我有以下结构的列表,myList <- replicate(5, data.frame(id = 1:10, mean = runif(10)), simplify =F)
我想通过合并来减少它
myList %>% reduce(function(x, y) merge(x, y, by = 'id'))
但是,这会导致以下列名:
id mean.x mean.y mean.x mean.y mean
虽然我想要类似的东西
id mean1 mean2 mean3 mean4 mean5
其中数字基于
myList
的顺序.显然我可以迭代
1:length(myList)
,但我觉得这个解决方案不优雅。另一种选择是在减少函数中引入检查,但这会导致对列表中的每个元素进行新的线性时间搜索,所以我认为它不是很有效。有没有另一种方法来实现这一目标?
最佳答案
新答案:
使用 rbindlist
和 dcast
来自 data.table
-包裹:
library(data.table)
mydata <- rbindlist(myList, idcol = 'df')
dcast(mydata, id ~ paste0('mean',df), value.var = 'mean')
或者使用 tidyverse 包:
library(dplyr)
library(tidyr)
myList %>%
bind_rows(., .id = 'df') %>%
spread(df, mean) %>%
rename_at(-1, funs(paste0('mean',.)))
两者都给出(显示
data.table
-output):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
当有duplicates in
id
在 myList
中的一个或多个数据帧中,你必须适应dcast
-步至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')
这也适用于
id
中没有重复项时.tidyverse
-code 然后必须适应: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):
一个可能的解决方案:
# 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))))
这使:
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
关于r - 遍历列表时获取列表的元素编号,我们在Stack Overflow上找到一个类似的问题: https://stackoverflow.com/questions/46404066/