如何将多个列转换为R中的各个行

I have a data frame in R that has many rows (over 3000) with F0 (fundamental frequency) tracks of an utterance in it. The rows have the following information in them: speaker ID, group #, repetition #, accent type, sex, and then 50 columns of F0 points. The data looks like this:

我在R中有一个数据帧，它有很多行（超过3000），其中有一个话语的F0（基频）轨道。行中包含以下信息：说话者ID，组＃，重复＃，重音类型，性别，然后50列F0点。数据如下所示：

Speaker Sex Group Repetition Accent    Word         1         2         3        4
    105   M     1          1      N AILMENT 102.31030 102.31030 102.31030 102.31127 
    105   M     1          1      N COLLEGE 111.80641 111.80313 111.68612 111.36020
    105   M     1          1      N  FATHER 124.06655 124.06655 124.06655 124.06655 

But instead of only going to X4, it has 50 points per row, so I have a 3562x56 data frame. I want to change it so each column of data in the F0 track (so after word, from 1:50) gets its own column, with the associated column number as another row. I want to keep all of the information in the first six columns with each data point as well, so it would look like this:

但是它不是仅仅转到X4，而是每行50个点，所以我有一个3562x56的数据帧。我想改变它，因此F0轨道中的每一列数据（因此，从1:50开始）得到它自己的列，相关的列号作为另一行。我想在每个数据点的前六列中保留所有信息，所以它看起来像这样：

Speaker Sex Group Repetition Accent    Word       Num        F0
    105   M     1          1      N AILMENT         1 102.31030
    105   M     1          1      N AILMENT         2 102.31030
    105   M     1          1      N AILMENT         3 102.31030
    105   M     1          1      N AILMENT         4 102.31127
    ...
    105   M     1          1      N COLLEGE         1 111.80641 
    105   M     1          1      N COLLEGE         1 111.80313 
    105   M     1          1      N COLLEGE         1 111.68612 
    105   M     1          1      N COLLEGE         1 111.36020 
    ...

The code I tried to use, while tedious, is as follows:

我尝试使用的代码虽然繁琐，但如下所示：

x = 1
for (i in 1:dim(normrangef0)[1])
{
     for (j in 1:50)
         {
             norm.all$Speaker[x] <- normrangef0$Speaker[i]
             norm.all$Sex[x] <- normrangef0$Sex[i]
             norm.all$Group[x] <- normrangef0$Group[i]
             norm.all$Repetition[x] <- normrangef0$Repetition[i]
             norm.all$Word[x] <- normrangef0$Word[i]
             norm.all$Accent[x] <- normrangef0$Accent[i]
             norm.all$Time[x] <- j
             norm.all$F0[x] <- normrangef0[i,j+6]
             x = x+1    
         }
}

However, when I do this with norm.all as a NULL object (just defined by norm.all = c() ), I end up with a list of over 200k items, many of which are NAs. When I define norm.all as a data frame (either an empty one or one of all 0s, in the 178100x8 data frame, I get an error:

但是，当我使用norm.all作为NULL对象（仅由norm.all = c（）定义）时，我最终得到了一个超过20万项的列表，其中许多是NA。当我将norm.all定义为数据帧（空数或全0中的一个，在178100x8数据帧中，我得到一个错误：

Error in `$<-.data.frame`(`*tmp*`, "Speaker", value = 105L) : replacement has 1 row, data has 0

Is my code just totally off? Is there another way to do this?

我的代码完全关闭了吗？还有另一种方法吗？

2 个解决方案

#1

Use melt from "reshape2"

使用来自“reshape2”的融化

library(reshape2)
melt(mydf, id.vars=c("Speaker", "Sex", "Group", "Repetition", "Accent", "Word"))
#    Speaker Sex Group Repetition Accent    Word variable    value
# 1      105   M     1          1      N AILMENT        1 102.3103
# 2      105   M     1          1      N COLLEGE        1 111.8064
# 3      105   M     1          1      N  FATHER        1 124.0666
# 4      105   M     1          1      N AILMENT        2 102.3103
# 5      105   M     1          1      N COLLEGE        2 111.8031
# 6      105   M     1          1      N  FATHER        2 124.0666
# 7      105   M     1          1      N AILMENT        3 102.3103
# 8      105   M     1          1      N COLLEGE        3 111.6861
# 9      105   M     1          1      N  FATHER        3 124.0666
# 10     105   M     1          1      N AILMENT        4 102.3113
# 11     105   M     1          1      N COLLEGE        4 111.3602
# 12     105   M     1          1      N  FATHER        4 124.0666

---

In base R, you can also use stack to stack the columns named 1 through 4, and cbind that with the first group of columns. Alternatively, unlist will also do this.

在基础R中，您还可以使用堆栈来堆叠名为1到4的列，并使用第一组列cbind。或者，unlist也会这样做。

---

You may also want to look into the "data.table" package to get a bit of a speed boost.

您可能还想查看“data.table”包以获得一点速度提升。

#2

With reshape:

重塑：

x <- read.table(header=T, text="Speaker Sex Group Repetition Accent    Word         1         2         3        4
105   M     1          1      N AILMENT 102.31030 102.31030 102.31030 102.31127
105   M     1          1      N COLLEGE 111.80641 111.80313 111.68612 111.36020
105   M     1          1      N  FATHER 124.06655 124.06655 124.06655 124.06655")

reshape(x, direction="long", sep='', varying=paste0('X', 1:4))
##     Speaker Sex Group Repetition Accent    Word time        X id
## 1.1     105   M     1          1      N AILMENT    1 102.3103  1
## 2.1     105   M     1          1      N COLLEGE    1 111.8064  2
## 3.1     105   M     1          1      N  FATHER    1 124.0666  3
## 1.2     105   M     1          1      N AILMENT    2 102.3103  1
## 2.2     105   M     1          1      N COLLEGE    2 111.8031  2
## 3.2     105   M     1          1      N  FATHER    2 124.0666  3
## 1.3     105   M     1          1      N AILMENT    3 102.3103  1
## 2.3     105   M     1          1      N COLLEGE    3 111.6861  2
## 3.3     105   M     1          1      N  FATHER    3 124.0666  3
## 1.4     105   M     1          1      N AILMENT    4 102.3113  1
## 2.4     105   M     1          1      N COLLEGE    4 111.3602  2
## 3.4     105   M     1          1      N  FATHER    4 124.0666  3