Python在数据框中扩展网络地址

Question

我有一个类似于以下的数据框：

df:

    ip          last_active
192.167.0.9     01/02/2012
226.90.2.12     05/06/2013
10.90.2.09      05/06/2014
12.60.2.80     
192.168.2.11-17 05/06/2016

有没有办法从df中的最后一行扩展IP地址？ 一个完美的解决方案是：

df:

    ip          last_active
192.167.0.9     01/02/2012
226.90.2.12     05/06/2013
10.90.2.09      05/06/2014
12.60.2.80     
192.168.2.11    05/06/2016
192.168.2.12    05/06/2016
192.168.2.13    05/06/2016
192.168.2.14    05/06/2016
192.168.2.15    05/06/2016
192.168.2.16    05/06/2016
192.168.2.17    05/06/2016

感谢任何指导！

Answer 1

您可以应用一个函数来列出范围元素中的 IP，如果您有最新的 pandas 版本，请使用 explode()

def ip_splitter(ip):
    if '-' in ip:
        last_octet_range=[int(i) for i in ip.split('.')[3].split('-')]
        new_ips = [i for i in range(last_octet_range[0],last_octet_range[1]+1)]
        expanded_range = ['.'.join(ip.split('.')[:3]+[str(i)]) for i in new_ips]
        return expanded_range
    return ip

df['ip']=df['ip'].apply(ip_splitter)

df

    ip                                                  last_active
0   192.167.0.9                                         01/02/2012
1   226.90.2.12                                         05/06/2013
2   10.90.2.09                                          05/06/2014
3   12.60.2.80                                          None
4   [192.168.2.11, 192.168.2.12, 192.168.2.13, 192...   05/06/2016

df.explode('ip')

    ip              last_active
0   192.167.0.9     01/02/2012
1   226.90.2.12     05/06/2013
2   10.90.2.09      05/06/2014
3   12.60.2.80      None
4   192.168.2.11    05/06/2016
4   192.168.2.12    05/06/2016
4   192.168.2.13    05/06/2016
4   192.168.2.14    05/06/2016
4   192.168.2.15    05/06/2016
4   192.168.2.16    05/06/2016
4   192.168.2.17    05/06/2016

Answer 2

一种可能的解决方案，基于命名的捕获组：

pat = re.compile(r'(?P<g1>(?:\d+\.){3})(?P<g2>\d+)-(?P<g3>\d+)')
outRows = []
for _, row in df.iterrows():
    ip = row.ip
    mtch = pat.match(ip)
    if mtch:
        n1 = mtch.group('g1')
        n2 = int(mtch.group('g2'))
        n3 = int(mtch.group('g3'))
        for n in range(n2, n3 + 1):
            outRows.append([ n1 + str(n), row.last_active])
    else:
        outRows.append([ ip, row.last_active])
result = pd.DataFrame(outRows, columns=df.columns)

Answer 3

这是我的方法：

# separate relevant parts
s = df['ip'].str.extract('(\d+\.\d+\.\d+.)(\d+)-?(\d+)?').ffill(1)

# convert last parts to int for easy manipulation
s[1] = s[1].astype(int)
s[2] = s[2].astype(int)

# masking the IP range
lowers = s[1].values[:,None] <= np.arange(256)
uppers =  np.arange(256)<= s[2].values[:,None]

# create new dataframe for the new IPs
u = pd.DataFrame(lowers & uppers, 
                 index=pd.MultiIndex.from_arrays([df['last_active'],s[0]])
                )

# final data
(u.where(u).stack().reset_index(name='dummy')
 .assign(ip=lambda x: x[0]+x['level_2'].astype(str))
 .drop([0,'level_2','dummy'], axis=1)
)   

输出：

   last_active            ip
0   01/02/2012   192.167.0.9
1   05/06/2013   226.90.2.12
2   05/06/2014     10.90.2.9
3          NaN    12.60.2.80
4   05/06/2016  192.168.2.11
5   05/06/2016  192.168.2.12
6   05/06/2016  192.168.2.13
7   05/06/2016  192.168.2.14
8   05/06/2016  192.168.2.15
9   05/06/2016  192.168.2.16
10  05/06/2016  192.168.2.17

Answer 4

我喜欢各种解决方案的不同之处。这是另一个改编自这里的（！）：https://stackoverflow.com/a/33291055/42346。

df2 = pd.DataFrame(columns=df.columns)

count = 0 
for idx,r in df.iterrows(): 
    data = r['ip'].split("-") 
    if len(data) > 1: 
        start = int(data[0].split('.')[-1]) 
        end = int(data[1]) + 1 
        for i in range(start,end): 
            df2.loc[count] = (data[0][:data[0].rfind('.')+1] + 
                              str(i),r['last_active']) 
            count += 1 
    else: 
        df2.loc[count] = (r['ip'],r['last_active']) 
        count += 1 

结果：

In [40]: df2
Out[40]: 
              ip last_active
0    192.167.0.9  01/02/2012
1    226.90.2.12  05/06/2013
2     10.90.2.09  05/06/2014
3     12.60.2.80        None
4   192.168.2.11  05/06/2016
5   192.168.2.12  05/06/2016
6   192.168.2.13  05/06/2016
7   192.168.2.14  05/06/2016
8   192.168.2.15  05/06/2016
9   192.168.2.16  05/06/2016
10  192.168.2.17  05/06/2016