在python中将十六进制转换为十进制

我正在尝试编写一个将十六进制转换为十进制的函数。 我有两个问题。我不能用数字来代替所有的字母。它替换一个字母，然后停止。第二，如何得到它，使它连续地将每个整数相加

 def toDecimal(hexidecimal):
    decimal=[hexidecimal[i:i+1] for i in range(0,len(hexidecimal), 1)]
    for i in range(0,len(decimal)):
            if 'a' in decimal:
                decimal[i]='10'
            if 'b' in decimal:
                decimal[i]='11'
            if 'c' in decimal:
                decimal[i]='12'
            if 'd' in decimal:
                decimal[i]='13'
            if 'e' in decimal:
                decimal[i]='14'
            if 'f' in decimal:
                decimal[i]='15'
            return decimal
      #Above I try to convert any letters into a number value 
    for i in range(0,len(decimal)):
        converted_decimal=decimal[i]*(16**i)
        total_decimal=converted_decimal+converted_decimal
    return total_decimal
     #Here I'm trying to add each converted decimal 

---

有一种更简单的方法可以做到这一点：

大的东西：

>>> s = "6f48f8248e828ce82f82"
>>> int(s, 16)
525528725744949635067778L

小一点的：

>>> int('BF', 16)
191

在数字前面添加一个

0x

，不会改变最终结果

作为一项功能：

def hex2dec(s):
    """return the integer value of a hexadecimal string s"""
    return int(s, 16)

---

有一种更简单的方法可以做到这一点：

大的东西：

>>> s = "6f48f8248e828ce82f82"
>>> int(s, 16)
525528725744949635067778L

小一点的：

>>> int('BF', 16)
191

在数字前面添加一个

0x

，不会改变最终结果

作为一项功能：

def hex2dec(s):
    """return the integer value of a hexadecimal string s"""
    return int(s, 16)

---

我只是在这里玩得很开心，但是要重写你的函数

def toDecimal(hexadecimal):
    decimal = int(hexadecimal, 16)
    return decimal

toDecimal('0xdeadbeef')

---

我猜你发明轮子只是为了看看它是怎么做的？：）

我只是在这里玩得很开心，但是要重写你的函数

def toDecimal(hexadecimal):
    decimal = int(hexadecimal, 16)
    return decimal

toDecimal('0xdeadbeef')

---

我猜你发明轮子只是为了看看它是怎么做的？：）

您的代码中有很多问题。让我们看一下：

hexidecimal= "7ac8965f" #hexadecimal value

decimal=[hexidecimal[i:i+1] for i in range(0,len(hexidecimal), 1)]
# >> decimal : ["7","a","c","8","9","6","5","f"]

for i in range(0,len(decimal)):
# first path : i = 0

        # First Error : 'in' is a array-wide search.
        # you want to use :'if decimal[i] == 'a' '
        if 'a' in decimal: # a is in decimal (second pos) so decimal[0] is modified !
            decimal[i]='10'
            # >> decimal : ["10","a","c","8","9","6","5","f"]

        if 'b' in decimal:
            decimal[i]='11'
        if 'c' in decimal:
            decimal[i]='12'
        if 'd' in decimal:
            decimal[i]='13'
        if 'e' in decimal:
            decimal[i]='14'
        if 'f' in decimal: # f is in decimal (last pos) so decimal[0] is modified !
            decimal[i]='15'
            # >> decimal : ["15","a","c","8","9","6","5","f"]

        #Second Error : anticipated return
        #Assuming the indentation is correct, the function exit here, on the 
        #first run of the function
        return decimal

现在有一个解决方案：

#dict for conversion (there are more elegant ways to do it, 
#                                                        but this is good enough)
conversion_table = { '0' : 0 ,
                     '1' : 1 ,
                     '2' : 2 ,
                     '3' : 3 ,
                      .......
                     'a' : 10 ,
                     'b' : 11 ,
                     ......
                     'f' : 15
                    }

hexidecimal= "7ac8965f" #hexadecimal value

hexa_list=[ digit for digit in hexidecimal]
# same thing as before, just more "elegant"

decimal = [ conversion_table[hex_digit] for hex_digit in hexa_list]
# convert everything in base10

# reduce the list
return reduce( lambda x,y : x*16 + y, decimal )

---

代码中有很多问题。让我们看一下：

hexidecimal= "7ac8965f" #hexadecimal value

decimal=[hexidecimal[i:i+1] for i in range(0,len(hexidecimal), 1)]
# >> decimal : ["7","a","c","8","9","6","5","f"]

for i in range(0,len(decimal)):
# first path : i = 0

        # First Error : 'in' is a array-wide search.
        # you want to use :'if decimal[i] == 'a' '
        if 'a' in decimal: # a is in decimal (second pos) so decimal[0] is modified !
            decimal[i]='10'
            # >> decimal : ["10","a","c","8","9","6","5","f"]

        if 'b' in decimal:
            decimal[i]='11'
        if 'c' in decimal:
            decimal[i]='12'
        if 'd' in decimal:
            decimal[i]='13'
        if 'e' in decimal:
            decimal[i]='14'
        if 'f' in decimal: # f is in decimal (last pos) so decimal[0] is modified !
            decimal[i]='15'
            # >> decimal : ["15","a","c","8","9","6","5","f"]

        #Second Error : anticipated return
        #Assuming the indentation is correct, the function exit here, on the 
        #first run of the function
        return decimal

现在有一个解决方案：

#dict for conversion (there are more elegant ways to do it, 
#                                                        but this is good enough)
conversion_table = { '0' : 0 ,
                     '1' : 1 ,
                     '2' : 2 ,
                     '3' : 3 ,
                      .......
                     'a' : 10 ,
                     'b' : 11 ,
                     ......
                     'f' : 15
                    }

hexidecimal= "7ac8965f" #hexadecimal value

hexa_list=[ digit for digit in hexidecimal]
# same thing as before, just more "elegant"

decimal = [ conversion_table[hex_digit] for hex_digit in hexa_list]
# convert everything in base10

# reduce the list
return reduce( lambda x,y : x*16 + y, decimal )

---

我认为阅读本惯用python教程会对您有所帮助：

---

我认为阅读本惯用python教程会对您有所帮助：

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这是任务吗？否则int（hexnumber，16）执行作业是的，这是一个赋值。请尝试

int（'0xA'，16）

或

eval（'0xA'）

这是一个赋值吗？否则int（hexnumber，16）会执行作业是的，这是一个赋值。请尝试

int（'0xA'，16）

或

eval（'0xA'）

这是一个赋值。因此我不允许使用int（）。这是一个赋值。因此我不允许使用int（）。