Python将3个列表转换为以list name为键名的字典列表

是这样还是你在寻找某种自动化来提取变量的别名?

source_ip_address = ["172.30.1.1","123.123.1.1","180.104.1.1"]
destination_ip_address = ["172.31.1.1","172.31.1.1","192.168.1.1"]
incident_name = ["Credential Sharing","Remote IP","OTP Failures"]


print([{'source_ip_address':x, 'destination_ip_address':y, 'incident_name':z} for x,y,z in zip(source_ip_address, destination_ip_address, incident_name)])

有很多不同的方法可以做到这一点,但这里是我要做的。

result = [{'source_ip_address': source,
           'destination_ip_address': dest,
           'incident_name': incident}
           for source, dest, incident in zip(source_ip_address,
                                             destination_ip_address,
                                             incident_name)]

('172.30.1.1','172.31.1.1','Credential Sharing') . for循环遍历所有元素并将它们分解为单独的位。所以在第一次迭代中, source='172.30.1.1' dest='172.31.1.1' ,和 incident='Credential Sharing' . 我认为很清楚字典是如何从一次迭代中构建出来的。for位于列表理解的内部,因此它遍历所有元素并生成包含每个字典的列表。

source_ip_address = ["172.30.1.1","123.123.1.1","180.104.1.1"]
destination_ip_address = ["172.31.1.1","172.31.1.1","192.168.1.1"]
incident_name = ["Credential Sharing","Remote IP","OTP Failures"]

ip_list = []
for a,b,c in zip(source_ip_address,destination_ip_address,incident_name):
    ip_list.append({'source_ip_address':a,
                          'destination_ip_address': b,
                          'incident_name': c})

print (ip_list)

其输出将是:

[{'source_ip_address': '172.30.1.1', 'destination_ip_address': '172.31.1.1', 'incident_name': 'Credential Sharing'}, {'source_ip_address': '123.123.1.1', 'destination_ip_address': '172.31.1.1', 'incident_name': 'Remote IP'}, {'source_ip_address': '180.104.1.1', 'destination_ip_address': '192.168.1.1', 'incident_name': 'OTP Failures'}]

如果您不熟悉zip并希望使用常规迭代,您也可以这样做。

n = len(source_ip_address)
ip_list = []

for i in range(n):
    ip_list.append({'source_ip_address':source_ip_address[i],
                          'destination_ip_address': destination_ip_address[i],
                          'incident_name': incident_name[i]})

print (ip_list)