我正在尝试创建一个输入名称并输出等级的函数。创建该功能的最佳方法是什么？

Question

我正在尝试创建输入名称并根据其具有的字母的顺序输出等级的

函数= ab
等级= 3，

import string

x = "richard"
y = "donald"
c = "Sam"
numbers = []


for i in range(1,27):
    numbers.append(i)

print(numbers)

alphabet_string = string.ascii_lowercase
alphabet_list = list(alphabet_string)

print(alphabet_list)

new_x = list(x)
new_y = list(y)
new_c = list(c)

zip_iterators = zip(alphabet_list,numbers)
dic = list(zip_iterators)

print(dic)

def rank(name):
    rank = 0
    for l in range(0,len(name)):
        for k,v in dic:
            if l == k:
                v += rank
        print(rank)

rank(new_c)

但我到目前为止失败了

Answer 1

letter_rank = {letter:rank for rank, letter in enumerate(string.ascii_lowercase, 1)}

def rank(name):
    return sum(letter_rank.get(c, 0) for c in name.lower())

Answer 2

您只需在ascii_lowercase String> String模块中常数

from string import ascii_lowercase

def rank(x):
    total = 0
    for char in x:
        if char in ascii_lowercase:
            total += ascii_lowercase.index(char) + 1
    return total

print(rank('abc'))

---

：6

Answer 3

您可以使用ascii_lowercase.index（Letter）或创建字典来查找字母的等级。 （我的示例不包括上限，但是如果您愿意，您可以这样做）

查找字典

from string import ascii_lowercase

alphabet_lookup = {letter:i+1 for i, letter in enumerate(ascii_lowercase)}

def rank(name):
    return sum(alphabet_lookup[c] for c in name.lower())


print(rank('baa')) # outputs 4

str.index

def rank(name):
    return sum(ascii_lowercase.index(c)+1 for c in name.lower())

Answer 4

您可以使用ord（）内置函数并列表理解，以创建所需的功能，如下所示：

x = "Villalobos Velasquez Santiago"

def fun(s):
    out = 0
    for i in s.lower():
        if i != " ":
            out += (ord(i)-96)
    return out

print(fun(x))

output：

333