并行问题
对 GDPR 维基百科条目运行一项包含 13 个问题的监管简报，展示把所有问题合并为一次 TypeSafe 调用可便宜 12.2 倍、快 10.0 倍，而答案完全不变。
你有一份文档和关于它的 N 个问题。你可以在一个请求里发送全部 N 个问题，也可以发 N 个请求、每个请求一个问题。使用 TypeSafe 时，两种方式的答案完全相同：每个问题都是独立地针对文档打分的，因此其答案不取决于请求里还包含什么其他内容。
为了验证这一点，实战指南用两种方式各把每个问题问上多次——N 个问题放进一个请求，以及每个请求一个问题——并比较多次运行之间的标准差：即答案从一次重复到下一次重复的变动幅度。一个问题带有什么样的噪声，在两种批量策略下就带有什么样的噪声。批量不增加任何噪声。无论哪种方式，大多数答案在全部 5 次重复中都完全一致，每次调用都返回同一个值，标准差恰好为 0.0。
成本和速度确实会变。文档在每个请求中都占大头。N 次单问题调用要为它付 N 次钱、跑 N 个来回；批量调用只付一次。文档越大，这笔节省就越接近完整的 N 倍。
这里的案例是一份监管简报。文档是维基百科上关于 GDPR 的条目（~54,000 字符，属于文档主导的工作负载，即文档占每个请求的大部分），合规团队要核查 13 件事：8 个 Noul 问题、2 个 Choice 问题，以及 3 个 Score 问题。
准备
bash
pip install ipython 'cooksafe>=0.2.0,<0.3.0'然后设置 TYPESAFE_API_KEY。
python
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：维基百科上关于 GDPR 的条目
以纯文本形式从该条目的固定修订版本抓取，并缓存在 API 调用旁边的 json_cache.json 中，因此即使线上条目继续被编辑，文档及其数字也保持不变。
python
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
每个答案按类型各跟踪一个数字：
Noul：“是”的概率。Choice：最大概率，即落在被选中标签上的概率。criteria将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。criteria从第 0 档起列出各档描述。
expandable
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",
}用两种方式各问 5 次
ask() 把问题的任意子集连同文档一起发送，并将每个答案归约为它那唯一被跟踪的数字。文档在每次调用中都逐字节相同。
两种批量策略各运行 RUNS = 5 次，即每个问题在每种策略下得到 5 个答案，足以比较均值（两者是否一致？）和标准差（批量是否引入噪声？）。调用会缓存到 json_cache.json，它随实战指南一起附带，因此重新渲染是免费的；删除该文件即可实时重跑。
expandable
@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批量不会改变答案
针对每个问题：在每种批量策略下，其被跟踪数字在 5 次运行中的均值和标准差。如果批量改变了答案，批量列就会与单问题列不同。均值偏移是偏差；标准差变大是噪声。
python
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按问题类型解读这张表：
Choice、Score 以及八个 Noul 中的六个在 5 次重复中返回的结果完全一致：在两种批量策略下标准差都恰好为 0.0，每一次批量调用和单问题调用都返回同一个数字。一次调用带 N 个问题，与 N 次调用各带一个问题的答案相同。
breach_72h和criminal_penalties带有一点运行间的采样噪声，而且它在两种批量策略下大小相同，均值也在该噪声范围内彼此一致。这种噪声是问题本身的属性，与你如何批量无关：批量既不移动答案，也不增加方差。
无论哪种方式，都不存在批量效应：没有任何一个问题的答案取决于与它共享同一请求的其他 12 个问题。
唯一的区别：成本与速度
答案相同，账单不同。这篇 ~54,000 字符的文章在每个请求中都占大头，因此：
成本：13 次单问题调用会把这篇文章重发 13 次；批量调用只发送一次。无论你以何种方式发起这些调用，这一节省都成立。
速度：图中数字是把 13 次单次调用的延迟加总而来的，因此假定它们一个接一个地运行。并发发起它们可以缩小差距，但 13 倍的 token 成本依然存在。
token 计数与延迟随答案一同缓存；成本是在其后计算的，两者都取 5 次运行的平均值。
python
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 中打开
同样的文章和同样的 13 个问题，打包进一个分享链接。打开它即可实时重跑这份简报；返回的将是同样的数字。
python
在 TypeSafe Playground 中打开这篇文章 + 问题 →
playground_link = make_playground_link(
{"article": DOCUMENT}, QUESTIONS, models=[TYPESAFE_MODEL]
)
display(
Markdown(
f"🔗 [Open this article + questions in the TypeSafe playground]({playground_link})"
)
)