Selecting lowest position from 'array'
$begingroup$
My goal is to move a 'monster' (mX, mY
) in a 2d grid towards the player (pX, pY
). The monster can move 8 directions.
I have working code for this, but I'm very new to Python and have a strong inclination it is awful and there is faster ways to do it.
I do this by creating a 3 x 3 array around the monsters position (array slot 4), and filling it with the distance from that array position to the player. Then I check if any are lower than the monsters current distance, and if so, move the monster to it.
Here is my current code, apologies if it makes you puke, I'm still learning the ropes.
# get the distance between the monster and player
dist = math.hypot(pX - mX, pY - mY)
if dist > 1.5 and dist < 10:
# make an 'array' grid to store updated distances in
goto = np.full((3, 3), 10, dtype=float)
# if each position in the array passes a
# collision check, add each new distance
if collisionCheck(mID, (mX-1), (mY-1), mMap) == 0:
goto[0][0] = round(math.hypot(pX - (mX-1), pY - (mY-1)), 1)
if collisionCheck(mID, mX, (mY-1), mMap) == 0:
goto[0][1] = round(math.hypot(pX - mX, pY - (mY-1)), 1)
if collisionCheck(mID, (mX+1), (mY-1), mMap) == 0:
goto[0][2] = round(math.hypot(pX - (mX+1), pY - (mY-1)), 1)
if collisionCheck(mID, (mX-1), mY, mMap) == 0:
goto[1][0] = round(math.hypot(pX - (mX-1), pY - mY), 1)
# goto[1][1] is skipped since that is the monsters current position
if collisionCheck(mID, (mX+1), mY, mMap) == 0:
goto[1][2] = round(math.hypot(pX - (mX+1), pY - mY), 1)
if collisionCheck(mID, (mX-1), (mY+1), mMap) == 0:
goto[2][0] = round(math.hypot(pX - (mX-1), pY - (mY+1)), 1)
if collisionCheck(mID, mX, (mY+1), mMap) == 0:
goto[2][1] = round(math.hypot(pX - mX, pY - (mY+1)), 1)
if collisionCheck(mID, (mX+1), (mY+1), mMap) == 0:
goto[2][2] = round(math.hypot(pX - (mX+1), pY - (mY+1)), 1)
# get the lowest distance, and its key
lowest = goto.min()
lowestKey = goto.argmin()
# if the lowest distance is lower than monsters current position, move
if lowest < dist:
if lowestKey == 0:
newX = mX - 1
newY = mY - 1
if lowestKey == 1:
newY = mY - 1
if lowestKey == 2:
newX = mX + 1
newY = mY - 1
if lowestKey == 3:
newX = mX - 1
if lowestKey == 5:
newX = mX + 1
if lowestKey == 6:
newY = mY + 1
newX = mX - 1
if lowestKey == 7:
newY = mY + 1
if lowestKey == 8:
newX = mX + 1
newY = mY + 1
What is the cleanest, simplest, and fastest way to do what I'm doing? This is going to loop through many monsters at once!
python array
New contributor
$endgroup$
add a comment |
$begingroup$
My goal is to move a 'monster' (mX, mY
) in a 2d grid towards the player (pX, pY
). The monster can move 8 directions.
I have working code for this, but I'm very new to Python and have a strong inclination it is awful and there is faster ways to do it.
I do this by creating a 3 x 3 array around the monsters position (array slot 4), and filling it with the distance from that array position to the player. Then I check if any are lower than the monsters current distance, and if so, move the monster to it.
Here is my current code, apologies if it makes you puke, I'm still learning the ropes.
# get the distance between the monster and player
dist = math.hypot(pX - mX, pY - mY)
if dist > 1.5 and dist < 10:
# make an 'array' grid to store updated distances in
goto = np.full((3, 3), 10, dtype=float)
# if each position in the array passes a
# collision check, add each new distance
if collisionCheck(mID, (mX-1), (mY-1), mMap) == 0:
goto[0][0] = round(math.hypot(pX - (mX-1), pY - (mY-1)), 1)
if collisionCheck(mID, mX, (mY-1), mMap) == 0:
goto[0][1] = round(math.hypot(pX - mX, pY - (mY-1)), 1)
if collisionCheck(mID, (mX+1), (mY-1), mMap) == 0:
goto[0][2] = round(math.hypot(pX - (mX+1), pY - (mY-1)), 1)
if collisionCheck(mID, (mX-1), mY, mMap) == 0:
goto[1][0] = round(math.hypot(pX - (mX-1), pY - mY), 1)
# goto[1][1] is skipped since that is the monsters current position
if collisionCheck(mID, (mX+1), mY, mMap) == 0:
goto[1][2] = round(math.hypot(pX - (mX+1), pY - mY), 1)
if collisionCheck(mID, (mX-1), (mY+1), mMap) == 0:
goto[2][0] = round(math.hypot(pX - (mX-1), pY - (mY+1)), 1)
if collisionCheck(mID, mX, (mY+1), mMap) == 0:
goto[2][1] = round(math.hypot(pX - mX, pY - (mY+1)), 1)
if collisionCheck(mID, (mX+1), (mY+1), mMap) == 0:
goto[2][2] = round(math.hypot(pX - (mX+1), pY - (mY+1)), 1)
# get the lowest distance, and its key
lowest = goto.min()
lowestKey = goto.argmin()
# if the lowest distance is lower than monsters current position, move
if lowest < dist:
if lowestKey == 0:
newX = mX - 1
newY = mY - 1
if lowestKey == 1:
newY = mY - 1
if lowestKey == 2:
newX = mX + 1
newY = mY - 1
if lowestKey == 3:
newX = mX - 1
if lowestKey == 5:
newX = mX + 1
if lowestKey == 6:
newY = mY + 1
newX = mX - 1
if lowestKey == 7:
newY = mY + 1
if lowestKey == 8:
newX = mX + 1
newY = mY + 1
What is the cleanest, simplest, and fastest way to do what I'm doing? This is going to loop through many monsters at once!
python array
New contributor
$endgroup$
$begingroup$
Could you add more description, how big is the board, is the player always in the 3x3 grid?
$endgroup$
– Peilonrayz
5 hours ago
$begingroup$
The board could be any size, this just creates a temporary array around the monster to decide which square it will move to
$endgroup$
– user1022585
5 hours ago
$begingroup$
Is there anything the monster can collide with?
$endgroup$
– Peilonrayz
5 hours ago
$begingroup$
The collisioncheck() routine returns a 0 or a 1. Its an external def that checks if any other monsters, players or walls are at that position.
$endgroup$
– user1022585
5 hours ago
$begingroup$
Hopefully I've not messed up my math when converting to this comment, but you can use:d = math.degrees(math.atan((pY - mY) / (pX - mY)))
and'N NE E SE S SW W NW'.split()[int((d + 22.5) // 45)]
to know which direction you want to go if there are no collisions.
$endgroup$
– Peilonrayz
4 hours ago
add a comment |
$begingroup$
My goal is to move a 'monster' (mX, mY
) in a 2d grid towards the player (pX, pY
). The monster can move 8 directions.
I have working code for this, but I'm very new to Python and have a strong inclination it is awful and there is faster ways to do it.
I do this by creating a 3 x 3 array around the monsters position (array slot 4), and filling it with the distance from that array position to the player. Then I check if any are lower than the monsters current distance, and if so, move the monster to it.
Here is my current code, apologies if it makes you puke, I'm still learning the ropes.
# get the distance between the monster and player
dist = math.hypot(pX - mX, pY - mY)
if dist > 1.5 and dist < 10:
# make an 'array' grid to store updated distances in
goto = np.full((3, 3), 10, dtype=float)
# if each position in the array passes a
# collision check, add each new distance
if collisionCheck(mID, (mX-1), (mY-1), mMap) == 0:
goto[0][0] = round(math.hypot(pX - (mX-1), pY - (mY-1)), 1)
if collisionCheck(mID, mX, (mY-1), mMap) == 0:
goto[0][1] = round(math.hypot(pX - mX, pY - (mY-1)), 1)
if collisionCheck(mID, (mX+1), (mY-1), mMap) == 0:
goto[0][2] = round(math.hypot(pX - (mX+1), pY - (mY-1)), 1)
if collisionCheck(mID, (mX-1), mY, mMap) == 0:
goto[1][0] = round(math.hypot(pX - (mX-1), pY - mY), 1)
# goto[1][1] is skipped since that is the monsters current position
if collisionCheck(mID, (mX+1), mY, mMap) == 0:
goto[1][2] = round(math.hypot(pX - (mX+1), pY - mY), 1)
if collisionCheck(mID, (mX-1), (mY+1), mMap) == 0:
goto[2][0] = round(math.hypot(pX - (mX-1), pY - (mY+1)), 1)
if collisionCheck(mID, mX, (mY+1), mMap) == 0:
goto[2][1] = round(math.hypot(pX - mX, pY - (mY+1)), 1)
if collisionCheck(mID, (mX+1), (mY+1), mMap) == 0:
goto[2][2] = round(math.hypot(pX - (mX+1), pY - (mY+1)), 1)
# get the lowest distance, and its key
lowest = goto.min()
lowestKey = goto.argmin()
# if the lowest distance is lower than monsters current position, move
if lowest < dist:
if lowestKey == 0:
newX = mX - 1
newY = mY - 1
if lowestKey == 1:
newY = mY - 1
if lowestKey == 2:
newX = mX + 1
newY = mY - 1
if lowestKey == 3:
newX = mX - 1
if lowestKey == 5:
newX = mX + 1
if lowestKey == 6:
newY = mY + 1
newX = mX - 1
if lowestKey == 7:
newY = mY + 1
if lowestKey == 8:
newX = mX + 1
newY = mY + 1
What is the cleanest, simplest, and fastest way to do what I'm doing? This is going to loop through many monsters at once!
python array
New contributor
$endgroup$
My goal is to move a 'monster' (mX, mY
) in a 2d grid towards the player (pX, pY
). The monster can move 8 directions.
I have working code for this, but I'm very new to Python and have a strong inclination it is awful and there is faster ways to do it.
I do this by creating a 3 x 3 array around the monsters position (array slot 4), and filling it with the distance from that array position to the player. Then I check if any are lower than the monsters current distance, and if so, move the monster to it.
Here is my current code, apologies if it makes you puke, I'm still learning the ropes.
# get the distance between the monster and player
dist = math.hypot(pX - mX, pY - mY)
if dist > 1.5 and dist < 10:
# make an 'array' grid to store updated distances in
goto = np.full((3, 3), 10, dtype=float)
# if each position in the array passes a
# collision check, add each new distance
if collisionCheck(mID, (mX-1), (mY-1), mMap) == 0:
goto[0][0] = round(math.hypot(pX - (mX-1), pY - (mY-1)), 1)
if collisionCheck(mID, mX, (mY-1), mMap) == 0:
goto[0][1] = round(math.hypot(pX - mX, pY - (mY-1)), 1)
if collisionCheck(mID, (mX+1), (mY-1), mMap) == 0:
goto[0][2] = round(math.hypot(pX - (mX+1), pY - (mY-1)), 1)
if collisionCheck(mID, (mX-1), mY, mMap) == 0:
goto[1][0] = round(math.hypot(pX - (mX-1), pY - mY), 1)
# goto[1][1] is skipped since that is the monsters current position
if collisionCheck(mID, (mX+1), mY, mMap) == 0:
goto[1][2] = round(math.hypot(pX - (mX+1), pY - mY), 1)
if collisionCheck(mID, (mX-1), (mY+1), mMap) == 0:
goto[2][0] = round(math.hypot(pX - (mX-1), pY - (mY+1)), 1)
if collisionCheck(mID, mX, (mY+1), mMap) == 0:
goto[2][1] = round(math.hypot(pX - mX, pY - (mY+1)), 1)
if collisionCheck(mID, (mX+1), (mY+1), mMap) == 0:
goto[2][2] = round(math.hypot(pX - (mX+1), pY - (mY+1)), 1)
# get the lowest distance, and its key
lowest = goto.min()
lowestKey = goto.argmin()
# if the lowest distance is lower than monsters current position, move
if lowest < dist:
if lowestKey == 0:
newX = mX - 1
newY = mY - 1
if lowestKey == 1:
newY = mY - 1
if lowestKey == 2:
newX = mX + 1
newY = mY - 1
if lowestKey == 3:
newX = mX - 1
if lowestKey == 5:
newX = mX + 1
if lowestKey == 6:
newY = mY + 1
newX = mX - 1
if lowestKey == 7:
newY = mY + 1
if lowestKey == 8:
newX = mX + 1
newY = mY + 1
What is the cleanest, simplest, and fastest way to do what I'm doing? This is going to loop through many monsters at once!
python array
python array
New contributor
New contributor
edited 13 mins ago
Jamal♦
30.3k11119227
30.3k11119227
New contributor
asked 5 hours ago
user1022585user1022585
1011
1011
New contributor
New contributor
$begingroup$
Could you add more description, how big is the board, is the player always in the 3x3 grid?
$endgroup$
– Peilonrayz
5 hours ago
$begingroup$
The board could be any size, this just creates a temporary array around the monster to decide which square it will move to
$endgroup$
– user1022585
5 hours ago
$begingroup$
Is there anything the monster can collide with?
$endgroup$
– Peilonrayz
5 hours ago
$begingroup$
The collisioncheck() routine returns a 0 or a 1. Its an external def that checks if any other monsters, players or walls are at that position.
$endgroup$
– user1022585
5 hours ago
$begingroup$
Hopefully I've not messed up my math when converting to this comment, but you can use:d = math.degrees(math.atan((pY - mY) / (pX - mY)))
and'N NE E SE S SW W NW'.split()[int((d + 22.5) // 45)]
to know which direction you want to go if there are no collisions.
$endgroup$
– Peilonrayz
4 hours ago
add a comment |
$begingroup$
Could you add more description, how big is the board, is the player always in the 3x3 grid?
$endgroup$
– Peilonrayz
5 hours ago
$begingroup$
The board could be any size, this just creates a temporary array around the monster to decide which square it will move to
$endgroup$
– user1022585
5 hours ago
$begingroup$
Is there anything the monster can collide with?
$endgroup$
– Peilonrayz
5 hours ago
$begingroup$
The collisioncheck() routine returns a 0 or a 1. Its an external def that checks if any other monsters, players or walls are at that position.
$endgroup$
– user1022585
5 hours ago
$begingroup$
Hopefully I've not messed up my math when converting to this comment, but you can use:d = math.degrees(math.atan((pY - mY) / (pX - mY)))
and'N NE E SE S SW W NW'.split()[int((d + 22.5) // 45)]
to know which direction you want to go if there are no collisions.
$endgroup$
– Peilonrayz
4 hours ago
$begingroup$
Could you add more description, how big is the board, is the player always in the 3x3 grid?
$endgroup$
– Peilonrayz
5 hours ago
$begingroup$
Could you add more description, how big is the board, is the player always in the 3x3 grid?
$endgroup$
– Peilonrayz
5 hours ago
$begingroup$
The board could be any size, this just creates a temporary array around the monster to decide which square it will move to
$endgroup$
– user1022585
5 hours ago
$begingroup$
The board could be any size, this just creates a temporary array around the monster to decide which square it will move to
$endgroup$
– user1022585
5 hours ago
$begingroup$
Is there anything the monster can collide with?
$endgroup$
– Peilonrayz
5 hours ago
$begingroup$
Is there anything the monster can collide with?
$endgroup$
– Peilonrayz
5 hours ago
$begingroup$
The collisioncheck() routine returns a 0 or a 1. Its an external def that checks if any other monsters, players or walls are at that position.
$endgroup$
– user1022585
5 hours ago
$begingroup$
The collisioncheck() routine returns a 0 or a 1. Its an external def that checks if any other monsters, players or walls are at that position.
$endgroup$
– user1022585
5 hours ago
$begingroup$
Hopefully I've not messed up my math when converting to this comment, but you can use:
d = math.degrees(math.atan((pY - mY) / (pX - mY)))
and 'N NE E SE S SW W NW'.split()[int((d + 22.5) // 45)]
to know which direction you want to go if there are no collisions.$endgroup$
– Peilonrayz
4 hours ago
$begingroup$
Hopefully I've not messed up my math when converting to this comment, but you can use:
d = math.degrees(math.atan((pY - mY) / (pX - mY)))
and 'N NE E SE S SW W NW'.split()[int((d + 22.5) // 45)]
to know which direction you want to go if there are no collisions.$endgroup$
– Peilonrayz
4 hours ago
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
Some thoughts from a non-game-developer:
- The second and subsequent
if lowestKey
should useelif
to short-circuit the evaluation when it finds a match. - I believe algorithms such as A* (A-star) are very well suited to find the shortest path between two points on a 2-dimensional map.
- Running this code through a linter such as
flake8
will show how to make your code more pythonic.
$endgroup$
add a comment |
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active
oldest
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oldest
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oldest
votes
$begingroup$
Some thoughts from a non-game-developer:
- The second and subsequent
if lowestKey
should useelif
to short-circuit the evaluation when it finds a match. - I believe algorithms such as A* (A-star) are very well suited to find the shortest path between two points on a 2-dimensional map.
- Running this code through a linter such as
flake8
will show how to make your code more pythonic.
$endgroup$
add a comment |
$begingroup$
Some thoughts from a non-game-developer:
- The second and subsequent
if lowestKey
should useelif
to short-circuit the evaluation when it finds a match. - I believe algorithms such as A* (A-star) are very well suited to find the shortest path between two points on a 2-dimensional map.
- Running this code through a linter such as
flake8
will show how to make your code more pythonic.
$endgroup$
add a comment |
$begingroup$
Some thoughts from a non-game-developer:
- The second and subsequent
if lowestKey
should useelif
to short-circuit the evaluation when it finds a match. - I believe algorithms such as A* (A-star) are very well suited to find the shortest path between two points on a 2-dimensional map.
- Running this code through a linter such as
flake8
will show how to make your code more pythonic.
$endgroup$
Some thoughts from a non-game-developer:
- The second and subsequent
if lowestKey
should useelif
to short-circuit the evaluation when it finds a match. - I believe algorithms such as A* (A-star) are very well suited to find the shortest path between two points on a 2-dimensional map.
- Running this code through a linter such as
flake8
will show how to make your code more pythonic.
answered 5 hours ago
l0b0l0b0
4,4141023
4,4141023
add a comment |
add a comment |
user1022585 is a new contributor. Be nice, and check out our Code of Conduct.
user1022585 is a new contributor. Be nice, and check out our Code of Conduct.
user1022585 is a new contributor. Be nice, and check out our Code of Conduct.
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$begingroup$
Could you add more description, how big is the board, is the player always in the 3x3 grid?
$endgroup$
– Peilonrayz
5 hours ago
$begingroup$
The board could be any size, this just creates a temporary array around the monster to decide which square it will move to
$endgroup$
– user1022585
5 hours ago
$begingroup$
Is there anything the monster can collide with?
$endgroup$
– Peilonrayz
5 hours ago
$begingroup$
The collisioncheck() routine returns a 0 or a 1. Its an external def that checks if any other monsters, players or walls are at that position.
$endgroup$
– user1022585
5 hours ago
$begingroup$
Hopefully I've not messed up my math when converting to this comment, but you can use:
d = math.degrees(math.atan((pY - mY) / (pX - mY)))
and'N NE E SE S SW W NW'.split()[int((d + 22.5) // 45)]
to know which direction you want to go if there are no collisions.$endgroup$
– Peilonrayz
4 hours ago