【问题标题】：How to remove those elements from a list for which the try function produced error?如何从 try 函数产生错误的列表中删除这些元素？
【发布时间】：2018-04-22 17:53:10
【问题描述】：

我想做什么

我有一个车辆运动数据集。因此，每个车辆 ID 都会重复多次。我使用purrr::map() 函数分别为每辆车拟合segmented::segmented() 回归模型。因为，模型可能不适合每辆车（可能是由于给定车辆的数据点很少），它可能会产生错误。所以，我将segmented() 函数包装在try() 中。

数据

下面是我data的结构：

data = my_df %>% split(., .$per.Vehicle.ID2)
   > str(data, max.level = 1)
List of 2
 $ 3.544.534:Classes ‘tbl_df’, ‘tbl’ and 'data.frame':  30 obs. of  4 variables:
 $ 3.553.545:Classes ‘tbl_df’, ‘tbl’ and 'data.frame':  72 obs. of  4 variables:

这里是data 重复性列表：

> dput(data)
structure(list(`3.544.534` = structure(list(per.Vehicle.ID2 = c("3.544.534", 
"3.544.534", "3.544.534", "3.544.534", "3.544.534", "3.544.534", 
"3.544.534", "3.544.534", "3.544.534", "3.544.534", "3.544.534", 
"3.544.534", "3.544.534", "3.544.534", "3.544.534", "3.544.534", 
"3.544.534", "3.544.534", "3.544.534", "3.544.534", "3.544.534", 
"3.544.534", "3.544.534", "3.544.534", "3.544.534", "3.544.534", 
"3.544.534", "3.544.534", "3.544.534", "3.544.534"), Time = c(307.1, 
307.7, 308.3, 308.9, 309.5, 310.1, 310.7, 311.3, 311.9, 312.5, 
313.1, 313.7, 314.3, 314.9, 315.5, 316.1, 316.7, 317.3, 317.9, 
318.5, 319.1, 319.7, 320.3, 320.9, 321.5, 322.1, 322.7, 323.3, 
323.9, 324.5), svel.mps_mean = c(NA, NA, NA, NA, NA, NA, NA, 
12.7755159281222, 12.5036616661267, 12.2395719427147, 11.9923745340627, 
11.7738694424139, 11.558525429244, 11.3191973673818, 11.0522994308264, 
10.7788324802049, 10.5051145516082, 10.2349319889114, 9.97501528086885, 
9.73263129457631, 9.49870470544252, 9.26388495185967, 9.03860711857004, 
NA, NA, NA, NA, NA, NA, NA), dssvel = c(NA, NA, NA, NA, NA, NA, 
NA, NA, NA, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, NA, NA, 
NA, NA, NA, NA, NA)), .Names = c("per.Vehicle.ID2", "Time", "svel.mps_mean", 
"dssvel"), row.names = c(NA, -30L), class = c("tbl_df", "tbl", 
"data.frame")), `3.553.545` = structure(list(per.Vehicle.ID2 = c("3.553.545", 
"3.553.545", "3.553.545", "3.553.545", "3.553.545", "3.553.545", 
"3.553.545", "3.553.545", "3.553.545", "3.553.545", "3.553.545", 
"3.553.545", "3.553.545", "3.553.545", "3.553.545", "3.553.545", 
"3.553.545", "3.553.545", "3.553.545", "3.553.545", "3.553.545", 
"3.553.545", "3.553.545", "3.553.545", "3.553.545", "3.553.545", 
"3.553.545", "3.553.545", "3.553.545", "3.553.545", "3.553.545", 
"3.553.545", "3.553.545", "3.553.545", "3.553.545", "3.553.545", 
"3.553.545", "3.553.545", "3.553.545", "3.553.545", "3.553.545", 
"3.553.545", "3.553.545", "3.553.545", "3.553.545", "3.553.545", 
"3.553.545", "3.553.545", "3.553.545", "3.553.545", "3.553.545", 
"3.553.545", "3.553.545", "3.553.545", "3.553.545", "3.553.545", 
"3.553.545", "3.553.545", "3.553.545", "3.553.545", "3.553.545", 
"3.553.545", "3.553.545", "3.553.545", "3.553.545", "3.553.545", 
"3.553.545", "3.553.545", "3.553.545", "3.553.545", "3.553.545", 
"3.553.545"), Time = c(262, 262.6, 263.2, 263.8, 264.4, 265, 
265.6, 266.2, 266.8, 267.4, 268, 268.6, 269.2, 269.8, 270.4, 
271, 271.6, 272.2, 272.8, 273.4, 274, 274.6, 275.2, 275.8, 276.4, 
277, 277.6, 278.2, 278.8, 279.4, 280, 280.6, 281.2, 281.8, 282.4, 
283, 283.6, 284.2, 284.8, 285.4, 286, 286.6, 287.2, 287.8, 288.4, 
289, 289.6, 290.2, 290.8, 291.4, 292, 292.6, 293.2, 293.8, 294.4, 
295, 295.6, 296.2, 296.8, 297.4, 298, 298.6, 299.2, 299.8, 300.4, 
301, 301.6, 302.2, 302.8, 303.4, 304, 304.6), svel.mps_mean = c(NA, 
NA, NA, NA, NA, NA, NA, 5.41298285821819, 5.48497881688925, 5.55898102091842, 
5.63821570373546, 5.73023228642822, 5.84505407541773, 5.98954476445736, 
6.1455976413909, 6.29775534569644, 6.4475118875263, 6.59939228553705, 
6.75929997962276, 6.92825864041472, 7.10600376881863, 7.29418216320438, 
7.48845217271764, 7.68381738580354, 7.87513283133227, 8.05995699864641, 
8.21465371209303, 8.31097200556874, 8.3417386030748, 8.32304537754036, 
8.26198297864187, 8.15886518084024, 8.02894718462323, 7.87911840872659, 
7.71538338260088, 7.54358017038221, 7.36910128510413, 7.1920560779047, 
7.00992171675244, 6.81783765068062, 6.61630770462671, 6.42117981828069, 
6.24687579703188, 6.09559517163776, 5.96909261287346, 5.87826537515735, 
5.83640038089119, 5.84922602270984, 5.9161965850754, 6.02778813388058, 
6.18611611187481, 6.38709031522456, 6.61991746112876, 6.88184116355984, 
7.1817199521547, 7.51057503223919, 7.8581088613562, 8.22211301486075, 
8.60478211935657, 9.01154624501708, 9.42860178480699, 9.83720909606077, 
10.2152191362441, 10.5568662978488, 10.8733070569773, NA, NA, 
NA, NA, NA, NA, NA), dssvel = c(NA, NA, NA, NA, NA, NA, NA, NA, 
NA, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 
0, -2, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 2, 
0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, NA, NA, NA, 
NA, NA, NA, NA)), .Names = c("per.Vehicle.ID2", "Time", "svel.mps_mean", 
"dssvel"), row.names = c(NA, -72L), class = c("tbl_df", "tbl", 
"data.frame"))), .Names = c("3.544.534", "3.553.545"))

功能

library(segmented)
segf2_1 <- function(df){
  try(segmented(lm(svel.mps_mean ~ Time, data=df), seg.Z = ~Time,
                psi = list(Time = df$Time[which(df$dssvel != 0)]),
                control = seg.control(seed=1, n.boot = 50)
  ),
  silent=TRUE)
}

在`data`上应用函数：

library(purrr)
model_subject = data %>% map(segf2_1)

它产生以下内容：

    > str(model_subject, max.level = 1)
List of 2
 $ 3.544.534:Class 'try-error'  atomic [1:1] Error in Z <= PSI : non-conformable arrays

  .. ..- attr(*, "condition")=List of 2
  .. .. ..- attr(*, "class")= chr [1:3] "simpleError" "error" "condition"
 $ 3.553.545:List of 23
  ..- attr(*, "class")= chr [1:2] "segmented" "lm"

您可以看到该模型不适合data 列表的第一个元素。 我想从model_subject 和data 列表中删除它。

删除有错误的元素：

我可以从model_subject 中删除产生错误的元素，如下所示：

# Removing the vehicles which have error:
model_subject[grep("Error", model_subject)] <- NULL

问题：从model_subject 列表中删除相应元素后，如何从data 中删除它们？ data 列表不包含 Error，但由于模型不适合 data 的第一个元素，我想删除它

手动，我可以做data[c(1)] <- NULL。但我想自动化这个（特别是如果有多个Error 元素）。请帮忙。

【问题讨论】：

标签： r purrr

【解决方案1】：

考虑使用purrr 包中的possibly()。

在拟合分段回归时，很多事情都可能出错。如果您要执行以下操作，则不必使用 try() 并最终得到 try-error 类。我还没有测试过你的函数，但我正在对我自己的一些代码做同样的事情。

library(segmented)
# your function but without try(), and better formatted for readibility
segf2_1 <- function(df){
  segmented(lm(svel.mps_mean ~ Time, data = df),
            seg.Z = ~ Time,
            psi = list(Time = df$Time[which(df$dssvel != 0)]),
            control = seg.control(seed=1, n.boot = 50),
            silent=TRUE)
}

library(purrr)
model_subject = map(data, possibly(segf2_1, otherwise = NULL))

【讨论】：

【解决方案2】：

我们可以使用Filter

Filter(function(x) length(x) > 1, model_subject)

或keep 来自purrr

library(purrr)
keep(model_subject, ~ length(.x) > 1)

或者另一种选择是使用class创建一个逻辑向量

model_subject[sapply(model_subject, function(x) !"try-error" %in% class(x))]

或者

model_subject[sapply(model_subject, function(x) !inherits(x, "try-error"))]

更新

如果 OP 已将这些错误元素分配给 NULL

model_subject[grep("Error", model_subject)] <- list(NULL)

然后，我们可以在is.null 上使用Filter 和Negate

Filter(Negate(is.null), model_subject)

或者

keep(model_subject, Negate(is.null))

如果我们需要得到一个逻辑索引

i1 <- !sapply(model_subject, is.null)

这可用于对“数据”进行子集化

data[i1]

更新2

使用 OP 的可重现示例

data[i1]
#$`3.553.545`
# A tibble: 72 x 4
#   per.Vehicle.ID2  Time svel.mps_mean dssvel
#   <chr>           <dbl>         <dbl>  <dbl>
# 1 3.553.545         262         NA        NA
# 2 3.553.545         263         NA        NA
# 3 3.553.545         263         NA        NA
# 4 3.553.545         264         NA        NA
# 5 3.553.545         264         NA        NA
# 6 3.553.545         265         NA        NA
# 7 3.553.545         266         NA        NA
# 8 3.553.545         266          5.41     NA
# 9 3.553.545         267          5.48     NA
#10 3.553.545         267          5.56      0
# ... with 62 more rows

【讨论】：

感谢您的回答。但我在问一个不同的问题。正如您所建议的，我可以在model_subject 列表中删除带有Error 的元素。我的解决方案是model_subject[grep("Error", model_subject)] <- NULL 。但是在这样做之后，我还想从data列表中删除相应的元素。既然data列表中没有Error，那么如何去掉对应的元素呢？
@umairdurrani 更新了答案
@umairdurrani 您没有提供可重现的示例。我认为您需要分配给list(NULL)，然后创建索引。
@umairdurrani 你测试过我更新的代码吗？它应该工作
谢谢！有用。你教我is.null 和keep。

我想做什么

数据

功能

在data上应用函数：

删除有错误的元素：

更新

更新2

在`data`上应用函数：