Mixing
Calculating lists bottom up
I'm trying calculate (using operators) the innermost list then the second innermost list etc. until there is no more lists. I'm also trying to do this so that it should be able to calculate the list no matter how many nested lists that's in it.
import operator
lst = [['x1', '*', ['x2', '*', 'x3']], '-', 'x3']
dictionary = {'x1': 4, 'x2': 5, 'x3': 7}
operator_dictionary = {"+": operator.add, "-": operator.sub, "*":
operator.mul, "/": operator.truediv}
def BINARYEXPR(program, dictionary):
for i in range(len(program)):
if isinstance(program[i],list):
return BINARYEXPR(program[i],dictionary)
operator = operator_dictionary[program[1]]
operand1 = dictionary[program[0]]
print("operand1: ", operand1)
operand2 = dictionary[program[2]]
print("operand2: ", operand2)
return operator(operand1,operand2)
print (BINARYEXPR(lst,dictionary))
So what I wanted to do here was to first calculate x2*x3 (5*7) which should give us 35, then calculate x1*35 (4*35) which should give us 140 and then finally take 140 - x3 (140-7) which should return 133. But instead I only managed to calculate the innermost list and hit return operator(operand1,operand2) which ends the function.
So what I'm stuck at is the recursion as I can't seem to figure out how to move on to the second innermost list whenever the innermost list has been calculated.
python list recursion
asked Nov 19 '18 at 20:56
Programming_ZeusProgramming_Zeus
134
add a comment |
I'm trying calculate (using operators) the innermost list then the second innermost list etc. until there is no more lists. I'm also trying to do this so that it should be able to calculate the list no matter how many nested lists that's in it.
import operator
lst = [['x1', '*', ['x2', '*', 'x3']], '-', 'x3']
dictionary = {'x1': 4, 'x2': 5, 'x3': 7}
operator_dictionary = {"+": operator.add, "-": operator.sub, "*":
operator.mul, "/": operator.truediv}
def BINARYEXPR(program, dictionary):
for i in range(len(program)):
if isinstance(program[i],list):
return BINARYEXPR(program[i],dictionary)
operator = operator_dictionary[program[1]]
operand1 = dictionary[program[0]]
print("operand1: ", operand1)
operand2 = dictionary[program[2]]
print("operand2: ", operand2)
return operator(operand1,operand2)
print (BINARYEXPR(lst,dictionary))
So what I wanted to do here was to first calculate x2*x3 (5*7) which should give us 35, then calculate x1*35 (4*35) which should give us 140 and then finally take 140 - x3 (140-7) which should return 133. But instead I only managed to calculate the innermost list and hit return operator(operand1,operand2) which ends the function.
So what I'm stuck at is the recursion as I can't seem to figure out how to move on to the second innermost list whenever the innermost list has been calculated.
python list recursion
asked Nov 19 '18 at 20:56
Programming_ZeusProgramming_Zeus
134
add a comment |
I'm trying calculate (using operators) the innermost list then the second innermost list etc. until there is no more lists. I'm also trying to do this so that it should be able to calculate the list no matter how many nested lists that's in it.
import operator
lst = [['x1', '*', ['x2', '*', 'x3']], '-', 'x3']
dictionary = {'x1': 4, 'x2': 5, 'x3': 7}
operator_dictionary = {"+": operator.add, "-": operator.sub, "*":
operator.mul, "/": operator.truediv}
def BINARYEXPR(program, dictionary):
for i in range(len(program)):
if isinstance(program[i],list):
return BINARYEXPR(program[i],dictionary)
operator = operator_dictionary[program[1]]
operand1 = dictionary[program[0]]
print("operand1: ", operand1)
operand2 = dictionary[program[2]]
print("operand2: ", operand2)
return operator(operand1,operand2)
print (BINARYEXPR(lst,dictionary))
So what I wanted to do here was to first calculate x2*x3 (5*7) which should give us 35, then calculate x1*35 (4*35) which should give us 140 and then finally take 140 - x3 (140-7) which should return 133. But instead I only managed to calculate the innermost list and hit return operator(operand1,operand2) which ends the function.
So what I'm stuck at is the recursion as I can't seem to figure out how to move on to the second innermost list whenever the innermost list has been calculated.
python list recursion
asked Nov 19 '18 at 20:56
Programming_ZeusProgramming_Zeus
134
I'm trying calculate (using operators) the innermost list then the second innermost list etc. until there is no more lists. I'm also trying to do this so that it should be able to calculate the list no matter how many nested lists that's in it.
import operator
lst = [['x1', '*', ['x2', '*', 'x3']], '-', 'x3']
dictionary = {'x1': 4, 'x2': 5, 'x3': 7}
operator_dictionary = {"+": operator.add, "-": operator.sub, "*":
operator.mul, "/": operator.truediv}
def BINARYEXPR(program, dictionary):
for i in range(len(program)):
if isinstance(program[i],list):
return BINARYEXPR(program[i],dictionary)
operator = operator_dictionary[program[1]]
operand1 = dictionary[program[0]]
print("operand1: ", operand1)
operand2 = dictionary[program[2]]
print("operand2: ", operand2)
return operator(operand1,operand2)
print (BINARYEXPR(lst,dictionary))
So what I wanted to do here was to first calculate x2*x3 (5*7) which should give us 35, then calculate x1*35 (4*35) which should give us 140 and then finally take 140 - x3 (140-7) which should return 133. But instead I only managed to calculate the innermost list and hit return operator(operand1,operand2) which ends the function.
So what I'm stuck at is the recursion as I can't seem to figure out how to move on to the second innermost list whenever the innermost list has been calculated.
python list recursion
python list recursion
asked Nov 19 '18 at 20:56
Programming_ZeusProgramming_Zeus
134
asked Nov 19 '18 at 20:56
Programming_ZeusProgramming_Zeus
134
asked Nov 19 '18 at 20:56
Programming_ZeusProgramming_Zeus
134
asked Nov 19 '18 at 20:56
Programming_ZeusProgramming_Zeus
134
asked Nov 19 '18 at 20:56
Programming_ZeusProgramming_Zeus
134
134
add a comment |
add a comment |
1 Answer
1
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A recursive function as follows should do the job:
import operator
lst = [['x1', '*', ['x2', '*', 'x3']], '-', 'x3']
dictionary = {'x1': 4, 'x2': 5, 'x3': 7}
operator_dictionary = {"+": operator.add, "-": operator.sub, "*":
operator.mul, "/": operator.truediv}
def calc(lst):
if type(lst) == str:
return dictionary[lst]
if len(lst) != 3:
raise ValueError("Incorrect expression: {}".format(lst))
op = operator_dictionary[lst[1]]
return op(calc(lst[0]), calc(lst[2]))
Since you are using infix notation, each expression has three components: expression, operator, expression. The function works by assuming the 0th and 2nd elements are operands and the 1st element is the operator. We calculate the sub-expressions recursively, and then apply the operation. Also, if at any point our function receives a list with length different than 3, we throw since this cannot be a well-formed expression.
answered Nov 19 '18 at 21:07
eozdeozd
8941515
I don't think giving the user a completely different solution is an appropriate answer to their question.
– ritlew
Nov 19 '18 at 21:09
@ritlew What do you mean by a completely different answer? The only main difference between this one and the one OP originally written is that they have forgotten the recursive call to the function, which I provide.
– eozd
Nov 19 '18 at 21:11
You didn't fix the users code, you provided your own to accomplish the same task that the user was trying to code for themselves. Your code looks nothing like OP's.
– ritlew
Nov 19 '18 at 21:15
Great solution, thank you!
– Programming_Zeus
Nov 19 '18 at 21:26
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
A recursive function as follows should do the job:
import operator
lst = [['x1', '*', ['x2', '*', 'x3']], '-', 'x3']
dictionary = {'x1': 4, 'x2': 5, 'x3': 7}
operator_dictionary = {"+": operator.add, "-": operator.sub, "*":
operator.mul, "/": operator.truediv}
def calc(lst):
if type(lst) == str:
return dictionary[lst]
if len(lst) != 3:
raise ValueError("Incorrect expression: {}".format(lst))
op = operator_dictionary[lst[1]]
return op(calc(lst[0]), calc(lst[2]))
Since you are using infix notation, each expression has three components: expression, operator, expression. The function works by assuming the 0th and 2nd elements are operands and the 1st element is the operator. We calculate the sub-expressions recursively, and then apply the operation. Also, if at any point our function receives a list with length different than 3, we throw since this cannot be a well-formed expression.
answered Nov 19 '18 at 21:07
eozdeozd
8941515
I don't think giving the user a completely different solution is an appropriate answer to their question.
– ritlew
Nov 19 '18 at 21:09
@ritlew What do you mean by a completely different answer? The only main difference between this one and the one OP originally written is that they have forgotten the recursive call to the function, which I provide.
– eozd
Nov 19 '18 at 21:11
You didn't fix the users code, you provided your own to accomplish the same task that the user was trying to code for themselves. Your code looks nothing like OP's.
– ritlew
Nov 19 '18 at 21:15
Great solution, thank you!
– Programming_Zeus
Nov 19 '18 at 21:26
add a comment |
A recursive function as follows should do the job:
import operator
lst = [['x1', '*', ['x2', '*', 'x3']], '-', 'x3']
dictionary = {'x1': 4, 'x2': 5, 'x3': 7}
operator_dictionary = {"+": operator.add, "-": operator.sub, "*":
operator.mul, "/": operator.truediv}
def calc(lst):
if type(lst) == str:
return dictionary[lst]
if len(lst) != 3:
raise ValueError("Incorrect expression: {}".format(lst))
op = operator_dictionary[lst[1]]
return op(calc(lst[0]), calc(lst[2]))
Since you are using infix notation, each expression has three components: expression, operator, expression. The function works by assuming the 0th and 2nd elements are operands and the 1st element is the operator. We calculate the sub-expressions recursively, and then apply the operation. Also, if at any point our function receives a list with length different than 3, we throw since this cannot be a well-formed expression.
answered Nov 19 '18 at 21:07
eozdeozd
8941515
I don't think giving the user a completely different solution is an appropriate answer to their question.
– ritlew
Nov 19 '18 at 21:09
@ritlew What do you mean by a completely different answer? The only main difference between this one and the one OP originally written is that they have forgotten the recursive call to the function, which I provide.
– eozd
Nov 19 '18 at 21:11
You didn't fix the users code, you provided your own to accomplish the same task that the user was trying to code for themselves. Your code looks nothing like OP's.
– ritlew
Nov 19 '18 at 21:15
Great solution, thank you!
– Programming_Zeus
Nov 19 '18 at 21:26
add a comment |
A recursive function as follows should do the job:
import operator
lst = [['x1', '*', ['x2', '*', 'x3']], '-', 'x3']
dictionary = {'x1': 4, 'x2': 5, 'x3': 7}
operator_dictionary = {"+": operator.add, "-": operator.sub, "*":
operator.mul, "/": operator.truediv}
def calc(lst):
if type(lst) == str:
return dictionary[lst]
if len(lst) != 3:
raise ValueError("Incorrect expression: {}".format(lst))
op = operator_dictionary[lst[1]]
return op(calc(lst[0]), calc(lst[2]))
Since you are using infix notation, each expression has three components: expression, operator, expression. The function works by assuming the 0th and 2nd elements are operands and the 1st element is the operator. We calculate the sub-expressions recursively, and then apply the operation. Also, if at any point our function receives a list with length different than 3, we throw since this cannot be a well-formed expression.
answered Nov 19 '18 at 21:07
eozdeozd
8941515
A recursive function as follows should do the job:
import operator
lst = [['x1', '*', ['x2', '*', 'x3']], '-', 'x3']
dictionary = {'x1': 4, 'x2': 5, 'x3': 7}
operator_dictionary = {"+": operator.add, "-": operator.sub, "*":
operator.mul, "/": operator.truediv}
def calc(lst):
if type(lst) == str:
return dictionary[lst]
if len(lst) != 3:
raise ValueError("Incorrect expression: {}".format(lst))
op = operator_dictionary[lst[1]]
return op(calc(lst[0]), calc(lst[2]))
Since you are using infix notation, each expression has three components: expression, operator, expression. The function works by assuming the 0th and 2nd elements are operands and the 1st element is the operator. We calculate the sub-expressions recursively, and then apply the operation. Also, if at any point our function receives a list with length different than 3, we throw since this cannot be a well-formed expression.
answered Nov 19 '18 at 21:07
eozdeozd
8941515
answered Nov 19 '18 at 21:07
eozdeozd
8941515
answered Nov 19 '18 at 21:07
eozdeozd
8941515
answered Nov 19 '18 at 21:07
eozdeozd
8941515
8941515
I don't think giving the user a completely different solution is an appropriate answer to their question.
– ritlew
Nov 19 '18 at 21:09
@ritlew What do you mean by a completely different answer? The only main difference between this one and the one OP originally written is that they have forgotten the recursive call to the function, which I provide.
– eozd
Nov 19 '18 at 21:11
You didn't fix the users code, you provided your own to accomplish the same task that the user was trying to code for themselves. Your code looks nothing like OP's.
– ritlew
Nov 19 '18 at 21:15
Great solution, thank you!
– Programming_Zeus
Nov 19 '18 at 21:26
add a comment |
I don't think giving the user a completely different solution is an appropriate answer to their question.
– ritlew
Nov 19 '18 at 21:09
@ritlew What do you mean by a completely different answer? The only main difference between this one and the one OP originally written is that they have forgotten the recursive call to the function, which I provide.
– eozd
Nov 19 '18 at 21:11
You didn't fix the users code, you provided your own to accomplish the same task that the user was trying to code for themselves. Your code looks nothing like OP's.
– ritlew
Nov 19 '18 at 21:15
Great solution, thank you!
– Programming_Zeus
Nov 19 '18 at 21:26
I don't think giving the user a completely different solution is an appropriate answer to their question.
– ritlew
Nov 19 '18 at 21:09
I don't think giving the user a completely different solution is an appropriate answer to their question.
– ritlew
Nov 19 '18 at 21:09
@ritlew What do you mean by a completely different answer? The only main difference between this one and the one OP originally written is that they have forgotten the recursive call to the function, which I provide.
– eozd
Nov 19 '18 at 21:11
@ritlew What do you mean by a completely different answer? The only main difference between this one and the one OP originally written is that they have forgotten the recursive call to the function, which I provide.
– eozd
Nov 19 '18 at 21:11
You didn't fix the users code, you provided your own to accomplish the same task that the user was trying to code for themselves. Your code looks nothing like OP's.
– ritlew
Nov 19 '18 at 21:15
You didn't fix the users code, you provided your own to accomplish the same task that the user was trying to code for themselves. Your code looks nothing like OP's.
– ritlew
Nov 19 '18 at 21:15
Great solution, thank you!
– Programming_Zeus
Nov 19 '18 at 21:26
Great solution, thank you!
– Programming_Zeus
Nov 19 '18 at 21:26
add a comment |
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