将数据帧导出到R中的文本文件时出错

Ako*_*kos 6 export r

我试图将R中的数据帧写入文本文件,但它返回到以下错误:

Error in if (inherits(X[[j]], "data.frame") && ncol(xj) > 1L)
             X[[j]] <- as.matrix(X[[j]]) : 
  missing value where TRUE/FALSE needed
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我使用以下命令进行导出:

write.table(df, file ='dfname.txt', sep='\t' )
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我不知道问题可能源于什么.至于"缺少需要TRUE/FALSE的数据",我只有一列包含TRUE/FALSE值,并且没有这些值丢失.

数据帧的内容:

> str(df)
'data.frame':   776 obs. of  15 variables:
 $ Age         : Factor w/ 4 levels "","A","J","SA": 2 2 2 2 2 2 2 2 2 2 ...
 $ Sex         : Factor w/ 2 levels "F","M": 1 1 1 1 2 2 2 2 2 2 ...
 $ Rep         : Factor w/ 11 levels "L","NR","NRF",..: 1 1 4 4 2 2 2 2 2 2 ...
 $ FA          : num  61.5 62.5 60.5 61 59.5 59.5 59.1 59.2 59.8 59.9 ...
 $ Mass        : num  20 19 16.5 17.5 NA 14 NA 23 19 18.5 ...
 $ Vir1        : num  999 999 999 999 999 999 999 999 999 999 ...
 $ Vir2        : num  999 999 999 999 999 999 999 999 999 999 ...
 $ Vir3        : num  40 999 999 999 999 999 999 999 999 999 ...
 $ Location    : Factor w/ 4 levels "Loc1",..: 4 4 4 4 4 4 2 2 2 2 ...
 $ Site        : Factor w/ 6 levels "A","B","C",..: 5 5 5 5 5 5 3 3 3 3 ...
 $ Date        : Date, format: "2010-08-30" "2010-08-30" ...
 $ Record      : int  35 34 39 49 69 38 145 112 125 140 ...
 $ SampleID    : Factor w/ 776 levels "AT1-A-F1","AT1-A-F10",..: 525 524 527 528
                                                                 529 526 111 78
                                                                 88 110 ...
 $ Vir1Inc     : logi  FALSE FALSE FALSE FALSE FALSE FALSE ...
 $ Month       :'data.frame':   776 obs. of  2 variables:
  ..$ Dates: Date, format: "2010-08-30" "2010-08-30" ...
  ..$ Month: Factor w/ 19 levels "Apr-2011","Aug-2010",..: 2 2 2 2 
                                                           2 2 18 18 18 18 ...
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我希望我已经给出了足够/正确的信息......

非常感谢,希瑟

ags*_*udy 5

重现错误的示例.我创建了一个嵌套的data.frame:

Month=data.frame(Dates= as.Date("2003-02-01") + 1:15,
                 Month=gl(12,2,15))
dd <- data.frame(Age=1:15)
dd$Month <- Month
str(dd)
'data.frame':   15 obs. of  2 variables:
 $ Age  : int  1 2 3 4 5 6 7 8 9 10 ...
 $ Month:'data.frame':  15 obs. of  2 variables:
  ..$ Dates: Date, format: "2003-02-02" "2003-02-03" "2003-02-04" ...
  ..$ Month: Factor w/ 12 levels "1","2","3","4",..: 1 1 2 2 3 3 4 4 5 5 ...
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不,我尝试保存它,我重现错误:

write.table(dd)
Error in if (inherits(X[[j]], "data.frame") && ncol(xj) > 1L) 
    X[[j]] <- as.matrix(X[[j]]) :  missing value where TRUE/FALSE needed
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没有反复,删除嵌套的一个选项data.frame:

write.table(data.frame(subset(dd,select=-c(Month)),unclass(dd$Month)))
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Flo*_*Flo 5

agstudy 的解决方案提供了一个很好的快速修复,但有一个 简单的替代/通用解决方案,您不必在 data.frame 中指定嵌套的元素:

以下位只是从agstudy 的解决方案中复制以获得嵌套的 data.frame dd

Month=data.frame(Dates= as.Date("2003-02-01") + 1:15,
                 Month=gl(12,2,15))
dd <- data.frame(Age=1:15)
dd$Month <- Month
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您可以使用akhilsbehl 的 LinearizeNestedList()函数(mrdwab 在此处提供)来展平(或线性化)嵌套级别:

library(devtools)
source_gist(4205477) #loads the function

ddf <- LinearizeNestedList(dd, LinearizeDataFrames = TRUE) 
# ddf is now a list with two elements (Age and Month)

ddf <- LinearizeNestedList(ddf, LinearizeDataFrames = TRUE) 
# ddf is now a list with 3 elements (Age, `Month/Dates` and `Month/Month`)

ddf <- as.data.frame.list(ddf)
# transforms the flattened/linearized list into a data.frame
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ddf现在是一个没有嵌套的 data.frame。但是,它的列名仍然反映了嵌套结构:

names(ddf)
[1] "Age"         "Month.Dates" "Month.Month"
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如果你想改变这个(在这种情况下,它似乎是多余的已经Month.之前被写入Dates,例如),您可以使用gsub一些正则表达式,我从萨沙Epskamp复制到删除前的列名的所有文本.

names(ddf) <- gsub(".*\\.","",names(ddf))  
names(ddf)
[1] "Age"   "Dates" "Month"
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现在唯一剩下的就是像往常一样导出 data.frame:

write.table(ddf, file="test.txt")
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