) not in edges:
edge = (key, adj_node)
edges.add(edge)
then when the edge a <-> b
is first encountered, it's as (key, adj_node) == (a, b)
and is added to the set. When it's encountered the second (and only other) time, it's as (key, adj_node) == (b, a)
, meaning (adj_node, key) == (a, b)
which is already in the set so (adj_node, key) not in edges
is false and (b, a)
doesn't get added to the set.
1
Looks like i got tuples and sets mixed up! thanks for pointing that out! It took me quite a while to understand your explanation on the if statement, but makes sense to me now; when it checks at "if (adj_node, key) not in edges" it still takes into account the tuple stored previously by the "edge = (key, adj_node)" statement, which is basically the same. Thanks!
– Francis Magnusson
2 days ago
add a comment |
When we say that sets have no order or that order doesn't matter, it means that {x, y} == {y, x}
. But (a, b)
and (b, a)
are tuples, order matters for them, so (a, b) != (b, a)
and therefore {(a, b), (b, a)}
is a set with two distinct elements, although it's equal to {(b, a), (a, b)}
.
When your code looks like this:
if (adj_node, key) not in edges:
edge = (key, adj_node)
edges.add(edge)
then when the edge a <-> b
is first encountered, it's as (key, adj_node) == (a, b)
and is added to the set. When it's encountered the second (and only other) time, it's as (key, adj_node) == (b, a)
, meaning (adj_node, key) == (a, b)
which is already in the set so (adj_node, key) not in edges
is false and (b, a)
doesn't get added to the set.
1
Looks like i got tuples and sets mixed up! thanks for pointing that out! It took me quite a while to understand your explanation on the if statement, but makes sense to me now; when it checks at "if (adj_node, key) not in edges" it still takes into account the tuple stored previously by the "edge = (key, adj_node)" statement, which is basically the same. Thanks!
– Francis Magnusson
2 days ago
add a comment |
When we say that sets have no order or that order doesn't matter, it means that {x, y} == {y, x}
. But (a, b)
and (b, a)
are tuples, order matters for them, so (a, b) != (b, a)
and therefore {(a, b), (b, a)}
is a set with two distinct elements, although it's equal to {(b, a), (a, b)}
.
When your code looks like this:
if (adj_node, key) not in edges:
edge = (key, adj_node)
edges.add(edge)
then when the edge a <-> b
is first encountered, it's as (key, adj_node) == (a, b)
and is added to the set. When it's encountered the second (and only other) time, it's as (key, adj_node) == (b, a)
, meaning (adj_node, key) == (a, b)
which is already in the set so (adj_node, key) not in edges
is false and (b, a)
doesn't get added to the set.
When we say that sets have no order or that order doesn't matter, it means that {x, y} == {y, x}
. But (a, b)
and (b, a)
are tuples, order matters for them, so (a, b) != (b, a)
and therefore {(a, b), (b, a)}
is a set with two distinct elements, although it's equal to {(b, a), (a, b)}
.
When your code looks like this:
if (adj_node, key) not in edges:
edge = (key, adj_node)
edges.add(edge)
then when the edge a <-> b
is first encountered, it's as (key, adj_node) == (a, b)
and is added to the set. When it's encountered the second (and only other) time, it's as (key, adj_node) == (b, a)
, meaning (adj_node, key) == (a, b)
which is already in the set so (adj_node, key) not in edges
is false and (b, a)
doesn't get added to the set.
answered 2 days ago
Alex HallAlex Hall
21.6k32053
21.6k32053
1
Looks like i got tuples and sets mixed up! thanks for pointing that out! It took me quite a while to understand your explanation on the if statement, but makes sense to me now; when it checks at "if (adj_node, key) not in edges" it still takes into account the tuple stored previously by the "edge = (key, adj_node)" statement, which is basically the same. Thanks!
– Francis Magnusson
2 days ago
add a comment |
1
Looks like i got tuples and sets mixed up! thanks for pointing that out! It took me quite a while to understand your explanation on the if statement, but makes sense to me now; when it checks at "if (adj_node, key) not in edges" it still takes into account the tuple stored previously by the "edge = (key, adj_node)" statement, which is basically the same. Thanks!
– Francis Magnusson
2 days ago
1
1
Looks like i got tuples and sets mixed up! thanks for pointing that out! It took me quite a while to understand your explanation on the if statement, but makes sense to me now; when it checks at "if (adj_node, key) not in edges" it still takes into account the tuple stored previously by the "edge = (key, adj_node)" statement, which is basically the same. Thanks!
– Francis Magnusson
2 days ago
Looks like i got tuples and sets mixed up! thanks for pointing that out! It took me quite a while to understand your explanation on the if statement, but makes sense to me now; when it checks at "if (adj_node, key) not in edges" it still takes into account the tuple stored previously by the "edge = (key, adj_node)" statement, which is basically the same. Thanks!
– Francis Magnusson
2 days ago
add a comment |
I think it just needs a little change, try this:
def get_edges(self):
edges = set()
for key in self.graph:
for adj_node in self.graph[key]:
if ((key, adj_node) not in edges) and ((adj_node, key) not in edges):
edge = (key, adj_node)
edges.add(edge)
else:
pass
return edges
Update:
So it is an Undigraph.
And it's me overcomplicated this.
And your way is actually better than my checking both ways.
The reason your code succeed, is that set
will only contain one instance of any value.
So each time do the add
, if there's already same tuple exists, it simply won't change the set.
And you already used the if
to check the tuple of opposite direction, so it won't create duplicate edges.
For example, when (a, b)
hits the if
checking, it will check (b,a)
exists in the set or not, if exists, then pass. If not, add (a, b) in the set, if (a, b) exists, the set won't change since only one instace will be in the set.
And later when looped to (b, a), since (a, b) already in the set, the if
will be false and passed.
So by this way, the set is safe, free of duplicate edges.
1
I still don't get the point of the check because the set will only contain one instance of any value
– roganjosh
2 days ago
So just add them to the set. What is theif
check preventing? "If it's not in the set, add it" is no different, from what I can see, to "add it to the set and let the set mechanics mean I only have one instance"
– roganjosh
2 days ago
Or, just add both? The question has 3 upvotes so I'm guessing it's me that's missing something here, but I genuinely don't see it :)
– roganjosh
2 days ago
1
@FrancisMagnusson that's not something to be sorry about. A big, and interesting, part of SO is helping people align their thinking :)
– roganjosh
2 days ago
1
oh is that so haha... thanks, then! :D
– Francis Magnusson
2 days ago
|
show 7 more comments
I think it just needs a little change, try this:
def get_edges(self):
edges = set()
for key in self.graph:
for adj_node in self.graph[key]:
if ((key, adj_node) not in edges) and ((adj_node, key) not in edges):
edge = (key, adj_node)
edges.add(edge)
else:
pass
return edges
Update:
So it is an Undigraph.
And it's me overcomplicated this.
And your way is actually better than my checking both ways.
The reason your code succeed, is that set
will only contain one instance of any value.
So each time do the add
, if there's already same tuple exists, it simply won't change the set.
And you already used the if
to check the tuple of opposite direction, so it won't create duplicate edges.
For example, when (a, b)
hits the if
checking, it will check (b,a)
exists in the set or not, if exists, then pass. If not, add (a, b) in the set, if (a, b) exists, the set won't change since only one instace will be in the set.
And later when looped to (b, a), since (a, b) already in the set, the if
will be false and passed.
So by this way, the set is safe, free of duplicate edges.
1
I still don't get the point of the check because the set will only contain one instance of any value
– roganjosh
2 days ago
So just add them to the set. What is theif
check preventing? "If it's not in the set, add it" is no different, from what I can see, to "add it to the set and let the set mechanics mean I only have one instance"
– roganjosh
2 days ago
Or, just add both? The question has 3 upvotes so I'm guessing it's me that's missing something here, but I genuinely don't see it :)
– roganjosh
2 days ago
1
@FrancisMagnusson that's not something to be sorry about. A big, and interesting, part of SO is helping people align their thinking :)
– roganjosh
2 days ago
1
oh is that so haha... thanks, then! :D
– Francis Magnusson
2 days ago
|
show 7 more comments
I think it just needs a little change, try this:
def get_edges(self):
edges = set()
for key in self.graph:
for adj_node in self.graph[key]:
if ((key, adj_node) not in edges) and ((adj_node, key) not in edges):
edge = (key, adj_node)
edges.add(edge)
else:
pass
return edges
Update:
So it is an Undigraph.
And it's me overcomplicated this.
And your way is actually better than my checking both ways.
The reason your code succeed, is that set
will only contain one instance of any value.
So each time do the add
, if there's already same tuple exists, it simply won't change the set.
And you already used the if
to check the tuple of opposite direction, so it won't create duplicate edges.
For example, when (a, b)
hits the if
checking, it will check (b,a)
exists in the set or not, if exists, then pass. If not, add (a, b) in the set, if (a, b) exists, the set won't change since only one instace will be in the set.
And later when looped to (b, a), since (a, b) already in the set, the if
will be false and passed.
So by this way, the set is safe, free of duplicate edges.
I think it just needs a little change, try this:
def get_edges(self):
edges = set()
for key in self.graph:
for adj_node in self.graph[key]:
if ((key, adj_node) not in edges) and ((adj_node, key) not in edges):
edge = (key, adj_node)
edges.add(edge)
else:
pass
return edges
Update:
So it is an Undigraph.
And it's me overcomplicated this.
And your way is actually better than my checking both ways.
The reason your code succeed, is that set
will only contain one instance of any value.
So each time do the add
, if there's already same tuple exists, it simply won't change the set.
And you already used the if
to check the tuple of opposite direction, so it won't create duplicate edges.
For example, when (a, b)
hits the if
checking, it will check (b,a)
exists in the set or not, if exists, then pass. If not, add (a, b) in the set, if (a, b) exists, the set won't change since only one instace will be in the set.
And later when looped to (b, a), since (a, b) already in the set, the if
will be false and passed.
So by this way, the set is safe, free of duplicate edges.
edited yesterday
answered 2 days ago
TiwTiw
1,570618
1,570618
1
I still don't get the point of the check because the set will only contain one instance of any value
– roganjosh
2 days ago
So just add them to the set. What is theif
check preventing? "If it's not in the set, add it" is no different, from what I can see, to "add it to the set and let the set mechanics mean I only have one instance"
– roganjosh
2 days ago
Or, just add both? The question has 3 upvotes so I'm guessing it's me that's missing something here, but I genuinely don't see it :)
– roganjosh
2 days ago
1
@FrancisMagnusson that's not something to be sorry about. A big, and interesting, part of SO is helping people align their thinking :)
– roganjosh
2 days ago
1
oh is that so haha... thanks, then! :D
– Francis Magnusson
2 days ago
|
show 7 more comments
1
I still don't get the point of the check because the set will only contain one instance of any value
– roganjosh
2 days ago
So just add them to the set. What is theif
check preventing? "If it's not in the set, add it" is no different, from what I can see, to "add it to the set and let the set mechanics mean I only have one instance"
– roganjosh
2 days ago
Or, just add both? The question has 3 upvotes so I'm guessing it's me that's missing something here, but I genuinely don't see it :)
– roganjosh
2 days ago
1
@FrancisMagnusson that's not something to be sorry about. A big, and interesting, part of SO is helping people align their thinking :)
– roganjosh
2 days ago
1
oh is that so haha... thanks, then! :D
– Francis Magnusson
2 days ago
1
1
I still don't get the point of the check because the set will only contain one instance of any value
– roganjosh
2 days ago
I still don't get the point of the check because the set will only contain one instance of any value
– roganjosh
2 days ago
So just add them to the set. What is the
if
check preventing? "If it's not in the set, add it" is no different, from what I can see, to "add it to the set and let the set mechanics mean I only have one instance"– roganjosh
2 days ago
So just add them to the set. What is the
if
check preventing? "If it's not in the set, add it" is no different, from what I can see, to "add it to the set and let the set mechanics mean I only have one instance"– roganjosh
2 days ago
Or, just add both? The question has 3 upvotes so I'm guessing it's me that's missing something here, but I genuinely don't see it :)
– roganjosh
2 days ago
Or, just add both? The question has 3 upvotes so I'm guessing it's me that's missing something here, but I genuinely don't see it :)
– roganjosh
2 days ago
1
1
@FrancisMagnusson that's not something to be sorry about. A big, and interesting, part of SO is helping people align their thinking :)
– roganjosh
2 days ago
@FrancisMagnusson that's not something to be sorry about. A big, and interesting, part of SO is helping people align their thinking :)
– roganjosh
2 days ago
1
1
oh is that so haha... thanks, then! :D
– Francis Magnusson
2 days ago
oh is that so haha... thanks, then! :D
– Francis Magnusson
2 days ago
|
show 7 more comments
Francis Magnusson is a new contributor. Be nice, and check out our Code of Conduct.
Francis Magnusson is a new contributor. Be nice, and check out our Code of Conduct.
Francis Magnusson is a new contributor. Be nice, and check out our Code of Conduct.
Francis Magnusson is a new contributor. Be nice, and check out our Code of Conduct.
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you mean to
sort
the data or use an helper set with sorted data to avoid insertion. What's the type of key and adj_node?– Jean-François Fabre
2 days ago
The check is irrelevant anyway because the set can only contain one of each value, so you can add the same value multiple times and you'll still only have one instance at the end in your set.
– roganjosh
2 days ago
sorry, i forgot to say (or emphasize, I suppose) that I'm not asking how to remove the duplicates! I was just wondering why it ignored the duplicates when I changed the condition on the if-statement
– Francis Magnusson
2 days ago
That's fine, my comment was somewhat tangential to what you're asking, but I still think it's valid. It's very possible I'm not seeing the full picture but I can't see the purpose of the check at all.
– roganjosh
2 days ago
So simple explaination is that
set
will only holds one instance of each value, and you usedif
to check opposite direction's tuple, so in effect both directions' duplicates are avoided.– Tiw
2 days ago