将 data.frame 从宽格式重塑为长格式

Question

我在将data.frame从宽表转换为长表时遇到了一些麻烦。 目前它看起来像这样：

Code Country        1950    1951    1952    1953    1954
AFG  Afghanistan    20,249  21,352  22,532  23,557  24,555
ALB  Albania        8,097   8,986   10,058  11,123  12,246

现在我想把这个data.frame变成一个长的data.frame 。 是这样的：

Code Country        Year    Value
AFG  Afghanistan    1950    20,249
AFG  Afghanistan    1951    21,352
AFG  Afghanistan    1952    22,532
AFG  Afghanistan    1953    23,557
AFG  Afghanistan    1954    24,555
ALB  Albania        1950    8,097
ALB  Albania        1951    8,986
ALB  Albania        1952    10,058
ALB  Albania        1953    11,123
ALB  Albania        1954    12,246

正如某些人在类似问题中所建议的那样，我已经查看并尝试使用melt()和reshape()函数。 但是，到目前为止，我只得到混乱的结果。

如果可能的话，我想用reshape() function 来做，因为它看起来更容易处理。

Answer 1

三种替代解决方案：

1）使用data.table ：

您可以使用与reshape2包中相同的melt功能（这是一个扩展和改进的实现）。 来自data.table的melt还具有比来自reshape2的melt -function 更多的参数。 例如，您还可以指定变量列的名称：

library(data.table)
long <- melt(setDT(wide), id.vars = c("Code","Country"), variable.name = "year")

这使：

 > long Code Country year value 1: AFG Afghanistan 1950 20,249 2: ALB Albania 1950 8,097 3: AFG Afghanistan 1951 21,352 4: ALB Albania 1951 8,986 5: AFG Afghanistan 1952 22,532 6: ALB Albania 1952 10,058 7: AFG Afghanistan 1953 23,557 8: ALB Albania 1953 11,123 9: AFG Afghanistan 1954 24,555 10: ALB Albania 1954 12,246

一些替代符号：

melt(setDT(wide), id.vars = 1:2, variable.name = "year")
melt(setDT(wide), measure.vars = 3:7, variable.name = "year")
melt(setDT(wide), measure.vars = as.character(1950:1954), variable.name = "year")

2）使用tidyr ：

library(tidyr)
long <- wide %>% gather(year, value, -c(Code, Country))

一些替代符号：

wide %>% gather(year, value, -Code, -Country)
wide %>% gather(year, value, -1:-2)
wide %>% gather(year, value, -(1:2))
wide %>% gather(year, value, -1, -2)
wide %>% gather(year, value, 3:7)
wide %>% gather(year, value, `1950`:`1954`)

3) 使用reshape2 ：

library(reshape2)
long <- melt(wide, id.vars = c("Code", "Country"))

给出相同结果的一些替代符号：

# you can also define the id-variables by column number
melt(wide, id.vars = 1:2)

# as an alternative you can also specify the measure-variables
# all other variables will then be used as id-variables
melt(wide, measure.vars = 3:7)
melt(wide, measure.vars = as.character(1950:1954))

---

笔记：

• reshape2已退休。 只有将其保留在 CRAN 上所需的更改才会进行。 （来源）
• 如果要排除NA值，可以将na.rm = TRUE添加到melt以及gather函数中。

---

数据的另一个问题是这些值将被 R 作为字符值读取（作为数字中的,的结果）。 您可以使用gsub和as.numeric修复它：

long$value <- as.numeric(gsub(",", "", long$value))

或直接使用data.table或dplyr ：

# data.table
long <- melt(setDT(wide),
             id.vars = c("Code","Country"),
             variable.name = "year")[, value := as.numeric(gsub(",", "", value))]

# tidyr and dplyr
long <- wide %>% gather(year, value, -c(Code,Country)) %>% 
  mutate(value = as.numeric(gsub(",", "", value)))

---

数据：

wide <- read.table(text="Code Country        1950    1951    1952    1953    1954
AFG  Afghanistan    20,249  21,352  22,532  23,557  24,555
ALB  Albania        8,097   8,986   10,058  11,123  12,246", header=TRUE, check.names=FALSE)

Answer 2

reshape()需要一段时间来适应，就像melt / cast一样。 这是一个重塑的解决方案，假设您的数据框称为d ：

reshape(d, 
        direction = "long",
        varying = list(names(d)[3:7]),
        v.names = "Value",
        idvar = c("Code", "Country"),
        timevar = "Year",
        times = 1950:1954)

Answer 3

使用tidyr_1.0.0 ，另一个选项是pivot_longer

library(tidyr)
pivot_longer(df1, -c(Code, Country), values_to = "Value", names_to = "Year")
# A tibble: 10 x 4
#   Code  Country     Year  Value 
#   <fct> <fct>       <chr> <fct> 
# 1 AFG   Afghanistan 1950  20,249
# 2 AFG   Afghanistan 1951  21,352
# 3 AFG   Afghanistan 1952  22,532
# 4 AFG   Afghanistan 1953  23,557
# 5 AFG   Afghanistan 1954  24,555
# 6 ALB   Albania     1950  8,097 
# 7 ALB   Albania     1951  8,986 
# 8 ALB   Albania     1952  10,058
# 9 ALB   Albania     1953  11,123
#10 ALB   Albania     1954  12,246

数据

df1 <- structure(list(Code = structure(1:2, .Label = c("AFG", "ALB"), class = "factor"), 
    Country = structure(1:2, .Label = c("Afghanistan", "Albania"
    ), class = "factor"), `1950` = structure(1:2, .Label = c("20,249", 
    "8,097"), class = "factor"), `1951` = structure(1:2, .Label = c("21,352", 
    "8,986"), class = "factor"), `1952` = structure(2:1, .Label = c("10,058", 
    "22,532"), class = "factor"), `1953` = structure(2:1, .Label = c("11,123", 
    "23,557"), class = "factor"), `1954` = structure(2:1, .Label = c("12,246", 
    "24,555"), class = "factor")), class = "data.frame", row.names = c(NA, 
-2L))

Answer 4

使用重塑包：

#data
x <- read.table(textConnection(
"Code Country        1950    1951    1952    1953    1954
AFG  Afghanistan    20,249  21,352  22,532  23,557  24,555
ALB  Albania        8,097   8,986   10,058  11,123  12,246"), header=TRUE)

library(reshape)

x2 <- melt(x, id = c("Code", "Country"), variable_name = "Year")
x2[,"Year"] <- as.numeric(gsub("X", "" , x2[,"Year"]))

Answer 5

由于这个答案被标记为r-faq ，我觉得从基础 R 分享另一个替代方案会很有用： stack 。

但是请注意，该stack不适用于factor s——它仅在is.vector为TRUE时才有效，并且从is.vector的文档中，我们发现：

如果 x 是指定模式的向量，除了 names 之外没有其他属性， is.vector返回TRUE 。 否则返回FALSE 。

我正在使用来自@Jaap's answer的样本数据，其中年份列中的值是factor s。

这是stack方法：

cbind(wide[1:2], stack(lapply(wide[-c(1, 2)], as.character)))
##    Code     Country values  ind
## 1   AFG Afghanistan 20,249 1950
## 2   ALB     Albania  8,097 1950
## 3   AFG Afghanistan 21,352 1951
## 4   ALB     Albania  8,986 1951
## 5   AFG Afghanistan 22,532 1952
## 6   ALB     Albania 10,058 1952
## 7   AFG Afghanistan 23,557 1953
## 8   ALB     Albania 11,123 1953
## 9   AFG Afghanistan 24,555 1954
## 10  ALB     Albania 12,246 1954

Answer 6

这是另一个示例，展示了使用tidyr gather的方法。 您可以选择要gather的列，方法是单独删除它们（就像我在这里所做的那样），或者明确包含您想要的年份。

请注意，为了处理逗号（如果未设置check.names = FALSE则添加 X），我还使用dplyr的 mutate 和parse_number中的readr将文本值转换回数字。 这些都是tidyverse的一部分，因此可以与library(tidyverse)一起加载

wide %>%
  gather(Year, Value, -Code, -Country) %>%
  mutate(Year = parse_number(Year)
         , Value = parse_number(Value))

回报：

   Code     Country Year Value
1   AFG Afghanistan 1950 20249
2   ALB     Albania 1950  8097
3   AFG Afghanistan 1951 21352
4   ALB     Albania 1951  8986
5   AFG Afghanistan 1952 22532
6   ALB     Albania 1952 10058
7   AFG Afghanistan 1953 23557
8   ALB     Albania 1953 11123
9   AFG Afghanistan 1954 24555
10  ALB     Albania 1954 12246

Answer 7

这是一个sqldf解决方案：

sqldf("Select Code, Country, '1950' As Year, `1950` As Value From wide
        Union All
       Select Code, Country, '1951' As Year, `1951` As Value From wide
        Union All
       Select Code, Country, '1952' As Year, `1952` As Value From wide
        Union All
       Select Code, Country, '1953' As Year, `1953` As Value From wide
        Union All
       Select Code, Country, '1954' As Year, `1954` As Value From wide;")

要在不输入所有内容的情况下进行查询，您可以使用以下命令：

^{感谢 G. Grothendieck 实施它。}

ValCol <- tail(names(wide), -2)

s <- sprintf("Select Code, Country, '%s' As Year, `%s` As Value from wide", ValCol, ValCol)
mquery <- paste(s, collapse = "\n Union All\n")

cat(mquery) #just to show the query
 #> Select Code, Country, '1950' As Year, `1950` As Value from wide
 #>  Union All
 #> Select Code, Country, '1951' As Year, `1951` As Value from wide
 #>  Union All
 #> Select Code, Country, '1952' As Year, `1952` As Value from wide
 #>  Union All
 #> Select Code, Country, '1953' As Year, `1953` As Value from wide
 #>  Union All
 #> Select Code, Country, '1954' As Year, `1954` As Value from wide

sqldf(mquery)

 #>    Code     Country Year  Value
 #> 1   AFG Afghanistan 1950 20,249
 #> 2   ALB     Albania 1950  8,097
 #> 3   AFG Afghanistan 1951 21,352
 #> 4   ALB     Albania 1951  8,986
 #> 5   AFG Afghanistan 1952 22,532
 #> 6   ALB     Albania 1952 10,058
 #> 7   AFG Afghanistan 1953 23,557
 #> 8   ALB     Albania 1953 11,123
 #> 9   AFG Afghanistan 1954 24,555
 #> 10  ALB     Albania 1954 12,246

不幸的是，我认为PIVOT和UNPIVOT不适用于R SQLite 。 如果您想以更复杂的方式编写查询，还可以查看以下帖子：

• 使用sprintf编写 sql 查询

• 将变量传递给sqldf

Answer 8

也可以使用cdata包，它使用（转换）控制表的概念：

# data
wide <- read.table(text="Code Country        1950    1951    1952    1953    1954
AFG  Afghanistan    20,249  21,352  22,532  23,557  24,555
ALB  Albania        8,097   8,986   10,058  11,123  12,246", header=TRUE, check.names=FALSE)

library(cdata)
# build control table
drec <- data.frame(
    Year=as.character(1950:1954),
    Value=as.character(1950:1954),
    stringsAsFactors=FALSE
)
drec <- cdata::rowrecs_to_blocks_spec(drec, recordKeys=c("Code", "Country"))

# apply control table
cdata::layout_by(drec, wide)

我目前正在探索该软件包并发现它很容易获得。 它是为更复杂的转换而设计的，包括反向转换。 有一个教程可用。

Answer 9

您还可以在R 食谱中看到许多示例

olddata_wide <- read.table(header=TRUE, text='
 subject sex control cond1 cond2
       1   M     7.9  12.3  10.7
       2   F     6.3  10.6  11.1
       3   F     9.5  13.1  13.8
       4   M    11.5  13.4  12.9
')
# Make sure the subject column is a factor
olddata_wide$subject <- factor(olddata_wide$subject)
olddata_long <- read.table(header=TRUE, text='
 subject sex condition measurement
       1   M   control         7.9
       1   M     cond1        12.3
       1   M     cond2        10.7
       2   F   control         6.3
       2   F     cond1        10.6
       2   F     cond2        11.1
       3   F   control         9.5
       3   F     cond1        13.1
       3   F     cond2        13.8
       4   M   control        11.5
       4   M     cond1        13.4
       4   M     cond2        12.9
')
# Make sure the subject column is a factor
olddata_long$subject <- factor(olddata_long$subject)