Function in a board game with chests and bandits
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This function is part of a python board game program. The game is a board game with chests and bandits hidden throughout the board. The function is dedicated to the "easy" section of the game (where it is a 8x8 grid).
def easy_level(Coins):
#This function is for the movement of the game in easy difficulty
while True:
oldcurrent=current
boardeasy[oldcurrent[0]][oldcurrent[1]]='*'
table_game_easy()
boardeasy[oldcurrent[0]][oldcurrent[1]]=' '
n = input('Enter the direction followed by the number Ex:Up 5 , Number should be < 8 n')
n=n.split()
if n[0].lower() not in ['up','left','down','right']:#Validates input
print('Wrong command, please input again')
continue
elif n[0].lower()=='up':
up(int(n[1].lower()),8)#Boundary is set to 8 as the 'easy' grid is a 8^8
elif n[0].lower()=='down':
down(int(n[1].lower()),8)
elif n[0].lower()=='left':
left(int(n[1].lower()),8)
elif n[0].lower()=='right':
right(int(n[1].lower()),8)
print("5 chests left")
print("8 bandits left")
print("Coins:",Coins)#Acts as a counter, displays the number of coins that the player has
if current[0] == Treasure1_Row and current[1] == Treasure1_Col
or current[0] == Treasure2_Row and current[1] == Treasure2_Col
or current[0] == Treasure3_Row and current[1] == Treasure3_Col
or current[0] == Treasure4_Row and current[1] == Treasure4_Col
or current[0] == Treasure5_Row and current[1] == Treasure5_Col
or current[0] == Treasure6_Row and current[1] == Treasure6_Col
or current[0] == Treasure7_Row and current[1] == Treasure7_Col
or current[0] == Treasure8_Row and current[1] == Treasure8_Col
or current[0] == Treasure9_Row and current[1] == Treasure9_Col
or current[0] == Treasure10_Row and current[1] == Treasure10_Col:
print("Hooray! You have found booty! +10 gold")
Coins = Coins+10 #Adds an additional 10 points
print("Coins:",Coins)
if current[0] == Bandit1_Row and current[1] == Bandit1_Col
or current[0] == Bandit2_Row and current[1] == Bandit2_Col
or current[0] == Bandit3_Row and current[1] == Bandit3_Col
or current[0] == Bandit4_Row and current[1] == Bandit4_Col
or current[0] == Bandit5_Row and current[1] == Bandit5_Col:
print("Oh no! You have landed on a bandit...they steal all your coins!")
Coins = Coins-Coins #Removes all coins
print("Coins:",Coins)
boardeasy[current[0]][current[1]]='*'#sets value to players position
python game
migrated from stackoverflow.com Nov 22 at 4:08
This question came from our site for professional and enthusiast programmers.
add a comment |
up vote
0
down vote
favorite
This function is part of a python board game program. The game is a board game with chests and bandits hidden throughout the board. The function is dedicated to the "easy" section of the game (where it is a 8x8 grid).
def easy_level(Coins):
#This function is for the movement of the game in easy difficulty
while True:
oldcurrent=current
boardeasy[oldcurrent[0]][oldcurrent[1]]='*'
table_game_easy()
boardeasy[oldcurrent[0]][oldcurrent[1]]=' '
n = input('Enter the direction followed by the number Ex:Up 5 , Number should be < 8 n')
n=n.split()
if n[0].lower() not in ['up','left','down','right']:#Validates input
print('Wrong command, please input again')
continue
elif n[0].lower()=='up':
up(int(n[1].lower()),8)#Boundary is set to 8 as the 'easy' grid is a 8^8
elif n[0].lower()=='down':
down(int(n[1].lower()),8)
elif n[0].lower()=='left':
left(int(n[1].lower()),8)
elif n[0].lower()=='right':
right(int(n[1].lower()),8)
print("5 chests left")
print("8 bandits left")
print("Coins:",Coins)#Acts as a counter, displays the number of coins that the player has
if current[0] == Treasure1_Row and current[1] == Treasure1_Col
or current[0] == Treasure2_Row and current[1] == Treasure2_Col
or current[0] == Treasure3_Row and current[1] == Treasure3_Col
or current[0] == Treasure4_Row and current[1] == Treasure4_Col
or current[0] == Treasure5_Row and current[1] == Treasure5_Col
or current[0] == Treasure6_Row and current[1] == Treasure6_Col
or current[0] == Treasure7_Row and current[1] == Treasure7_Col
or current[0] == Treasure8_Row and current[1] == Treasure8_Col
or current[0] == Treasure9_Row and current[1] == Treasure9_Col
or current[0] == Treasure10_Row and current[1] == Treasure10_Col:
print("Hooray! You have found booty! +10 gold")
Coins = Coins+10 #Adds an additional 10 points
print("Coins:",Coins)
if current[0] == Bandit1_Row and current[1] == Bandit1_Col
or current[0] == Bandit2_Row and current[1] == Bandit2_Col
or current[0] == Bandit3_Row and current[1] == Bandit3_Col
or current[0] == Bandit4_Row and current[1] == Bandit4_Col
or current[0] == Bandit5_Row and current[1] == Bandit5_Col:
print("Oh no! You have landed on a bandit...they steal all your coins!")
Coins = Coins-Coins #Removes all coins
print("Coins:",Coins)
boardeasy[current[0]][current[1]]='*'#sets value to players position
python game
migrated from stackoverflow.com Nov 22 at 4:08
This question came from our site for professional and enthusiast programmers.
2
what do you mean more efficient?
– SuperStew
Nov 14 at 0:00
1
Reduce the amount of code needed perhaps?
– J.Peggy
Nov 14 at 0:01
1
How can I make the code in this function more efficient?isn't a good title for Code Review because stateing your concerns about the code, applies to too many questions on this site to be useful. The site standard is for the title to simply state the task accomplished by the code. Please see How to Ask for examples, and revise the title accordingly.
– Heslacher
Nov 22 at 5:30
add a comment |
up vote
0
down vote
favorite
up vote
0
down vote
favorite
This function is part of a python board game program. The game is a board game with chests and bandits hidden throughout the board. The function is dedicated to the "easy" section of the game (where it is a 8x8 grid).
def easy_level(Coins):
#This function is for the movement of the game in easy difficulty
while True:
oldcurrent=current
boardeasy[oldcurrent[0]][oldcurrent[1]]='*'
table_game_easy()
boardeasy[oldcurrent[0]][oldcurrent[1]]=' '
n = input('Enter the direction followed by the number Ex:Up 5 , Number should be < 8 n')
n=n.split()
if n[0].lower() not in ['up','left','down','right']:#Validates input
print('Wrong command, please input again')
continue
elif n[0].lower()=='up':
up(int(n[1].lower()),8)#Boundary is set to 8 as the 'easy' grid is a 8^8
elif n[0].lower()=='down':
down(int(n[1].lower()),8)
elif n[0].lower()=='left':
left(int(n[1].lower()),8)
elif n[0].lower()=='right':
right(int(n[1].lower()),8)
print("5 chests left")
print("8 bandits left")
print("Coins:",Coins)#Acts as a counter, displays the number of coins that the player has
if current[0] == Treasure1_Row and current[1] == Treasure1_Col
or current[0] == Treasure2_Row and current[1] == Treasure2_Col
or current[0] == Treasure3_Row and current[1] == Treasure3_Col
or current[0] == Treasure4_Row and current[1] == Treasure4_Col
or current[0] == Treasure5_Row and current[1] == Treasure5_Col
or current[0] == Treasure6_Row and current[1] == Treasure6_Col
or current[0] == Treasure7_Row and current[1] == Treasure7_Col
or current[0] == Treasure8_Row and current[1] == Treasure8_Col
or current[0] == Treasure9_Row and current[1] == Treasure9_Col
or current[0] == Treasure10_Row and current[1] == Treasure10_Col:
print("Hooray! You have found booty! +10 gold")
Coins = Coins+10 #Adds an additional 10 points
print("Coins:",Coins)
if current[0] == Bandit1_Row and current[1] == Bandit1_Col
or current[0] == Bandit2_Row and current[1] == Bandit2_Col
or current[0] == Bandit3_Row and current[1] == Bandit3_Col
or current[0] == Bandit4_Row and current[1] == Bandit4_Col
or current[0] == Bandit5_Row and current[1] == Bandit5_Col:
print("Oh no! You have landed on a bandit...they steal all your coins!")
Coins = Coins-Coins #Removes all coins
print("Coins:",Coins)
boardeasy[current[0]][current[1]]='*'#sets value to players position
python game
This function is part of a python board game program. The game is a board game with chests and bandits hidden throughout the board. The function is dedicated to the "easy" section of the game (where it is a 8x8 grid).
def easy_level(Coins):
#This function is for the movement of the game in easy difficulty
while True:
oldcurrent=current
boardeasy[oldcurrent[0]][oldcurrent[1]]='*'
table_game_easy()
boardeasy[oldcurrent[0]][oldcurrent[1]]=' '
n = input('Enter the direction followed by the number Ex:Up 5 , Number should be < 8 n')
n=n.split()
if n[0].lower() not in ['up','left','down','right']:#Validates input
print('Wrong command, please input again')
continue
elif n[0].lower()=='up':
up(int(n[1].lower()),8)#Boundary is set to 8 as the 'easy' grid is a 8^8
elif n[0].lower()=='down':
down(int(n[1].lower()),8)
elif n[0].lower()=='left':
left(int(n[1].lower()),8)
elif n[0].lower()=='right':
right(int(n[1].lower()),8)
print("5 chests left")
print("8 bandits left")
print("Coins:",Coins)#Acts as a counter, displays the number of coins that the player has
if current[0] == Treasure1_Row and current[1] == Treasure1_Col
or current[0] == Treasure2_Row and current[1] == Treasure2_Col
or current[0] == Treasure3_Row and current[1] == Treasure3_Col
or current[0] == Treasure4_Row and current[1] == Treasure4_Col
or current[0] == Treasure5_Row and current[1] == Treasure5_Col
or current[0] == Treasure6_Row and current[1] == Treasure6_Col
or current[0] == Treasure7_Row and current[1] == Treasure7_Col
or current[0] == Treasure8_Row and current[1] == Treasure8_Col
or current[0] == Treasure9_Row and current[1] == Treasure9_Col
or current[0] == Treasure10_Row and current[1] == Treasure10_Col:
print("Hooray! You have found booty! +10 gold")
Coins = Coins+10 #Adds an additional 10 points
print("Coins:",Coins)
if current[0] == Bandit1_Row and current[1] == Bandit1_Col
or current[0] == Bandit2_Row and current[1] == Bandit2_Col
or current[0] == Bandit3_Row and current[1] == Bandit3_Col
or current[0] == Bandit4_Row and current[1] == Bandit4_Col
or current[0] == Bandit5_Row and current[1] == Bandit5_Col:
print("Oh no! You have landed on a bandit...they steal all your coins!")
Coins = Coins-Coins #Removes all coins
print("Coins:",Coins)
boardeasy[current[0]][current[1]]='*'#sets value to players position
python game
python game
edited Nov 22 at 9:07
200_success
127k15148412
127k15148412
asked Nov 13 at 23:52
J.Peggy
migrated from stackoverflow.com Nov 22 at 4:08
This question came from our site for professional and enthusiast programmers.
migrated from stackoverflow.com Nov 22 at 4:08
This question came from our site for professional and enthusiast programmers.
2
what do you mean more efficient?
– SuperStew
Nov 14 at 0:00
1
Reduce the amount of code needed perhaps?
– J.Peggy
Nov 14 at 0:01
1
How can I make the code in this function more efficient?isn't a good title for Code Review because stateing your concerns about the code, applies to too many questions on this site to be useful. The site standard is for the title to simply state the task accomplished by the code. Please see How to Ask for examples, and revise the title accordingly.
– Heslacher
Nov 22 at 5:30
add a comment |
2
what do you mean more efficient?
– SuperStew
Nov 14 at 0:00
1
Reduce the amount of code needed perhaps?
– J.Peggy
Nov 14 at 0:01
1
How can I make the code in this function more efficient?isn't a good title for Code Review because stateing your concerns about the code, applies to too many questions on this site to be useful. The site standard is for the title to simply state the task accomplished by the code. Please see How to Ask for examples, and revise the title accordingly.
– Heslacher
Nov 22 at 5:30
2
2
what do you mean more efficient?
– SuperStew
Nov 14 at 0:00
what do you mean more efficient?
– SuperStew
Nov 14 at 0:00
1
1
Reduce the amount of code needed perhaps?
– J.Peggy
Nov 14 at 0:01
Reduce the amount of code needed perhaps?
– J.Peggy
Nov 14 at 0:01
1
1
How can I make the code in this function more efficient? isn't a good title for Code Review because stateing your concerns about the code, applies to too many questions on this site to be useful. The site standard is for the title to simply state the task accomplished by the code. Please see How to Ask for examples, and revise the title accordingly.– Heslacher
Nov 22 at 5:30
How can I make the code in this function more efficient? isn't a good title for Code Review because stateing your concerns about the code, applies to too many questions on this site to be useful. The site standard is for the title to simply state the task accomplished by the code. Please see How to Ask for examples, and revise the title accordingly.– Heslacher
Nov 22 at 5:30
add a comment |
1 Answer
1
active
oldest
votes
up vote
0
down vote
We're definitely missing some context here to run your code or understand what it is supposed to do, but there is enough to detect a few things that could be easily improved.
Coins = Coins-Coins #Removes all coins
This should be:
Coins = 0 #Removes all coins
Also
Coins = Coins+10 #Adds an additional 10 points
can be written
Coins += 10 #Adds an additional 10 points
Don't perform the same operations more than needed. In particular when handling the user input, you can limit the number of index accesses, to call to lower function, to call to int function:
user_input = input('Enter the direction followed by the number Ex:Up 5 , Number should be < 8 n').split()
if len(user_input) != 2:
print('Wrong command, please input again')
continue
direct, number = user_input
direct = direct.lower()
number = int(number.lower())
if direct not in ['up','left','down','right']:#Validates input
print('Wrong command, please input again')
continue
elif direct == 'up':
up(number, 8) #Boundary is set to 8 as the 'easy' grid is a 8^8
elif direct == 'down':
down(number, 8)
elif direct == 'left':
left(number, 8)
elif direct == 'right':
right(number, 8)
Then, you can actually change the condition order so that you don't need to list twice the valid directions:
if direct == 'up':
up(number, 8) #Boundary is set to 8 as the 'easy' grid is a 8^8
elif direct == 'down':
down(number, 8)
elif direct == 'left':
left(number, 8)
elif direct == 'right':
right(number, 8)
else:
print('Wrong command, please input again')
continue
You could probably rewrite the comparisons to have something like:
if current == Treasure1_Pos
or current == Treasure2_Pos
or current == Treasure3_Pos
or current == Treasure4_Pos
or current == Treasure5_Pos
or current == Treasure6_Pos
or current == Treasure7_Pos
or current == Treasure8_Pos
or current == Treasure9_Pos
or current == Treasure10_Pos:
print("Hooray! You have found booty! +10 gold")
Coins += 10 #Adds an additional 10 points
print("Coins:",Coins)
if current == Bandit1_Pos
or current == Bandit2_Pos
or current == Bandit3_Pos
or current == Bandit4_Pos
or current == Bandit5_Pos:
print("Oh no! You have landed on a bandit...they steal all your coins!")
Coins = 0 #Removes all coins
print("Coins:",Coins)
And you could even define a data structure (list, set) to hold all the relevant positions and write something like:
if current in Treasure_Positions:
print("Hooray! You have found booty! +10 gold")
Coins += 10 #Adds an additional 10 points
print("Coins:",Coins)
if current in Bandit_Positions:
print("Oh no! You have landed on a bandit...they steal all your coins!")
Coins = 0 #Removes all coins
print("Coins:",Coins)
Then more things look wrong/improvable about boardeasy but we'd need to see what it does.
add a comment |
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
0
down vote
We're definitely missing some context here to run your code or understand what it is supposed to do, but there is enough to detect a few things that could be easily improved.
Coins = Coins-Coins #Removes all coins
This should be:
Coins = 0 #Removes all coins
Also
Coins = Coins+10 #Adds an additional 10 points
can be written
Coins += 10 #Adds an additional 10 points
Don't perform the same operations more than needed. In particular when handling the user input, you can limit the number of index accesses, to call to lower function, to call to int function:
user_input = input('Enter the direction followed by the number Ex:Up 5 , Number should be < 8 n').split()
if len(user_input) != 2:
print('Wrong command, please input again')
continue
direct, number = user_input
direct = direct.lower()
number = int(number.lower())
if direct not in ['up','left','down','right']:#Validates input
print('Wrong command, please input again')
continue
elif direct == 'up':
up(number, 8) #Boundary is set to 8 as the 'easy' grid is a 8^8
elif direct == 'down':
down(number, 8)
elif direct == 'left':
left(number, 8)
elif direct == 'right':
right(number, 8)
Then, you can actually change the condition order so that you don't need to list twice the valid directions:
if direct == 'up':
up(number, 8) #Boundary is set to 8 as the 'easy' grid is a 8^8
elif direct == 'down':
down(number, 8)
elif direct == 'left':
left(number, 8)
elif direct == 'right':
right(number, 8)
else:
print('Wrong command, please input again')
continue
You could probably rewrite the comparisons to have something like:
if current == Treasure1_Pos
or current == Treasure2_Pos
or current == Treasure3_Pos
or current == Treasure4_Pos
or current == Treasure5_Pos
or current == Treasure6_Pos
or current == Treasure7_Pos
or current == Treasure8_Pos
or current == Treasure9_Pos
or current == Treasure10_Pos:
print("Hooray! You have found booty! +10 gold")
Coins += 10 #Adds an additional 10 points
print("Coins:",Coins)
if current == Bandit1_Pos
or current == Bandit2_Pos
or current == Bandit3_Pos
or current == Bandit4_Pos
or current == Bandit5_Pos:
print("Oh no! You have landed on a bandit...they steal all your coins!")
Coins = 0 #Removes all coins
print("Coins:",Coins)
And you could even define a data structure (list, set) to hold all the relevant positions and write something like:
if current in Treasure_Positions:
print("Hooray! You have found booty! +10 gold")
Coins += 10 #Adds an additional 10 points
print("Coins:",Coins)
if current in Bandit_Positions:
print("Oh no! You have landed on a bandit...they steal all your coins!")
Coins = 0 #Removes all coins
print("Coins:",Coins)
Then more things look wrong/improvable about boardeasy but we'd need to see what it does.
add a comment |
up vote
0
down vote
We're definitely missing some context here to run your code or understand what it is supposed to do, but there is enough to detect a few things that could be easily improved.
Coins = Coins-Coins #Removes all coins
This should be:
Coins = 0 #Removes all coins
Also
Coins = Coins+10 #Adds an additional 10 points
can be written
Coins += 10 #Adds an additional 10 points
Don't perform the same operations more than needed. In particular when handling the user input, you can limit the number of index accesses, to call to lower function, to call to int function:
user_input = input('Enter the direction followed by the number Ex:Up 5 , Number should be < 8 n').split()
if len(user_input) != 2:
print('Wrong command, please input again')
continue
direct, number = user_input
direct = direct.lower()
number = int(number.lower())
if direct not in ['up','left','down','right']:#Validates input
print('Wrong command, please input again')
continue
elif direct == 'up':
up(number, 8) #Boundary is set to 8 as the 'easy' grid is a 8^8
elif direct == 'down':
down(number, 8)
elif direct == 'left':
left(number, 8)
elif direct == 'right':
right(number, 8)
Then, you can actually change the condition order so that you don't need to list twice the valid directions:
if direct == 'up':
up(number, 8) #Boundary is set to 8 as the 'easy' grid is a 8^8
elif direct == 'down':
down(number, 8)
elif direct == 'left':
left(number, 8)
elif direct == 'right':
right(number, 8)
else:
print('Wrong command, please input again')
continue
You could probably rewrite the comparisons to have something like:
if current == Treasure1_Pos
or current == Treasure2_Pos
or current == Treasure3_Pos
or current == Treasure4_Pos
or current == Treasure5_Pos
or current == Treasure6_Pos
or current == Treasure7_Pos
or current == Treasure8_Pos
or current == Treasure9_Pos
or current == Treasure10_Pos:
print("Hooray! You have found booty! +10 gold")
Coins += 10 #Adds an additional 10 points
print("Coins:",Coins)
if current == Bandit1_Pos
or current == Bandit2_Pos
or current == Bandit3_Pos
or current == Bandit4_Pos
or current == Bandit5_Pos:
print("Oh no! You have landed on a bandit...they steal all your coins!")
Coins = 0 #Removes all coins
print("Coins:",Coins)
And you could even define a data structure (list, set) to hold all the relevant positions and write something like:
if current in Treasure_Positions:
print("Hooray! You have found booty! +10 gold")
Coins += 10 #Adds an additional 10 points
print("Coins:",Coins)
if current in Bandit_Positions:
print("Oh no! You have landed on a bandit...they steal all your coins!")
Coins = 0 #Removes all coins
print("Coins:",Coins)
Then more things look wrong/improvable about boardeasy but we'd need to see what it does.
add a comment |
up vote
0
down vote
up vote
0
down vote
We're definitely missing some context here to run your code or understand what it is supposed to do, but there is enough to detect a few things that could be easily improved.
Coins = Coins-Coins #Removes all coins
This should be:
Coins = 0 #Removes all coins
Also
Coins = Coins+10 #Adds an additional 10 points
can be written
Coins += 10 #Adds an additional 10 points
Don't perform the same operations more than needed. In particular when handling the user input, you can limit the number of index accesses, to call to lower function, to call to int function:
user_input = input('Enter the direction followed by the number Ex:Up 5 , Number should be < 8 n').split()
if len(user_input) != 2:
print('Wrong command, please input again')
continue
direct, number = user_input
direct = direct.lower()
number = int(number.lower())
if direct not in ['up','left','down','right']:#Validates input
print('Wrong command, please input again')
continue
elif direct == 'up':
up(number, 8) #Boundary is set to 8 as the 'easy' grid is a 8^8
elif direct == 'down':
down(number, 8)
elif direct == 'left':
left(number, 8)
elif direct == 'right':
right(number, 8)
Then, you can actually change the condition order so that you don't need to list twice the valid directions:
if direct == 'up':
up(number, 8) #Boundary is set to 8 as the 'easy' grid is a 8^8
elif direct == 'down':
down(number, 8)
elif direct == 'left':
left(number, 8)
elif direct == 'right':
right(number, 8)
else:
print('Wrong command, please input again')
continue
You could probably rewrite the comparisons to have something like:
if current == Treasure1_Pos
or current == Treasure2_Pos
or current == Treasure3_Pos
or current == Treasure4_Pos
or current == Treasure5_Pos
or current == Treasure6_Pos
or current == Treasure7_Pos
or current == Treasure8_Pos
or current == Treasure9_Pos
or current == Treasure10_Pos:
print("Hooray! You have found booty! +10 gold")
Coins += 10 #Adds an additional 10 points
print("Coins:",Coins)
if current == Bandit1_Pos
or current == Bandit2_Pos
or current == Bandit3_Pos
or current == Bandit4_Pos
or current == Bandit5_Pos:
print("Oh no! You have landed on a bandit...they steal all your coins!")
Coins = 0 #Removes all coins
print("Coins:",Coins)
And you could even define a data structure (list, set) to hold all the relevant positions and write something like:
if current in Treasure_Positions:
print("Hooray! You have found booty! +10 gold")
Coins += 10 #Adds an additional 10 points
print("Coins:",Coins)
if current in Bandit_Positions:
print("Oh no! You have landed on a bandit...they steal all your coins!")
Coins = 0 #Removes all coins
print("Coins:",Coins)
Then more things look wrong/improvable about boardeasy but we'd need to see what it does.
We're definitely missing some context here to run your code or understand what it is supposed to do, but there is enough to detect a few things that could be easily improved.
Coins = Coins-Coins #Removes all coins
This should be:
Coins = 0 #Removes all coins
Also
Coins = Coins+10 #Adds an additional 10 points
can be written
Coins += 10 #Adds an additional 10 points
Don't perform the same operations more than needed. In particular when handling the user input, you can limit the number of index accesses, to call to lower function, to call to int function:
user_input = input('Enter the direction followed by the number Ex:Up 5 , Number should be < 8 n').split()
if len(user_input) != 2:
print('Wrong command, please input again')
continue
direct, number = user_input
direct = direct.lower()
number = int(number.lower())
if direct not in ['up','left','down','right']:#Validates input
print('Wrong command, please input again')
continue
elif direct == 'up':
up(number, 8) #Boundary is set to 8 as the 'easy' grid is a 8^8
elif direct == 'down':
down(number, 8)
elif direct == 'left':
left(number, 8)
elif direct == 'right':
right(number, 8)
Then, you can actually change the condition order so that you don't need to list twice the valid directions:
if direct == 'up':
up(number, 8) #Boundary is set to 8 as the 'easy' grid is a 8^8
elif direct == 'down':
down(number, 8)
elif direct == 'left':
left(number, 8)
elif direct == 'right':
right(number, 8)
else:
print('Wrong command, please input again')
continue
You could probably rewrite the comparisons to have something like:
if current == Treasure1_Pos
or current == Treasure2_Pos
or current == Treasure3_Pos
or current == Treasure4_Pos
or current == Treasure5_Pos
or current == Treasure6_Pos
or current == Treasure7_Pos
or current == Treasure8_Pos
or current == Treasure9_Pos
or current == Treasure10_Pos:
print("Hooray! You have found booty! +10 gold")
Coins += 10 #Adds an additional 10 points
print("Coins:",Coins)
if current == Bandit1_Pos
or current == Bandit2_Pos
or current == Bandit3_Pos
or current == Bandit4_Pos
or current == Bandit5_Pos:
print("Oh no! You have landed on a bandit...they steal all your coins!")
Coins = 0 #Removes all coins
print("Coins:",Coins)
And you could even define a data structure (list, set) to hold all the relevant positions and write something like:
if current in Treasure_Positions:
print("Hooray! You have found booty! +10 gold")
Coins += 10 #Adds an additional 10 points
print("Coins:",Coins)
if current in Bandit_Positions:
print("Oh no! You have landed on a bandit...they steal all your coins!")
Coins = 0 #Removes all coins
print("Coins:",Coins)
Then more things look wrong/improvable about boardeasy but we'd need to see what it does.
edited Nov 22 at 9:07
Toby Speight
22.6k537109
22.6k537109
answered Nov 22 at 8:44
Josay
24.5k13783
24.5k13783
add a comment |
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2
what do you mean more efficient?
– SuperStew
Nov 14 at 0:00
1
Reduce the amount of code needed perhaps?
– J.Peggy
Nov 14 at 0:01
1
How can I make the code in this function more efficient?isn't a good title for Code Review because stateing your concerns about the code, applies to too many questions on this site to be useful. The site standard is for the title to simply state the task accomplished by the code. Please see How to Ask for examples, and revise the title accordingly.– Heslacher
Nov 22 at 5:30