两种备选解决方案：

1）使用data.table：

可以使用melt函数：

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

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其给出：

> 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

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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")

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2）使用tidyr：

使用pivot_longer()：

library(tidyr)

long <- wide %>% 
  pivot_longer(
    cols = `1950`:`1954`, 
    names_to = "year",
    values_to = "value"
)

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注意事项：

• names_to和values_to分别默认为"name"和"value"，因此您可以将其非常简洁地写成wide %>% pivot_longer(1950:1954)。
• cols参数使用高度灵活的tidyselect DSL，因此您可以使用否定选择（!c(Code, Country)）、选择帮助器（starts_with("19"); matches("^\\d{4}$")）、数字索引（3:7）等来选择相同的列。
• tidyr::pivot_longer()是tidyr::gather()和reshape2::melt()的继任者，这两个版本已不再开发。
  转换值

数据的另一个问题是，这些值将被R读取为字符值（由于数字中的,）。您可以在整形之前使用gsub和as.numeric进行修复：

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

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或者在整形期间，使用data.table或tidyr：

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

# tidyr
long <- wide %>%
  pivot_longer(
    cols = `1950`:`1954`, 
    names_to = "year",
    values_to = "value",
    values_transform = ~ as.numeric(gsub(",", "", .x))
  )

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)

reshape()需要一段时间来适应，就像melt/cast一样。这里是一个整形的解决方案，假设你的 Dataframe 被称为d：

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

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对于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

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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))

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使用 reshape 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"]))

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由于这个答案被标记为r-faq，我觉得分享另一个基于R的替代方案会很有用：stack。
但是，请注意，stack不适用于factor s-它只适用于is.vector是TRUE，并且从is.vector的文档中，我们发现：
is.vector返回TRUE，如果x是一个指定模式的向量，除了名称之外没有任何属性 *。否则返回FALSE。
我使用的是示例数据from @Jaap's answer，其中year列中的值为factor s。
以下是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

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下面是另一个例子，展示了从tidyr中使用gather。您可以通过单独删除它们（就像我在这里做的那样）来选择gather的列，或者通过显式包含您想要的年份。
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readr的parse_number将文本值转换回数字。

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

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回报率：

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

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以下是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;")

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要在不键入所有内容的情况下进行查询，可以使用以下命令：
感谢G. Grothendieck的实施。

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，我不认为PIVOT和UNPIVOT适用于RSQLite。如果你想以更复杂的方式编写你的查询，你也可以看看这些帖子：

• Using sprintf writing up sql queries的
• Pass variables to sqldf的

您还可以使用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)

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我目前正在探索这个包，发现它很容易访问。它是为更复杂的转换而设计的，包括反向转换。有a tutorial可用。

这里有两个基本R中的选项（当输入是一个矩阵而不是一个矩阵时，使用x=unlist(df)而不是x=c(m)）：

> m=matrix(sample(1:100,6),3,dimnames=list(2021:2023,c("male","female")))
> m
     male female
2021   89     42
2022   39     96
2023   26     40
> cbind(expand.grid(dimnames(m)),x=c(m))
  Var1   Var2  x
1 2021   male 89
2 2022   male 39
3 2023   male 26
4 2021 female 42
5 2022 female 96
6 2023 female 40
> data.frame(row=rownames(m),col=colnames(m)[col(m)],x=c(m))
   row    col  x
1 2021   male 89
2 2022   male 39
3 2023   male 26
4 2021 female 42
5 2022 female 96
6 2023 female 40

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第三种选择是使用as.table后跟as.data.frame，但它将行和列名转换为因子，如果输入是一个矩阵，则必须首先将其转换为矩阵：

> as.data.frame(as.table(m))
  Var1   Var2 Freq
1 2021   male   89
2 2022   male   39
3 2023   male   26
4 2021 female   42
5 2022 female   96
6 2023 female   40
> as.data.frame(as.table(m))|>sapply(class)
     Var1      Var2      Freq
 "factor"  "factor" "integer"
> d=as.data.frame(m)
> as.data.frame(as.table(d))
Error in h(simpleError(msg, call)) :
  error in evaluating the argument 'x' in selecting a method for function 'as.data.frame': cannot coerce to a table
> as.data.frame(as.table(as.matrix(d)))
  Var1   Var2 Freq
1 2021   male   89
2 2022   male   39
3 2023   male   26
4 2021 female   42
5 2022 female   96
6 2023 female   40

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第四个选项是使用stack，但它将行名称和列名转换为因子，并且当输入是矩阵时列名转换为Rle因子（但当输入是矩阵时则不是）：

> stack(m)
DataFrame with 6 rows and 3 columns
       row    col     value
  <factor>  <Rle> <integer>
1     2021   male        89
2     2022   male        39
3     2023   male        26
4     2021 female        42
5     2022 female        96
6     2023 female        40

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当stack的输入是一个字符串时，行名称不会作为列包含在内，所以你必须cbind它们：

> d=as.data.frame(m);cbind(row=rownames(d),stack(d))
   row values    ind
1 2021     89   male
2 2022     39   male
3 2023     26   male
4 2021     42 female
5 2022     96 female
6 2023     40 female

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