你有一份文档和关于它的 N 个问题。你可以发送一个包含全部 N 个问题的请求，也可以发送 N 个每次只含一个问题的请求。用 TypeSafe 时，两种方式得到的答案完全相同：每个问题都独立地对照文档评分，因此它的答案不依赖于请求中还有什么其他内容。
为验证这一点，本实战指南以两种方式——一个请求包含全部 N 个问题，以及每个请求一个问题——把每个问题各问若干次，并比较运行之间的标准差：一个答案在相邻两次重复之间会移动多远。一个问题无论带有什么噪声，在两种批处理策略下都同样存在；批处理不会增加噪声。无论哪种方式，大多数答案在全部 5 次重复中都完全一致——每次调用都是同一个值，标准差恰为 0.0。
成本和速度确实会变。文档占据了每个请求的绝大部分。N 次单问题调用要为它付 N 次费用、经历 N 个往返；批量调用只付一次。文档越大，这项节省就越接近整整 N 倍。
这里的案例是一份监管简报。文档是关于 GDPR 的 Wikipedia 条目（~54,000 字符，属于文档占主导的工作负载——文档占每个请求的绝大部分），合规团队要检查 13 项内容：8 个 Noul 问题、2 个 Choice 问题和 3 个 Score 问题。
环境准备
bash
pip install ipython "typesafe-sdk>=0.5.7" cooksafe --extra-index-url https://pypi.typesafe.ai/
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：关于 GDPR 的 Wikipedia 条目
从该条目的一个固定修订版本以纯文本形式抓取，并随 API 调用一起缓存在 json_cache.json 中，这样即使线上条目被编辑，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
按类型，为每个答案跟踪一个数字：
Noul：“是”的概率。Choice：最大概率，即选中标签上的概率。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。criteria列出各档位的描述，从 0 档向上。
python
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() 连同文档发送问题的任意子集，并把每个答案归约为它被跟踪的那一个数字。每次调用中的文档在字节层面完全相同。
两种批处理策略各运行 RUNS = 5 次，每种策略下每个问题得到 5 个答案，足以比较均值（两种方式一致吗？）和标准差（批处理会增加噪声吗？）。调用被缓存到 json_cache.json，该文件随本实战指南一起发布，因此重新渲染不花费任何费用；删除它即可重新实跑。
python
@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
批处理不会改变答案
每个问题对应：它的跟踪数字在两种批处理策略下各 5 次运行的均值和标准差。如果批处理改变了答案，批量列就会与单次列不同。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
按问题类型阅读该表：
- Choice、Score 以及八个 Noul 中的六个在 5 次重复中返回完全相同的结果：两种批处理策略下标准差都恰为 0.0，每次批量调用和单次调用都返回同一个数。一次包含 N 个问题的调用给出的答案与 N 次每次一个问题的调用相同。
breach_72h和criminal_penalties带有少量运行之间的采样噪声，两种批处理策略下噪声大小相同，均值的一致性也在该噪声范围之内。噪声是问题自身的属性，而不是批处理方式的属性：批处理既不移动答案，也不增加方差。
无论哪种方式，都不存在批处理效应：没有任何问题的答案依赖于与其共享请求的另外 12 个问题。
唯一的差别：成本和速度
答案相同，账单不同。这篇 ~54,000 字符的文章占据了每个请求的绝大部分，因此：
- 成本：13 次单问题调用要重发 13 次文章；批量调用只发送一次。无论你怎样发起这些调用，这项节省都成立。
- 速度：该数字是 13 次单次调用延迟的总和，因此假设它们一个接一个地运行。并发发起这些调用会缩小差距，但 13 倍的 token 成本保持不变。
token 数和延迟随答案一起缓存；成本在其后套用，二者都是在 5 次运行上取的平均值。
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
同一篇文章和同样的 13 个问题被打包进一个分享链接。打开它即可实时重跑这份简报；返回的数字相同。
python
在 TypeSafe Playground 中打开此文章 + 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})"
)
)