Question

我有一个如下数据框，

  id  sex age trt.1 response.1 trt.2 response.2
1  1 <NA>  NA     A          1     B          1
2  2 <NA>  NA     A          1     B          1
3  3 <NA>  NA     A          1     B          1
4  4    M  28     A          1     B          1
5  5    F  39     A          1     B          1
6  6    M  47     A          1     B          1

我想将其更改为

  id  sex age times response
1  1 <NA>  NA  A      1
2  1 <NA>  NA  B      1
3  2 <NA>  NA  A      1
4  2 <NA>  NA  B      1
.
.
.
.

我尝试了以下内容，

reshape(merged, idvar = "id", varying = list(4:7), v.names="response", times=c("A","B"), direction="long")

但是我收到了这个错误，

Error in reshapeLong(data, idvar = idvar, timevar = timevar, varying = varying,  : 
  'times' is wrong length

我尝试了时间= C（“A”，“B”，“A”，“B”）那时我得到输出，但原始数据帧是70行，输出应该是140行，但是当我用这个我得到280行是错的。任何人都可以帮我解决我的错误吗？

Answer 1

这是你期待的吗？您只需要将Dim UserInput UserInput = InputChecker ' You can reuse the accepted user input If Length = 8 Then Wscript.Echo "User input: """ & UserInput & """" Function InputChecker Dim Input, Length Do Input = InputBox("Enter your name","Enter your name (must be 8 characters)") If Input = False Then Exit Do Input = Trim(Input) ' Remove spaces at front and back Length = Len(Input) If Length <> 8 Then If Msgbox("Input must be 8 characters, not " & Len(Input) & "!",5+48,"INPUT LENGTH ERROR") = 2 Then Input = "" ' Reset it to empty string Exit Do End If End If Loop Until Length = 8 InputChecker = Input End Function更改为varying = list(4:7)，因为您实际上正在同时融化两件。

varying = list(c(4,6), c(5,7))

Answer 2

重塑非常棒，但我真的不明白为什么你需要在这种情况下使用它。

text = "id  sex age trt.1 response.1 trt.2 response.2
1  1 <NA>  NA     A          1     B          1
2  2 <NA>  NA     A          1     B          1
3  3 <NA>  NA     A          1     B          1
4  4    M  28     A          1     B          1
5  5    F  39     A          1     B          1
6  6    M  47     A          1     B          1"

data <- read.table(text = text)

result1 <- data[, c("id", "sex", "age", "trt.1", "response.1")]
result2 <- data[, c("id", "sex", "age", "trt.2", "response.2")]

names(result1) <- c("id", "sex", "age", "trt", "response")
names(result2) <- c("id", "sex", "age", "trt", "response")

result <- rbind(result1, result2)

这是输出：

id  sex age trt response
1   1 <NA>  NA   A        1
2   2 <NA>  NA   A        1
3   3 <NA>  NA   A        1
4   4    M  28   A        1
5   5    F  39   A        1
6   6    M  47   A        1
11  1 <NA>  NA   B        1
...

Answer 3

这是最简单的方法：

library(dplyr)
library(tidyr)

data %>%
  gather(variable, value,
         trt.1:response.2) %>%
  separate(variable, c("variable", "number")) %>%
  spread(variable, value)

Answer 4

我们可以使用melt中的library(data.table) measure，其中pattern个参数可以带有多个library(data.table)#v1.9.6+ melt(setDT(df2), measure=patterns('^trt', 'response'), value.name=c('times', 'response')) # id sex age variable times response # 1: 1 <NA> NA 1 A 1 # 2: 2 <NA> NA 1 A 1 # 3: 3 <NA> NA 1 A 1 # 4: 4 M 28 1 A 1 # 5: 5 F 39 1 A 1 # 6: 6 M 47 1 A 1 # 7: 1 <NA> NA 2 B 1 # 8: 2 <NA> NA 2 B 1 # 9: 3 <NA> NA 2 B 1 #10: 4 M 28 2 B 1 #11: 5 F 39 2 B 1 #12: 6 M 47 2 B 1

Width:50%;

在R中使用重塑功能时出错

4 个答案: