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我正在尝试将许多文件读入 R 以将每个文件的第 14 列中的数据评估为自己的直方图。在我尝试访问实际数据（列名除外）之前，它似乎可以正常工作。

该代码似乎在一定程度上起作用：

    test<-sapply(1:Num.Files,function(x){readLines(File.names[x])})
    head(test)
    data<-read.table(header=TRUE,text=test[1])
    head(data)
    names(data)[14]<-'column14'
    names(data)
    data$column14

但结果如下：

test<-sapply(1:Num.Files,function(x){readLines(File.names[x])}) #进入数据头(test) #这里可以看到信息都在[,1]
[ 1，]“目标cov av_cov 87A_cvg 87Ag 87Agr 87Agr 87A_gra 87A%_1 87A%_3 87A%_5 87A%_10 87A%_20 87A%_30 87A%_40\t87A%_50\t87A%_75\t87A%_100”[2]， "1:028 400\t0.42\t400\t0.42\t1\t1\t2\t41.8\t0.0\t0.0\t0.0\t0.0\t0.0\t0.0\ t0.0\t0.0\t0.0"
[3,] "1:296 400\t0.42\t400\t0.42\t1\t1\t2\t41.8\t0.0\t0.0\ t0.0\t0.0\t0.0\t0.0\t0.0\t0.0\t0.0"
[4,] "1:453 1646\t8.11\t1646\t8.11\t7\ t8\t13\t100.0\t100.0\t87.2\t32.0\t0.0\t0.0\t0.0\t0.0\t0.0\t0.0"
[5,] "1 :427 1646\t8.11\t1646\t8.11\t7\t8\t13\t100.0\t100.0\t87.2\t32.0\t0.0\t0.0\t0.0\t0。 0\t0.0\t0.0"
[6,] "1:736 5105\t29.68\t5105\t29.68\t14\t29\t48\t100.0\t100.0\t100.0\t86.0\t65.7\t49.4\ t35.5\t16.9\t0.0\t0.0"
data<-read.table(header=TRUE,text=test[1]) head(data) #所有正确的列名 [1] Targ cov av_cov X87A_cvg X87Ag X87Agr X87Agr.1 [8] X87A_gra X87A._1 X87A._3 X87A._5 X87A._10 X87A._20 X87A._30 [15] X87A._40 X87A._50 X87A._75 X87A._100 <0 行>（或 0 -length row.names) names(data)[14]<-'column14' names(data) #column 14 被改变了 [1] "Targ" "cov" "av_cov" "X87A_cvg" "X87Ag" "X87Agr"
[7] “X87Agr.1” “X87A_gra” “X87A._1” “X87A._3” “X87A._5” “X87A._10” [13] “X87A._20” “column14” “X87A._40” “X87A. _50" "X87A._75" "X87A._100" data$column14 #但逻辑列中没有任何内容(0)

我是否错误地使用了 read.table？有人有建议吗？

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2 回答 2

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why not just use read.table to begin with instead of the readLines business? try:

test <- lapply(File.names, read.table, header=TRUE, sep='\t')`.

Also, your variable test is a vector of rows. the first row is the header. you're passing only the first row test[1] as the argument to read.table(text=...).

using read.table(text=test) should work better. However, you should probably use some variation on the lapply call that I wrote above instead of all this extra monkeying around.

于 2012-08-14T19:31:01.373 回答
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是的，您错误地使用了 read.table，您只传递了标题行 ( text[1])。

我会read.table直接调用文件名。如果你坚持传递 的结果readLines，你可能想重新加入一切，比如paste(test, collapse="\n").

于 2012-08-14T19:17:41.710 回答