如何使用 Python 将八进制转换为十进制
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How to use Python to convert an octal to a decimal
提问by Austin Wildgrube
I have this little homework assignment and I needed to convert decimal to octal and then octal to decimal. I did the first part and can not figure out the second to save my life. The first part went like this:
我有这个小小的家庭作业，我需要将十进制转换为八进制，然后将八进制转换为十进制。我做了第一部分，无法弄清楚第二部分来挽救我的生命。第一部分是这样的：
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)
I read how to convert it here: Octal to Decimalbut have no clue how to turn it into code. Any help is appreciated. Thank you.
我在这里阅读了如何转换它：八进制到十进制，但不知道如何将其转换为代码。任何帮助表示赞赏。谢谢你。
采纳答案by martin
From decimal to octal:
从十进制到八进制：
oct(42) # '052'
Octal to decimal
八进制转十进制
int('052', 8) # 42
If you want to return octal as a string then you might want to wrap it in str.
如果要将八进制作为字符串返回，则可能需要将其包装在str.
回答by martin
Someone might find these useful
有人可能会发现这些有用
These first lines take any decimal number and convert it to any desired number base
这些第一行采用任何十进制数并将其转换为任何所需的数字基数
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)
The second lines do the same but for a given range and a given decimal
第二行做同样的事情，但对于给定的范围和给定的小数
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)
The third lines convert from any base back to decimal
第三行从任何基数转换回十进制
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)
Note: I also have the GUI version if anyone needs them
注意：如果有人需要，我也有 GUI 版本
回答by Edi Gogidze
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))
回答by Edi Gogidze
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))
回答by Jagath Gowda
I am not an expert in Python.. but I wrote this logic.. it works fine. Feel free to suggest to me if this is not the right way to do it.
我不是 Python 专家.. 但我写了这个逻辑.. 它工作正常。如果这不是正确的方法，请随时向我建议。
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)
