python - 模式计算器Python(Pattern Calculator Python)

Question

I had a friend of mine show me an interesting problem he had on a coding challange at an interview.

(我有一个朋友向我展示了一个有趣的问题，他在一次采访中遇到了编码挑战。)

So you have a number like n = 3413289830 and a pattern like p = a-bcdefghij, you need to create a function that takes this input and outputs -413289827.

(因此，您有一个像n = 3413289830的数字和一个像p = a-bcdefghij的模式，您需要创建一个接受此输入并输出-413289827的函数。)

Obviously it should work for any combination of numbers and letters for addition and subtraction.

(显然，它应该适用于数字和字母的任何组合以进行加减。)

I worked out this code but I am pretty sure it can be improved as I think it's a bit inefficient.

(我编写了这段代码，但是我很确定可以改进它，因为我认为它效率不高。)

pattern = 'a-bcdefghij'       
n = '3413289830'
lst = []
def splitnb(n, pattern):
    save = dict()
    if(len(n) != len(pattern) - 1):
        print('Pattern needs to match number')
    else:
        if( '-' in pattern):
            patlst = pattern.split('-')
        elif('+' in pattern):
            patlst = pattern.split('+')
        for char in n:
            a = list(n)
        for pat in patlst:
            first  = patlst[0].split()
            rest = pat.split()
        for l in first[0]:
            f1 = l
            lst.append(f1)
        for l2 in rest[0]:
            f2 = l2
            lst.append(f2)
        save = dict(zip(lst, a))
        nb = ''
        if( '-' in pattern):
            for i in first[0]:
                numb = int(save.get(i))
            for j in rest[0]:
                nb += save.get(j)
                numb1 = numb - int(nb)
        elif('+' in pattern):
            for i in first[0]:
                numb = int(save.get(i))
            for j in rest[0]:
                nb += save.get(j)
                numb1 = numb + int(nb)
    return numb1

f1 = splitnb(n, pattern)
f2 = splitnb('1234', 'ab+cd')
f3 = splitnb('22', 'a-b')

  ask by thouxanstx translate from so

Answer 1

One way to do this is to take the pattern and replace each char with the number that should be there and then eval the result

(一种方法是采用模式并将每个字符替换为应该存在的数字，然后eval结果)

string.ascii_letters is a string of all ascii characters in alphabetical order starting with lowercase.

(string.ascii_letters是所有ASCII字符的字符串，按字母顺序string.ascii_letters写开始。)

This can be used to convert the char into the index of the digit that should be extracted from n

(这可用于将char转换为应从n提取的数字的索引)

>>> [n[string.ascii_letters.index(x)] if x in string.ascii_letters else x for x in pattern]
['3', '-', '4', '1', '3', '2', '8', '9', '8', '3', '0']

We add if x in string.ascii_letters else x so that the operators are not converted.

(我们if x in string.ascii_letters else x添加if x in string.ascii_letters else x以便不转换运算符。)

Then you join the resulting list to get the string

(然后，您加入结果列表以获取字符串)

>>> ''.join(n[string.ascii_letters.index(x)] if x in string.ascii_letters else x for x in pattern)
'3-413289830'

Removing the brackets turns the list comprehension into a generator which should be slightly more performant.

(删除括号将列表理解变成一个生成器，它应该会稍微好一些。)

You can then use eval to run this string as if it was python code

(然后，您可以使用eval来运行此字符串，就好像它是python代码一样)

>>> eval(''.join(n[string.ascii_letters.index(x)] if x in string.ascii_letters else x for x in pattern))
-413289827

You should only use eval if you trust the input that you are being given as it can execute arbitrary code

(仅当您信任输入信息时，才应使用eval，因为它可以执行任意代码)