[英]How can I shorten this substitution cipher code?
输入字符串将仅包含字母字符。 该函数应返回一个字符串,其中所有字符都已在字母表中“向上”移动了两个位置。
例如:
- “a”会变成“c”
- “z”会变成“b”
我写了这段代码,但我觉得它太长了。 我怎样才能让它更短更好?
def encrypt_message(strings: str) -> str:
our_al = ["a", "b", "c", "d", "e", "f", "g", "h", "i", 'j', 'k', 'l', 'm', 'n', 'o', 'p', 'q', 'r', 's', 't', 'u',
'v', 'w', 'x', 'y', 'z']
new_s = ""
for character in strings:
index = our_al.index(character)
if index <= 23:
index += 2
new_s += our_al[index]
elif index == 24:
new_s += our_al[0]
elif index == 25:
new_s += our_al[1]
return new_s
print(encrypt_message("abc"))
print(encrypt_message("xyz"))
print(encrypt_message(""))
一些实用程序将是有益的。 如果您重复使用此函数,您并不总是希望迭代字符以查找索引,因此查找dict
。
from string import ascii_lowercase as al
index = {c: i for i, c in enumerate(al)}
def encrypt_message(s: str) -> str:
return ''.join(al[(index[c] + 2) % 26] for c in s)
>>> encrypt_message('xyz')
'zab'
您可以使用itertools.islice
和itertools.cycle
来获取下一个字符(位于前 2 个位置):
from itertools import islice, cycle
from string import ascii_lowercase
def get_next_2(c):
index = ascii_lowercase.index(c)
return next(islice(cycle(ascii_lowercase), index + 2, None))
def encrypt_message(strings):
return ''.join(map(get_next_2, strings))
如果您喜欢单行解决方案,您可以使用:
from string import ascii_lowercase as al
def encrypt_message(strings):
return ''.join([al[(al.index(c) + 2) % 26] for c in strings])
您可以进行两项改进:
%
) 来简化计算
(25 + 2) % 26
计算结果为1
def encrypt_message(strings: str) -> str:
new_s = ""
for character in strings:
if character not in string.ascii_lowercase:
continue
index = string.ascii_lowercase.index(character)
new_index = (index + 2) % len(string.ascii_lowercase)
new_s += string.ascii_lowercase[new_index]
return new_s
您还可以将字符转换为它们的 ASCII 值,添加两个并将它们转换回来。
for character in strings:
if character not in string.ascii_lowercase:
continue
index = ord(character) - 96
new_index = (index + 2) % 26 + 96 # a is 97
return chr(new_index)
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