R 在遍历列表时获取列表的元素号

我有以下结构的列表

---

myList新答案：

使用
数据中的
rbindlist
和
dcast
。表
-包：

library(data.table)
mydata <- rbindlist(myList, idcol = 'df')
dcast(mydata, id ~ paste0('mean',df), value.var = 'mean')

两者都给出（
数据。表
-显示输出）：

当
myList
中的一个或多个数据帧存在时，您必须将
dcast
-步骤调整为
dcast（mydata，id+rowid（id，df）~paste0（'mean'，df），value.var='mean'）
，以获得正确的结果。检查以下示例以查看结果：

myList <- replicate(5, data.frame(id = sample(1:10, 10, TRUE), mean = runif(10)), simplify = FALSE)
mydata <- rbindlist(myList, idcol = 'df')
dcast(mydata, id + rowid(id,df) ~ paste0('mean',df), value.var = 'mean')


旧答案（仍然有效）：

一种可能的解决办法：

# option 1
myList <- mapply(function(x,y) {names(x)[2] = paste0('mean',y); x}, myList, 1:length(myList), SIMPLIFY = FALSE)
Reduce(function(x, y) merge(x, y, by = 'id'), myList)

# option 2 (quite similar to @zx8754's solution)
mydata <- Reduce(function(x, y) merge(x, y, by = 'id'), myList)
setNames(mydata, c('id', paste0('mean', seq_along(myList))))

与Reduce合并，然后更新列名：

res <- Reduce(function(...) merge(..., all = TRUE, by = "id"), myList)
colnames(res)[2:ncol(res)] <- paste0("mean", 1:length(myList))

res我们可以使用
set\u名称

library(tidyverse)
myList %>%
      reduce(merge, by = 'id') %>% 
      set_names(c("id", paste0("mean", 1:5)))
#   id      mean1       mean2      mean3      mean4     mean5
#1   1 0.07122593 0.480300675 0.34944190 0.48718226 0.9118796
#2   2 0.18375430 0.850652470 0.24780063 0.45148232 0.2587470
#3   3 0.18617054 0.526188340 0.48716956 0.53354343 0.9057241
#4   4 0.87838756 0.811985522 0.49024819 0.10412944 0.7830501
#5   5 0.29287646 0.974811919 0.31413846 0.01508965 0.4587954
#6   6 0.62304018 0.004421152 0.81053625 0.80032467 0.7630185
#7   7 0.78445890 0.006362844 0.73643248 0.15952795 0.4386658
#8   8 0.71568076 0.081139996 0.36933728 0.31771823 0.2794372
#9   9 0.25523328 0.081603285 0.00298272 0.33698950 0.2413859
#10 10 0.86274552 0.432177738 0.26064580 0.75639537 0.3125151

您还可以尝试修改
Reduce
（或
Reduce
）调用中的
函数，以自动添加索引：

Reduce(function(x, y){

         # get indices of columns that are not the common one, in x and y
         col_noby_x <- which(colnames(x) != "id")
         col_noby_y <- which(colnames(y) != "id")

         # maximum of indices in x (at the end of the column names)
         ind_x <- max(as.numeric(sub(".+(\\d+)$", "\\1", colnames(x)[col_noby_x]))) 

         # if there is no indice yet, put 1 and 2, else modify names only in y, taking the max value of indices in x plus one.
         if(!is.na(ind_x)) colnames(y)[col_noby_y] <- paste0(colnames(y)[col_noby_y], ind_x +1) else {colnames(x)[col_noby_x] <- paste0(colnames(x)[col_noby_x], 1); colnames(y)[col_noby_y] <- paste0(colnames(y)[col_noby_y], 2)}  

         # finally merge
         merge(x, y, by="id")}, myList)      

#   id      mean1     mean2     mean3     mean4      mean5
#1   1 0.10698388 0.0277198 0.5109345 0.8885772 0.79983437
#2   2 0.29750846 0.7951743 0.9558739 0.9691619 0.31805857
#3   3 0.07115142 0.2401011 0.8106464 0.5101563 0.78697618
#4   4 0.39564336 0.7225532 0.7583893 0.4275574 0.77151883
#5   5 0.55860511 0.4111913 0.8403031 0.4284490 0.51489116
#6   6 0.92191777 0.9142926 0.4708712 0.2451099 0.84142501
#7   7 0.08218166 0.2741819 0.6772842 0.7939364 0.86930336
#8   8 0.35392512 0.2088531 0.0801731 0.2734870 0.62963218
#9   9 0.64068537 0.8427225 0.1904426 0.2389339 0.73145206
#10 10 0.31304719 0.9898133 0.8173664 0.2013031 0.04658273

Reduce（函数（x，y）{
#获取非常用列的索引（在x和y中）
col_noby_x这里有两个一行

使用
purrr:reduce2
和
dplyr:：inner_join
代替
merge
：

library(dplyr)
library(purrr)
myList %>% reduce2(map(2:length(.),~c("",.x)), inner_join, by = 'id',copy=F)

#    id       mean     mean2     mean3      mean4      mean5
# 1   1 0.44560715 0.4575765 0.6075921 0.06504922 0.90410342
# 2   2 0.60606716 0.5004711 0.7866959 0.89632285 0.09890028
# 3   3 0.59928281 0.4894146 0.4495071 0.66090212 0.56046997
# 4   4 0.55630819 0.4166869 0.1984523 0.08040737 0.18375885
# 5   5 0.97714203 0.1223497 0.7923596 0.53054508 0.93747149
# 6   6 0.07751312 0.6217220 0.3861749 0.30062805 0.03177210
# 7   7 0.22839323 0.3994350 0.6382234 0.98578452 0.27032222
# 8   8 0.73628572 0.8804618 0.8240999 0.44205508 0.73901477
# 9   9 0.81894510 0.2186181 0.9317510 0.60035660 0.65002083
# 10 10 0.26197059 0.5569660 0.9167330 0.58912675 0.81367176

或者使用
plyr:：join_all
和
tibble:：repair_name
（相同的输出）：

什么是
a
，什么是
reduce
？可能
a
实际上是
id
？而
%%>%%>%%
将是
%%>%
？reduce和merging是否保证在结束数据帧中id之后的第一列是列表的第一个元素？我知道我可以做到这一点，我在任务中提到了这个解决方案ion。由于可能存在性能问题，我不想这样做。我真的不想迭代1:length（myList）。这是一个可行的解决方案，但我觉得它并不优雅。@Joaquin为什么不觉得它优雅？无论如何，我找到了另一个不使用
1:length（myList）的解决方案
。查看更新，HTHI不相信您的第一个解决方案有任何问题，它是我在新解决方案之前使用的解决方案。我只是想知道是否有一种方法可以不迭代索引就完成它。这实际上不是一个关于如何使事情运行的问题，但是否有其他方法可以完成它。@Jaap，dcast answer对样本数据没有问题，但在以下方面失败：
myList@Moody_Mudskipper只需在
dcast
中使用
rowid
答案：
dcast（mydata，id+rowid（id，df）~paste0（'mean'，df），value.var='mean'）
看到你的评论后，我测试了你的答案。它适用于OP发布的原始示例，但不适用于我的答案中提到的你。这可能解释了否决票（不是我的）。@Jaap，我不明白它怎么不起作用，我得到的输出与OP的代码
myList%>%reduce（函数（x，y）merge相同（x，y，by='id'））
（除了列名）。也许我们的理解有所不同，我认为您的解决方案将均值的第一个实例放在第一行（对于给定的id），然后在下一行添加一个，用NAs填充每个缺少的值。我所做的是连续的内部联接，只保留公共的
id
值，我的第二个答案使用
tibble
包进行
repair\u names
，而不是我之前提到的
tidyr
，现在已更正，可能是导致id被否决的原因已标识：）。
myList %>% 
  bind_rows(., .id = 'df') %>% 
  group_by(df, id) %>% 
  mutate(ri = row_number()) %>% 
  ungroup() %>% 
  spread(df, mean) %>% 
  rename_at(3:7, funs(paste0('mean',.)))

# option 1
myList <- mapply(function(x,y) {names(x)[2] = paste0('mean',y); x}, myList, 1:length(myList), SIMPLIFY = FALSE)
Reduce(function(x, y) merge(x, y, by = 'id'), myList)

# option 2 (quite similar to @zx8754's solution)
mydata <- Reduce(function(x, y) merge(x, y, by = 'id'), myList)
setNames(mydata, c('id', paste0('mean', seq_along(myList))))

   id     mean1     mean2      mean3      mean4      mean5
1   1 0.1119114 0.4193226 0.86619590 0.52543072 0.52879193
2   2 0.4630863 0.8786721 0.02012432 0.77274088 0.09227344
3   3 0.9832522 0.4687838 0.49074271 0.01611625 0.69919423
4   4 0.7017467 0.7845002 0.44692958 0.64485570 0.40808345
5   5 0.6204856 0.1687563 0.54407165 0.54236973 0.09947167
6   6 0.1480965 0.7654041 0.43591864 0.22468554 0.84557988
7   7 0.0179509 0.3610114 0.45420122 0.20612154 0.76899342
8   8 0.9862083 0.5579173 0.13540519 0.97311401 0.13947602
9   9 0.3140737 0.2213044 0.05187671 0.07870425 0.23880332
10 10 0.4515313 0.2367271 0.65728768 0.22149073 0.90578043

res <- Reduce(function(...) merge(..., all = TRUE, by = "id"), myList)
colnames(res)[2:ncol(res)] <- paste0("mean", 1:length(myList))

library(tidyverse)
myList %>%
      reduce(merge, by = 'id') %>% 
      set_names(c("id", paste0("mean", 1:5)))
#   id      mean1       mean2      mean3      mean4     mean5
#1   1 0.07122593 0.480300675 0.34944190 0.48718226 0.9118796
#2   2 0.18375430 0.850652470 0.24780063 0.45148232 0.2587470
#3   3 0.18617054 0.526188340 0.48716956 0.53354343 0.9057241
#4   4 0.87838756 0.811985522 0.49024819 0.10412944 0.7830501
#5   5 0.29287646 0.974811919 0.31413846 0.01508965 0.4587954
#6   6 0.62304018 0.004421152 0.81053625 0.80032467 0.7630185
#7   7 0.78445890 0.006362844 0.73643248 0.15952795 0.4386658
#8   8 0.71568076 0.081139996 0.36933728 0.31771823 0.2794372
#9   9 0.25523328 0.081603285 0.00298272 0.33698950 0.2413859
#10 10 0.86274552 0.432177738 0.26064580 0.75639537 0.3125151

Reduce(function(x, y){

         # get indices of columns that are not the common one, in x and y
         col_noby_x <- which(colnames(x) != "id")
         col_noby_y <- which(colnames(y) != "id")

         # maximum of indices in x (at the end of the column names)
         ind_x <- max(as.numeric(sub(".+(\\d+)$", "\\1", colnames(x)[col_noby_x]))) 

         # if there is no indice yet, put 1 and 2, else modify names only in y, taking the max value of indices in x plus one.
         if(!is.na(ind_x)) colnames(y)[col_noby_y] <- paste0(colnames(y)[col_noby_y], ind_x +1) else {colnames(x)[col_noby_x] <- paste0(colnames(x)[col_noby_x], 1); colnames(y)[col_noby_y] <- paste0(colnames(y)[col_noby_y], 2)}  

         # finally merge
         merge(x, y, by="id")}, myList)      

#   id      mean1     mean2     mean3     mean4      mean5
#1   1 0.10698388 0.0277198 0.5109345 0.8885772 0.79983437
#2   2 0.29750846 0.7951743 0.9558739 0.9691619 0.31805857
#3   3 0.07115142 0.2401011 0.8106464 0.5101563 0.78697618
#4   4 0.39564336 0.7225532 0.7583893 0.4275574 0.77151883
#5   5 0.55860511 0.4111913 0.8403031 0.4284490 0.51489116
#6   6 0.92191777 0.9142926 0.4708712 0.2451099 0.84142501
#7   7 0.08218166 0.2741819 0.6772842 0.7939364 0.86930336
#8   8 0.35392512 0.2088531 0.0801731 0.2734870 0.62963218
#9   9 0.64068537 0.8427225 0.1904426 0.2389339 0.73145206
#10 10 0.31304719 0.9898133 0.8173664 0.2013031 0.04658273

library(dplyr)
library(purrr)
myList %>% reduce2(map(2:length(.),~c("",.x)), inner_join, by = 'id',copy=F)

#    id       mean     mean2     mean3      mean4      mean5
# 1   1 0.44560715 0.4575765 0.6075921 0.06504922 0.90410342
# 2   2 0.60606716 0.5004711 0.7866959 0.89632285 0.09890028
# 3   3 0.59928281 0.4894146 0.4495071 0.66090212 0.56046997
# 4   4 0.55630819 0.4166869 0.1984523 0.08040737 0.18375885
# 5   5 0.97714203 0.1223497 0.7923596 0.53054508 0.93747149
# 6   6 0.07751312 0.6217220 0.3861749 0.30062805 0.03177210
# 7   7 0.22839323 0.3994350 0.6382234 0.98578452 0.27032222
# 8   8 0.73628572 0.8804618 0.8240999 0.44205508 0.73901477
# 9   9 0.81894510 0.2186181 0.9317510 0.60035660 0.65002083
# 10 10 0.26197059 0.5569660 0.9167330 0.58912675 0.81367176

myList %>% join_all('id','inner') %>% repair_names