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Stop using litellm._should_retry

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Paul Gauthier 2024-08-06 14:32:52 -03:00
parent f4d0864bcb
commit 1e232d4db6
1 changed files with 0 additions and 22 deletions
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22
aider/sendchat.py
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@ -18,27 +18,6 @@ def lazy_litellm_retry_decorator(func):
def wrapper(*args, **kwargs): def wrapper(*args, **kwargs):
import httpx import httpx
def should_giveup(e):
if not hasattr(e, "status_code"):
return False
if type(e) in (
httpx.ConnectError,
httpx.RemoteProtocolError,
httpx.ReadTimeout,
):
return False
# These seem to return .status_code = ""
# litellm._should_retry() expects an int and throws a TypeError
#
# litellm.llms.anthropic.AnthropicError
# litellm.exceptions.APIError
if not e.status_code:
return False
return not litellm._should_retry(e.status_code)
decorated_func = backoff.on_exception( decorated_func = backoff.on_exception(
backoff.expo, backoff.expo,
( (
@ -53,7 +32,6 @@ def lazy_litellm_retry_decorator(func):
litellm.exceptions.InternalServerError, litellm.exceptions.InternalServerError,
litellm.llms.anthropic.AnthropicError, litellm.llms.anthropic.AnthropicError,
), ),
giveup=should_giveup,
max_time=60, max_time=60,
on_backoff=lambda details: print( on_backoff=lambda details: print(
f"{details.get('exception', 'Exception')}\nRetry in {details['wait']:.1f} seconds." f"{details.get('exception', 'Exception')}\nRetry in {details['wait']:.1f} seconds."