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我正在尝试（有效地）重新排列 R 中的数据框。

我的数据是从两个参与者群体（1 或 0，即疾病组和对照组）的四个不同实验中收集的实验数据。

示例数据框：

Subject type    Experiment 1    Experiment 2    Experiment 3    Experiment 4
           0             4.6             2.5             1.4             5.3
           0             4.7             2.4             1.8             5.1
           1             3.5             1.2             5.6             7.5
           1             3.8             1.7             6.2             8.1

我想重新排列我的数据框，使其结构如下（原因是，当它们在 R 中的结构如下时，它使我更容易在数据上运行函数）：

Subject type    Experiment    Measure
           0             1        4.6
           0             2        2.5
           0             3        1.4
           0             4        5.3
           0             1        4.7
           0             2        2.4
           0             3        1.8
           0             4        5.1
           1             1        3.5
           1             2        1.2
           1             3        5.6
           1             4        7.5
           1             1        3.8
           1             2        1.7
           1             3        6.2
           1             4        8.1

如您所见，发生的情况是每个主题现在占据了四行；现在，每一行都与单个测量有关，而不是单个主题。这（至少现在）对我来说更方便插入 R 函数。也许及时我会想出一种完全跳过这一步的方法，但我是 R 新手，这似乎是最好的做事方式。

无论如何-问题是，进行此数据框转换的最有效方法是什么？目前我正在这样做：

# Input dframe1
dframe1 <- structure(list(subject_type = c(0L, 0L, 1L, 1L), experiment_1 = c(4.6, 
4.7, 3.5, 3.8), experiment_2 = c(2.5, 2.4, 1.2, 1.7), experiment_3 = c(1.4, 
1.8, 5.6, 6.2), experiment_4 = c(5.3, 5.1, 7.5, 8.1)), .Names = c("subject_type", 
"experiment_1", "experiment_2", "experiment_3", "experiment_4"
), class = "data.frame", row.names = c(NA, -4L))

# Create a matrix
temporary_matrix <- matrix(ncol=3, nrow=nrow(dframe1) * 4)
colnames(temporary_matrix) <- c("subject_type","experiment","measure")

# Rearrange dframe1 so that a different measure is in each column
for(i in 1:nrow(dframe1)) {
  temporary_matrix[i*4-3,"subject_type"] <- dframe1$subject_type[i]
  temporary_matrix[i*4-3,"experiment"] <- 1
  temporary_matrix[i*4-3,"measure"] <- dframe1$experiment_1[i]
  temporary_matrix[i*4-2,"subject_type"] <- dframe1$subject_type[i]
  temporary_matrix[i*4-2,"experiment"] <- 2
  temporary_matrix[i*4-2,"measure"] <- dframe1$experiment_2[i]
  temporary_matrix[i*4-1,"subject_type"] <- dframe1$subject_type[i]
  temporary_matrix[i*4-1,"experiment"] <- 3
  temporary_matrix[i*4-1,"measure"] <- dframe1$experiment_3[i]
  temporary_matrix[i*4-0,"subject_type"] <- dframe1$subject_type[i]
  temporary_matrix[i*4-0,"experiment"] <- 4
  temporary_matrix[i*4-0,"measure"] <- dframe1$experiment_4[i]
}

# Convert matrix to a data frame
dframe2 <- data.frame(temporary_matrix)

# NOTE: For some reason, this has to be converted back into a double (at some point above it becomes a factor)
dframe2$measure <- as.double(as.character(dframe2$measure))

当然有更好的方法来做到这一点？！

4

3 回答 3

5

使用该reshape2软件包，这非常简单。

library(reshape2)

# assuming your data.frame is called `dat`
melt(dat, id.vars=c("Subject type"))

如果你愿意，你可以让它变得更漂亮：

newdat <- melt(dat, id.vars=c("Subject type"), variable.name="Experiment", value.name="Measure")

# remove "experiment " from the names, and convert to numeric
newdat$Experiment <- as.numeric(gsub("Experiment\\s*", "", as.character(newdat$Experiment)))
于 2013-02-25T01:23:44.730 回答
4

基础reshape方法：

获取数据：

dframe1 <- structure(list(subject_type = c(0L, 0L, 1L, 1L), experiment_1 = c(4.6, 
4.7, 3.5, 3.8), experiment_2 = c(2.5, 2.4, 1.2, 1.7), experiment_3 = c(1.4, 
1.8, 5.6, 6.2), experiment_4 = c(5.3, 5.1, 7.5, 8.1)), .Names = c("subject_type", 
"experiment_1", "experiment_2", "experiment_3", "experiment_4"
), class = "data.frame", row.names = c(NA, -4L))

将变量设置为堆栈：

expandvars <- paste('experiment',1:4,sep='_')

重塑远方！

dfrm1res <- reshape(
                   dframe1,
                   idvar="subject_type",
                   varying=list(expandvars),
                   v.names=c("value"),
                   direction="long",
                   new.row.names=1:16
                    )

结果：

> dfrm1res
   subject_type time value
1             0    1   4.6
2             0    1   4.7
3             1    1   3.5
4             1    1   3.8
5             0    2   2.5
6             0    2   2.4
7             1    2   1.2
8             1    2   1.7
9             0    3   1.4
10            0    3   1.8
11            1    3   5.6
12            1    3   6.2
13            0    4   5.3
14            0    4   5.1
15            1    4   7.5
16            1    4   8.1
于 2013-02-25T01:54:23.717 回答
4 
data.frame(subject_type=dframe1$subject_type, stack(dframe1[2:5] )  )
   subject_type values          ind
1             0    4.6 experiment_1
2             0    4.7 experiment_1
3             1    3.5 experiment_1
4             1    3.8 experiment_1
5             0    2.5 experiment_2
6             0    2.4 experiment_2
7             1    1.2 experiment_2
8             1    1.7 experiment_2
9             0    1.4 experiment_3
10            0    1.8 experiment_3
11            1    5.6 experiment_3
12            1    6.2 experiment_3
13            0    5.3 experiment_4
14            0    5.1 experiment_4
15            1    7.5 experiment_4
16            1    8.1 experiment_4

或使用 base reshape（尽管我的偏好似乎与 thelatemal 的使用不同。）：

dframe1$subject=1:4
reshape(dframe1, direction="long", idvar=c("subject_type", "subject"),
                 varying=2:5, sep="_", v.names="exp_value")
 #--------------------------
      subject_type subject time exp_value
0.1.1            0       1    1       4.6
0.2.1            0       2    1       4.7
1.3.1            1       3    1       3.5
1.4.1            1       4    1       3.8
0.1.2            0       1    2       2.5
0.2.2            0       2    2       2.4
1.3.2            1       3    2       1.2
1.4.2            1       4    2       1.7
0.1.3            0       1    3       1.4
0.2.3            0       2    3       1.8
1.3.3            1       3    3       5.6
1.4.3            1       4    3       6.2
0.1.4            0       1    4       5.3
0.2.4            0       2    4       5.1
1.3.4            1       3    4       7.5
1.4.4            1       4    4       8.1
于 2013-02-25T02:03:29.263 回答