Question

我试图将字符分配给值线的最小量可能

这就是我所知道的全部方法。我查看了其他解决方案，但似乎无法使它们工作

letter = random.randint(1,26)
if letter == 1:
    actualLetter = "A"
elif letter == 2:
    actualLetter = "B"
elif letter == 3:
    actualLetter = "C"
elif letter == 4:
    actualLetter = "D"
elif letter == 5:
    actualLetter = "E"
elif letter == 6:
    actualLetter = "F"
elif letter == 7:
    actualLetter = "G"
elif letter == 8:
    actualLetter = "H"
elif letter == 9:
    actualLetter = "I"
elif letter == 10:
    actualLetter = "J"
elif letter == 11:
    actualLetter = "K"
elif letter == 12:
    actualLetter = "L"
elif letter == 13:
    actualLetter = "M"
elif letter == 14:
    actualLetter = "N"
elif letter == 15:
    actualLetter = "O"
elif letter == 16:
    actualLetter = "P"
elif letter == 17:
    actualLetter = "Q"
elif letter == 18:
    actualLetter = "R"
elif letter == 19:
    actualLetter = "S"
elif letter == 20:
    actualLetter = "T"
elif letter == 21:
    actualLetter = "U"
elif letter == 22:
    actualLetter = "V"
elif letter == 23:
    actualLetter = "W"
elif letter == 24:
    actualLetter = "X"
elif letter == 25:
    actualLetter = "Y"
elif letter == 26:
    actualLetter = "Z"

actualLetter被分配了与该值相对应的字母。

Answer 1

您可以使用string.ascii_uppercase：

import random
from string import ascii_uppercase:

letter = random.randint(1, 26)
actualLetter = ascii_uppercase[letter - 1]

更好的是，使用random.choice（优选溶液）：

actualLetter = random.choice(ascii_uppercase)

或者，而不string.ascii_uppercase：

letter = random.randint(1, 26)
actualLetter = chr(letter + ord('A') - 1)

Answer 2

您可以使用'ord（）'和'chr（）'依靠内置的ascii映射

在[8]中：ord（'A'）
出[8]：65

在[9]中：chr（65）
出[9]：“ A”

尝试简化为数字分配字符

2 个答案: