You don't need to hard-code the header values, just set a variable to the header and test if the new header is different.

last_header = '  '
for value, header in list1:
    if header != last_header:
        print(header, '--------------')
        last_header = header
    print value

Update: Based on comment of OP, that only the response headers are needed. Even more easy as written in below documentation of Requests module:

We can view the server's response headers using a Python dictionary:

>>> r.headers
{
    'content-encoding': 'gzip',
    'transfer-encoding': 'chunked',
    'connection': 'close',
    'server': 'nginx/1.0.4',
    'x-runtime': '148ms',
    'etag': '"e1ca502697e5c9317743dc078f67693f"',
    'content-type': 'application/json'
}

And especially the documentation notes:

The dictionary is special, though: it's made just for HTTP headers. According to RFC 7230, HTTP Header names are case-insensitive.

So, we can access the headers using any capitalization we want:

and goes on to explain even more cleverness concerning RFC compliance.

The Requests documentation states:

Using Response.iter_content will handle a lot of what you would otherwise have to handle when using Response.raw directly. When streaming a download, the above is the preferred and recommended way to retrieve the content.

It offers as example:

>>> r = requests.get('https://api.github.com/events', stream=True)
>>> r.raw
<requests.packages.urllib3.response.HTTPResponse object at 0x101194810>
>>> r.raw.read(10)
'\x1f\x8b\x08\x00\x00\x00\x00\x00\x00\x03'

But also offers advice on how to do it in practice by redirecting to a file etc. and using a different method:

Using Response.iter_content will handle a lot of what you would otherwise have to handle when using Response.raw directly

LAYOUT = "{!s:10} {!s:14} {!s:4} {!s:10}"

yourList = ["Ford","Mustang",1966,"Red"]

print LAYOUT.format("Brand","Modell","Year","Color")
print LAYOUT.format(*yourList)

produces output in columns which will line up (so long as you don't exceed the field width specifiers!), ie

Brand      Modell         Year Color     
Ford       Mustang        1966 Red       
Mercedes   F-Cell         2011 White

If you have your HTTP response packet data in http_response_data variable as python bytes type you can get response headers just in one line:

headers_text = http_response_data.partition(b'\r\n\r\n')[0].decode('utf-8')
print(headers_text)

Also you might need to use 'cp852' instead of 'utf-8' if HTTP headers are not UTF-8 encoded.

This takes into account the fact that HTTP headers are separated from HTTP content body by two newlines (all header lines are separated just by one newline).

Next is small example of usage solution above by receiving HTTP bytes response from TCP port 80 of Google's server using standard socket library.

Try it online!

import socket
s = socket.socket()
s.connect(('google.com', 80))
s.send(b'GET / HTTP/1.1\r\n\r\n')
s.shutdown(socket.SHUT_WR)
http_response_data = s.recv(8192) # TCP response stored as bytes
s.close()
headers_text = http_response_data.partition(b'\r\n\r\n')[0].decode('utf-8')
print(headers_text)

PS:

1. Your headers might be not UTF-8 encoded then instead of .decode('utf-8') try other encoding/code-page like this .decode('cp852').
2. headers_text will contain also status line like HTTP/1.1 200 OK if you don't need it (if important to have only key: value lines) you can use next code instead:

headers_text = http_response_data.partition(b'\r\n\r\n')[0].partition(b'\r\n')[2].decode('utf-8')
print(headers_text)

3. According to your tutorial TCP data as bytes type is available at the end of body of if protocol == 6 : block as data variable, use it as http_response_data in my solution.

Use DataFrame.dot with all columns without first for matrix multiplication, add separator and last remove last character (separator) by indexing:

df['Result'] = df.iloc[:, 1:].dot(df.columns[1:] + ',').str[:-1]
print (df)
    Id  A  B  C  D Result
0  101  0  0  0  1      D
1  102  0  0  0  0       
2  103  1  0  1  0    A,C
3  104  1  0  1  1  A,C,D

Your solution should be changed with extract matched index with join:

df['Result'] = df.apply(lambda x: ','.join(x.index[1:][x.iloc[1:] == 1]), axis=1)
print (df)
    Id  A  B  C  D Result
0  101  0  0  0  1      D
1  102  0  0  0  0       
2  103  1  0  1  0    A,C
3  104  1  0  1  1  A,C,D