並行提問
在 GDPR 維基百科文章上做一次 13 個問題的合規簡報，證明把所有問題合併進一次 TypeSafe 呼叫要便宜 12.2 倍、快 10.0 倍，而答案不變。
你有一份文件，還有關於它的 N 個問題。你可以發一個包含全部 N 個問題的請求，也可以發 N 個各含一個問題的請求。用 TypeSafe，兩種方式得到的答案一樣：每個問題都是獨立針對文件打分的，所以它的答案不取決於請求裡還有什麼別的東西。
為了驗證這一點，這份 cookbook 把每個問題用兩種方式各問好幾次 —— 全部 N 個放一個請求，以及每次請求只放一個問題 —— 然後比較逐次執行的標準差：一個答案從一次重複到下一次會波動多遠。一個問題有什麼噪聲，在兩種分批策略下都有，分批本身不增添噪聲。多數答案在兩種方式下、全部 5 次重複裡都完全一致，每次呼叫的值都一樣，標準差恰好為 0.0。
成本和速度確實會變。文件在每次請求裡都佔大頭。N 次單問題呼叫要為它付 N 次費，走 N 次往返；合批的呼叫只付一次。文件越大，這個節省就越接近整整 N 倍。
這裡的案例是一份合規簡報。文件是維基百科上的 GDPR 條目（約 54,000 字元，屬於文件主導型負載 —— 文件佔每次請求的大部分），一個合規團隊想核查 13 件事：8 個 Noul 問題、2 個 Choice 問題，以及 3 個 Score 問題。
準備
pip install ipython 'cooksafe>=0.2.0,<0.3.0'然後設定 TYPESAFE_API_KEY。
import json
import os
import urllib.request
from pathlib import Path
from statistics import mean, stdev
from time import perf_counter
from cooksafe import JsonCache, make_playground_link
from IPython.display import Markdown, display
from typesafe_sdk import Choice, ChoiceAnswer, Noul, NoulAnswer, Score, TypeSafeClient
TYPESAFE_MODEL = "jev-1.12"
PRICE = (
0.042,
0.00,
) # $ per 1M tokens (input, output); TypeSafe jev-1.12 as of 2026-09, see README
RUNS = 5 # repeats per batching strategy, to estimate each answer's run-to-run std dev
client = TypeSafeClient(api_key=os.environ["TYPESAFE_API_KEY"], timeout=120.0)
json_cache = JsonCache(Path("json_cache.json"))文件：維基百科上的 GDPR 條目
以純文本形式從該條目的一個固定修訂版抓取，並和 API 呼叫一起快取進 json_cache.json，所以即便線上條目被編輯，文件和它的數字也保持不變。
WIKIPEDIA_REVISION = 1363040264 # "General Data Protection Regulation", as of 2026-07
@json_cache
def fetch_article(revision_id: int) -> str:
url = (
"https://en.wikipedia.org/w/api.php?action=query&format=json"
f"&prop=extracts&explaintext=1&revids={revision_id}"
)
request = urllib.request.Request(
url, headers={"User-Agent": "typesafe-cookbook/1.0"}
)
with urllib.request.urlopen(request) as response:
pages = json.loads(response.read())["query"]["pages"]
return next(iter(pages.values()))["extract"]
DOCUMENT = {
"source": f"https://en.wikipedia.org/?oldid={WIKIPEDIA_REVISION}",
"text": fetch_article(WIKIPEDIA_REVISION),
}
print(f"{len(DOCUMENT['text']):,} characters")
display(Markdown(f"📄 [Read the pinned Wikipedia revision]({DOCUMENT['source']})"))53,777 characters問題：8 個 noul + 2 個 choice + 3 個 score
按型別，每個答案追蹤一個數字：
Noul：答案為「是」的機率。Choice：最大機率，即落在所選中標籤上的機率。criteria把每個標籤對映到它的含義。Score：歸一化到 0-1 的分數，即分數除以最高檔位。criteria從檔位 0 往上列出各檔描述。
QUESTIONS = {
"breach_72h": Noul(
instructions="Must a personal data breach be reported to the supervisory authority within 72 hours?"
),
"applies_non_eu": Noul(
instructions="Does the regulation apply to organisations established outside the EU that offer goods or services to people in the EU?"
),
"dpo_all_orgs": Noul(
instructions="Must every organisation appoint a Data Protection Officer, regardless of what data it processes?"
),
"pre_ticked_consent": Noul(
instructions="Can valid consent be obtained through pre-ticked boxes or inactivity?"
),
"right_erasure": Noul(
instructions="Does the regulation grant individuals a right to erasure of their personal data?"
),
"data_portability": Noul(
instructions="Does the regulation include a right to data portability?"
),
"us_federal_law": Noul(instructions="Is the GDPR a United States federal law?"),
"criminal_penalties": Noul(
instructions="Does the GDPR itself impose criminal penalties such as imprisonment?"
),
"instrument_type": Choice(
instructions="What kind of EU legal instrument is the GDPR?",
criteria={
"Regulation": "Directly binding law in all member states, no national implementation needed.",
"Directive": "Sets goals that member states implement through national law.",
"Treaty": "An international treaty between states.",
"Recommendation": "Non-binding guidance.",
},
),
"max_fine": Choice(
instructions="What is the maximum administrative fine for the most serious infringements?",
criteria={
"TwentyM_or_4pct": "Up to EUR 20 million or 4% of annual worldwide turnover, whichever is greater.",
"TenM_or_2pct": "Up to EUR 10 million or 2% of annual worldwide turnover, whichever is greater.",
"FixedCap": "A fixed amount not tied to turnover.",
"NoFines": "The GDPR provides no administrative fines.",
},
),
"individual_rights": Score(
instructions="How strong are the rights the GDPR grants to individuals over their data?",
criteria=[
"None: individuals get no rights over their data.",
"Weak: a right to be informed, but little control.",
"Moderate: access and correction rights, but limited means to act on them.",
"Strong: access, erasure, portability, and objection rights, with enforcement behind them.",
],
),
"penalty_severity": Score(
instructions="How severe are the penalties the GDPR provides for non-compliance?",
criteria=[
"None: no penalties of any kind.",
"Symbolic: small fixed fines unlikely to change behavior.",
"Substantial: fines large enough to matter to most companies.",
"Severe: fines scaled to global revenue, material even to the largest companies.",
],
),
"compliance_burden": Score(
instructions="How heavy is the compliance burden the GDPR places on organisations?",
criteria=[
"Negligible: no meaningful obligations.",
"Light: a few notices and disclosures.",
"Moderate: documented processes and some dedicated roles for larger processors.",
"Heavy: records, impact assessments, officers, and breach procedures for many organisations.",
"Extreme: obligations so demanding that ordinary organisations cannot fully comply.",
],
),
}
N = len(QUESTIONS)
METRIC = { # question type -> the one number we track per answer
Noul: "p(yes)",
Choice: "max prob",
Score: "normalized score",
}兩種方式各問 5 次
ask() 把問題的任意子集連同文件一起傳送，並把每個答案歸約成它那一個被追蹤的數字。每次呼叫裡文件都逐位元組相同。
兩種分批策略各執行 RUNS = 5 次，於是每個問題在每種策略下都有 5 個答案，足夠比較均值（兩者一致嗎？）和標準差（分批會引入噪聲嗎？）。呼叫結果被快取到 json_cache.json，它隨 cookbook 一起釋出，所以重新渲染不花錢；把它刪掉就能重跑即時呼叫。
@json_cache
def ask(keys: tuple[str, ...], run: int):
"""One TypeSafe call -> ({key: tracked metric}, input_tokens, output_tokens, latency_s);
``run`` only forces a distinct live call per repeat."""
started = perf_counter()
response = client.system_one(
state={"article": DOCUMENT},
questions={key: QUESTIONS[key] for key in keys},
model=TYPESAFE_MODEL,
)
values = {}
for key in keys:
answer = response.answers[key]
if isinstance(answer, NoulAnswer):
values[key] = answer.noul
elif isinstance(answer, ChoiceAnswer):
values[key] = max(answer.probabilities.values())
else:
values[key] = answer.score / (len(QUESTIONS[key].criteria) - 1)
return (
values,
response.usage.input_tokens,
response.usage.output_tokens,
perf_counter() - started,
)
def priced(result):
"""({key: metric}, in_tokens, out_tokens, latency) -> ({key: metric}, cost_usd, latency)."""
values, input_tokens, output_tokens, latency = result
return values, input_tokens / 1e6 * PRICE[0] + output_tokens / 1e6 * PRICE[1], latency
# Price after cache retrieval, so a price change needs no new calls.
batched = [
priced(ask(tuple(QUESTIONS), run)) for run in range(RUNS)
] # all N in one call, x RUNS
singles = [
{key: priced(ask((key,), run)) for key in QUESTIONS} for run in range(RUNS)
] # N x 1, x RUNS分批不改變答案
逐個問題：在每種分批策略下，它那個被追蹤的數字在 5 次執行中的均值和標準差。如果分批改變了答案，合批那幾列就會和單問那幾列不同。均值偏移是偏差，標準差變大是噪聲。
print(
f"{'question':<22}{'metric':<18}{'batched mean':>13}{'single mean':>12}"
f"{'batched std':>13}{'single std':>12}"
)
for key, question in QUESTIONS.items():
batched_values = [values[key] for values, _cost, _latency in batched]
single_values = [singles[run][key][0][key] for run in range(RUNS)]
print(
f"{key:<22}{METRIC[type(question)]:<18}{mean(batched_values):>13.3f}"
f"{mean(single_values):>12.3f}{stdev(batched_values):>13.4f}{stdev(single_values):>12.4f}"
)question metric batched mean single mean batched std single std
breach_72h p(yes) 0.804 0.814 0.0055 0.0055
applies_non_eu p(yes) 0.990 0.990 0.0000 0.0000
dpo_all_orgs p(yes) 0.030 0.030 0.0000 0.0000
pre_ticked_consent p(yes) 0.040 0.040 0.0000 0.0000
right_erasure p(yes) 0.990 0.990 0.0000 0.0000
data_portability p(yes) 0.990 0.990 0.0000 0.0000
us_federal_law p(yes) 0.010 0.010 0.0000 0.0000
criminal_penalties p(yes) 0.108 0.108 0.0045 0.0084
instrument_type max prob 1.000 1.000 0.0000 0.0000
max_fine max prob 1.000 1.000 0.0000 0.0000
individual_rights normalized score 1.000 1.000 0.0000 0.0000
penalty_severity normalized score 1.000 1.000 0.0000 0.0000
compliance_burden normalized score 0.750 0.750 0.0000 0.0000按問題型別來讀這張表：
- choice、score，以及 8 個 noul 裡的 6 個，在全部 5 次重複中都完全一致：兩種分批策略下標準差都恰好為 0.0，每次合批呼叫和單問呼叫返回的數字都一樣。一次包含 N 個問題的呼叫，和 N 次各含一個問題的呼叫，給出相同答案。
breach_72h和criminal_penalties帶一點逐次執行的取樣噪聲，而且在兩種分批策略下大小相同，均值也在這個噪聲範圍內一致。這個噪聲是問題本身的屬性，而不是你如何分批的屬性：分批既不偏移答案，也不增加方差。
無論哪種方式，都不存在分批效應：沒有任何問題的答案取決於和它共享請求的那另外 12 個問題。
唯一的區別：成本和速度
答案相同，賬單不同。約 54,000 字元的條目在每次請求裡都佔大頭，所以：
- 成本：13 次單問題呼叫會把條目重發 13 次；合批呼叫只發一次。這個節省不管你怎樣發起呼叫都成立。
- 速度：該數字是把 13 次單問呼叫的延遲加總，所以它假設這些呼叫一個接一個地跑。併發發起會縮小差距，但 13 倍的 token 成本依然在。
token 數和延遲和答案一起被快取；成本是在之後套上去的，兩者都對 5 次執行取平均。
batched_cost = mean(cost for _values, cost, _latency in batched)
batched_latency = mean(latency for _values, _cost, latency in batched)
singles_cost = mean(
sum(singles[run][key][1] for key in QUESTIONS) for run in range(RUNS)
)
singles_latency = mean(
sum(singles[run][key][2] for key in QUESTIONS) for run in range(RUNS)
)
print(f"{'batching':<24}{'calls':>6}{'cost':>12}{'total time':>12}")
print(
f"{f'one call, all {N}':<24}{1:>6}{'$' + format(batched_cost, '.6f'):>12}{format(batched_latency, '.2f') + 's':>12}"
)
print(
f"{f'{N} calls, one each':<24}{N:>6}{'$' + format(singles_cost, '.6f'):>12}{format(singles_latency, '.2f') + 's':>12}"
)
print(
f"\nbatching: {singles_cost / batched_cost:.1f}x cheaper, {singles_latency / batched_latency:.1f}x faster"
)batching calls cost total time
one call, all 13 1 $0.000497 0.27s
13 calls, one each 13 $0.006090 2.71s
batching: 12.2x cheaper, 10.0x faster在 TypeSafe playground 裡開啟
同一篇文章、同樣的 13 個問題，打包成一個分享連結。開啟它就能即時重跑這份簡報；返回同樣的數字。
playground_link = make_playground_link(
{"article": DOCUMENT}, QUESTIONS, models=[TYPESAFE_MODEL]
)
display(
Markdown(
f"🔗 [Open this article + questions in the TypeSafe playground]({playground_link})"
)
)