Question

我有这段代码，用于创建一个数据框，用于保存多个分类模型的性能矩阵：

eval_table <- data.frame(norm_method_1=c("","","","",""),norm_method_2=c("","","","",""),norm_method_3=c("","","","",""),norm_method_4=c("","","","",""),norm_method_5=c("","","","",""),norm_method_6=c("","","","",""),norm_method_7=c("","","","",""),norm_method_8=c("","","","",""))
    rownames(eval_table) <- c("Log. Model Coeff","Log. Model Confusion Matrix","Log. Model AUC","Linear Modl Coeff","Linear Model overfit ratio")

    col <- list(mnm_coef,cm,auc,lm_summary$coefficients,lm_summary$training_testing_error_ratio)
    eval_table[,c("norm_method_1")] <- col

当尝试使用我预先准备的5个元素的列表更新数据帧的第一列（最后一行）时（数据帧有5行），我收到以下错误：

Error in `[<-.data.frame`(`*tmp*`, , c("norm_method_1"), value = list( : 
  replacement element 1 is a matrix/data frame of 2 rows, need 5

知道为什么吗？

更新：

dim(col)
NULL
str(col)
> str(col)
List of 5
 $ : num [1:2, 1:30] 21.436 40.828 -0.785 0.54 -1.375 ...
  ..- attr(*, "dimnames")=List of 2
  .. ..$ : chr [1:2] "2" "3"
  .. ..$ : chr [1:30] "(Intercept)" "is_top_rated_listing1" "seller_is_top_rated_seller" "is_auto_pay1" ...
 $ :List of 5
  ..$ positive: NULL
  ..$ table   : 'table' int [1:3, 1:3] 414 73 433 101 329 1186 147 60 10546
  .. ..- attr(*, "dimnames")=List of 2
  .. .. ..$ Prediction: chr [1:3] "1" "2" "3"
  .. .. ..$ Reference : chr [1:3] "1" "2" "3"
  ..$ overall : Named num [1:7] 0.849 0.402 0.843 0.856 0.809 ...
  .. ..- attr(*, "names")= chr [1:7] "Accuracy" "Kappa" "AccuracyLower" "AccuracyUpper" ...
  ..$ byClass : num [1:3, 1:8] 0.45 0.204 0.981 0.98 0.989 ...
  .. ..- attr(*, "dimnames")=List of 2
  .. .. ..$ : chr [1:3] "Class: 1" "Class: 2" "Class: 3"
  .. .. ..$ : chr [1:8] "Sensitivity" "Specificity" "Pos Pred Value" "Neg Pred Value" ...
  ..$ dots    : list()
  ..- attr(*, "class")= chr "confusionMatrix"
 $ :Classes 'auc', 'numeric'  atomic [1:1] 0.592
  .. ..- attr(*, "partial.auc")= logi FALSE
  .. ..- attr(*, "percent")= logi FALSE
  .. ..- attr(*, "roc")=List of 8
  .. .. ..$ percent      : logi FALSE
  .. .. ..$ sensitivities: num [1:1021] 1 0.998 0.998 0.998 0.996 ...
  .. .. ..$ specificities: num [1:1021] 0 0 0.00151 0.00302 0.00302 ...
  .. .. ..$ thresholds   : num [1:1021] -Inf 0.358 0.359 0.36 0.361 ...
  .. .. ..$ direction    : chr "<"
  .. .. ..$ cases        : num [1:462] 0.51 0.727 0.725 0.667 0.667 ...
  .. .. ..$ controls     : num [1:662] 0.903 0.967 0.756 0.966 0.981 ...
  .. .. ..$ fun.sesp     :function (thresholds, controls, cases, direction)  
  .. .. ..- attr(*, "class")= chr "roc"
 $ : num [1:28, 1:4] 3.18 3.49 -3.55 -1.16 4.1 ...
  ..- attr(*, "dimnames")=List of 2
  .. ..$ : chr [1:28] "(Intercept)" "is_top_rated_listing1" "seller_is_top_rated_seller" "is_auto_pay1" ...
  .. ..$ : chr [1:4] "Estimate" "Std. Error" "t value" "Pr(>|t|)"
 $ : num 0.819
>

Answer 1

如果遇到这种情况，您需要检查数据的大小是否符合您的想法。使用

length(col)

验证您对col的假设。消息是说尺寸不兼容。

如果是，那么作业将起作用 - 例如，

col <- c(1,2,3,4,5)
eval_table[,c("norm_method_1")] <- col

在这种情况下，length(col)返回5.

另一个有用的命令是str - 即str(eval_table)和str(col)。这将帮助您查看大小不兼容。

在这种情况下，您的第一个元素是矩阵。所以我们可以这样嘲笑：

mat <- matrix(1,2)
col <- list(mat, 2, 3, 4, 5)

然后你会看到你得到的错误。要解决这个问题，请使用矩阵的第一个元素作为第一个元素构建列表，并将其构建为向量：

 col <- c(mat[1][1], 2,3, 4,5)

然后这个有效：

eval_table[,c("norm_method_1")] <- col

R数据帧更新整列

1 个答案: