如何将一列数据框写入一个文件列表？

这里是我的虚拟数据集：如何将一列数据框写入一个文件列表？

dataset<-data.frame(a=c(1,2,3,4),b=c('a','b','c','d'), c=c("HI","DD","gg","ff")) 
g=list(c("a","b"),c(2,3,4), c(44,33,11,22),c("chr","ID","i","II")) 
dataset$l<-g 
dataset 

a b c    l 
1 1 a HI   a, b 
2 2 b DD  2, 3, 4 
3 3 c gg 44, 33, 11, 22 
4 4 d ff chr, ID, i, II 

> mode(dataset$l) 
[1] "list"

，当我尝试将数据集写入文件：

> write.table(dataset, "dataset.txt", quote=F, sep="\t") 
Error in write.table(x, file, nrow(x), p, rnames, sep, eol, na, dec, as.integer(quote), : 
    unimplemented type 'list' in 'EncodeElement'

我怎样才能解决这个问题？

来源

2012-11-24 Wayne Zhang

有保存输出作为表的特定需求？如果是这样，您希望/期望您的表格中的列表能够被处理或代表吗？ –

真正的数据是相当大的我最好把它们放在一个Excel数据表具有良好的外观。但他们中的一列，这使得我无法使用“wirte.table”或“wirte.csv” –

如果你想有它在Excel中工作，你需要做的事情列出来适应他们到不错的列表列。我下面贴的选项使用function'listFlatten' –

我能想到的几个选项，这取决于你想要达到的目的。

如果仅用于显示，那么您可能只需要capture.output()或sink();这些都不是回读成R非常方便：

capture.output(dataset, file="myfile.txt") 
### Result is a text file that looks like this: 
# a b c    l 
# 1 1 a HI   a, b 
# 2 2 b DD  2, 3, 4 
# 3 3 c gg 44, 33, 11, 22 
# 4 4 d ff chr, ID, i, II 
sink("myfile.txt") 
dataset 
sink() 
## Same result as `capture.output()` approach

如果你希望能够读取结果表返回到R（虽然没有保留的事实，列“l”是一个列表），你可以采取类似于@DWin所建议的方法。

在下面的代码中，dataset2[sapply...行标识哪些变量是列表并将它们连接成单个字符串。因此，它们变成简单的字符变量，允许您使用write.table()。

dataset2 <- dataset # make a copy just to be on the safe side 
dataset2[sapply(dataset2, is.list)] <- apply(dataset2[sapply(dataset2, is.list)], 
              1, function(x) 
               paste(unlist(x), 
                 sep=", ", collapse=", ")) 
str(dataset2) 
# 'data.frame': 4 obs. of 4 variables: 
# $ a: num 1 2 3 4 
# $ b: Factor w/ 4 levels "a","b","c","d": 1 2 3 4 
# $ c: Factor w/ 4 levels "DD","ff","gg",..: 4 1 3 2 
# $ l: chr "a, b" "2, 3, 4" "44, 33, 11, 22" "chr, ID, i, II" 
write.table(dataset2, "myfile.txt", quote=FALSE, sep="\t") 
# can be read back in with: dataset3 <- read.delim("myfile.txt")

来源

2012-11-25 02:51:34 A5C1D2H2I1M1N2O1R2T1

谢谢mrdwab！你的方法奏效！ –

有帮助的“应用”和“sapply”。这是一个奇妙的把戏！我在这里是新来的，所以我正在寻找要投票的地方。 –

我在哪里可以将答案标记为已接受的答案？ –

你可以使用dput来做到这一点。

dput(dataset, "dataset.txt")

来源

2012-11-24 21:05:11

你也可以使用保存（）

save(dataset, file="dataset.RData")

来源

2012-11-24 23:59:08

输出保存不可读。转储或输入的输出是ASCII，对于理解R对象结构的人是可读的，但我猜你希望它更传统地排列。

> apply(dataset, 1, function(x) paste(x, sep=",", collapse=",")) 
[1] "1,a,HI,c(\"a\", \"b\")"     
[2] "2,b,DD,c(2, 3, 4)"      
[3] "3,c,gg,c(44, 33, 11, 22)"     
[4] "4,d,ff,c(\"chr\", \"ID\", \"i\", \"II\")"

反斜杠不会出现在文本文件输出：

writeLines(con="test.txt", apply(dataset, 1, function(x) paste(x, sep=",", collapse=","))) 
#-------output----- 
1,a,HI,c("a", "b") 
2,b,DD,c(2, 3, 4) 
3,c,gg,c(44, 33, 11, 22) 
4,d,ff,c("chr", "ID", "i", "II")

来源

2012-11-25 00:44:20

如果要求一个是保存为Excel等格式，这可能帮助：

writableTable <- tableFlatten(dataset, filler="") 
    # a b c l.01 l.02 l.03 l.04 
    # 1 a HI a b   
    # 2 b DD 2 3 4  
    # 3 c gg 44 33 11 22 
    # 4 d ff chr ID i II 

    write.csv(writableTable, "myFile.csv")

tableFlatten使用功能listFlatten其中，顾名思义，采用嵌套列表并将它们弄平。但是，如果列表中的元素大小不同，则会添加填充符（它可以是NA s，空格或任何其他用户定义的选项）

其代码如下。

tableFlatten <- function(tableWithLists, filler="") { 
# takes as input a table with lists and returns a flat table 
# empty spots in lists are filled with value of `filler` 
# 
# depends on: listFlatten(.), findGroupRanges(.), fw0(.) 

    # index which columns are lists 
    listCols <- sapply(tableWithLists, is.list) 

    tableWithLists[listCols] 
    tableWithLists[!listCols] 

    # flatten lists into table 
    flattened <- sapply(tableWithLists[listCols], listFlatten, filler=filler, simplify=FALSE) 

    # fix names 
    for (i in 1:length(flattened)) colnames(flattened[[i]]) <- fw0(ncol(flattened[[i]]), 2) 

    # REASSEMBLE, IN ORDER 
    # find pivot point counts 
    pivots <- sapply(findGroupRanges(listCols), length) 

    #index markers 
    indNonList <- indList <- 1 

    # nonListGrp <- (0:(length(pivots)/2)) * 2 + 1 
    # ListGrp <- (1:(length(pivots)/2)) * 2 
    final <- data.frame(row.names=row.names(tableWithLists)) 
    for (i in 1:length(pivots)) { 
     if(i %% 2 == 1) { 
      final <- cbind(final, 
         tableWithLists[!listCols][indNonList:((indNonList<-indNonList+pivots[[i]])-1)] 
         ) 
     } else { 
      final <- cbind(final, 
         flattened[indList:((indList<-indList+pivots[[i]])-1)] 
         ) 
     } 
    } 

    return(final) 
} 


#===================================== 

listFlatten <- function(obj, filler=NA) { 
## Flattens obj like rbind, but if elements are of different length, plugs in value filler 

    # Initialize Vars 
    bind <- FALSE 

    # IF ALL ELEMENTS ARE MATRIX-LIKE OR VECTORS, MAKE SURE SAME NUMBER OF COLUMNS 
    matLike <- sapply(obj, function(x) !is.null(dim(x))) 
    vecLike <- sapply(obj, is.vector) 

    # If all matrix-like. 
    if (all(matLike)) { 
    maxLng <- max(sapply(obj[matLike], ncol)) 
    obj[matLike] <- lapply(obj[matLike], function(x) t(apply(x, 1, c, rep(filler, maxLng - ncol(x))))) 
    bind <- TRUE 

    # If all vector-like 
    } else if (all(vecLike)) { 
    maxLng <- max(sapply(obj[vecLike], length)) 
    obj[vecLike] <- lapply(obj[vecLike], function(x) c(x, rep(filler, maxLng - length(x)))) 
    bind <- TRUE 

    # If all are either matrix- or vector-like 
    } else if (all(matLike & vecLike)) { 

    maxLng <- max(sapply(obj[matLike], ncol), sapply(obj[vecLike], length)) 

    # Add in filler's as needed 
    obj[matLike] <- 
     lapply(obj[matLike], function(x) t(apply(x, 1, c, rep(filler, maxLng - ncol(x))))) 
    obj[vecLike] <- 
     lapply(obj[vecLike], function(x) c(x, rep(filler, maxLng - length(x)))) 
    bind <- TRUE 
    } 

    # If processed and ready to be returned, then just clean it up 
    if(bind) { 
    ret <- (do.call(rbind, obj)) 
    colnames(ret) <- paste0("L", fw0(1:ncol(ret), digs=2)) 
    return(ret) 
    } 

    # Otherwise, if obj is sitll a list, continue recursively  
    if (is.list(obj)) { 
     return(lapply(obj, listFlatten)) 
    } 

    # If none of the above, return an error. 
    stop("Unknown object type") 
} 
#-------------------------------------------- 

findGroupRanges <- function(booleanVec) { 
# returns list of indexes indicating a series of identical values 
    pivots <- which(sapply(2:length(booleanVec), function(i) booleanVec[[i]] != booleanVec[[i-1]])) 

    ### THIS ISNT NEEDED... 
    # if (identical(pivots, numeric(0))) 
    # pivots <- length(booleanVec) 

    pivots <- c(0, pivots, length(booleanVec)) 
    lapply(seq(2, length(pivots)), function(i) 
    seq(pivots[i-1]+1, pivots[i]) 
) 
} 

#-------------------------------------------- 



fw0 <- function(num, digs=NULL, mkSeq=TRUE) { 
    ## formats digits with leading 0's. 
    ## num should be an integer or range of integers. 
    ## if mkSeq=T, then an num of length 1 will be expanded to seq(1, num). 

    # TODO 1: put more error check 
    if (is.list(num)) 
    lapply(num, fw0) 

    if (!is.vector(num)) { 
    stop("num should be integer or vector") 
    } 

    # convert strings to numbers 
    num <- as.numeric(num) 

    # If num is a single number and mkSeq is T, expand to seq(1, num) 
    if(mkSeq && !length(num)>1) 
    num <- (1:num) 

    # number of digits is that of largest number or digs, whichever is max 
    digs <- max(nchar(max(abs(num))), digs) 

    # if there are a mix of neg & pos numbers, add a space for pos numbs 
    posSpace <- ifelse(sign(max(num)) != sign(min(num)), " ", "") 

    # return: paste appropriate 0's and preface neg/pos mark 
    sapply(num, function(x) ifelse(x<0, 
    paste0("-", paste0(rep(0, max(0, digs-nchar(abs(x)))), collapse=""), abs(x)), 
    paste0(posSpace, paste0(rep(0, max(0, digs-nchar(abs(x)))), collapse=""), x) 
    )) 
} 

#-----------------------------------------------

来源

2012-11-25 22:01:23

有趣的功能（+1）。我将不得不更仔细地查看它，看看你在做什么。但事实上我有工作[连接的字符串（https://github.com/mrdwab/2657-R-Functions/blob/master/docs/Part-1-Concat-Split.md）和分裂他们相关的功能草案分成不同的栏目（因为实际上似乎发生了很多 - 谷歌表格等常常在同一个单元格中给多个回答问题提供答案）。 – A5C1D2H2I1M1N2O1R2T1

由@Ananda提供的answer是优秀的，但是，我遇到了一个问题，当我有一个数据帧时都列出了两列。

dataset<-data.frame(a=c(1,2,3,4),b=c('a','b','c','d'), c=c("HI","DD","gg","ff")) 
g=list(c("a","b"),c(2,3,4), c(44,33,11,22),c("chr","ID","i","II")) 
dataset$l<-g 
dataset$l2<-g 
dataset 

    a b c    l    l2 
1 1 a HI   a, b   a, b 
2 2 b DD  2, 3, 4  2, 3, 4 
3 3 c gg 44, 33, 11, 22 44, 33, 11, 22 
4 4 d ff chr, ID, i, II chr, ID, i, II

使用原始答案，两个列表都包含两列的连接内容。

a b c        l        l2 
1 1 a HI      a, b, a, b      a, b, a, b 
2 2 b DD    2, 3, 4, 2, 3, 4    2, 3, 4, 2, 3, 4 
3 3 c gg 44, 33, 11, 22, 44, 33, 11, 22 44, 33, 11, 22, 44, 33, 11, 22 
4 4 d ff chr, ID, i, II, chr, ID, i, II chr, ID, i, II, chr, ID, i, II

相反，试试这个修改后的版本：

dataset2 <- dataset # make a copy just to be on the safe side 
dataset2[sapply(dataset2, is.list)] <- 
    sapply(dataset2[sapply(dataset2, is.list)], 
      function(x)sapply(x, function(y) paste(unlist(y),collapse=", "))) 
dataset2 

    a b c    l    l2 
1 1 a HI   a, b   a, b 
2 2 b DD  2, 3, 4  2, 3, 4 
3 3 c gg 44, 33, 11, 22 44, 33, 11, 22 
4 4 d ff chr, ID, i, II chr, ID, i, II

来源

2014-07-17 17:27:04 Lance

-1

我偶然发现这一点，同时也有很多伟大的答案，我结束了别人做一些事情。为后人分享。

library(dplyr) 

flatten_list = function(x){ 
    if (typeof(x) != "list") { 
     return(x) 
    } 
    sapply(x, function(y) paste(y, collapse = " | ")) 
} 

data %>% 
    mutate_each(funs(flatten_list)) -> 
    write_csv("data.csv")

来源

2016-05-28 17:31:09

如何将一列数据框写入一个文件列表？

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