python - Parsing maths expression -

so following code function: "eqn" converts string: "d" mathematical expression, list of integers , operators according reverse polish notation, "add", "subtract", "multiply" , "divide" represent +, -, * , / respectively. there "negate" operator takes positive integer , turns negative integer. example inputs , outputs:

• eqn("10 5 add") should produce output of [10, 5 , "add"]
• eqn("5 6 add 2 negate divide") should produce output of [5, 6, "add", -2, "divide"]

with negative function though, if not used correctly (not after number) function should return index number of first incorrect "negate".

for example:

• eqn("negate 2 add 2") should produce output of 0

• eqn("2 4 add negate") should produce output of 3

  def eqn(d):  extrac = []  e in eqnstr.split():  if len(e) == 0:     continue   try:     extrac.append(int(e))     continue  except valueerror:     pass   try:     extrac.append(float(e))     continue  except valueerror:     pass  extrac.append(e)  return extrac 

the problem negate function won't work. instead of changing integer negative comes out "negate" in output when should 'disappear'. example:

eqn("5 6 add 2 negate divide") should produce output of [5, 6, "add", -2, "divide"] instead produces [5, 6, "add", 2, "negate", "divide"] how fix , how show index number of first invalid negate?

you need add special case handle negate. try this:

def eqn(d):     extrac = []      e in d.split():         if len(e) == 0:             continue          try:             extrac.append(int(e))             continue         except valueerror:             pass          try:             extrac.append(float(e))             continue         except valueerror:             pass          if e == 'negate':             if isinstance(extrac[-1], str):                 return len(extrac)             else:                 extrac[-1] *= -1                 continue          extrac.append(e)      return extrac 

output:

   eqn("2 4 add negate") => 3    eqn("5 6 add 2 negate divide") => [5, 6, 'add', -2, 'divide']