如何使用 Python 將八進制轉換為十進制

Question

我有這個小小的家庭作業，我需要將十進制轉換為八進制，然后將八進制轉換為十進制。 我做了第一部分，無法弄清楚第二部分來挽救我的生命。 第一部分是這樣的：

decimal = int(input("Enter a decimal integer greater than 0: "))

print("Quotient Remainder Octal")
bstring = " "
while decimal > 0:
    remainder = decimal % 8
    decimal = decimal // 8
    bstring = str(remainder) + bstring
    print ("%5d%8d%12s" % (decimal, remainder, bstring))
print("The octal representation is", bstring)

我在這里閱讀了如何轉換它： Octal to Decimal ，但我不知道如何將其轉換為代碼。

Answer 1

從十進制到八進制：

oct(42) # '052'

八進制轉十進制

int('052', 8) # 42

如果要將八進制作為字符串返回，則可能需要將其包裝在str 。

Answer 2

這些第一行采用任何十進制數並將其轉換為任何所需的數字基數

def dec2base():
    a = int(input('Enter decimal number: \t'))
    d = int(input('Enter expected base: \t'))
    b = ""
    while a != 0:
        x = '0123456789ABCDEF'
        c = a % d
        c1 = x[c]
        b = str(c1) + b
        a = int(a // d)
    return (b)

第二行做同樣的事情，但對於給定的范圍和給定的小數

def dec2base_R():
    a = int(input('Enter start decimal number:\t'))
    e = int(input('Enter end decimal number:\t'))
    d = int(input('Enter expected base:\t'))
    for i in range (a, e):
        b = ""
        while i != 0:
            x = '0123456789ABCDEF'
            c = i % d
            c1 = x[c]
            b = str(c1) + b
            i = int(i // d)
    return (b)

第三行從任何基數轉換回十進制

def todec():
    c = int(input('Enter base of the number to convert to decimal:\t'))
    a = (input('Then enter the number:\t ')).upper()
    b = list(a)
    s = 0
    x = ['0', '1', '2', '3', '4', '5', '6', '7', '8', '9', 'A', 'B', 'C', 'D', 'E', 'F']
    for pos, digit in enumerate(b[-1::-1]):
        y = x.index(digit)
        if int(y)/c >= 1:
            print('Invalid input!!!')
            break
        s = (int(y) * (c**pos)) + s
    return (s)

注意：如果有人需要，我也有 GUI 版本

Answer 3

def decimal_to_octal(num1):
  new_list = []
  while num1 >= 1:
    num1 = num1/8
    splited = str(num1).split('.')
    num1 = int(splited[0])
    appendednum = float('0.'+splited[1])*8
    new_list.append(int(appendednum))
    decimal_to_octal(num1)    
  return "your number in octal: "+''.join(str(v) for v in new_list[::-1])

print(decimal_to_octal(384))

Answer 4

我不是 Python 專家......但我寫了這個邏輯......它工作正常。

def octToDec(oct):
    lenOct = str(oct)
    le = len(lenOct)
    octal = 0
    for i in (range(le)):
        octal = octal + int(lenOct[i])* pow(8, le-1)
        le -= 1
    print(octal)

octToDec(number)

Answer 5

def decimalToOctal(num1):
  new_list = []
  while num1 >= 1:
    num1 = num1/8
    splited = str(num1).split('.')
    num1 = int(splited[0])
    appendednum = float('0.' + splited[1])*8
    new_list.append(int(appendednum))
    decimalToOctal(num1)
  return "your number in octal: " + ''.join(str(v) for v in new_list[::-1])

print(decimalToOctal(384))

一個更簡單的版本：

a = input("Enter a string of octal digits: ")
while a != 0:
    de = str(int(a,8));
    print("The integer value is", de);
    break