python 转化16进制 python 16进制转10进制

"""
16进制转10进制
"""
# str="A5 42 D2 00 4A 00 01 00 01 F4 00 32 00 32 00 50 01 0C 00 01 00 01 F4 00 32 00 32 00 50 01 01 00 01 00 01 F4 01 F4 01 F4 01 F4 01 F4 00 32 00 32 00 32 00 32 00 32 00 32 00 32 00 32 00 32 00 32 00 50 01 00 00 01 F4 00 32 00 32 50 00 00 00 D2 41 5A "
str=""
with open("bitChange",mode="r",encoding="UTF-8") as f: #从文件里读数据
   str= f.read()
lst=str.split(" ")#根据空格切割
lst=lst[5:len(lst)-4]#根据自己实际情况来做，我这里是根据我这边的通讯协议格式来的
dic={}
count=0
for c in lst:
    dic[count]=c
    count+=1
print(dic)
print(lst)
#16进制转10进制
new_Lst=[]#得到需要的16进制
starW= int(input("请输入起始位置:"))-1
lenW=int(input("请输入长度："))
new_Lst=lst[starW:lenW+starW]
new_Str="".join(new_Lst)
print("16进制：{}".format(" ".join(new_Lst)))
new_Str=new_Str[::-1]
resoult={}
resoultCount=0
strs=""
while resoultCount<len(new_Str)-1:
    if new_Str[resoultCount]=="F":
        resoult[resoultCount]=15
    elif new_Str[resoultCount]=="E":
        resoult[resoultCount] = 14
    elif new_Str[resoultCount] == "D":
        resoult[resoultCount] = 13
    elif new_Str[resoultCount] == "C":
        resoult[resoultCount] = 12
    elif new_Str[resoultCount] == "B":
        resoult[resoultCount] = 11
    elif new_Str[resoultCount] == "A":
        resoult[resoultCount] = 10
    else:
        resoult[resoultCount] = new_Str[resoultCount]
    resoultCount+=1
sum=0
for k,v in resoult.items():
    sum+=int(v)*16**k
print("10进制：{}".format(sum))

进制之间的转换可以访问