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我有一个需要更改为长格式的宽格式 csv 文件。我刚刚给出了前 3 行。

CODEA   C45 ragek   ra80    ra98    ... Obese14 Overweight14 Obese21 hibp14 hibp21 Overweight21
1   1   NA  3   4   1   NA  NA  NA  NA  NA  NA      NA  NA
2   3   2   3   3   1   0   0   0   0   1   0   0   0
3   4   2   3   6   1   NA  NA  NA  NA  NA  NA  NA  NA

这样下去。Obese 14 (Yes/No); Overweight(yes/no)等等

> names(Copy.of.BP_2)

 [1] "CODEA"  "C45"                     "ragek"                   "ra80"              
 [5] "ra98"   "CBCLAggressionAt1410"    "CBCLInternalisingAt1410" "Obese14"              
 [9] "Overweight14"   "Overweight21"    "Obese21"                 "hibp14"               
[13] "hibp21"

它有 6898 个观测值和 13 个变量

我正在尝试以堆叠格式组织这些数据；我认为以下一个将是一个不错的选择。我不确定如何组合obese和分类，overweight因为原始长版本有obese14、overweight14和4 个不同的类别。obese 21overweight21

CODEA ...  time         Obese        Overweight      HiBP

           14 
           21
           14
           21 ... etc

我给出的语法为：

BP.stack1=reshape(Copy.of.BP_2, 
   timevar="time",direction="long",
   varying=list(names(Copy.of.BP_2[8:13]),
   v.names="Obese","Overweight","HiBP",idvar=c("CODEA")

它似乎不起作用，它给出一个+标志并等待进一步的命令。

我应该使用melt和cast吗？我阅读了reshape包装手册，但无法理解。

编辑：问题重组

4

1 回答 1

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坚持使用基础 R reshape()，请尝试以下操作。

我认为我已经使用以下内容重新创建了您的示例数据：

Copy.of.BP_2 <- 
structure(list(CODEA = c(1, 3, 4), C45 = c(NA, 2, 2), ragek = c(3, 
3, 3), ra80 = c(4, 3, 6), ra98 = c(1, 1, 1), CBCLAggressionAt1410 = c(NA, 
0, NA), CBCLInternalisingAt1410 = c(NA, 0, NA), Obese14 = c(NA, 
0, NA), Overweight14 = c(NA, 0, NA), Overweight21 = c(NA, 1, 
NA), Obese21 = c(NA, 0, NA), hibp14 = c(NA, 0, NA), hibp21 = c(NA, 
0, NA)), .Names = c("CODEA", "C45", "ragek", "ra80", "ra98", 
"CBCLAggressionAt1410", "CBCLInternalisingAt1410", "Obese14", 
"Overweight14", "Overweight21", "Obese21", "hibp14", "hibp21"
), row.names = c(NA, -3L), class = "data.frame")

Copy.of.BP_2
#   CODEA C45 ragek ra80 ra98 CBCLAggressionAt1410 CBCLInternalisingAt1410
# 1     1  NA     3    4    1                   NA                      NA
# 2     3   2     3    3    1                    0                       0
# 3     4   2     3    6    1                   NA                      NA
#   Obese14 Overweight14 Overweight21 Obese21 hibp14 hibp21
# 1      NA           NA           NA      NA     NA     NA
# 2       0            0            1       0      0      0
# 3      NA           NA           NA      NA     NA     NA

首先，为方便起见，让我们创建一个度量变量的向量——我们想要从宽格式到长格式“堆叠”的变量。

measurevars <- names(Copy.of.BP_2)[grepl("Obese|Overweight|hibp", 
                                         names(Copy.of.BP_2))]

接下来，使用reshape()，指定方向、标识变量以及哪些变量随时间“变化”（measurevars，从上面）。

BP_2_long <- reshape(Copy.of.BP_2, direction = "long", idvar="CODEA",
                     varying = measurevars, sep = "")
BP_2_long
#      CODEA C45 ragek ra80 ra98 CBCLAggressionAt1410 CBCLInternalisingAt1410
# 1.14     1  NA     3    4    1                   NA                      NA
# 3.14     3   2     3    3    1                    0                       0
# 4.14     4   2     3    6    1                   NA                      NA
# 1.21     1  NA     3    4    1                   NA                      NA
# 3.21     3   2     3    3    1                    0                       0
# 4.21     4   2     3    6    1                   NA                      NA
#      time Obese Overweight hibp
# 1.14   14    NA         NA   NA
# 3.14   14     0          0    0
# 4.14   14    NA         NA   NA
# 1.21   21    NA         NA   NA
# 3.21   21     0          1    0
# 4.21   21    NA         NA   NA

如果您只对 id 列和 measure 列感兴趣，您还可以在命令中添加一个drop参数：reshape()

BP_2_long_2 <- reshape(
  Copy.of.BP_2, direction = "long", idvar="CODEA",
  varying = measurevars, sep = "",
  drop = !names(Copy.of.BP_2) %in% c(measurevars, "CODEA"))
BP_2_long_2
#      CODEA time Obese Overweight hibp
# 1.14     1   14    NA         NA   NA
# 3.14     3   14     0          0    0
# 4.14     4   14    NA         NA   NA
# 1.21     1   21    NA         NA   NA
# 3.21     3   21     0          1    0
# 4.21     4   21    NA         NA   NA

更新：为什么您的代码不起作用

这是您尝试过的内容的逐个参数细分，并附有有关如何尝试修复它的评论。

BP.stack1 = 
reshape(Copy.of.BP_2,                    # Fine
timevar="time",                          # Fine
direction="long",                        # Fine
varying=list(names(Copy.of.BP_2)[8:13]), # Wrong. Use "varying = 8:13" instead
v.names="Obese","Overweight","HiBP",     # Wrong. This needs to be in c()
idvar=c("CODEA")                         # Almost... missing your closing ")"

因此，要获得完整的工作命令：

BP.stack1 = reshape(
  Copy.of.BP_2, 
  timevar="time", 
  direction="long", 
  varying=8:13, 
  v.names=c("Obese", "Overweight", "HiBP"),
  idvar=c("CODEA"))

我通常尽量不要过多地依赖列号，因为与要重命名的列相比，这些列更有可能被重新排列。因此，我使用grepl()根据某种模式匹配名称。

于 2012-12-27T16:43:27.543 回答