Python 插入字母和数字-将字母向右移动';编号';时代 def char(): 字母=str(输入(“输入字母表中的字母:”) x=int(输入(“将字母向右移动的输入值x次数:”) 新字母=ord(字母)+x 如果newLetter>90且newLetter122: 余数=新字母%122 newLetterI=97+余数 印刷品(chr(新字母I)) char()
这是我的代码,它将字母向右移动了“多次”。Python 插入字母和数字-将字母向右移动';编号';时代 def char(): 字母=str(输入(“输入字母表中的字母:”) x=int(输入(“将字母向右移动的输入值x次数:”) 新字母=ord(字母)+x 如果newLetter>90且newLetter122: 余数=新字母%122 newLetterI=97+余数 印刷品(chr(新字母I)) char(),python,python-3.x,Python,Python 3.x,这是我的代码,它将字母向右移动了“多次”。 这是非常冗长的,但这是唯一的方法,我发现如何做到这一点,没有很多错误,我真的不明白。我只是想知道它是否正常,我只是想知道一旦字母到达Z或Z,包装是否正常。请检查一下: def char(): letter = str(input("Input Letter from alphabet: ")) x = int(input("Input Value to shift letter to the right x number of time
这是非常冗长的,但这是唯一的方法,我发现如何做到这一点,没有很多错误,我真的不明白。我只是想知道它是否正常,我只是想知道一旦字母到达Z或Z,包装是否正常。请检查一下:
def char():
letter = str(input("Input Letter from alphabet: "))
x = int(input("Input Value to shift letter to the right x number of times: : "))
newLetter = ord(letter) + x
if newLetter > 90 and newLetter < 97:
remainder = newLetter % 90
newLetterI = 65 + remainder
print(chr(newLetterI))
elif newLetter > 122:
remainder = newLetter % 122
newLetterI = 97 + remainder
print(chr(newLetterI))
char()
输出:
#!/usr/bin/python
def next_nchar(char, n):
# 97, 122
upper = None
if char.isupper():
upper = True
lw = char.lower()
start = ord(lw)
num = ord(lw) + n
if num > 122:
ex = num - 122
else:
ex = None
if not ex:
r = range(start, num)
else:
r = range(start, 123) + range(97, 97+ex-1)
if upper:
return [chr(x).upper() for x in r]
return [chr(x) for x in r]
for i in ['a', 'k', 'y', 'P']:
print next_nchar(i, 5)
你应该用这个
['a', 'b', 'c', 'd', 'e']
['k', 'l', 'm', 'n', 'o']
['y', 'z', 'a', 'b', 'c']
['P', 'Q', 'R', 'S', 'T']
导入系统
def char():
字母=sys.stdin.readline().strip()
x=int(sys.stdin.readline())
oldLetter=ord(字母)
如果oldLetter>64且oldLetter<91:
如果(oldLetter+x)96和oldLetter<123:
如果(oldLetter+x)我可以从您的代码中找到一个问题,那就是概括似乎没有多大意义。例如,如果将新字母模化122,当移位为26时会遇到问题,而且由于必须记住ascii值,因此读取起来有点困难 我将尝试解释一个稍微改进的版本:
import sys
def char():
letter = sys.stdin.readline().strip()
x = int(sys.stdin.readline())
oldLetter = ord(letter)
if oldLetter > 64 and oldLetter < 91:
if (oldLetter+x) <= 90:
remainder = (oldLetter + x) % 65
newLetter = 65 + remainder
else:
remainder = (oldLetter + x) % 90
newLetter = 64 + remainder
print(chr(newLetter))
elif oldLetter > 96 and oldLetter < 123:
if (oldLetter+x) <= 122:
remainder = (oldLetter + x) % 97
newLetter = 97 + remainder
else:
remainder = (oldLetter + x) % 122
newLetter = 96 + remainder
print(chr(newLetter))
char()
以下是一个更简洁的版本,没有任何注释:
letter = input("Enter a letter from the alphabet: ") # No need for str, it is already a string
shift = int(input("Enter your letter shift: "))
if letter.islower(): # Much better than worrying about ascii values
initial = 'a'
else:
initial = 'A'
letterIndex = (ord(letter) - ord(initial)) # So 'a' would have index 0, 'b' 1, 'z' 25
newIndex = (letterIndex + shift) % 26 # Making sure our index is from 0 to 25
newLetter = chr(ord(initial) + newIndex) # after 'a' or 'A', increment by newIndex
print newLetter
这是最具可读性的方法,IMO:
letter = input("Enter a letter from the alphabet: ")
shift = int(input("Enter your letter shift: "))
initial = 'a' if letter.islower() else 'A'
newIndex = (ord(letter) - ord(initial) + shift) % 26
print(chr(ord(initial) + newIndex))
可能是重复的谢谢。我看到了这个,但我不太明白它的代码。
import string
def shiftchar(char, n):
if char.isupper():
letters = string.uppercase # all uppercase letters in alphabetical order
elif char.islower():
letters = string.lowercase # all lowercase letters in alphabetical order
try:
initcharpos = letters.index(char)
newchar = letters[(initcharpos + n) % 26]
# get the letter n characters later, with wraparound
return newchar
except (NameError, ValueError):
# NameError: char is mixed-case or non-English, so letters is never defined
# ValueError: char not found in letters; probably a multi-char string
raise TypeError("Expected a single English char as first argument")
print shiftchar('A', 5) # F
print shiftchar('z', 1) # a
print shiftchar('foo', 5) # TypeError