Python从base64转换为二进制 [英] Python converting from base64 to binary

问题描述

我有一个关于将base64编码的字符串转换为二进制的问题.我正在以下链接中收集Fingerprint2D，

I have a problem about converting a base64 encoded string into binary. I am collecting the Fingerprint2D in the following link,

url = "https://pubchem.ncbi.nlm.nih.gov/rest/pug/compound/cid/108770/property/Fingerprint2D/xml"

Fingerprint2D=AAADccB6OAAAAAAAAAAAAAAAAAAAAAAAAAA8WIEAAAAAAACxAAAAHgAACAAADAzBmAQwzoMABgCI AiTSSACCCAAhIAAAiAEMTMgMJibMsZuGeijn4BnI+YeQ0OMOKAACAgAKAABQAAQEABQAAAAAAAAA AA==

Pubchem中的说明g说这是115字节的字符串，转换为二进制时应为920位.我尝试使用以下命令将其转换为二进制文件，

The descriptiong in the Pubchem says that this is 115 byte string, and it should be 920 bits when converted into binary. I try to convert it to the binary with the following,

    response = requests.get(url)
    tree = ET.fromstring(response.text)

    for el in tree[0]:
        if "Fingerprint2D" in el.tag:
            fpp = bin(int(el.text, 16))
            print(len(fpp))

如果我使用上面的代码，则会收到以下错误，值错误:base16的int()的无效文字:

If I use the code above, I'm getting the following error, "Value error: invalid literal for int() with base16:

如果我使用下面的代码，则fpp(二进制)的长度等于1278，这不是我期望的.

And if I use the code below, length of fpp (binary) is equal to 1278 which is not what I expected.

    response = requests.get(url)
    tree = ET.fromstring(response.text)

    for el in tree[0]:
        if "Fingerprint2D" in el.tag:
            fpp = bin(int(hexlify(el.text), 16))
            print(len(fpp))

非常感谢！

推荐答案

要解码base64格式，您需要将bytes对象传递给base64.decodebytes函数:

To decode base64 format you need to pass a bytes object to the base64.decodebytes function:

import base64

t = "AAADccB6OAAAAAAAAAAAAAAAAAAAAAAAAAA8WIEAAAAAAACxAAAAHgAACAAADAzBmAQwzoMABgCI AiTSSACCCAAhIAAAiAEMTMgMJibMsZuGeijn4BnI+YeQ0OMOKAACAgAKAABQAAQEABQAAAAAAAAA AA==".encode("ascii")

decoded = base64.decodebytes(t)

print(decoded)
print(len(decoded)*8)

我得到以下信息:

b'\x00\x00\x03q\xc0z8\x00\x00\x00\x00\x00\x00\x00\x00\x00\x00\x00\x00\x00\x00\x00\x00\x00\x00\x00<X\x81\x00\x00\x00\x00\x00\x00\xb1\x00\x00\x00\x1e\x00\x00\x08\x00\x00\x0c\x0c\xc1\x98\x040\xce\x83\x00\x06\x00\x88\x02$\xd2H\x00\x82\x08\x00! \x00\x00\x88\x01\x0cL\xc8\x0c&&\xcc\xb1\x9b\x86z(\xe7\xe0\x19\xc8\xf9\x87\x90\xd0\xe3\x0e(\x00\x02\x02\x00\n\x00\x00P\x00\x04\x04\x00\x14\x00\x00\x00\x00\x00\x00\x00\x00'
920

那么920位就可以了.

So 920 bits as expected.

要以二进制形式获取数据，只需对字节进行迭代，然后使用format并将其填充为8位零填充(bin添加0b标头，因此不合适)和join字符串，即可转换为二进制一起:

To get data as binary just iterate on the bytes and convert to binary using format and zero-padding to 8 digits (bin adds a 0b header so it's not suitable), and join the strings together:

print("".join(["{:08b}".format(x) for x in decoded]))

导致:

00000000000000000000001101110001110000000111101000111000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000011110001011000100000010000000000000000000000000000000000000000000000001011000100000000000000000000000000011110000000000000000000001000000000000000000000001100000011001100000110011000000001000011000011001110100000110000000000000110000000001000100000000010001001001101001001001000000000001000001000001000000000000010000100100000000000000000000010001000000000010000110001001100110010000000110000100110001001101100110010110001100110111000011001111010001010001110011111100000000110011100100011111001100001111001000011010000111000110000111000101000000000000000001000000010000000000000101000000000000000000101000000000000000001000000010000000000000101000000000000000000000000000000000000000000000000000000000000000000

(预期为920个字符)

(which is 920 chars, as expected)

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