Question

我有以下列表清单：

my_lol <- structure(list(coolfactor_score = list(structure(c(0.164477631065473, 
0.198253819406019, 0.396414447052519, 0.133118603987442, 0.107735498488546
), .Names = c("B", "Mac", "NK", "Neu", "Stro")), structure(c(0.186215537135912, 
0.18408529174803, 0.375349920115798, 0.247664923324821, 0.006684327675438
), .Names = c("B", "Mac", "NK", "Neu", "Stro"))), sr_crt = list(
    structure(list(crt = 0.133118603987442, sr = 0.407076876403305), .Names = c("crt", 
    "sr")), structure(list(crt = 0.18408529174803, sr = 0.0829181742326453), .Names = c("crt", 
    "sr"))), sample_names = c("Sample1", "Sample2")), .Names = c("coolfactor_score", 
"sr_crt", "sample_names"))

看起来像这样：

> my_lol
$coolfactor_score
$coolfactor_score[[1]]
        B       Mac        NK       Neu      Stro 
0.1644776 0.1982538 0.3964144 0.1331186 0.1077355 

$coolfactor_score[[2]]
          B         Mac          NK         Neu        Stro 
0.186215537 0.184085292 0.375349920 0.247664923 0.006684328 


$sr_crt
$sr_crt[[1]]
$sr_crt[[1]]$crt
[1] 0.1331186

$sr_crt[[1]]$sr
[1] 0.4070769


$sr_crt[[2]]
$sr_crt[[2]]$crt
[1] 0.1840853

$sr_crt[[2]]$sr
[1] 0.08291817



$sample_names
[1] "Sample1" "Sample2"
# Note that the number of samples can be more than 2 and cell type more than 5.

如何将其整理到此数据框（tibbles）

CellType    Sample    CoolFactorScore  SR            CRT
B           Sample1   0.1644776        0.4070769     0.1331186
Mac         Sample1   0.1982538        0.4070769     0.1331186
NK          Sample1   0.3964144        0.4070769     0.1331186
Neu         Sample1   0.1331186        0.4070769     0.1331186
Stro        Sample1   0.1077355        0.4070769     0.1331186
B           Sample2   0.186215537      0.08291817    0.1840853
Mac         Sample2   0.184085292      0.08291817    0.1840853
NK          Sample2   0.375349920      0.08291817    0.1840853
Neu         Sample2   0.247664923      0.08291817    0.1840853
Stro        Sample2   0.006684328      0.08291817    0.1840853

Answer 1

使用基础R的一种方式：

mylist <- lapply(1:2, function(i) {
  #this is the important bit where you extract the corresponding elements
  #of sample 1 first and sample 2 second.
  df <- data.frame(lapply(my_lol, '[', i))
  names(df) <- c('CoolFactorScore', 'CRT', 'SR', 'Sample')
  df$CellType <- rownames(df)
  row.names(df) <- NULL
  df
})

do.call(rbind, mylist)

输出：

  CoolFactorScore       CRT         SR  Sample CellType
1      0.164477631 0.1331186 0.40707688 Sample1        B
2      0.198253819 0.1331186 0.40707688 Sample1      Mac
3      0.396414447 0.1331186 0.40707688 Sample1       NK
4      0.133118604 0.1331186 0.40707688 Sample1      Neu
5      0.107735498 0.1331186 0.40707688 Sample1     Stro
6      0.186215537 0.1840853 0.08291817 Sample2        B
7      0.184085292 0.1840853 0.08291817 Sample2      Mac
8      0.375349920 0.1840853 0.08291817 Sample2       NK
9      0.247664923 0.1840853 0.08291817 Sample2      Neu
10     0.006684328 0.1840853 0.08291817 Sample2     Stro

Answer 2

这是一个不那么优雅的方法：

int <- lapply(1:2, function(x) do.call(data.frame, 
              c(list(CoolFactorScore=my_lol[[1]][[x]]), 
                my_lol[[2]][[x]], 
                list(Sample=my_lol[[3]][[x]])))) 
do.call(rbind, int)

      CoolFactorScore       crt         sr  Sample
B         0.164477631 0.1331186 0.40707688 Sample1
Mac       0.198253819 0.1331186 0.40707688 Sample1
NK        0.396414447 0.1331186 0.40707688 Sample1
Neu       0.133118604 0.1331186 0.40707688 Sample1
Stro      0.107735498 0.1331186 0.40707688 Sample1
B1        0.186215537 0.1840853 0.08291817 Sample2
Mac1      0.184085292 0.1840853 0.08291817 Sample2
NK1       0.375349920 0.1840853 0.08291817 Sample2
Neu1      0.247664923 0.1840853 0.08291817 Sample2
Stro1     0.006684328 0.1840853 0.08291817 Sample2

Answer 3

这是一个使用data.table包函数的无循环解决方案。

library(data.table)

步骤1：打开列表

unlist(my_lol) -> tmp1

第2步：转置并将其转换为data.table
这样，您将获得可以从原始数据组成的最宽的表。它应该根据要求转换（在后续步骤中）到长表。

as.data.table(t(tmp1)) -> tmp2

第3步：＆＃39; sample_names1＆＃39;和＆＃39; sample_names2＆＃39;有必要转变为样本＆＃39;手动。
如果您想要推广到多个sample_names值，那么您应该根据可能值的语法修改此步骤。（此版本适用于以下sample_names值语法：＆＃39; Sample1＆＃39;，＆＃39; Sample2＆＃39;，＆＃39; Sample3＆＃39;等等。）

names(tmp2) <- gsub('sample_names\\d+', 'Sample', names(tmp2))

步骤4：根据tmp2表的字段名称

创建度量字段名称

measure <- unique(names(tmp2))

步骤5：从宽表（tmp2）

创建更长的表（tmp3）

tmp3 <- melt(tmp2, 
             measure.vars = patterns(measure), 
             value.name = measure)

步骤6：根据请求重命名列

names(tmp3) <- gsub('coolfactor_score.', '', names(tmp3))
names(tmp3) <- gsub('sr_crt.', '', names(tmp3))
setnames(tmp3, 'crt', 'CRT')
setnames(tmp3, 'sr', 'SR')

步骤7：从tmp3中创建更长的表（mylist）

mylist <- melt(tmp3,
               id.vars = c('Sample',
                           'CRT',
                           'SR'),
               measure.vars = c('B', 
                                'Mac',  
                                'NK',   
                                'Neu',
                                'Stro'),
               value.name = 'CoolFactorScore',
               variable.name = 'CellType')

步骤8：根据请求重新排序列

setcolorder(mylist, c('CellType', 'Sample', 'CoolFactorScore', 'SR', 'CRT'))

步骤9：根据请求重新排序行

mylist <- mylist[order(Sample, CellType)]

如何将列表列表转换为R中的整洁tibble或data.frame

3 个答案: