response=craft.generate_code(language="python",description="实现一个快速排序算法,对整数列表进行升序排列")# 输出生成的代码print(response.code)""" defquick_sort(arr):iflen(arr)<=1:returnarr pivot=arr[len(arr)//2]left=[xforxinarrifx<pivot]middle=[xforxinarrifx==pivot]right=[xforxinarrifx...
这种能力基于先进的LLM(大语言模型)技术,能够准确捕捉开发意图。 # 示例:使用Craft API生成Python排序代码from codebuddy.craft import CodeGenerator# 初始化Craft客户端craft = CodeGenerator(api_key="your_api_key")# 用自然语言描述需求response = craft.generate_code(language="python",description="实现一个快...
Optionally, the software can also return the individual values of the fictive temperature components using a Prony series fit to approximate the stretched exponential relaxation form. RelaxPy aims to provide a flexible, open-source framework for glass relaxation modeling where new advances in glass ...
for faster recovery "llm_retry_backoff": 2.0, # Keep exponential backoff factor "llm_retry_jitter": 0.2, # Increased jitter for better distribution "llm_rate_limit": 12, # Set below the 15 RPM limit to provide buffer "llm_batch_size": 1, # Keep batch size at 1 for better control...
有打不开的情况都会多刷新几次。因此，我们也需要让 python 进行重试。而 retrying 模块应运而生 ...
toExponential() - Round off numbers in exponential notation @goblindoom95 var x = 1.234; var y = x.toExponential(2); // returns y = 1.23e+0 var y = x.toExponential(5); // returns y = 1.23450e+0 content_copy #javascript#jsfunctions#numbers ...
You will get remarks on generation of gather/scatter prefetches for indirect memory reference, similar to Figure 5.Figure 5 - Gather/scatter prefetch remarks shown in compiler optimization report3. AVX-512ERMAVX-512ER provides fast and high-precision approximation instructions for exponential, ...
5. 给出实例代码或伪代码,展示如何实现“exponential backoff”重试策略 以下是一个使用Python实现的简单“exponential backoff”重试策略的伪代码示例: python import time import random def exponential_backoff(initial_delay, max_delay, retries): delay = initial_delay for attempt in range(retries): try: ...
RegOT is a collection of state-of-the-art solvers for regularized optimal transport (OT) problems, implemented in efficient C++ code. This repository is the Python interface to RegOT. 📝 Formulation RegOT mainly solves two types of regularized OT problems: the entropic-regularized OT (EROT) ...
[sign]integeral-digits[.[fractional-digits]][eE[sign]exponential-digits] 1 那么,一路传下来就行了。不在分析线上的,或者在分析线上就有问题的,就是不符合标准的。 classSolution:# s字符串defisNumeric(self,s):lens=len(s)iflens==0:#空字符串直接返回FalsereturnFalseidx=0# 首位是符号位ifs[0]i...