Debugging 没有抛出错误的错误

Debugging 没有抛出错误的错误,debugging,r,Debugging,R,我试图对一些数据进行乘法插补,但它以一种非常奇怪的方式失败了。我设置了选项(error=recover),mi()不会抛出错误,但在第一次尝试后会停止插补。它也不会返回值。因此,我甚至不知道从哪里开始调试。下面是最小的可复制示例 > library(mi) > temp <- mi(dat) Beginning Multiple Imputation ( Wed Dec 14 10:44:44 2011 ): Iteration 1 Chain 1 : HLTHA5.fac

我试图对一些数据进行乘法插补,但它以一种非常奇怪的方式失败了。我设置了选项(error=recover),mi()不会抛出错误,但在第一次尝试后会停止插补。它也不会返回值。因此,我甚至不知道从哪里开始调试。下面是最小的可复制示例

> library(mi)
> temp <- mi(dat)
Beginning Multiple Imputation ( Wed Dec 14 10:44:44 2011 ):
Iteration 1 
 Chain 1 : HLTHA5.fac*  BMI*  INCOME*  
> temp
Error: object 'temp' not found


dat<-structure(list(treat = c(FALSE, FALSE, TRUE, TRUE, TRUE, FALSE, 
FALSE, FALSE, TRUE, FALSE, FALSE, TRUE, FALSE, TRUE, TRUE, TRUE, 
FALSE, TRUE, TRUE, TRUE, FALSE, TRUE, TRUE, FALSE, TRUE), NUMADULT = c(2, 
1, 2, 1, 2, 1, 2, 1, 2, 4, 1, 1, 2, 2, 1, 2, 1, 1, 2, 1, 2, 2, 
3, 3, 1), HLTHA5.fac = structure(c(3L, NA, 3L, 2L, 4L, 5L, 5L, 
4L, 4L, 3L, 3L, 5L, 3L, 4L, 5L, 4L, 2L, 2L, 3L, 5L, 4L, 5L, 4L, 
3L, 3L), .Label = c("0", "1", "2", "3", "4"), class = "factor"), 
    SOURCEA = structure(c(1L, 2L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 
    2L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 
    1L), .Label = c("Yes", "No", "Don't know", "Refused"), class = "factor"), 
    BMI = c(27.363941459046, 24.0265857515842, 34.3236939308346, 
    27.0907152026518, 32.6101901381975, 34.1643655360753, 21.4628674188624, 
    29.1751398412094, 22.5924920198551, 39.6719545438681, 38.5220574557939, 
    20.1156133421915, 30.6612391698034, 35.7332536282609, 26.5664872147956, 
    25.6016897082437, 19.3649931598758, 28.1868713091175, NA, 
    32.4438116170843, 32.5507197719099, 21.1090717674633, 32.2340044872853, 
    24.3699149340904, 27.3369153440247), SMOKE2 = structure(c(2L, 
    1L, 2L, 2L, 2L, 2L, 2L, 1L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 
    1L, 2L, 2L, 2L, 1L, 2L, 2L, 2L, 1L), .Label = c("Yes", "No"
    ), class = "factor"), INCOME = structure(c(16L, 4L, 13L, 
    11L, 13L, 7L, 22L, 6L, NA, 1L, 13L, 18L, 12L, 20L, NA, NA, 
    2L, 13L, 17L, NA, 12L, 21L, 9L, 15L, 13L), .Label = c("Less than $10,800", 
    "$10,800-$14,600", "$14,601-$16,250", "$16,251-$18,300", 
    "$18,301-$21,800", "$21,801-$25,000", "$25,001-$27,500", 
    "$27,501-$29,300", "$29,301-$33,100", "$33,101-$36,700", 
    "$36,701-$38,700", "$38,701-$44,200", "$44,201-$50,000", 
    "$50,001-$58,000", "$58,001-$66,500", "$66,501-$73,500", 
    "$73,501-$80,000", "$80,001-88,200", "$88,201-$100,000", 
    "$100,001-$120,000", "$120,001-$130,000", "$130,001-$150,000", 
    "$150,001-$250,000", "Over $250,000", "Don't know", "Refused"
    ), class = "factor"), RESPMAR = structure(c(1L, 5L, 1L, 4L, 
    3L, 6L, 1L, 1L, 1L, 1L, 4L, 4L, 1L, 1L, 4L, 1L, 3L, 3L, 1L, 
    3L, 1L, 1L, 1L, 1L, 1L), .Label = c("Married", "Living w partner", 
    "Widowed", "Divorced", "Separated", "Single", "Other", "Don't know", 
    "Refused"), class = "factor"), RESPGRAD = structure(c(5L, 
    1L, 2L, 5L, 3L, 3L, 5L, 2L, 4L, 2L, 4L, 3L, 4L, 4L, 2L, 3L, 
    2L, 5L, 2L, 4L, 4L, 5L, 2L, 2L, 3L), .Label = c("< HS 0-11", 
    "HS graduate", "Some colge 13-15", "Collge grad 16", "Post college 16+", 
    "Don't know", "Refused"), class = "factor"), RACEA2 = structure(c(1L, 
    1L, 1L, 2L, 2L, 2L, 1L, 1L, 1L, 1L, 2L, 1L, 1L, 1L, 2L, 1L, 
    3L, 1L, 1L, 1L, 2L, 1L, 1L, 1L, 1L), .Label = c("White (Not-Latino)", 
    "Black (Not-Latino)", "Latino (total)", "Asian", "Biracial/Multi", 
    "Native American", "Other", "Don't know", "Refused"), class = "factor"), 
    INSUREDA = structure(c(1L, 2L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 
    1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 2L, 1L, 
    1L), .Label = c("Insured", "Not insured", "Don't know", "Refused"
    ), class = "factor"), PAP.adj = c(TRUE, FALSE, FALSE, FALSE, 
    FALSE, TRUE, TRUE, FALSE, FALSE, FALSE, TRUE, FALSE, TRUE, 
    TRUE, TRUE, TRUE, FALSE, TRUE, TRUE, FALSE, TRUE, TRUE, TRUE, 
    TRUE, TRUE)), .Names = c("treat", "NUMADULT", "HLTHA5.fac", 
"SOURCEA", "BMI", "SMOKE2", "INCOME", "RESPMAR", "RESPGRAD", 
"RACEA2", "INSUREDA", "PAP.adj"), row.names = c(1L, 13L, 15L, 
23L, 26L, 33L, 38L, 53L, 56L, 60L, 62L, 85L, 109L, 116L, 138L, 
217L, 240L, 262L, 264L, 269L, 277L, 295L, 328L, 334L, 338L), class = "data.frame")
>库(mi)
>温度
错误:找不到对象“temp”

dat由于
error=recover
选项不起作用,一种可行的替代方法是设置
选项(error=dump.frames)
。这将为您获取有关错误的一些信息,您可以打印出来,或者更有效地使用
debugger()

ls()
#[1]“dat”
选项(错误=dump.frames)
mi(dat)
ls()
#[1]“dat”“last.dump”#显然有错误
#使用调试器进行调查()
调试器(dump=last.dump)
#或者,将last.dump打印到控制台
最后一个垃圾场
$`mi(dat)`
$`mi(dat)`
$`.local(对象,…)`
$`mi.default(对象、信息、n.imp、n.iter、R.hat、max.minutes、rand.imp.method`
属性(,“error.message”)

[1] “aveVar[s,i,]
debug(mi)
中的错误并逐步执行代码?不管怎样,
mi
是S4通用的,并且
debug(mi)
没有提供任何有用的东西。@PaulHiemstra和gsk3——是的,我在几年前就读过,但从来没有发现需要它。即使是Chambers(2008)也说,“通常,转储.frames()没有好处。”,因为recover()的行为类似于dump.frames(),如果计算不是交互式的。“好吧,现在它有一个用途了!我把它添加到我的调试建议列表中:这是有问题的。值得注意的是,相关的帮助文件和Chambers(2008)都有轻微的警告,提示
调试器(dump=last.dump)
并不总是完全等同于
recover()
。promise对象、连接上的I/O和交互式图形等内容保存不好,因此使用
dump.frames
进行调试比较困难。
ls()
# [1] "dat"
options(error=dump.frames)
mi(dat)
ls()
# [1] "dat"       "last.dump"  # Apparently there WAS an error

# INVESTIGATE WITH debugger()
debugger(dump=last.dump)

# ALTERNATIVELY, PRINT last.dump TO CONSOLE
last.dump

$`mi(dat)`
<environment: 0x05155c44>

$`mi(dat)`
<environment: 0x05158f30>

$`.local(object, ...)`
<environment: 0x05158cac>

$`mi.default(object, info, n.imp, n.iter, R.hat, max.minutes, rand.imp.method`
<environment: 0x047dc3a0>

attr(,"error.message")
[1] "Error in aveVar[s, i, ] <- c(avevar.mean, avevar.sd) :
\n  number of items to replace is not a multiple of replacement length\n"
attr(,"class")
[1] "dump.frames"