它使得突然出现了大量的垃圾对象,空闲空间必然减少,从而大大增加了下一次创建新对象时强制主 GC 的机会。...(7) 使用软引用类型如果需要使用经常用到的图片, 可以使用软引用类型, 它可以尽可能将图片保存在内存中, 供程序调用, 而不引起 OutOfMemory。 数据挖掘中的利器--XGBoost理论篇 XGBoost是各种数据挖掘或...
Spillingis the technique that moves data automatically when an operation that would otherwise succeed runs out of memory due to other dataframes or series taking up needed space in GPU memory. It enables out-of-core computations on datasets that don’t fit into memory. RAPIDS cuDF and dask-cu...
@throws(classOf[XGBoostError]) def trainDistributed( trainingDataIn: RDD[XGBLabeledPoint], params: Map[String, Any], round: Int, nWorkers: Int, obj: ObjectiveTrait = null, eval: EvalTrait = null, useExternalMemory: Boolean = false, //NaN作为missing的值 missing: Float = Float.NaN, hasG...
实际中可能还需要考虑block size的设置,文中提出可通过实验确定,在此不展开。 4.3 out of core 当数据集非常大时,不可能一次性全部载入内存,此时就要使用out of core机制。这个机制会启动一个独立的线程来预取数据,这样在计算的同时这个线程可以把数据预取然后放置在memory buffer中供计算使用。从实现方案来说有两种...
报错:RuntimeError: CUDA out of memory. Tried to allocate 46.00 MiB (GPU 0; 2.00 GiB total capacity; 54.79 MiB already allocated; 39.30 MiB free; 74.00 MiB reserved in total by PyTorch) 原因:可以看出在GPU充足的情况下无法使用,本机有两个GPU,其中一个GPU的内存不可用?
XGBoost is designed to be memory efficient. Usually it can handle problems as long as the data fit into your memory (This usually means millions of instances). If you are running out of memory, checkout external memory version or distributed version of xgboost. Running xgboost on Platform X ...
XGBoost 算法运行效率优化 分块并行学习 缓存优化 Out-of-core Computation XGBoost 与 GBDT 的区别 XGBoost 库安装 XGBoost 算法 XGBoost 损失函数 假设输入训练数据集 \(T = \{ (\mathbf{x}_1, y_1), (\mathbf{x}_2, y_2), \cdots, (\mathbf{x}_N, y_N) \}\), \(\mathbf{x}_i \in ...
Optimize and bound the size of the histogram on CPU, to control memory footprint XGBoost has a new parametermax_cached_hist_nodefor users to limit the CPU cache size for histograms. It can help prevent XGBoost from caching histograms too aggressively. Without the cache, performance is likely ...
*/@throws(classOf[XGBoostError])deftrainDistributed( trainingDataIn:RDD[XGBLabeledPoint], params:Map[String,Any], round:Int, nWorkers:Int, obj:ObjectiveTrait=null, eval:EvalTrait=null, useExternalMemory:Boolean=false,//NaN作为missing的值missing:Float=Float.NaN, ...
具体地,给定一个 Q(\mathcal{D}) = (S, \widetilde{r}^+_\mathcal{D}, \widetilde{r}_\mathcal{D}^-, \widetilde{\omega}_\mathcal{D}), \; S = \{x_1, x_2, \cdots, x_k\} ,以及一个 memory budget(内存预算)b,那么prune操作的结果就是返回一个新的 Q^\prime(\mathcal{D}) ...