Question

我有以下“ arp -a”命令输出列表，只要检测到网络中的设备，该输出就会连续。

['XX.1.60.XX                ether   XX:16:35:b2:c9:XX   C                     wlan0', 'XXX.168.42.XXX           ether   XX:35:cd:e8:a5:XX   C                     usb0', 'XX.1.62.XXX              ether   XX:9f:db:72:61:XX   C                     wlan0', '']

我希望生成一个以IP地址为键，硬件地址为值的字典，即：

{ 'XX.1.60.XX': XX:16:35:b2:c9:XX, 'XXX.168.42.XXX': 'XX:35:cd:e8:a5:XX'}

Answer 1

使用split()将行切成单词

data = ['XX.1.60.XX                ether   XX:16:35:b2:c9:XX   C                     wlan0', 'XXX.168.42.XXX           ether   XX:35:cd:e8:a5:XX   C                     usb0', 'XX.1.62.XXX              ether   XX:9f:db:72:61:XX   C                     wlan0', '']

result = {}
for line in data:
    try:
        ip, _, hardware, _, _ = line.split()
        result[ip] = hardware
    except ValueError:
        # bad line format
        pass

print(result)
# {'XX.1.60.XX': 'XX:16:35:b2:c9:XX', 'XXX.168.42.XXX': 'XX:35:cd:e8:a5:XX', 'XX.1.62.XXX': 'XX:9f:db:72:61:XX'}

Answer 2

使用字典理解：

o = ['XX.1.60.XX                ether   XX:16:35:b2:c9:XX   C                     wlan0', 'XXX.168.42.XXX           ether   XX:35:cd:e8:a5:XX   C                     usb0', 'XX.1.62.XXX              ether   XX:9f:db:72:61:XX   C                     wlan0', '']

o = {v[0]: v[2] for v in (v.split() for v in o) if len(v) >= 3}
print(o)

打印：

{'XX.1.60.XX': 'XX:16:35:b2:c9:XX', 'XXX.168.42.XXX': 'XX:35:cd:e8:a5:XX', 'XX.1.62.XXX': 'XX:9f:db:72:61:XX'}

Answer 3

这可以做到：

out = {re.findall(r'X.*?(?= )',i)[0]: re.findall(r'X.*?(?= )',i)[1] for i in x if i!=''}

收益：

{'XX.1.60.XX': 'XX:16:35:b2:c9:XX', 'XXX.168.42.XXX': 'XX:35:cd:e8:a5:XX', 'XX.1.62.XXX': 'XX:9f:db:72:61:XX'}

Answer 4

dict(re.findall(r'([\dX.]+)\s.*?([\dX:a-f]+)\s', line)[0] for line in data if line)
# {'XX.1.60.XX': 'XX:16:35:b2:c9:XX', 'XXX.168.42.XXX': 'XX:35:cd:e8:a5:XX', 'XX.1.62.XXX': 'XX:9f:db:72:61:XX'}

我如何从列表中生成字典并删除所有空格

4 个答案: