Question

我一直试图通过查看其他帖子来查看我的数据，但我一直收到错误。我的数据new如下所示：

id  year    name    gdp
1   1980    Jamie   45
1   1981    Jamie   60
1   1982    Jamie   70
2   1990    Kate    40
2   1991    Kate    25
2   1992    Kate    67
3   1994    Joe     35
3   1995    Joe     78
3   1996    Joe     90

我想通过id选择年份值最高的行。所以想要的输出是：

id  year    name    gdp
1   1982    Jamie   70
2   1992    Kate    67
3   1996    Joe     90

从Selecting Rows which contain daily max value in R开始，我尝试了以下操作，但无效

ddply(new,~id,function(x){x[which.max(new$year),]})

我也试过

tapply(new$year, new$id, max)

但这并没有给我想要的输出。

任何建议都会有所帮助！

Answer 1

另一个适用于大型表的选项是使用data.table。

DT <- read.table(text = "id  year    name    gdp
                          1   1980    Jamie   45
                          1   1981    Jamie   60
                          1   1982    Jamie   70
                          2   1990    Kate    40
                          2   1991    Kate    25
                          2   1992    Kate    67
                          3   1994    Joe     35
                          3   1995    Joe     78
                          3   1996    Joe     90",
                 header = TRUE)

require("data.table")
DT <- as.data.table(DT)

setkey(DT,id,year)
res = DT[,j=list(year=year[which.max(gdp)]),by=id]
res

setkey(res,id,year)
DT[res]
# id year  name gdp
# 1:  1 1982 Jamie  70
# 2:  2 1992  Kate  67
# 3:  3 1996   Joe  90

Answer 2

ave再次在这里工作，并将解释最多一年中有多行的情况。

new[with(new, year == ave(year,id,FUN=max) ),]

#  id year  name gdp
#3  1 1982 Jamie  70
#6  2 1992  Kate  67
#9  3 1996   Joe  90

Answer 3

只需使用split：

df <- do.call(rbind, lapply(split(df, df$id),
  function(subdf) subdf[which.max(subdf$year)[1], ]))

例如，

df <- data.frame(id = rep(1:10, each = 3), year = round(runif(30,0,10)) + 1980, gdp = round(runif(30, 40, 70)))
print(head(df))
#   id year gdp
# 1  1 1990  49
# 2  1 1981  47
# 3  1 1987  69
# 4  2 1985  57
# 5  2 1989  41
# 6  2 1988  54

df <- do.call(rbind, lapply(split(df, df$id), function(subdf) subdf[which.max(subdf$year)[1], ]))
print(head(df))
#    id year gdp
# 1   1 1990  49
# 2   2 1989  41
# 3   3 1989  55
# 4   4 1988  62
# 5   5 1989  48
# 6   6 1990  41

Answer 4

您可以使用duplicated

执行此操作

# your data
 df <- read.table(text="id  year    name    gdp
1   1980    Jamie   45
1   1981    Jamie   60
1   1982    Jamie   70
2   1990    Kate    40
2   1991    Kate    25
2   1992    Kate    67
3   1994    Joe     35
3   1995    Joe     78
3   1996    Joe     90" , header=TRUE)

# Sort by id and year (latest year is last for each id)
df <- df[order(df$id , df$year), ]

# Select the last row by id
df <- df[!duplicated(df$id, fromLast=TRUE), ]

Answer 5

您的ddply努力对我来说很好，但您在回调函数中引用了原始数据集。

ddply(new,~id,function(x){x[which.max(new$year),]})
# should be
ddply(new,.(id),function(x){x[which.max(x$year),]})

按组选择最大行值

5 个答案: