如何将DF列中的列表转换为逗号分隔的值？

Question

现在，我有一个输出到CSV文件的DataFrame。如果我要打印“受影响的IP地址”列，它将如下所示：

['10.0.7.248']
['10.0.7.248', '10.0.8.56']
['10.0.6.72']
['10.0.6.72', '10.0.5.46']
['10.0.9.126']
['10.0.9.126', '10.0.7.248']
['10.0.9.126', '10.0.7.248', '10.0.8.56']
['10.0.6.72']
['10.0.6.72', '10.0.5.46']
['10.0.9.126']
['10.0.9.126', '10.0.7.248']
['10.0.9.126', '10.0.7.248', '10.0.8.56']

该列中的每个值都是一个列表，有些只是单个IP，有些是多个。有没有办法从输出中省略括号和撇号？如果可能的话，我更希望它的输出是这样的：

10.0.7.248
10.0.7.248, 10.0.8.56
10.0.6.72
10.0.6.72, 10.0.5.46
10.0.9.126
10.0.9.126, 10.0.7.248
10.0.9.126, 10.0.7.248, 10.0.8.56

这个get是写到一个CSV文件中的，所以我不知道如何省略这些字符，所以它只是it，它们之间用逗号分隔。

这是我的剧本：

def main():
    csv_data = open_csv()
    get_scan_results(csv_data)

def open_csv():
    #Opens CSV file
    with open(f"{csv_filename}.csv", newline='') as f:
        reader = csv.reader(f)
        data = list(reader)    
    return data

def get_scan_results(data):
    # New dictionary to be created
    new_dict = {}
    for ip, host, os, vuln_title, vuln_id, cvss2, cvss3, descr, proof, solu, cves in data[1:]:
        # Converts CVSSv3 score into a 'Risk Exposure' metric, blank values return 'Null'
        if len(cvss3.strip()):
            converted_cvss3 = float(cvss3)
            if converted_cvss3 < 4.0:
                s = "Low"
            elif converted_cvss3 >= 4 and converted_cvss3 < 7:
                s = "Moderate"
            else:
                s = "High"
        elif len(cvss2.strip()):
            converted_cvss2 = float(cvss2)
            if converted_cvss2 < 4.0:
                s = "Low"
            elif converted_cvss2 >= 4 and converted_cvss2 < 7:
                s = "Moderate"
            else:
                s = "High"
        else:
            s = "Null"
        # Populates 'new_dict' with values, the keys will also be the column names in CSV/Excel
        vuln_data = new_dict.setdefault(vuln_id, {"Name": vuln_title, "Description": descr, "Source of Discovery": csv_filename, "Vulnerability ID": vuln_id, "Affected IP Address": [], "Solution": solu, "Risk Exposure": s })
        vuln_data["Affected IP Address"].append(ip)
        print (vuln_data["Affected IP Address"])


    # Creates DF object and exports to CSV
    new_list = new_dict.values()
    df = pd.DataFrame(new_list)
    df.to_csv(f"{exported_csv_filename}.csv", index=False)

if __name__ == "__main__":
    main()

Answer 1

以下是几种方法

如果您将数据作为代码发布，它将删除我在创建数据框架以回答问题时所做的假设。

# if its of type list, then iterate through and concat with ','
df['IP'].apply(lambda x:  ', '.join(x))

0                          10.0.7.248
1                10.0.7.248,10.0.8.56
2                           10.0.6.72
3                 10.0.6.72,10.0.5.46
4                          10.0.9.126
5               10.0.9.126,10.0.7.248
6     10.0.9.126,10.0.7.248,10.0.8.56
7                           10.0.6.72
8                 10.0.6.72,10.0.5.46
9                          10.0.9.126
10              10.0.9.126,10.0.7.248
11    10.0.9.126,10.0.7.248,10.0.8.56

或

# if its of type string, replace brackets and apostrophes
df['IP'].replace(r"'|\]|\[","", regex=True) 

0                            10.0.7.248
1                 10.0.7.248, 10.0.8.56
2                             10.0.6.72
3                  10.0.6.72, 10.0.5.46
4                            10.0.9.126
5                10.0.9.126, 10.0.7.248
6     10.0.9.126, 10.0.7.248, 10.0.8.56
7                             10.0.6.72
8                  10.0.6.72, 10.0.5.46
9                            10.0.9.126
10               10.0.9.126, 10.0.7.248
11    10.0.9.126, 10.0.7.248, 10.0.8.56