将 IP 地址与 IP 网络匹配并返回关联的列

Question

>我在熊猫中有两个数据帧。

import pandas as pd
inp1 = [{'network':'1.0.0.0/24', 'A':1, 'B':2}, {'network':'5.46.8.0/23', 'A':3, 'B':4}, {'network':'78.212.13.0/24', 'A':5, 'B':6}]
df1 = pd.DataFrame(inp)
print("df1", df1)
inp2 = [{'ip':'1.0.0.10'}, {'ip':'blahblahblah'}, {'ip':'78.212.13.249'}]
df2 = pd.DataFrame(inp2)
print("df2", df2)

输出：

network  A  B
0      1.0.0.0/24  1  2
1     5.46.8.0/23  3  4
2  78.212.13.0/24  5  6
ip
0       1.0.0.10
1   blahblahblah
2  78.212.13.249

我想要的最终输出如下所示：

ip              A           B
0       1.0.0.10         1           2
1   blahblahblah         NaN         Nan
2  78.212.13.249         5           6

我想遍历df2['ip']中的每个单元格，并检查它是否属于df1['network']中的网络。如果它属于网络，它将返回特定 IP 地址的相应 A 和 B 列。我已经参考了这篇文章并考虑了netaddr, IPNetwork, IPAddress, ipaddress但无法完全弄清楚。

感谢帮助！

Answer 1

您可以使用netaddr+apply()来做到这一点。下面是一个示例：

from netaddr import IPNetwork, IPAddress, AddrFormatError

network_df = pd.DataFrame([
{'network': '1.0.0.0/24', 'A': 1, 'B': 2},
{'network': '5.46.8.0/23', 'A': 3, 'B': 4},
{'network': '78.212.13.0/24', 'A': 5, 'B': 6}
])
ip_df = pd.DataFrame([{'ip': '1.0.0.10'}, {'ip': 'blahblahblah'}, {'ip': '78.212.13.249'}])
# create all networks using netaddr
networks = (IPNetwork(n) for n in network_df.network.to_list())

def find_network(ip):
# return empty string when bad/wrong IP
try:
ip_address = IPAddress(ip)
except AddrFormatError:
return ''
# return network name as string if we found network
for network in networks:
if ip_address in network:
return str(network.cidr)
return ''

# add network column. set network names by ip column
ip_df['network'] = ip_df['ip'].apply(find_network)
# just merge by network columns(str in both dataframes)
result = pd.merge(ip_df, network_df, how='left', on='network')
# you don't need network column in expected output...
result = result.drop(columns=['network'])
print(result)
#               ip    A    B
# 0       1.0.0.10  1.0  2.0
# 1   blahblahblah  NaN  NaN
# 2  78.212.13.249  5.0  6.0

请参阅评论。希望这有帮助。

Answer 2

如果你愿意使用R而不是Python，我写了一个IP地址包可以解决这个问题。仍然有一个底层循环，但它是在C++中实现的(更快！

library(tibble)
library(ipaddress)
library(fuzzyjoin)
addr <- tibble(
address = ip_address(c("1.0.0.10", "blahblahblah", "78.212.13.249"))
)
#> Warning: Problem on row 2: blahblahblah
nets <- tibble(
network = ip_network(c("1.0.0.0/24", "5.46.8.0/23", "78.212.13.0/24")),
A = c(1, 3, 5),
B = c(2, 4, 6)
)
fuzzy_left_join(addr, nets, c("address" = "network"), is_within)
#> # A tibble: 3 x 4
#>         address        network     A     B
#>       <ip_addr>     <ip_netwk> <dbl> <dbl>
#> 1      1.0.0.10     1.0.0.0/24     1     2
#> 2            NA             NA    NA    NA
#> 3 78.212.13.249 78.212.13.0/24     5     6

^{创建于 2020-09-02 由 reprex 软件包 (v0.3.0(}