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我正在尝试使用 ROCR 包从分析中导出生物特征数据。这是我到目前为止所做的代码:

pred = performance(Matching.Score,Distribution)
perf = prediction(pred,"fnr", "fpr")

An object of class “performance”

Slot "x.name":

[1] "False positive rate"

Slot "y.name":

[1] "False negative rate"

Slot "alpha.name":

[1] "Cutoff"

Slot "x.values":

[[1]]

[1] 0.00000 0.00000 0.00000 0.00000 0.00000 0.00000
[15] 0.00000 0.00000 0.00000 0.00000 0.00000 0.00000
  ......

Slot "y.values":

[[1]]

[1] 1.00000 0.99999 0.99998 0.99997 0.99996 0.99995
[15] 0.99986 0.99985 0.99984 0.99983 0.99982 0.99981
    ......

Slot "alpha.values":

[[1]]

[1]  Inf      1.0427800 1.0221150 1.0056240 1.0032630 0.9999599
[12] 0.9644779 0.9633058 0.9628996 0.9626501 0.9607665 0.9605930
    .......

这会产生几个插槽。我想使用以下命令将结果值导出到文本文件中以进行 Excel 修改:

write(pred, "filename")

但是,当我尝试写入文件时,我收到一条错误消息:

Error in cat(list(...), file, sep, fill, labels, append) : 
  argument 1 (type 'S4') cannot be handled by 'cat'

有没有办法解决?

我会很感激任何建议。谢谢!

马特·彼得森

4

1 回答 1

7

检查生成的S4对象的类结构str,提取相关变量以构建数据框并使用write.table/write.csv导出结果。例如,对于预测pred

R> library("ROCR")
R> data(ROCR.simple)
R> pred <- prediction(ROCR.simple$predictions, ROCR.simple$labels)
R> perf <- performance(pred, "fnr", "fpr")
R> str(pred)
Formal class 'prediction' [package "ROCR"] with 11 slots
  ..@ predictions:List of 1
  .. ..$ : num [1:200] 0.613 0.364 0.432 0.14 0.385 ...
  ..@ labels     :List of 1
  .. ..$ : Ord.factor w/ 2 levels "0"<"1": 2 2 1 1 1 2 2 2 2 1 ...
  ..@ cutoffs    :List of 1
  .. ..$ : num [1:201] Inf 0.991 0.985 0.985 0.983 ...
  ..@ fp         :List of 1
  .. ..$ : num [1:201] 0 0 0 0 1 1 2 3 3 3 ...
  ..@ tp         :List of 1
  .. ..$ : num [1:201] 0 1 2 3 3 4 4 4 5 6 ...
  ..@ tn         :List of 1
  .. ..$ : num [1:201] 107 107 107 107 106 106 105 104 104 104 ...
  ..@ fn         :List of 1
  .. ..$ : num [1:201] 93 92 91 90 90 89 89 89 88 87 ...
  ..@ n.pos      :List of 1
  .. ..$ : int 93
  ..@ n.neg      :List of 1
  .. ..$ : int 107
  ..@ n.pos.pred :List of 1
  .. ..$ : num [1:201] 0 1 2 3 4 5 6 7 8 9 ...
  ..@ n.neg.pred :List of 1
  .. ..$ : num [1:201] 200 199 198 197 196 195 194 193 192 191 ...

R> write.csv(data.frame(fp=pred@fp, fn=pred@fn), file="result_pred.csv")

和性能perf

R> str(perf)
Formal class 'performance' [package "ROCR"] with 6 slots
  ..@ x.name      : chr "False positive rate"
  ..@ y.name      : chr "False negative rate"
  ..@ alpha.name  : chr "Cutoff"
  ..@ x.values    :List of 1
  .. ..$ : num [1:201] 0 0 0 0 0.00935 ...
  ..@ y.values    :List of 1
  .. ..$ : num [1:201] 1 0.989 0.978 0.968 0.968 ...
  ..@ alpha.values:List of 1
  .. ..$ : num [1:201] Inf 0.991 0.985 0.985 0.983 ...

R> write.csv(data.frame(fpr=perf@x.values,
                        fnr=perf@y.values, 
                        alpha.values=perf@alpha.values), 
             file="result_perf.csv")
于 2009-11-11T12:31:52.933 回答