我有两个数据集
df1 <- structure(list(V1 = structure(1:12, .Label = c("A0A0A6YXQ7",
"A0A0A6YXS5", "A0A0A6YXW8", "A0A0A6YXX6", "A0A0A6YXZ1", "A0A0A6YY28",
"A0A0A6YY43", "A0A0A6YY47", "A0A0A6YY78", "A0A0A6YY89", "A0A0A6YY91",
"A0A0A7NQN9"), class = "factor")), .Names = "from", class = "data.frame", row.names = c(NA,
-12L))
和
df2 <- structure(list(from = structure(1:8, .Label = c("A0A0A6YXQ7",
"A0A0A6YXW8", "A0A0A6YXX6", "A0A0A6YXZ1", "A0A0A6YY28", "A0A0A6YY47",
"A0A0A6YY78", "A0A0A6YY91"), class = "factor"), to = structure(c(4L,
5L, 1L, 2L, 6L, 3L, 3L, 3L), .Label = c("Arhgap15", "Igtp", "MumuTL",
"Myo1f", "Pak2", "pol"), class = "factor")), .Names = c("from",
"to"), class = "data.frame", row.names = c(NA, -8L))
df1和df2有一个名为from的列
df1中的所有字符串都应该在df2中。如果不是，我想将它们显示在df1中的确切顺序，对于列to，它们会得到NA
例如，在df2中缺少以下字符串
A0A0A6YXS5和A0A0A6YY43以及A0A0A6YY89和A0A0A7NQN9
所以输出看起来像这样
From To
A0A0A6YXQ7 Myo1f
A0A0A6YXS5 NA
A0A0A6YXW8 Pak2
A0A0A6YXX6 Arhgap15
A0A0A6YXZ1 Igtp
A0A0A6YY28 pol
A0A0A6YY43 NA
A0A0A6YY47 MumuTL
A0A0A6YY78 MumuTL
A0A0A6YY89 NA
A0A0A6YY91 MumuTL
A0A0A7NQN9 NA
一个不太好的解决方案是合并这两个数据框
merge(df1, df2, by = "from", all.x = TRUE)
适用于某些数据但不适用于其他数据
让我们看看更大的数据
df1<- structure(list(from = structure(c(10L, 11L, 12L, 13L, 14L, 15L,
16L, 17L, 18L, 19L, 20L, 21L, 1L, 2L, 3L, 4L, 5L, 6L, 7L, 8L,
9L), .Label = c("A0A023J6K5", "A0A023J6L7", "A0A023J6M1", "A0A067XG53",
"A0A067XKM5", "A0A067XKP8", "A0A067XKR4", "A0A067XKW4", "A0A067XKW7",
"A0A0A6YXQ7", "A0A0A6YXS5", "A0A0A6YXW8", "A0A0A6YXX6", "A0A0A6YXZ1",
"A0A0A6YY28", "A0A0A6YY43", "A0A0A6YY47", "A0A0A6YY78", "A0A0A6YY89",
"A0A0A6YY91", "A0A0A7NQN9"), class = "factor")), .Names = "from", class = "data.frame", row.names = c(NA,
-21L))
和
df2 <- structure(list(from = structure(c(10L, 11L, 12L, 13L, 14L, 15L,
16L, 17L, 1L, 2L, 3L, 4L, 5L, 6L, 7L, 8L, 9L), .Label = c("A0A023J6K5",
"A0A023J6L7", "A0A023J6M1", "A0A023T778", "A0A067XG53", "A0A067XKM5",
"A0A067XKP8", "A0A067XKR4", "A0A067XKW4", "A0A0A6YXQ7", "A0A0A6YXW8",
"A0A0A6YXX6", "A0A0A6YXZ1", "A0A0A6YY28", "A0A0A6YY47", "A0A0A6YY78",
"A0A0A6YY91"), class = "factor"), to = structure(c(7L, 10L, 1L,
4L, 11L, 6L, 6L, 6L, 9L, 8L, 9L, 5L, 2L, 8L, 3L, 9L, 8L), .Label = c("Arhgap15",
"Cask", "COXI", "Igtp", "Magohb", "MumuTL", "Myo1f", "ND1", "ND4",
"Pak2", "pol"), class = "factor")), .Names = c("from", "to"), class = "data.frame", row.names = c(NA,
-17L))
然后我需要获得以下输出
from to
A0A0A6YXQ7 Myo1f
A0A0A6YXS5
A0A0A6YXW8 Pak2
A0A0A6YXX6 Arhgap15
A0A0A6YXZ1 Igtp
A0A0A6YY28 pol
A0A0A6YY43
A0A0A6YY47 MumuTL
A0A0A6YY78 MumuTL
A0A0A6YY89
A0A0A6YY91 MumuTL
A0A0A7NQN9
A0A023J6K5 ND4
A0A023J6L7 ND1
A0A023J6M1 ND4
A0A067XG53 Cask
A0A067XKM5 ND1
A0A067XKP8 COXI
A0A067XKR4 ND4
A0A067XKW4 ND1
A0A067XKW7
但如果我这样做
#> merge(df1, df2, by = "from", all.x = TRUE)
# from to
#1 A0A023J6K5 ND4
#2 A0A023J6L7 ND1
#3 A0A023J6M1 ND4
#4 A0A067XG53 Cask
#5 A0A067XKM5 ND1
#6 A0A067XKP8 COXI
#7 A0A067XKR4 ND4
#8 A0A067XKW4 ND1
#9 A0A067XKW7 <NA>
#10 A0A0A6YXQ7 Myo1f
#11 A0A0A6YXS5 <NA>
#12 A0A0A6YXW8 Pak2
#13 A0A0A6YXX6 Arhgap15
#14 A0A0A6YXZ1 Igtp
#15 A0A0A6YY28 pol
#16 A0A0A6YY43 <NA>
#17 A0A0A6YY47 MumuTL
#18 A0A0A6YY78 MumuTL
#19 A0A0A6YY89 <NA>
#20 A0A0A6YY91 MumuTL
#21 A0A0A7NQN9 <NA>
我基本上希望保持df1的顺序相同，只需在to中将值粘贴到from相似的地方。
答案 0 :(得分：0)
dplyr::left_join不同， merge保留了行顺序。使用更大的示例数据：
dplyr::left_join(df1, df2)
# Joining, by = "from"
# from to
# 1 A0A0A6YXQ7 Myo1f
# 2 A0A0A6YXS5 <NA>
# 3 A0A0A6YXW8 Pak2
# 4 A0A0A6YXX6 Arhgap15
# 5 A0A0A6YXZ1 Igtp
# 6 A0A0A6YY28 pol
# 7 A0A0A6YY43 <NA>
# 8 A0A0A6YY47 MumuTL
# 9 A0A0A6YY78 MumuTL
# 10 A0A0A6YY89 <NA>
# 11 A0A0A6YY91 MumuTL
# 12 A0A0A7NQN9 <NA>
# 13 A0A023J6K5 ND4
# 14 A0A023J6L7 ND1
# 15 A0A023J6M1 ND4
# 16 A0A067XG53 Cask
# 17 A0A067XKM5 ND1
# 18 A0A067XKP8 COXI
# 19 A0A067XKR4 ND4
# 20 A0A067XKW4 ND1
# 21 A0A067XKW7 <NA>
# Warning message:
# Column `from` joining factors with different levels, coercing to character vector