并行提问
并行提问
在 GDPR 维基百科文章上做一次 13 个问题的合规简报，证明把所有问题合并进一次 TypeSafe 调用要便宜 12.2 倍、快 10.0 倍，而答案不变。
你有一份文档，还有关于它的 N 个问题。你可以发一个包含全部 N 个问题的请求，也可以发 N 个各含一个问题的请求。用 TypeSafe，两种方式得到的答案一样：每个问题都是独立针对文档打分的，所以它的答案不取决于请求里还有什么别的东西。
为了验证这一点，这份 cookbook 把每个问题用两种方式各问好几次 —— 全部 N 个放一个请求，以及每次请求只放一个问题 —— 然后比较逐次运行的标准差：一个答案从一次重复到下一次会波动多远。一个问题有什么噪声，在两种分批策略下都有，分批本身不增添噪声。多数答案在两种方式下、全部 5 次重复里都完全一致，每次调用的值都一样，标准差恰好为 0.0。
成本和速度确实会变。文档在每次请求里都占大头。N 次单问题调用要为它付 N 次费，走 N 次往返；合批的调用只付一次。文档越大，这个节省就越接近整整 N 倍。
这里的案例是一份合规简报。文档是维基百科上的 GDPR 条目（约 54,000 字符，属于文档主导型负载 —— 文档占每次请求的大部分），一个合规团队想核查 13 件事：8 个 Noul 问题、2 个 Choice 问题，以及 3 个 Score 问题。
准备
pip install ipython 'cooksafe>=0.2.0,<0.3.0'然后设置 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"))文档：维基百科上的 GDPR 条目
以纯文本形式从该条目的一个固定修订版抓取，并和 API 调用一起缓存进 json_cache.json，所以即便线上条目被编辑，文档和它的数字也保持不变。
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问题：8 个 noul + 2 个 choice + 3 个 score
按类型，每个答案追踪一个数字：
Noul：答案为「是」的概率。Choice：最大概率，即落在所选中标签上的概率。criteria把每个标签映射到它的含义。Score：归一化到 0-1 的分数，即分数除以最高档位。criteria从档位 0 往上列出各档描述。
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",
}两种方式各问 5 次
ask() 把问题的任意子集连同文档一起发送，并把每个答案归约成它那一个被追踪的数字。每次调用里文档都逐字节相同。
两种分批策略各运行 RUNS = 5 次，于是每个问题在每种策略下都有 5 个答案，足够比较均值（两者一致吗？）和标准差（分批会引入噪声吗？）。调用结果被缓存到 json_cache.json，它随 cookbook 一起发布，所以重新渲染不花钱；把它删掉就能重跑实时调用。
@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分批不改变答案
逐个问题：在每种分批策略下，它那个被追踪的数字在 5 次运行中的均值和标准差。如果分批改变了答案，合批那几列就会和单问那几列不同。均值偏移是偏差，标准差变大是噪声。
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按问题类型来读这张表：
- choice、score，以及 8 个 noul 里的 6 个，在全部 5 次重复中都完全一致：两种分批策略下标准差都恰好为 0.0，每次合批调用和单问调用返回的数字都一样。一次包含 N 个问题的调用，和 N 次各含一个问题的调用，给出相同答案。
breach_72h和criminal_penalties带一点逐次运行的采样噪声，而且在两种分批策略下大小相同，均值也在这个噪声范围内一致。这个噪声是问题本身的属性，而不是你如何分批的属性：分批既不偏移答案，也不增加方差。
无论哪种方式，都不存在分批效应：没有任何问题的答案取决于和它共享请求的那另外 12 个问题。
唯一的区别：成本和速度
答案相同，账单不同。约 54,000 字符的条目在每次请求里都占大头，所以：
- 成本：13 次单问题调用会把条目重发 13 次；合批调用只发一次。这个节省不管你怎样发起调用都成立。
- 速度：该数字是把 13 次单问调用的延迟加总，所以它假设这些调用一个接一个地跑。并发发起会缩小差距，但 13 倍的 token 成本依然在。
token 数和延迟和答案一起被缓存；成本是在之后套上去的，两者都对 5 次运行取平均。
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 里打开
同一篇文章、同样的 13 个问题，打包成一个分享链接。打开它就能实时重跑这份简报；返回同样的数字。
playground_link = make_playground_link(
{"article": DOCUMENT}, QUESTIONS, models=[TYPESAFE_MODEL]
)
display(
Markdown(
f"🔗 [Open this article + questions in the TypeSafe playground]({playground_link})"
)
)