我有以下结构的列表，
myList <- replicate(5, data.frame(id = 1:10, mean = runif(10)), simplify =F)
我希望通过合并来减少它
myList %>% reduce(function(x, y) merge(x, y, by = 'id'))
然而，这会导致以下姓氏：
id mean.x mean.y mean.x mean.y mean
虽然我想要像
这样的东西 id mean1 mean2 mean3 mean4 mean5
数字基于myList的顺序。
显然我可以迭代1:length(myList)，但我发现这个解决方案不够优雅。其他选择是在reduce函数中引入一个检查，但这会导致对列表中每个元素进行新的线性时间搜索，所以我认为它不是非常有效。
还有另一种方法可以达到这个目的吗？
答案 0 :(得分：6)
新答案：
使用rbindlist包中的dcast和data.table：
library(data.table)
mydata <- rbindlist(myList, idcol = 'df')
dcast(mydata, id ~ paste0('mean',df), value.var = 'mean')
或者使用tidyverse软件包：
library(dplyr)
library(tidyr)
myList %>%
bind_rows(., .id = 'df') %>%
spread(df, mean) %>%
rename_at(-1, funs(paste0('mean',.)))
，它们都显示（data.table - 输出）：
id mean1 mean2 mean3 mean4 mean5 1: 1 0.6937674 0.005642891 0.4155868 0.74184186 0.54513885 2: 2 0.3602352 0.569412043 0.8018570 0.29177043 0.34521060 3: 3 0.6353133 0.512876032 0.8711914 0.44660086 0.16338451 4: 4 0.2106574 0.555638598 0.8240744 0.37495213 0.57443740 5: 5 0.9530160 0.059930577 0.0930678 0.39862717 0.91568414 6: 6 0.3723244 0.598526326 0.4970844 0.01978011 0.07832631 7: 7 0.2923137 0.712971846 0.3805590 0.25676592 0.11682605 8: 8 0.6208868 0.426853621 0.5533876 0.64054247 0.78949419 9: 9 0.9032609 0.274705843 0.3525957 0.46994429 0.32883110 10: 10 0.9707088 0.351394642 0.1089803 0.97969335 0.77791085
当myList中的一个或多个数据框中有duplicates in id时，您必须调整dcast - 步骤dcast(mydata, id + rowid(id,df) ~ paste0('mean',df), value.var = 'mean')以获得正确的结果。请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')
当id中没有重复项时，这也有效。
tidyverse - 代码随后适用于：
myList %>%
bind_rows(., .id = 'df') %>%
group_by(df, id) %>%
mutate(ri = row_number()) %>%
ungroup() %>%
spread(df, mean) %>%
rename_at(3:7, funs(paste0('mean',.)))
旧答案（仍然有效）：
可能的解决方案：
# 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))))
给出：
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
答案 1 :(得分：4)
您还可以尝试修改function（或Reduce）调用中的reduce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
答案 2 :(得分：3)
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))
答案 3 :(得分：2)
我们可以使用set_names
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
答案 4 :(得分：0)
这是两个一个衬垫
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
或使用plyr::join_all和tibble::repair_names（相同的输出）：
myList %>% join_all('id','inner') %>% repair_names