重塑宽格式，以多列长格式

I want to reshape a wide format dataset that has multiple tests which are measured at 3 time points:

我想重塑一个具有多个测试的宽格式数据集，这些测试在3个时间点测量:

   ID   Test Year   Fall Spring Winter
    1   1   2008    15      16      19
    1   1   2009    12      13      27
    1   2   2008    22      22      24
    1   2   2009    10      14      20
    2   1   2008    12      13      25
    2   1   2009    16      14      21
    2   2   2008    13      11      29
    2   2   2009    23      20      26
    3   1   2008    11      12      22
    3   1   2009    13      11      27
    3   2   2008    17      12      23
    3   2   2009    14      9       31

into a data set that separates the tests by column but converts the measurement time into long format, for each of the new columns like this:

将测试按列分离的数据集，但将度量时间转换为长格式，对于每一列这样的新列:

    ID  Year    Time        Test1 Test2
    1   2008    Fall        15      22
    1   2008    Spring      16      22
    1   2008    Winter      19      24
    1   2009    Fall        12      10
    1   2009    Spring      13      14
    1   2009    Winter      27      20
    2   2008    Fall        12      13
    2   2008    Spring      13      11
    2   2008    Winter      25      29
    2   2009    Fall        16      23
    2   2009    Spring      14      20
    2   2009    Winter      21      26
    3   2008    Fall        11      17
    3   2008    Spring      12      12
    3   2008    Winter      22      23
    3   2009    Fall        13      14
    3   2009    Spring      11      9
    3   2009    Winter      27      31

I have unsuccessfully tried to use reshape and melt. Existing posts address transforming to single column outcome.

我尝试过重新塑造和融化，但没有成功。现有员额的地址改为单一栏的结果。

4 个解决方案

#1

Using reshape2:

使用reshape2:

# Thanks to Ista for helping with direct naming using "variable.name"
df.m <- melt(df, id.var = c("ID", "Test", "Year"), variable.name = "Time")
df.m <- transform(df.m, Test = paste0("Test", Test))
dcast(df.m, ID + Year + Time ~ Test, value.var = "value")

---

Update: Using data.table melt/cast from versions >= 1.9.0:

data.table from versions 1.9.0 imports reshape2 package and implements fast melt and dcast methods in C for data.tables. A comparison of speed on bigger data is shown below.

数据。来自版本1.9.0的表导入reshape2包，并在C中实现数据表的快速熔融和dcast方法。比较大数据的速度如下所示。

For more info regarding NEWS, go here.

有关新闻的更多信息，请访问这里。

require(data.table) ## ver. >=1.9.0
require(reshape2)

dt <- as.data.table(df, key=c("ID", "Test", "Year"))
dt.m <- melt(dt, id.var = c("ID", "Test", "Year"), variable.name = "Time")
dt.m[, Test := paste0("Test", Test)]
dcast.data.table(dt.m, ID + Year + Time ~ Test, value.var = "value")

At the moment, you'll have to write dcast.data.table explicitly as it's not a S3 generic in reshape2 yet.

目前，您必须编写dcast.data。表是显式的，因为它还不是reshape2中的S3泛型。

---

Benchmarking on bigger data:

# generate data:
set.seed(45L)
DT <- data.table(ID = sample(1e2, 1e7, TRUE), 
        Test = sample(1e3, 1e7, TRUE), 
        Year = sample(2008:2014, 1e7,TRUE), 
        Fall = sample(50, 1e7, TRUE), 
        Spring = sample(50, 1e7,TRUE), 
        Winter = sample(50, 1e7, TRUE))
DF <- as.data.frame(DT)

reshape2 timings:

reshape2_melt <- function(df) {
    df.m <- melt(df, id.var = c("ID", "Test", "Year"), variable.name = "Time")
}
# min. of three consecutive runs
system.time(df.m <- reshape2_melt(DF))
#   user  system elapsed 
# 43.319   4.909  48.932 

df.m <- transform(df.m, Test = paste0("Test", Test))

reshape2_cast <- function(df) {
    dcast(df.m, ID + Year + Time ~ Test, value.var = "value")
}
# min. of three consecutive runs
system.time(reshape2_cast(df.m))
#   user  system elapsed 
# 57.728   9.712  69.573 

data.table timings:

DT_melt <- function(dt) {
    dt.m <- melt(dt, id.var = c("ID", "Test", "Year"), variable.name = "Time")
}
# min. of three consecutive runs
system.time(dt.m <- reshape2_melt(DT))
#   user  system elapsed 
#  0.276   0.001   0.279 

dt.m[, Test := paste0("Test", Test)]

DT_cast <- function(dt) {
    dcast.data.table(dt.m, ID + Year + Time ~ Test, value.var = "value")
}
# min. of three consecutive runs
system.time(DT_cast(dt.m))
#   user  system elapsed 
# 12.732   0.825  14.006 

melt.data.table is ~175x faster than reshape2:::melt and dcast.data.table is ~5x than reshape2:::dcast.

melt.data。表比reshape2快175x:熔体和dcast.data。表是~5x比reshape2:: dcast。

#2

Base reshape function alternative method is below. Though this required using reshape twice, there might be a simpler way.

基本重塑函数替代方法如下。尽管这需要使用两次整形，但可能有一种更简单的方法。

Assuming your dataset is called df1

假设您的数据集被称为df1

tmp <- reshape(df1,idvar=c("ID","Year"),timevar="Test",direction="wide")
result <- reshape(
   tmp,
   idvar=c("ID","Year"),
   varying=list(3:5,6:8),
   v.names=c("Test1","Test2"),
   times=c("Fall","Spring","Winter"),
   direction="long"
)

Which gives:

这使:

> result
              ID Year   time Test1 Test2
1.2008.Fall    1 2008   Fall    15    22
1.2009.Fall    1 2009   Fall    12    10
2.2008.Fall    2 2008   Fall    12    13
2.2009.Fall    2 2009   Fall    16    23
3.2008.Fall    3 2008   Fall    11    17
3.2009.Fall    3 2009   Fall    13    14
1.2008.Spring  1 2008 Spring    16    22
1.2009.Spring  1 2009 Spring    13    14
2.2008.Spring  2 2008 Spring    13    11
2.2009.Spring  2 2009 Spring    14    20
3.2008.Spring  3 2008 Spring    12    12
3.2009.Spring  3 2009 Spring    11     9
1.2008.Winter  1 2008 Winter    19    24
1.2009.Winter  1 2009 Winter    27    20
2.2008.Winter  2 2008 Winter    25    29
2.2009.Winter  2 2009 Winter    21    26
3.2008.Winter  3 2008 Winter    22    23
3.2009.Winter  3 2009 Winter    27    31

#3

Sticking with base R, this is another good candidate for the "stack + reshape" routine. Assuming our dataset is called "mydf":

坚持使用基本的R，这是“堆栈+重塑”例程的另一个很好的候选。假设我们的数据集被称为“mydf”:

mydf.temp <- data.frame(mydf[1:3], stack(mydf[4:6]))
mydf2 <- reshape(mydf.temp, direction = "wide", 
                 idvar=c("ID", "Year", "ind"), 
                 timevar="Test")
names(mydf2) <- c("ID", "Year", "Time", "Test1", "Test2")
mydf2
#    ID Year   Time Test1 Test2
# 1   1 2008   Fall    15    22
# 2   1 2009   Fall    12    10
# 5   2 2008   Fall    12    13
# 6   2 2009   Fall    16    23
# 9   3 2008   Fall    11    17
# 10  3 2009   Fall    13    14
# 13  1 2008 Spring    16    22
# 14  1 2009 Spring    13    14
# 17  2 2008 Spring    13    11
# 18  2 2009 Spring    14    20
# 21  3 2008 Spring    12    12
# 22  3 2009 Spring    11     9
# 25  1 2008 Winter    19    24
# 26  1 2009 Winter    27    20
# 29  2 2008 Winter    25    29
# 30  2 2009 Winter    21    26
# 33  3 2008 Winter    22    23
# 34  3 2009 Winter    27    31

#4

tidyverse/tidyr solution:

tidyverse / tidyr解决方案:

library(dplyr)
library(tidyr)

df %>% 
  gather("Time", "Value", Fall, Spring, Winter) %>% 
  spread(Test, Value, sep = "")