Grok 4.7 vs Grok 4.6: Which Should You Use?

TL;DR: Pick Grok 4.7 for complex coding and agentic work; pick Grok 4.6 for lightweight tasks where you want to spend fewer tokens. The two generations cost exactly the same on HiAPI ($0.84 input / $2.52 output per million tokens) with the same 500K context, so price and window size won't decide this for you. Grok 4.7 is live on HiAPI — open the model page to try it, and new accounts get 200 free credits.

Grok 4.7 vs Grok 4.6 at a glance

Price, context, and reasoning levels are identical. The real dividing line is coding ability and output habits. Here is what two same-prompt runs actually look like.

Same prompt, test 1: an LRU cache in Python

We ran both models on HiAPI's production API on 2026-10-29, identical parameters (default high reasoning level, max_tokens capped), one run each:

Prompt:

Write a Python LRU cache class with O(1) get and put, TTL-based expiry, and thread safety. Code only, plus usage notes of no more than 5 lines.

Grok 4.7's answer (39.7s, 3285 output tokens, ~$0.011):

class LRUCache:
    def __init__(self, capacity: int, default_ttl: Optional[float] = None):
        if capacity <= 0:
            raise ValueError("capacity must be positive")
        self.capacity = capacity
        self.default_ttl = default_ttl
        self._data: OrderedDict[Hashable, tuple[Any, Optional[float]]] = OrderedDict()
        self._lock = threading.Lock()

    def get(self, key: Hashable, default: Any = None) -> Any:
        with self._lock:
            item = self._data.get(key)
            if item is None:
                return default
            value, expire_at = item
            if expire_at is not None and time.monotonic() >= expire_at:
                del self._data[key]
                return default
            self._data.move_to_end(key)
            return value

    def put(self, key: Hashable, value: Any, ttl: Optional[float] = None) -> None:
        with self._lock:
            ttl = self.default_ttl if ttl is None else ttl
            if ttl is not None and ttl <= 0:
                self._data.pop(key, None)
                return
            expire_at = None if ttl is None else time.monotonic() + ttl
            if key in self._data:
                self._data.move_to_end(key)
            elif len(self._data) >= self.capacity:
                self._data.popitem(last=False)
            self._data[key] = (value, expire_at)

(Full output includes imports and a usage example; 4.7 tracks TTL with time.monotonic() and lets get take a default value.)

Grok 4.6's answer (61.9s, 3915 output tokens, ~$0.012):

class LRUCache:
    def __init__(self, capacity, default_ttl=None):
        if not isinstance(capacity, int) or capacity < 1:
            raise ValueError("capacity must be positive int")
        self.capacity = capacity
        self.default_ttl = default_ttl
        self._cache = OrderedDict()
        self._lock = threading.Lock()

    def get(self, key):
        with self._lock:
            if key not in self._cache:
                return None
            value, expire_at = self._cache[key]
            if expire_at is not None and time.time() > expire_at:
                del self._cache[key]
                return None
            self._cache.move_to_end(key)
            return value

    def put(self, key, value, ttl=None):
        with self._lock:
            now = time.time()
            expire_at = None
            effective_ttl = ttl if ttl is not None else self.default_ttl
            if effective_ttl is not None:
                expire_at = now + effective_ttl
                if expire_at <= now:
                    if key in self._cache:
                        del self._cache[key]
                    return
            if key in self._cache:
                self._cache.move_to_end(key)
            elif len(self._cache) >= self.capacity:
                self._cache.popitem(last=False)
            self._cache[key] = (value, expire_at)

Both implementations are complete and runnable: OrderedDict for O(1), lazy TTL expiry, and a threading lock.

Same prompt, test 2: a pool filling problem

A pool fills in 6 hours with the inlet pipe alone and empties in 8 hours with the drain alone. Starting from empty, both pipes are opened at once. How many hours until the pool is exactly full? Show your reasoning, then give the answer.

Both wrote the same LaTeX working — 1/6 − 1/8 = 1/24 → 1 ÷ 1/24 = 24 — and reached the correct conclusion with equivalent reasoning.

The differences you can actually see

• On the coding task, 4.7 finished 22 seconds faster (39.7s vs 61.9s) with a shorter answer (3285 vs 3915 output tokens)
• On the reasoning task, both answered correctly with equivalent working
• In output style, 4.7 favors structured headings and LaTeX and writes more compactly; 4.6 also uses LaTeX but is a bit more verbose (637 vs 573 tokens)
• These two head-to-head runs plus 5 more scenario runs (debugging, code review, system design, and more) make 9 requests in total, all finished cleanly (finish=stop) with no truncation
• In our capped tests (max_tokens set), 4.7 did not consume more output tokens than 4.6. The traictory.com figure of ~81K tokens per task, roughly 2x 4.6, comes from high-effort runs with no cap — a different setup. Neither number cancels the other, and neither should be quoted as the whole story
• xAI's own Terminal-Bench 4.0 numbers move from 20.3% to 38.0%, which points the same direction as the coding test above

Which one for which task

• Complex coding, multi-step agentic work: pick Grok 4.7
• Everyday Q&A, format conversions, small scripts: pick Grok 4.6 — output quality holds up and you spend fewer tokens
• Happy with 4.6 in production: don't migrate for migration's sake. Route new tasks to 4.7 first and fall back if it disappoints
• Budget-sensitive but want 4.7's ability: both generations cost the same per token, so running lightweight tasks on 4.6 is where the saving comes from

Switching costs: one field

Both generations use the same /v1/chat/completions endpoint with identical parameters. Change model from grok-4.6 to grok-4.7 and everything else stays put:

import requests

resp = requests.post(
    "https://api.hiapi.ai/v1/chat/completions",
    headers={"Authorization": "Bearer $HIAPI_API_KEY"},
    json={
        "model": "grok-4.7",  # change this
        "messages": [{"role": "user", "content": "your task"}],
    },
)

FAQ

Is Grok 4.7 more expensive than 4.6? Not per token: both cost $0.84 input / $2.52 output per million tokens on HiAPI. traictory.com reports that 4.7 uses about 81K tokens per task at high effort with no cap, roughly 2x 4.6. That is consumption, not unit price — and in our own capped runs, 4.7 did not outspend 4.6.

What are 4.7's known weaknesses? Some Cursor users report occasional hallucinations, bad code merges, and heavy token usage, and a portion have rolled back to 4.6. For simple tasks, staying on 4.6 is a safe alternative.

Any difference in context or reasoning levels? No. Both offer a 500K context window and four reasoning levels (low / medium / high / xhigh) with high as the default.

Where can I try it? The Grok 4.7 model page has a live playground, and new accounts get 200 free credits. For pricing details see Grok 4.7 API pricing, for getting started see the Grok 4.7 API guide, for ready-to-use prompts see Grok 4.7 prompt recipes, and if you're coming from the older generation, the Grok 4.6 how-to still applies.