如果我不依赖外部库，则可以这样做：
def parse(x):
operators = set('+-*/')
op_out = [] #This holds the operators that are found in the string (left to right)
num_out = [] #this holds the non-operators that are found in the string (left to right)
buff = []
for c in x: #examine 1 character at a time
if c in operators:
#found an operator. Everything we've accumulated in `buff` is
#a single "number". Join it together and put it in `num_out`.
num_out.append(''.join(buff))
buff = []
op_out.append(c)
else:
#not an operator. Just accumulate this character in buff.
buff.append(c)
num_out.append(''.join(buff))
return num_out,op_out
print parse('3/2*15')
这不是最优雅的方法，但是它可以为您提供合理的数据结构（就我而言仍然如此）
现在编写代码以实际解析和评估数字-这将在浮点数中完成所有操作，但是更改起来很容易…
import operator
def my_eval(nums,ops):
nums = list(nums)
ops = list(ops)
operator_order = ('*/','+-') #precedence from left to right. operators at same index have same precendece.
#map operators to functions.
op_dict = {'*':operator.mul,
'/':operator.div,
'+':operator.add,
'-':operator.sub}
Value = None
for op in operator_order: #Loop over precedence levels
while any(o in ops for o in op): #Operator with this precedence level exists
idx,oo = next((i,o) for i,o in enumerate(ops) if o in op) #Next operator with this precedence
ops.pop(idx) #remove this operator from the operator list
values = map(float,nums[idx:idx+2]) #here I just assume float for everything
value = op_dict[oo](*values)
nums[idx:idx+2] = [value] #clear out those indices
return nums[0]
print my_eval(*parse('3/2*15'))
602392714
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