问题描述
RServe产生的进程是否可能共享一些一次加载到内存中的通用库?想象一下，我需要同时在100个不同的RConnections上执行以下代码.
Is it possible that processes spawned by RServe share some common libraries loaded once into memory?Imagine that I need to execute bellow code on 100 different RConnections concurrently.
library(libraryOfSize40MB)
fun()
这意味着我仅需要3.9GB的内存即可加载库.我希望先加载一次库，然后执行fun()一百次，以便可以在便宜的主机上运行它.
It means that I need about 3.9GB of memory just to load library. I would prefer to load library once and then execute fun() one hundred times, so that I can run this on cheap host.
这可能有用吗? https://github.com/su/Rserve/blob/master/NEWS #L40-L48
推荐答案
有可能.您必须先使用run.serve从R Shell运行RServe，然后再加载库:
It is possible. You have to run RServe from R shell using run.serve preceded by loaded libraries:
library(Rserve)
#load libraries so all connections will share them
library("yaml")
library("reshape")
library("rjson")
library("zoo")
(...)
library("stringr")
run.Rserve(debug = TRUE, port = 6311, remote=TRUE, auth=FALSE, args="--no-save", config.file = "/etc/Rserve.conf")
每个新的连接都将能够看到该库
Every new connection will be able to see this libraries
library(RSclient)
con = RS.connect(host='10.1.2.3')
RS.eval(con, quote(search()))
> #lots of libraries available
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