突然停止执行命令并继续执行下一个命令

I have a set of files on which I need to apply rpart algorithm. Some of these files takes too long for computation. How can I skip such cases (eg. cases that take more than an hour) and continue on to the next one?

我有一组文件，我需要在其上应用rpart算法。其中一些文件的计算时间太长。我怎样才能跳过这种情况（例如，需要一个多小时的案例）并继续下一个案例？

for (i in num)
{
print(i)
infilename  = filenames[i]
tmpData = read.table(infilename, header = TRUE,  sep= "\t")
retval = rpart(fmla[i],  dat=tmpData, method = "class")
print (retval)
}

Edit: Based on suggestin from @Dwin, I am doing the following but it does not work. Where I am doing wrong?

编辑：基于来自@Dwin的suggestin，我正在做以下但是它不起作用。我在哪里做错了？

for (i in num)
{
print(i)
infilename  = filenames[i]
tmpData = read.table(infilename, header = TRUE,  sep= "\t")
retVal= NULL
setTimeLimit(cpu=10)
retval = try(rpart(fmla,  dat=tmpData, method = "class") )
print (retval)
}

1 个解决方案

#1

Because you are just using regular R functions (and not coding this from scratch), you will need to come up with some way to estimate the conditions leading to excessive times. This might be a test that looks at the dimensions of a dataframe and skips the next rpart computation if the product of dim(dfrm) exceeds a certain threshold.

因为您只是使用常规R函数（而不是从头开始编码），所以您需要想出一些方法来估计导致过多时间的条件。这可能是一个测试，它会查看数据帧的维度，如果dim（dfrm）的乘积超过某个阈值，则会跳过下一个rpart计算。

retval = if(prod(dim(tmpData)) < 1e6) {
                rpart(fmla[i],  dat=tmpData, method = "class") }

Notice that at the moment you are overwriting retval with every loop iteration rather than storing it to a durable object.

请注意，此时您将使用每次循环迭代覆盖retval，而不是将其存储到持久对象中。

You could also try using the functions setTimeLimit and setSessionLimit but these will throw an error condition and you may need to put your code inside a try function to recover gracefully:

您也可以尝试使用函数setTimeLimit和setSessionLimit但这些会抛出一个错误条件，您可能需要将您的代码放在try函数中以优雅地恢复：

 setTimeLimit(cpu=2)
for (i in 4:8) {x <- 1:10^i;x=x^3}
 max(x)
#[1] 1e+24
# did not exceed the limits
x^(1/3)
#[1]     1     2     3     4     5     6     7Error: reached CPU time limit

#1

retval = if(prod(dim(tmpData)) < 1e6) {
                rpart(fmla[i],  dat=tmpData, method = "class") }