Parse formula into a dictionary
    formula_dict[variable_name] = [encoding, dropped_name]
    Parse interactions into a list
    inter_list = [[A,B], [A,C], [A,B,C]]

    formula = "A + beta + A*beta | Dev(A,drop=1), Poly(beta,degree=3)"

def _parse_formula(formula):
    """
    Parse formula into a dictionary
    formula_dict[variable_name] = [encoding, dropped_name]
    Parse interactions into a list
    inter_list = [[A,B], [A,C], [A,B,C]]

    formula = "A + beta + A*beta | Dev(A,drop=1), Poly(beta,degree=3)"
    """
    #TODO: DEAL WITH BSPLINE HAS MULTIPLE ARGUEMENTS (NOT JUST ONE)

    # break formula apart from encoding instructions
    formula,instr = formula.replace(' ','').split('|')
    # break formula apart at + sign
    formula_terms = formula.split('+') 
    # examine the instructions term, first split by ),
    instr = instr.split('),')
    # elements in the instr list will match the number of non interaction 
    #      elements in formula
    # go through each formula term, making a dictionary whose key is variable name
    formula_dict = {}
    encoding     = []
    other_arg    = []
    inter_list   = []
    for term in formula_terms:
        if "*" in term:
            # then this is an interaction term, make 'list of lists'
            term = term.split('*')
            inter_list.append(term)
        else:
            # then this is not an interaction term, make blank dictionary
            formula_dict[term] = [encoding, other_arg]
    # loop through instructions, parse each term
    for term in instr:
        # remove punctuation in term
        term = re.sub('[()]','',term)
        # check for each encoding type
        if "Dev" in term:
            # remove first three letters (Dev)
            term = term[3:]
            # split on comma
            var_name,arg = term.split(',')
            # split on equals sign in arg, take part after
            drop_name = arg.split('=')[1]
            # put var_name and drop_name into proper key in formula_dict
            formula_dict[var_name] = ["Dev",drop_name]
        elif "Dum" in term:
            # remove first three letters (Dum)
            term = term[3:]
            # split on comma
            var_name,arg = term.split(',')
            # split on equals sign in arg, take part after
            ref_name = arg.split('=')[1]
            # put var_name and drop_name into proper key in formula_dict
            formula_dict[var_name] = ["Dum",ref_name]
        elif "Poly" in term:
            # remove first four letters (Poly)
            term = term[4:]
            # split on comma
            var_name,arg = term.split(',')
            # split on equals sign in arg, take part after
            degree = arg.split('=')[1]
            # put var_name and drop_name into proper key in formula_dict
            formula_dict[var_name] = ["Poly",degree]
        else:
            raise Exception("Unknown Encoding")
    return formula_dict,inter_list