纯Python实现字典列表按用户分组的数值求和与报告计数

Hey, I've got a clean pure-Python solution for your problem—grouping those report dictionaries by user, summing the value and commission fields, and counting the total reports per user. Here's how to do it:

Step-by-Step Implementation #

# Your original list of report dictionaries
reports = [
    {"title": "Report 1", "value": 1000, "commission": 2000, "user": "user_1"},
    {"title": "Report 2", "value": 500, "commission": 300, "user": "user_1"},
    {"title": "Report 3", "value": 1500, "commission": 300, "user": "user_2"}
]

# Create a temporary dictionary to track aggregated data per user
grouped = {}

for report in reports:
    user_id = report["user"]
    
    # Initialize the user's entry if they're not already in the tracker
    if user_id not in grouped:
        grouped[user_id] = {
            "user": user_id,
            "value": 0,
            "commission": 0,
            "total": 0
        }
    
    # Update sums and report count
    grouped[user_id]["value"] += report["value"]
    grouped[user_id]["commission"] += report["commission"]
    grouped[user_id]["total"] += 1

# Convert the tracker's values into the final list format
result = list(grouped.values())
print(result)

What This Does #

• Initialization: We use a dictionary grouped to keep each user's aggregated data organized by their user ID. This makes lookups fast and avoids duplicates.
• Iterate & Aggregate: For each report, we check if the user exists in our tracker. If not, we set up their entry with starting values. Then we add the report's numeric values to their totals and increment the report count.
• Final Format: Converting the dictionary's values to a list gives you exactly the output structure you're looking for.

Output #

Running this code will produce:

[{"value": 1500, "commission": 2300, "user": "user_1", "total": 2}, {"value": 1500, "commission": 300, "user": "user_2", "total": 1}]

内容的提问来源于stack exchange，提问作者Douglas Laiber