Question

我有以下数据集（除了Weight = W和Height = H以外，还包含25个左右的变量），都跨越了10年。

当前，它具有以下格式，并且否时间索引。

df <- structure(list(data = structure(1:4, .Label = c("Ind_1", "Ind_2", 
"Ind_3", "Ind_4"), class = "factor"), r1weight = c(56, 76, 87, 64
), r2weight = c(57, 75, 88, 66), r3weight = c(56, 76, 87, 65), r4weight = c(56L, 
73L, 85L, 63L), r5weight = c(55L, 77L, 84L, 65L), r1height = c(151L, 163L, 
173L, 153L), r2height = c(154L, 164L, NA, 154L), r3height = c(NA, 165L, NA, 
152L), r4height = c(153L, 162L, 172L, 154L), r5height = c(152,161,171,154)), class = 
"data.frame", row.names = c(NA, 
 -4L)) 

  data  r1w r2w r3w r4w r5w r1h r2h r3h r4h r5h
1 Ind_1  56  57  56  56  55 151 154  NA 153 152
2 Ind_2  76  75  76  73  77 163 164 165 162 161
3 Ind_3  87  88  87  85  84 173  NA  NA 172 171
4 Ind_4  64  66  65  63  65 153 154 152 154 154`

我需要添加时间变量并将其整形为长格式，希望能得到类似的东西。

dflong <- structure(list(time = structure(1:20, .Label = c("1", "2", 
     "3", "4", "5", "1","2","3","4","5", "1","2","3","4","5","1","2","3","4","5"), 
     class = "factor"), Ind = c(1,1,1,1,1,2,2,2,2,2,3,3,3,3,3,4,4,4,4,4), W = c(56,57,56,56,55,76,75,76,73,77,87,88,87,85,84,64,66,65,63,65),
     H = c(151,154,NA,153,152,163,164,165,162,161,173,NA,NA,172,171,153,154,152,154,154)), class = "data.frame", row.names = c(NA, -20L))

看起来

   time Ind  W   H
1     1   1 56 151
2     2   1 57 154
3     3   1 56  NA
4     4   1 56 153
5     5   1 55 152
6     1   2 76 163
7     2   2 75 164
8     3   2 76 165
9     4   2 73 162
10    5   2 77 161
11    1   3 87 173
12    2   3 88  NA
13    3   3 87  NA
14    4   3 85 172
15    5   3 84 171
16    1   4 64 153
17    2   4 66 154
18    3   4 65 152
19    4   4 63 154
20    5   4 65 154`

我尝试使用reshape2命令，到目前为止，我已经知道：

library(reshape2)
dflong <- melt(df,id.vars = c("idhhpn",r1w-r10w, r1h-r10h (help writing compactly),
     time(needs help constructing) )`

我不想写“ r1w，r2w，r3w”，但更像是r1weight-r10weight，因此我不必为所有25个变量编写所有10个时间实例。

到目前为止，我已经掌握了这一点

使用以下代码

melt <- melt(setDT(HRSdata), measure = patterns("idhhpn", "srhlt", "highbp", "diabetes", "cancer", "lungev", "heartp", "strokev", "psychev", "arth", "obese", "agey", "marpart", "male", "black", "hispan", "logass", "logdebt", "atotal", "debt", "lths", "hsorged", "somehs", "scorAA", "bachelor", "graduate", "works62", "works65", "momagey", "dadagey", "dadalive", "momalive", "vigact3", "smokesn"), 
     value.name = c("idhhpn", "srhlt", "highbp", "diabetes", "cancer", "lungev", "heartp", "strokev", "psychev", "arth", "obese", "agey", "marpart", "male", "black", "hispan", "logass", "logdebt", "atotal", "debt", "lths", "hsorged", "somehs", "scorAA", "bachelor", "graduate", "works62", "works65", "momagey", "dadagey", "dadalive", "momalive", "vigact3", "smokesn"), 
     variable.name = "time")[, 
      idhhpn := as.integer(sub("\\D+", "", HRSdata))][order(idhhpn)][, .(time, idhhpn, srhlt, highbp, diabetes, cancer, lungev, heartp, strokev, psychev, arth, obese, agey, marpart, male, black, hispan, logass, logdebt, atotal, debt, lths, hsorged, somehs, scorAA, bachelor, graduate, works62, works65, momagey, dadagey, dadalive, momalive, vigact3, smokesn        )]

Answer 1

使用tidyverse和gather的{{1}}方法是

spread

相同的解决方案，但修改后的变量名称为“ r [num] [varname]”（@ iod提供）：

library(tidyverse)

df %>%
  gather(time, ind, -data) %>%
  separate(time, into = c("indName", "time")) %>%
  spread(indName, ind)


#    data time  H  W
#1  Ind_1   1 151 56
#2  Ind_1   2 154 57
#3  Ind_1   3  NA 56
#4  Ind_1   4 153 56
#5  Ind_1   5 152 55
#6  Ind_2   1 163 76
#7  Ind_2   2 164 75
#8  Ind_2   3 165 76
#9  Ind_2   4 162 73
#10 Ind_2   5 161 77
#11 Ind_3   1 173 87
#12 Ind_3   2  NA 88
#13 Ind_3   3  NA 87
#14 Ind_3   4 172 85
#15 Ind_3   5 171 84
#16 Ind_4   1 153 64
#17 Ind_4   2 154 66
#18 Ind_4   3 152 65
#19 Ind_4   4 154 63
#20 Ind_4   5 154 65

Answer 2

您可以使用melt中的data.table函数，然后使用cbind-

setDT(df)
df <- cbind(setnames(melt(df)[grep("^H_",variable),],"value","H"),
            setnames(melt(df)[grep("^W_",variable),],"value","W"))
df <- df[,Ind:=gsub(".*_","",data)] ##cleaning Ind_
df <- df[, time:=1:.N, by = .(Ind)]
df <- df[,.(time,W,H,Ind)]

输出-

> df
    time  W   H Ind
 1:    1 56 151   1
 2:    1 76 163   2
 3:    1 87 173   3
 4:    1 64 153   4
 5:    2 57 154   1
 6:    2 75 164   2
 7:    2 88  NA   3
 8:    2 66 154   4
 9:    3 56  NA   1
10:    3 76 165   2
11:    3 87  NA   3
12:    3 65 152   4
13:    4 56 153   1
14:    4 73 162   2
15:    4 85 172   3
16:    4 63 154   4
17:    5 55 152   1
18:    5 77 161   2
19:    5 84 171   3
20:    5 65 154   4

Answer 3

使用Sub loop_test() Dim i As Long 'use Long, integer only goes up to c32k Dim ws_num As Long Dim starting_ws As Worksheet Set starting_ws = ActiveSheet ws_num = ThisWorkbook.Worksheets.Count Dim lastrow As Long Dim LastColumn As Long Dim StartCell As Range, r As Range Dim objTable As ListObject For i = 1 To ws_num With ThisWorkbook.Worksheets(i) 'don't need to activate Set StartCell = .Range("A1") lastrow = .Cells(.Rows.Count, StartCell.Column).End(xlUp).Row Set r = .Range(StartCell, .Cells(lastrow, LastColumn)) Set objTable = .ListObjects.Add(xlSrcRange, r, , xlYes) End With Next i End Sub的{{1}}选项可以使用data.table。在示例中，列名具有通用的measure/patterns作为'weight'，'height'，我们在melt参数中将其指定为将其转换为'long'格式，然后使用{{提取数字部分1}}创建“ Ind”

patterns

将HRS数据从宽格式重整为长格式并创建时间变量

3 个答案: