R:gather（）用于整理具有两个列标题的数据

我想整理一下前两行中不幸设置了两列标题的一些数据：

• 第一行（标题）：实际上是度量的类型（例如。 估计，标准误差，上限，下限）

• 第二行（也是标题）：是测量年份

是否有某种方法可以使用

gather（）

或其他任何方法来整理此数据

此外，当一个度量值被重复时（例如，Rank，Rank.1），它实际上应该只读取Rank，并且由于年份的不同而有所不同。有没有办法解决这个问题

  Country_Territory WBCode Estimate  StdErr NumSrc    Rank   Lower
1              Year   <NA>  1996.00 1996.00   1996 1996.00 1996.00
2           Andorra    ADO     1.32    0.48      1   87.10   72.04
3       Afghanistan    AFG    -1.29    0.34      2    4.30    0.00
4            Angola    AGO    -1.17    0.26      4    9.68    0.54
    Upper Estimate.1 StdErr.1 NumSrc.1  Rank.1 Lower.1 Upper.1
1 1996.00    1998.00  1998.00     1998 1998.00 1998.00 1998.00
2   96.77       1.38     0.46        1   89.18   74.74   96.91
3   27.42      -1.18     0.33        2    9.79    0.00   31.44
4   27.42      -1.41     0.21        6    1.55    0.00   13.40

没有给我想要的东西：

   Country_Territory WBCode    measure  number
1               Year   <NA>   Estimate 1996.00
2            Andorra    ADO   Estimate    1.32
3        Afghanistan    AFG   Estimate   -1.29
4             Angola    AGO   Estimate   -1.17
5               Year   <NA>     StdErr 1996.00
6            Andorra    ADO     StdErr    0.48

国家/地区WBCode度量值编号
1年估计数1996.00
2安道尔ADO估计数1.32
3阿富汗空军估计数-1.29
4安哥拉前估计数-1.17
5年标准1996.00
6安道尔ADO标准差0.48

因为年份与国家/地区混在一起。

这是一种选择：

library(tidyverse)

# get unique Year values and create column names (to add later)
df %>%
  filter(Country_Territory == "Year") %>%
  gather() %>%
  filter(value != "Year" & !is.na(value)) %>%
  pull(value) %>%
  unique() %>%
  paste0("Year_",.) -> col_years

# reshape data (excluding the Year row)
df %>%
  filter(Country_Territory != "Year") %>%
  gather(key,y,-Country_Territory, -WBCode) %>%
  separate(key, c("measure","v")) %>%
  group_by(v = ifelse(is.na(v), 0, v)) %>%
  nest() -> df_info

reduce(df_info$data, function(x,y) left_join(x,y,by=c("Country_Territory","WBCode","measure"))) %>%
  setNames(c("Country_Territory", "WBCode", "measure", col_years))

# # A tibble: 18 x 5
#   Country_Territory WBCode measure  Year_1996 Year_1998
#   <chr>             <chr>  <chr>        <dbl>     <dbl>
# 1 Andorra           ADO    Estimate      1.32      1.38
# 2 Afghanistan       AFG    Estimate     -1.29     -1.18
# 3 Angola            AGO    Estimate     -1.17     -1.41
# 4 Andorra           ADO    StdErr        0.48      0.46
# 5 Afghanistan       AFG    StdErr        0.34      0.33
# 6 Angola            AGO    StdErr        0.26      0.21
# 7 Andorra           ADO    NumSrc        1         1   
# 8 Afghanistan       AFG    NumSrc        2         2   
# 9 Angola            AGO    NumSrc        4         6   
# 10 Andorra           ADO    Rank         87.1      89.2 
# 11 Afghanistan       AFG    Rank          4.3       9.79
# 12 Angola            AGO    Rank          9.68      1.55
# 13 Andorra           ADO    Lower        72.0      74.7 
# 14 Afghanistan       AFG    Lower         0         0   
# 15 Angola            AGO    Lower         0.54      0   
# 16 Andorra           ADO    Upper        96.8      96.9 
# 17 Afghanistan       AFG    Upper        27.4      31.4 
# 18 Angola            AGO    Upper        27.4      13.4 

这是一种选择：

library(tidyverse)

# get unique Year values and create column names (to add later)
df %>%
  filter(Country_Territory == "Year") %>%
  gather() %>%
  filter(value != "Year" & !is.na(value)) %>%
  pull(value) %>%
  unique() %>%
  paste0("Year_",.) -> col_years

# reshape data (excluding the Year row)
df %>%
  filter(Country_Territory != "Year") %>%
  gather(key,y,-Country_Territory, -WBCode) %>%
  separate(key, c("measure","v")) %>%
  group_by(v = ifelse(is.na(v), 0, v)) %>%
  nest() -> df_info

reduce(df_info$data, function(x,y) left_join(x,y,by=c("Country_Territory","WBCode","measure"))) %>%
  setNames(c("Country_Territory", "WBCode", "measure", col_years))

# # A tibble: 18 x 5
#   Country_Territory WBCode measure  Year_1996 Year_1998
#   <chr>             <chr>  <chr>        <dbl>     <dbl>
# 1 Andorra           ADO    Estimate      1.32      1.38
# 2 Afghanistan       AFG    Estimate     -1.29     -1.18
# 3 Angola            AGO    Estimate     -1.17     -1.41
# 4 Andorra           ADO    StdErr        0.48      0.46
# 5 Afghanistan       AFG    StdErr        0.34      0.33
# 6 Angola            AGO    StdErr        0.26      0.21
# 7 Andorra           ADO    NumSrc        1         1   
# 8 Afghanistan       AFG    NumSrc        2         2   
# 9 Angola            AGO    NumSrc        4         6   
# 10 Andorra           ADO    Rank         87.1      89.2 
# 11 Afghanistan       AFG    Rank          4.3       9.79
# 12 Angola            AGO    Rank          9.68      1.55
# 13 Andorra           ADO    Lower        72.0      74.7 
# 14 Afghanistan       AFG    Lower         0         0   
# 15 Angola            AGO    Lower         0.54      0   
# 16 Andorra           ADO    Upper        96.8      96.9 
# 17 Afghanistan       AFG    Upper        27.4      31.4 
# 18 Angola            AGO    Upper        27.4      13.4 

或许（如果您只有两个系列的措施）：

或许（如果您只有两个系列的措施）：

数据表解决方案

需要做一些准备（设置colnames和创建一个唯一名称表），但是速度非常快

该解决方案也可以使用两年以上

library( data.table )
dt <- as.data.table( df )  #or use setDT( df )

#extract unique years from the first row from the thirs column untill end of dt
dt.years <- as.data.table ( unique( t( (dt[1, 3:ncol(dt)]) ) ) )
dt.years[, year_id := 1:.N ]
setnames(dt.years, c("year", "year_id" ) )

#melt row 2:n of the data.table
dt.melt <- melt( dt[2:nrow(dt)], 
                 id.vars = c( "Country_Territory", "WBCode"),
                 measure = patterns( "Estimate", "StdErr", "NumSrc", "Rank", "Lower", "Upper"),
                 value.name = c( "Estimate", "StdErr", "NumSrc", "Rank", "Lower", "Upper" ),
                 variable.name = "year")

#left join both datatables
result <- dt.years[dt.melt, on = c( year_id = "year")]

#cleaning and renaming
result[, year_id := NULL]

数据表解决方案

需要做一些准备（设置colnames和创建一个唯一名称表），但是速度非常快

该解决方案也可以使用两年以上

library( data.table )
dt <- as.data.table( df )  #or use setDT( df )

#extract unique years from the first row from the thirs column untill end of dt
dt.years <- as.data.table ( unique( t( (dt[1, 3:ncol(dt)]) ) ) )
dt.years[, year_id := 1:.N ]
setnames(dt.years, c("year", "year_id" ) )

#melt row 2:n of the data.table
dt.melt <- melt( dt[2:nrow(dt)], 
                 id.vars = c( "Country_Territory", "WBCode"),
                 measure = patterns( "Estimate", "StdErr", "NumSrc", "Rank", "Lower", "Upper"),
                 value.name = c( "Estimate", "StdErr", "NumSrc", "Rank", "Lower", "Upper" ),
                 variable.name = "year")

#left join both datatables
result <- dt.years[dt.melt, on = c( year_id = "year")]

#cleaning and renaming
result[, year_id := NULL]

---

melt（）

的

data.table

方法能够同时重塑多个测量列的形状。不需要使用

patterns（）

函数重命名列

library(data.table)
# reshape multiple measure columns simultaneously from wide to long format
cols <- c("Estimate", "StdErr", "NumSrc", "Rank", "Lower", "Upper")
long <- melt(setDT(df), measure.vars = patterns(cols), value.name = cols)
# extract years
yrs <- long[Country_Territory == "Year", .(variable, Year = as.integer(Estimate))]
# join to get a separate Year column, remove Year rows and helper column 
result <- yrs[long[Country_Territory != "Year"], on = "variable"][, variable := NULL][]
result

---

重塑后，

变量

列表示宽格式中属于一个列子集的行，即属于一个特定年份的行。

数据。

melt（）

的表方法能够同时重塑多个度量列。不需要使用

patterns（）

函数重命名列

library(data.table)
# reshape multiple measure columns simultaneously from wide to long format
cols <- c("Estimate", "StdErr", "NumSrc", "Rank", "Lower", "Upper")
long <- melt(setDT(df), measure.vars = patterns(cols), value.name = cols)
# extract years
yrs <- long[Country_Territory == "Year", .(variable, Year = as.integer(Estimate))]
# join to get a separate Year column, remove Year rows and helper column 
result <- yrs[long[Country_Territory != "Year"], on = "variable"][, variable := NULL][]
result

---

重塑后，

变量

列表示宽格式中属于一个列子集的行，即属于一个特定年份的行。

如@Uwe所述，您不需要首先重命名列。。我删除了我答案中的（多余的）部分。正如@Uwe提到的，您不需要首先重命名列。。我删除了答案中的（多余的）部分。在完整的数据集上工作得绝对完美！我想知道你的这部分代码是做什么的？看起来你们在分离“度量”，实际上并没有两个部分可以分开

separate（key，c（“measure”，“v”））%%>%group_by（v=ifelse（is.na（v），0，v））%%>%

在完整的数据集上工作得绝对完美！我想知道你的这部分代码是做什么的？看起来你们在分离“度量”，实际上并没有两个部分可以分开<代码>分开（键，c（“测量”，“v”））%%>%group_by（v=ifelse（is.na（v），0，v））%%>%

#    year Country_Territory WBCode Estimate StdErr NumSrc  Rank Lower Upper
# 1: 1996           Andorra    ADO     1.32   0.48      1 87.10 72.04 96.77
# 2: 1996       Afghanistan    AFG    -1.29   0.34      2  4.30  0.00 27.42
# 3: 1996            Angola    AGO    -1.17   0.26      4  9.68  0.54 27.42
# 4: 1998           Andorra    ADO     1.38   0.46      1 89.18 74.74 96.91
# 5: 1998       Afghanistan    AFG    -1.18   0.33      2  9.79  0.00 31.44
# 6: 1998            Angola    AGO    -1.41   0.21      6  1.55  0.00 13.40

library(data.table)
# reshape multiple measure columns simultaneously from wide to long format
cols <- c("Estimate", "StdErr", "NumSrc", "Rank", "Lower", "Upper")
long <- melt(setDT(df), measure.vars = patterns(cols), value.name = cols)
# extract years
yrs <- long[Country_Territory == "Year", .(variable, Year = as.integer(Estimate))]
# join to get a separate Year column, remove Year rows and helper column 
result <- yrs[long[Country_Territory != "Year"], on = "variable"][, variable := NULL][]
result

   Year Country_Territory WBCode Estimate StdErr NumSrc  Rank Lower Upper
1: 1996           Andorra    ADO     1.32   0.48      1 87.10 72.04 96.77
2: 1996       Afghanistan    AFG    -1.29   0.34      2  4.30  0.00 27.42
3: 1996            Angola    AGO    -1.17   0.26      4  9.68  0.54 27.42
4: 1998           Andorra    ADO     1.38   0.46      1 89.18 74.74 96.91
5: 1998       Afghanistan    AFG    -1.18   0.33      2  9.79  0.00 31.44
6: 1998            Angola    AGO    -1.41   0.21      6  1.55  0.00 13.40