在 Python 中用 Pandas 确定奇数和偶数

Question

我有以下一些 IP 地址，我想根据它们的最后一位数字对它们进行分类。

一个 IPv4 地址由四个数字组成：

每个包含一到三个数字(0-255)
用一个点 (.) 分隔每个数字或一组数字

现在我想引用 IP 地址的最后一位数字，如果它是 [Odd]，则相关列用奇数填充，如果它是 [Even]，它会用偶数填充。

IP Address          
192.168.1.1         #last digit is 1 and consider it as odd
192.168.1.2         #last digit is 2 and even
192.168.152.200     #last digit is 200 and is even
192.168.54.98       #last digit is 98 and is even
192.168.98.93       #last digit is 93 and is odd
 .....
 ......

预期结果：

IP Address                 Status
192.168.1.1                Odd
192.168.1.2                Even
192.168.152.200            Even
192.168.54.98              Even
192.168.98.93              Odd
........ 
........

Answer 1

数据：

df = pd.DataFrame({"IP Address" : 
["192.168.1.1",
"192.168.1.2",
"192.168.152.200",
"192.168.54.98",
"192.168.98.93"]})

df:

        IP Address
0      192.168.1.1
1      192.168.1.2
2  192.168.152.200
3    192.168.54.98
4    192.168.98.93

df['New-variable'] = df['IP Address'].apply(lambda x:"Odd" if int(x.split(".")[-1]) % 2 else  "Even")

df:

        IP Address New-variable
0      192.168.1.1          Odd
1      192.168.1.2         Even
2  192.168.152.200         Even
3    192.168.54.98         Even
4    192.168.98.93          Odd

Answer 2

你可以试试这个

IPList = ["192.168.1.1",
    "192.168.1.1" ,       
    "192.168.1.2",         
    "192.168.152.200",   
    "192.168.54.98",      
    "192.168.98.93" ]      
final_list = []
for ip in IPList:
    _ip = ip.split(".")
    if int(_ip[-1]) %2==0:
        final_list.append([ip,'Even'])
        continue
    final_list.append([ip,'Odd'])
df = pd.DataFrame(final_list, columns =['IP Address', 'Type']) 
df

输出：

    IP Address      Type
0   192.168.1.1     Odd
1   192.168.1.1     Odd
2   192.168.1.2     Even
3   192.168.152.200 Even
4   192.168.54.98   Even
5   192.168.98.93   Odd

Answer 3

假设你的df叫x，你可以这样做：

import numpy as np
# First remove the '.' so that you can convert to float
x['IP_num'] = (x['IP'].apply(lambda x: ''.join([ch for ch in x if ch.isdigit()]))).astype(float)
# Then create a new column if the IP is odd / even
x['even_odd'] = np.where(x['IP_num'] % 2 == 0,'Even','Odd')

输出：

                IP             IP_num even_odd
0      192.168.1.1      19,216,811.00      Odd
1      192.168.1.1      19,216,811.00      Odd
2  192.168.152.200 192,168,152,200.00     Even
3    192.168.54.98   1,921,685,498.00     Even
4    192.168.98.93   1,921,689,893.00      Odd

如果需要，您可以删除 'IP_num' 列。

Answer 4

这只是显示输入：

import pandas as pd

adr_df = pd.DataFrame(['192.168.1.1', '192.168.1.2', '192.168.152.200', 
                        '192.168.54.98', '192.168.98.93'], columns=['IP Adress'])

一个例子，如果你真的喜欢使用 regex 来捕获最后一位数字，你可以使用以下命令（小心使用 [=13= 转义点字符]):

adr_df['Last Nr'] = adr_df['IP Adress'].str.extract(r'.*\..*\..*\.(.*)').astype(int)

当然，可以有更精确的正则表达式字符串来匹配 ip，但这个对我有用。

检查您可以使用一个小的 lambda 函数达到的奇数：

adr_df.loc['Status'] = adr_df['Last Nr'].apply(lambda x: 'Odd' if x%2 else 'Even')

Answer 5

此处不需要 for 或 apply 循环 - 在最后 . 之后提取值，转换为整数，使用 % 2 并最后传递给 numpy.where:

df['new'] = np.where(df['IP Address'].str.split('.').str[-1].astype(int) % 2,'Odd','Even')
    
print (df)
        IP Address   new
0      192.168.1.1   Odd
1      192.168.1.2  Even
2  192.168.152.200  Even
3    192.168.54.98  Even
4    192.168.98.93   Odd

Answer 6

df

df['Status'] = [int(str(x).strip()[-1]) for x in df['IP Address']]

df['Status'] = np.where(df['Status']%2, 'Odd', 'Even')

Answer 7

您可以 extract 最后一位数字，然后使用方法 divmod 和 map:

df['IP Adress'].str.extract('(\d+)$', expand=False).astype(int)\
.divmod(2)[1].map({1: 'odd', 0: 'even'})

在 Python 中用 Pandas 确定奇数和偶数

Detemining Odd and Even values with Pandas in Python

python

dataframe

python-2.7

pandas