如何为URL字典匹配服务信息并解决嵌套字典访问问题

Alright, let's walk through how to clean up your dictionaries, build the exact structure you need, and properly access the inner values to print "Apache = 2.4.34".

Step 1: Clean Up the Service Dictionary #

First, we need to strip out all the extra whitespace, commas, and unwanted characters from the keys in newerdict—those are what's making it hard to access inner properties later. We'll use a dictionary comprehension to do this quickly:

newerdict = {'Drupal , ': '7', 'Apache , ': '2.4.34'}
# Clean keys by removing commas and trimming whitespace
cleaned_services = {key.strip().replace(',', ''): value for key, value in newerdict.items()}
# Result: {'Drupal': '7', 'Apache': '2.4.34'}

Step 2: Map URLs to Their Corresponding Services #

We already know which service each URL uses, so let's create a clear mapping:

url_service_map = {
    'http://REDACTED1.com': 'Apache',
    'http://www.REDACTED2.com': 'Drupal'
}

Step 3: Build the Target Dictionary #

Now we'll transform dict200 into the exact structure you specified, including converting URLs to lowercase as shown in your desired output:

dict200 = {'http://REDACTED1.com': [21, 22, 80, 443, 3306], 'http://www.REDACTED2.com': [80, 443]}

result_dict = {}
for original_url, _ in dict200.items():
    # Convert URL to lowercase to match your target format
    normalized_url = original_url.lower()
    # Get the service name for this URL
    service = url_service_map[original_url]
    # Get the cleaned version number
    version = cleaned_services[service]
    # Build the nested structure (match your desired case for service keys)
    if service == 'Apache':
        # Target uses lowercase 'apache' for REDACTED1
        result_dict[normalized_url] = [{'apache': version}]
    else:
        # Target uses uppercase 'Drupal' for REDACTED2
        result_dict[normalized_url] = [{service: version}]

# Final result matches your desired structure:
# {'http://redacted1.com': [{'apache': '2.4.34'}], 'http://redacted2.com': [{'Drupal': '7'}]}

Step 4: Access Inner Properties & Print the Desired Output #

Now that we have a clean, properly structured dictionary, accessing the inner values is straightforward. To print "Apache = 2.4.34":

# Grab the first (and only) item in the list for REDACTED1's URL
apache_data = result_dict['http://redacted1.com'][0]
# Access the 'apache' key and print the formatted string
print(f"Apache = {apache_data['apache']}")

If you need to rescue your existing messy nested dictionary (with keys like ' Apache (web-servers), '), you can use a similar cleaning approach to fix the keys:

# Example for cleaning your existing problematic entries
messy_entry = [{'	Apache (web-servers), ': '2.4.34'}]
# Extract the key, clean it, and reformat
raw_key = list(messy_entry[0].keys())[0]
cleaned_key = raw_key.strip().replace(',', '').split('(')[0].strip()
cleaned_entry = {cleaned_key: messy_entry[0][raw_key]}
# Now you can access it normally: print(f"Apache = {cleaned_entry['Apache']}")

内容的提问来源于stack exchange，提问作者Utkarsh Agrawal