r-lapply循环，根据经过的时间跳到下一次迭代

Question

我想创建一个lapply循环，当当前迭代需要60秒以上才能完成时，该循环将跳到下一次迭代。该脚本包含一个trycatch，用于处理函数可能返回的错误。我尝试过使用withTimeout((，但没有成功。

reg.out <- lapply(1:nrow(dt), function(i) {     
reg.mets <- withTimeout(
tryCatch(
1 * dt$dim1[i]),
error = function(e) NULL), 
timeout = 60, 
onTimeout = "silent")      
})

Answer 1

您已经翻转了withTimeout和tryCatch。

这里是一个基于以下简单函数的最小可重复示例：我们有一个data.frame，delay次(以秒为单位(；然后我们循环通过CCD_ 5次并且用CCD_。

如果执行时间超过2秒，我们使用withTimeout跳过一个步骤。

# Delay times from 1 to 10 seconds in 1 second steps
df <- data.frame(delay = seq(1:10))
library(R.utils)
lst <- lapply(1:nrow(df), function(i) {
cat(sprintf("Processing row %i/%i...", i, nrow(df)))
tryCatch(
withTimeout( {
Sys.sleep(df$delay[i]);
cat("[Done]n") },
timeout = 2),
TimeoutException = function(ex) cat("[Skipped due to timeout]n"))
})
#Processing row 1/10...[Done]
#Processing row 2/10...[Skipped due to timeout]
#Processing row 3/10...[Skipped due to timeout]
#Processing row 4/10...[Skipped due to timeout]
#Processing row 5/10...[Skipped due to timeout]
#Processing row 6/10...[Skipped due to timeout]
#Processing row 7/10...[Skipped due to timeout]
#Processing row 8/10...[Skipped due to timeout]
#Processing row 9/10...[Skipped due to timeout]
#Processing row 10/10...[Skipped due to timeout]

请注意，我们如何将withTimeout封装在tryCatch中以捕获(默认(超时错误，并在不终止lapply循环的情况下打印自定义错误消息。