Hey there! Great question—you’re right that dictionary comprehensions (you mentioned list comprehensions, but we’re building dictionaries here) are clean, and avoiding repeated iterations over the same list is a smart call for efficiency, especially with larger datasets.

Let’s break down your options:

Option 1: Clean Nested Dictionary Comprehensions (Multiple Iterations) #

If your roomStayInfos list is small and performance isn’t a critical concern, this is the most concise way to build your nested dictionary. It uses three separate dictionary comprehensions to populate each sub-dictionary:

import json

jsn = """{ "availResponse": { "roomStayInfos": [ { "rateCodeID": 400166, "amount": 9000, "depositAmount": 0, "baseAmt": 9000 }, { "rateCodeID": 402451, "amount": 96000, "depositAmount": 0, "baseAmt": 96000 }, { "rateCodeID": 400164, "amount": 9000, "depositAmount": 0, "baseAmt": 9000 }, { "rateCodeID": 402598, "amount": 5100, "depositAmount": 0, "baseAmt": 9000 } ] } }"""
availability = json.loads(jsn, strict=False)

room_stays = availability["availResponse"]["roomStayInfos"]

result = {
    "depositAmounts": {str(rs["rateCodeID"]): rs["depositAmount"] for rs in room_stays},
    "amount": {str(rs["rateCodeID"]): rs["amount"] for rs in room_stays},
    "baseAmt": {str(rs["rateCodeID"]): rs["baseAmt"] for rs in room_stays}
}

print(json.dumps(result, indent=2))

This will output exactly what you want, but it does iterate over the list three times.

Option 2: Single Iteration (Efficient, Readable) #

If you want to iterate only once (the main goal you mentioned), your original for-loop approach is already excellent. We can make it a bit cleaner using collections.defaultdict to avoid manually initializing empty sub-dictionaries:

import json
from collections import defaultdict

jsn = """{ "availResponse": { "roomStayInfos": [ { "rateCodeID": 400166, "amount": 9000, "depositAmount": 0, "baseAmt": 9000 }, { "rateCodeID": 402451, "amount": 96000, "depositAmount": 0, "baseAmt": 96000 }, { "rateCodeID": 400164, "amount": 9000, "depositAmount": 0, "baseAmt": 9000 }, { "rateCodeID": 402598, "amount": 5100, "depositAmount": 0, "baseAmt": 9000 } ] } }"""
availability = json.loads(jsn, strict=False)

result = defaultdict(dict)
for rs in availability["availResponse"]["roomStayInfos"]:
    rc_id = str(rs["rateCodeID"])
    result["depositAmounts"][rc_id] = rs["depositAmount"]
    result["amount"][rc_id] = rs["amount"]
    result["baseAmt"][rc_id] = rs["baseAmt"]

# Convert defaultdict to regular dict if needed
result = dict(result)

print(json.dumps(result, indent=2))

This iterates over the list once, updates all three sub-dictionaries in one go, and keeps the code readable.

Can We Do This with a Single Comprehension? #

Short answer: Not really in a way that’s more readable or efficient than the loop. Dictionary comprehensions are designed to generate a single dictionary at a time. To build multiple sub-dictionaries in one comprehension, you’d have to use a workaround (like generating an intermediate structure and then splitting it), which ends up being less clear and may still involve multiple iterations.

For example, here’s a workaround that uses a single comprehension to build an intermediate map, then splits it into your target structure—but it still iterates twice (once for the intermediate map, once for each sub-dictionary):

room_stays = availability["availResponse"]["roomStayInfos"]
intermediate = {str(rs["rateCodeID"]): {
    "deposit": rs["depositAmount"],
    "amount": rs["amount"],
    "base": rs["baseAmt"]
} for rs in room_stays}

result = {
    "depositAmounts": {k: v["deposit"] for k, v in intermediate.items()},
    "amount": {k: v["amount"] for k, v in intermediate.items()},
    "baseAmt": {k: v["base"] for k, v in intermediate.items()}
}

This works, but it’s not better than your original loop for clarity or efficiency.

Final Recommendation #

Stick with your original loop (or the defaultdict version) for single-iteration efficiency and readability. The nested comprehension approach is great for small lists where you prioritize brevity over minimal iterations.

Expected Output #

{
  "depositAmounts": {
    "400166": 0,
    "402451": 0,
    "400164": 0,
    "402598": 0
  },
  "amount": {
    "400166": 9000,
    "402451": 96000,
    "400164": 9000,
    "402598": 5100
  },
  "baseAmt": {
    "400166": 9000,
    "402451": 96000,
    "400164": 9000,
    "402598": 9000
  }
}

内容的提问来源于stack exchange，提问作者A l w a y s S u n n y