Question

我的数据框排列如下

df <- structure(list(NAME1    =  c("AAA","CCC", "BBB", "BBB"), 
             NAME2            =  c("BBB", "AAA","DDD", "AAA"),
             ID1              =  c(1,3,2,2),
             ID2              =  c(2,1,4,1)),
        .Names                =  c("NAME1","NAME2","ID1","ID2"), 
        row.names             =  c("1", "2", "3", "4"), class =("data.frame"))

我有另一个数据框（df1）并想添加ID列。 ID值应与df中的相同。所需的数据框应如下所示。

df1 <- structure(list(NAME         =  c("AAA","BBB", "CCC", "DDD"), 
                      SIZE         =  c(0.9, 1.7, 1.4, 1.1),                     
                       ID          =  c(1,2,3,4)),
                .Names             =  c("NAME","SIZE", "ID"), 
                row.names          =  c("1", "2", "3", "4"), class =("data.frame"))

任何建议将不胜感激。欢呼声。

Answer 1

您应该重新格式化“密钥”，此时，应该很容易合并以获取新ID。

示例：

library(data.table)
setDT(df)
setDT(df1)
df1[, ID := NULL][] ## I assume you're starting without an ID
df1
#    NAME SIZE
# 1:  AAA  0.9
# 2:  BBB  1.7
# 3:  CCC  1.4
# 4:  DDD  1.1

idkey <- unique(melt(df, measure.vars = patterns("NAME", "ID"),
                     value.name = c("NAME", "ID")), 
                by = c("NAME", "ID"))[, c("NAME", "ID"), with = FALSE]

idkey
#    NAME ID
# 1:  AAA  1
# 2:  CCC  3
# 3:  BBB  2
# 4:  DDD  4

df1[idkey, on = "NAME"]
#    NAME SIZE ID
# 1:  AAA  0.9  1
# 2:  CCC  1.4  3
# 3:  BBB  1.7  2
# 4:  DDD  1.1  4

基础R方法可能类似于：

idkey <- unique(
  data.frame(NAME = unlist(df[grep("NAME", names(df))], use.names = FALSE), 
             ID = unlist(df[grep("ID", names(df))], use.names = FALSE)))
merge(df1, idkey, by = "NAME")
#   NAME SIZE ID
# 1  AAA  0.9  1
# 2  BBB  1.7  2
# 3  CCC  1.4  3
# 4  DDD  1.1  4

从另一个数据框添加ID列匹配值

1 个答案: