R 合并不同列的数据帧以创建新的数据帧，然后创建一个列表_R_List_Merge_Dataframe

R 合并不同列的数据帧以创建新的数据帧，然后创建一个列表

r list merge dataframe

R 合并不同列的数据帧以创建新的数据帧，然后创建一个列表,r,list,merge,dataframe,R,List,Merge,Dataframe,我有两个带有全局列ID的数据帧 df1: structure(list(ID = c("ILMN_1343291", "ILMN_1343292", "ILMN_1343293", "ILMN_1343294", "ILMN_1651209", "ILMN_1651217"), sample1 = c(15.8694849642673, 11.012283668738, 14.995174021271, 15.5837972369044, 8.08964880277197, 7.83406

我有两个带有全局列

ID

的数据帧

df1:

structure(list(ID = c("ILMN_1343291", "ILMN_1343292", "ILMN_1343293", 
"ILMN_1343294", "ILMN_1651209", "ILMN_1651217"), sample1 = c(15.8694849642673, 
11.012283668738, 14.995174021271, 15.5837972369044, 8.08964880277197, 
7.83406791970529), sample2 = c(15.8286939941996, 10.8407625194439, 
14.8818391783215, 15.5951964773238, 8.07397790649028, 7.85669590957356
), sample3 = c(15.8694849642673, 11.453825129904, 14.764465720783, 
15.6917233494908, 8.12884795713027, 7.86320578481668), sample4 = c(15.8694849642673, 
13.1291568945281, 14.8250076629948, 15.7193468523012, 8.07217210194779, 
7.84280084254162), sample5 = c(15.8694849642673, 10.223217362866, 
14.9595743905588, 15.362353159395, 8.12336672246356, 7.80055774746275
), sample6 = c(15.8694849642673, 11.3892037666505, 15.3837334429764, 
15.65923530912, 8.05738048722836, 7.81925084187836)), .Names = c("ID", 
"sample1", "sample2", "sample3", "sample4", "sample5", "sample6"
), class = "data.frame", row.names = c(NA, 6L))

structure(list(ID = c("ILMN_1343291", "ILMN_1343292", "ILMN_1343293", 
"ILMN_1343294", "ILMN_1651209", "ILMN_1651217"), PVALUE1 = c(0, 
0, 0, 0, 0, 0.4675325), PVALUE2 = c(0, 0, 0, 0, 0, 0.3290415), 
    PVALUE3 = c(0, 0, 0, 0, 0, 0.3074713), PVALUE4 = c(0, 0, 
    0, 0, 0, 0.3543759), PVALUE5 = c(0, 0, 0, 0, 0, 0.6465517
    ), PVALUE6 = c(0, 0, 0, 0, 0.001445087, 0.5390173)), .Names = c("ID", 
"PVALUE1", "PVALUE2", "PVALUE3", "PVALUE4", "PVALUE5", "PVALUE6"
), class = "data.frame", row.names = c(NA, 6L))

  [[1]]
      ID           sample1   PVALUE1
    1 ILMN_1343291 15.869485 0.0000000
    2 ILMN_1343292 11.012284 0.0000000
    3 ILMN_1343293 14.995174 0.0000000
    4 ILMN_1343294 15.583797 0.0000000
    5 ILMN_1651209  8.089649 0.0000000
    6 ILMN_1651217  7.834068 0.4675325
[[2]]
  ID           sample2   PVALUE2
1 ILMN_1343291 15.828694 0.0000000
2 ILMN_1343292 10.840763 0.0000000
3 ILMN_1343293 14.881839 0.0000000
4 ILMN_1343294 15.595196 0.0000000
5 ILMN_1651209  8.073978 0.0000000
6 ILMN_1651217  7.856696 0.3290415

[[3]]
  ID           sample3   PVALUE3
1 ILMN_1343291 15.869485 0.0000000
2 ILMN_1343292 11.453825 0.0000000
3 ILMN_1343293 14.764466 0.0000000
4 ILMN_1343294 15.691723 0.0000000
5 ILMN_1651209  8.128848 0.0000000
6 ILMN_1651217  7.863206 0.3074713

[[4]]
  ID           sample4   PVALUE4
1 ILMN_1343291 15.869485 0.0000000
2 ILMN_1343292 13.129157 0.0000000
3 ILMN_1343293 14.825008 0.0000000
4 ILMN_1343294 15.719347 0.0000000
5 ILMN_1651209  8.072172 0.0000000
6 ILMN_1651217  7.842801 0.3543759

[[5]]
  ID           sample5   PVALUE5
1 ILMN_1343291 15.869485 0.0000000
2 ILMN_1343292 10.223217 0.0000000
3 ILMN_1343293 14.959574 0.0000000
4 ILMN_1343294 15.362353 0.0000000
5 ILMN_1651209  8.123367 0.0000000
6 ILMN_1651217  7.800558 0.6465517

[[6]]
  ID           sample6     PVALUE6
1 ILMN_1343291 15.869485 0.000000000
2 ILMN_1343292 11.389204 0.000000000
3 ILMN_1343293 15.383733 0.000000000
4 ILMN_1343294 15.659235 0.000000000
5 ILMN_1651209  8.057380 0.001445087
6 ILMN_1651217  7.819251 0.539017300

df2:

structure(list(ID = c("ILMN_1343291", "ILMN_1343292", "ILMN_1343293", 
"ILMN_1343294", "ILMN_1651209", "ILMN_1651217"), sample1 = c(15.8694849642673, 
11.012283668738, 14.995174021271, 15.5837972369044, 8.08964880277197, 
7.83406791970529), sample2 = c(15.8286939941996, 10.8407625194439, 
14.8818391783215, 15.5951964773238, 8.07397790649028, 7.85669590957356
), sample3 = c(15.8694849642673, 11.453825129904, 14.764465720783, 
15.6917233494908, 8.12884795713027, 7.86320578481668), sample4 = c(15.8694849642673, 
13.1291568945281, 14.8250076629948, 15.7193468523012, 8.07217210194779, 
7.84280084254162), sample5 = c(15.8694849642673, 10.223217362866, 
14.9595743905588, 15.362353159395, 8.12336672246356, 7.80055774746275
), sample6 = c(15.8694849642673, 11.3892037666505, 15.3837334429764, 
15.65923530912, 8.05738048722836, 7.81925084187836)), .Names = c("ID", 
"sample1", "sample2", "sample3", "sample4", "sample5", "sample6"
), class = "data.frame", row.names = c(NA, 6L))

structure(list(ID = c("ILMN_1343291", "ILMN_1343292", "ILMN_1343293", 
"ILMN_1343294", "ILMN_1651209", "ILMN_1651217"), PVALUE1 = c(0, 
0, 0, 0, 0, 0.4675325), PVALUE2 = c(0, 0, 0, 0, 0, 0.3290415), 
    PVALUE3 = c(0, 0, 0, 0, 0, 0.3074713), PVALUE4 = c(0, 0, 
    0, 0, 0, 0.3543759), PVALUE5 = c(0, 0, 0, 0, 0, 0.6465517
    ), PVALUE6 = c(0, 0, 0, 0, 0.001445087, 0.5390173)), .Names = c("ID", 
"PVALUE1", "PVALUE2", "PVALUE3", "PVALUE4", "PVALUE5", "PVALUE6"
), class = "data.frame", row.names = c(NA, 6L))

  [[1]]
      ID           sample1   PVALUE1
    1 ILMN_1343291 15.869485 0.0000000
    2 ILMN_1343292 11.012284 0.0000000
    3 ILMN_1343293 14.995174 0.0000000
    4 ILMN_1343294 15.583797 0.0000000
    5 ILMN_1651209  8.089649 0.0000000
    6 ILMN_1651217  7.834068 0.4675325
[[2]]
  ID           sample2   PVALUE2
1 ILMN_1343291 15.828694 0.0000000
2 ILMN_1343292 10.840763 0.0000000
3 ILMN_1343293 14.881839 0.0000000
4 ILMN_1343294 15.595196 0.0000000
5 ILMN_1651209  8.073978 0.0000000
6 ILMN_1651217  7.856696 0.3290415

[[3]]
  ID           sample3   PVALUE3
1 ILMN_1343291 15.869485 0.0000000
2 ILMN_1343292 11.453825 0.0000000
3 ILMN_1343293 14.764466 0.0000000
4 ILMN_1343294 15.691723 0.0000000
5 ILMN_1651209  8.128848 0.0000000
6 ILMN_1651217  7.863206 0.3074713

[[4]]
  ID           sample4   PVALUE4
1 ILMN_1343291 15.869485 0.0000000
2 ILMN_1343292 13.129157 0.0000000
3 ILMN_1343293 14.825008 0.0000000
4 ILMN_1343294 15.719347 0.0000000
5 ILMN_1651209  8.072172 0.0000000
6 ILMN_1651217  7.842801 0.3543759

[[5]]
  ID           sample5   PVALUE5
1 ILMN_1343291 15.869485 0.0000000
2 ILMN_1343292 10.223217 0.0000000
3 ILMN_1343293 14.959574 0.0000000
4 ILMN_1343294 15.362353 0.0000000
5 ILMN_1651209  8.123367 0.0000000
6 ILMN_1651217  7.800558 0.6465517

[[6]]
  ID           sample6     PVALUE6
1 ILMN_1343291 15.869485 0.000000000
2 ILMN_1343292 11.389204 0.000000000
3 ILMN_1343293 15.383733 0.000000000
4 ILMN_1343294 15.659235 0.000000000
5 ILMN_1651209  8.057380 0.001445087
6 ILMN_1651217  7.819251 0.539017300

所需输出：

structure(list(ID = c("ILMN_1343291", "ILMN_1343292", "ILMN_1343293", 
"ILMN_1343294", "ILMN_1651209", "ILMN_1651217"), sample1 = c(15.8694849642673, 
11.012283668738, 14.995174021271, 15.5837972369044, 8.08964880277197, 
7.83406791970529), sample2 = c(15.8286939941996, 10.8407625194439, 
14.8818391783215, 15.5951964773238, 8.07397790649028, 7.85669590957356
), sample3 = c(15.8694849642673, 11.453825129904, 14.764465720783, 
15.6917233494908, 8.12884795713027, 7.86320578481668), sample4 = c(15.8694849642673, 
13.1291568945281, 14.8250076629948, 15.7193468523012, 8.07217210194779, 
7.84280084254162), sample5 = c(15.8694849642673, 10.223217362866, 
14.9595743905588, 15.362353159395, 8.12336672246356, 7.80055774746275
), sample6 = c(15.8694849642673, 11.3892037666505, 15.3837334429764, 
15.65923530912, 8.05738048722836, 7.81925084187836)), .Names = c("ID", 
"sample1", "sample2", "sample3", "sample4", "sample5", "sample6"
), class = "data.frame", row.names = c(NA, 6L))

structure(list(ID = c("ILMN_1343291", "ILMN_1343292", "ILMN_1343293", 
"ILMN_1343294", "ILMN_1651209", "ILMN_1651217"), PVALUE1 = c(0, 
0, 0, 0, 0, 0.4675325), PVALUE2 = c(0, 0, 0, 0, 0, 0.3290415), 
    PVALUE3 = c(0, 0, 0, 0, 0, 0.3074713), PVALUE4 = c(0, 0, 
    0, 0, 0, 0.3543759), PVALUE5 = c(0, 0, 0, 0, 0, 0.6465517
    ), PVALUE6 = c(0, 0, 0, 0, 0.001445087, 0.5390173)), .Names = c("ID", 
"PVALUE1", "PVALUE2", "PVALUE3", "PVALUE4", "PVALUE5", "PVALUE6"
), class = "data.frame", row.names = c(NA, 6L))

  [[1]]
      ID           sample1   PVALUE1
    1 ILMN_1343291 15.869485 0.0000000
    2 ILMN_1343292 11.012284 0.0000000
    3 ILMN_1343293 14.995174 0.0000000
    4 ILMN_1343294 15.583797 0.0000000
    5 ILMN_1651209  8.089649 0.0000000
    6 ILMN_1651217  7.834068 0.4675325
[[2]]
  ID           sample2   PVALUE2
1 ILMN_1343291 15.828694 0.0000000
2 ILMN_1343292 10.840763 0.0000000
3 ILMN_1343293 14.881839 0.0000000
4 ILMN_1343294 15.595196 0.0000000
5 ILMN_1651209  8.073978 0.0000000
6 ILMN_1651217  7.856696 0.3290415

[[3]]
  ID           sample3   PVALUE3
1 ILMN_1343291 15.869485 0.0000000
2 ILMN_1343292 11.453825 0.0000000
3 ILMN_1343293 14.764466 0.0000000
4 ILMN_1343294 15.691723 0.0000000
5 ILMN_1651209  8.128848 0.0000000
6 ILMN_1651217  7.863206 0.3074713

[[4]]
  ID           sample4   PVALUE4
1 ILMN_1343291 15.869485 0.0000000
2 ILMN_1343292 13.129157 0.0000000
3 ILMN_1343293 14.825008 0.0000000
4 ILMN_1343294 15.719347 0.0000000
5 ILMN_1651209  8.072172 0.0000000
6 ILMN_1651217  7.842801 0.3543759

[[5]]
  ID           sample5   PVALUE5
1 ILMN_1343291 15.869485 0.0000000
2 ILMN_1343292 10.223217 0.0000000
3 ILMN_1343293 14.959574 0.0000000
4 ILMN_1343294 15.362353 0.0000000
5 ILMN_1651209  8.123367 0.0000000
6 ILMN_1651217  7.800558 0.6465517

[[6]]
  ID           sample6     PVALUE6
1 ILMN_1343291 15.869485 0.000000000
2 ILMN_1343292 11.389204 0.000000000
3 ILMN_1343293 15.383733 0.000000000
4 ILMN_1343294 15.659235 0.000000000
5 ILMN_1651209  8.057380 0.001445087
6 ILMN_1651217  7.819251 0.539017300

我想合并

df1

和

df2

的不同列，然后创建所有数据帧的列表。我想通过

ID

合并

df1

和

df2

的第二列，这是数据帧中的第一列并存储在列表中，然后合并

df1

和

df2

存储在列表中的第三列，依此类推，直到所有列都完成。我知道我可以使用

merge（）

来合并两个数据帧，并使用

list（）

来存储。在这里，我需要在循环中这样做。如何才能做到这一点与最低限度的代码，否则我需要写一个漫长的代码来实现这一切。谢谢

这里有一个尝试

Res <- Map(function(x, y) cbind(df1[1], x, y), df1[-1], df2[-1])
tempnames <- rbind.data.frame(names(df1)[-1], names(df2)[-1])
Map(function(x, y) `names<-`(x, c("ID", as.character(y))), Res, tempnames)
# $sample1
#             ID   sample1   PVALUE1
# 1 ILMN_1343291 15.869485 0.0000000
# 2 ILMN_1343292 11.012284 0.0000000
# 3 ILMN_1343293 14.995174 0.0000000
# 4 ILMN_1343294 15.583797 0.0000000
# 5 ILMN_1651209  8.089649 0.0000000
# 6 ILMN_1651217  7.834068 0.4675325
# 
# $sample2
#             ID   sample2   PVALUE2
# 1 ILMN_1343291 15.828694 0.0000000
# 2 ILMN_1343292 10.840763 0.0000000
# 3 ILMN_1343293 14.881839 0.0000000
# 4 ILMN_1343294 15.595196 0.0000000
# 5 ILMN_1651209  8.073978 0.0000000
# 6 ILMN_1651217  7.856696 0.3290415
...

Res您想合并
还是cbind
？换句话说，这两个数据集是否每行都有相同的ID
s？是的，这里两个数据集都有相同的ID
sWillMap（data.frame，Map（cbind，df1[-1]，df2[-1]），ID=df1[1]）
或Map（data.frame，ID=df1[1]，Map（cbind，df1[-1]，df2[-1]））
为您工作？似乎很酷，除了删除原始列名并最后而不是首先分配列ID
。需要保留原始列名和列ID
，谢谢您的回答。我可以知道它是如何工作的吗？如果ID
不一致，我可以使用Map（）
合并吗？如果我有很多文件，并且创建所有文件的列表，那么merge
可能会更复杂。上面的案例考虑了一个文件。@ AgHaZuSain我不明白你的意思。上面提到的数据帧<代码> DF1和<代码> DF2，这是在组合创建多个文件的列表之后。现在有一个例子，我有很多文件，比如df1
和df2
，我想创建一个所有文件的列表Map（）
创建一个包含两个文件的列表，但我担心是否可以使用list[[length（list）+1]]
来存储文件。