转换数据帧列表：不是简单的rbind，第二行到新列

转换数据帧列表：不是简单的rbind，第二行到新列

我有一张单子

employeesList = list(data.frame(first = ("Al"), second = "Jones"), 
                 data.frame(first = c("Al", "Barb"), second = c("Jones",       "Smith")),
             data.frame(first = c("Al", "Barb", "Carol"), second =   c("Jones", "Smith", "Adams")),
             data.frame(first = ("Al"), second = "Jones"))

我期待着生产这个

employeesDF = data.frame(first = c("Al", "Al", "Al", "Al"), second = c("Jones", "Jones", "Jones", "Jones"),
                        first2 = c(NA, "Barb", "Barb", NA), second2 = c(NA, "Smith", "Smith", NA),
                        first3 = c(NA, NA, "Carol", NA), second3 = c(NA, NA, "Adams", NA))

我希望每个数据框都是结果数据框中的一行。 请注意，转换时的第一个数据帧将有两列，转换时的第二个df将有四列，转换时的第三个df将产生6列，转换时的第四个df将产生2列，依此类推。我意识到必须有一个NA值填充 我做了一些研究，如果目标只是rbind，问题就解决了。我看不到解决我问题的办法。

看完这篇文章,， 我开始学英语了

res1 = cbind(t(employeesList[[1]][1]), t(employeesList[[1]][2]))
res2 = cbind(t(employeesList[[2]][1]), t(employeesList[[2]][2]))
res3 = cbind(t(employeesList[[3]][1]), t(employeesList[[3]][2]))
res4 = cbind(t(employeesList[[4]][1]), t(employeesList[[4]][2]))

然后

---

但是我可能有大量的数据帧--res1，…，resn。列数事先未指定，但可能少于10列。我的过程没有命名列，我认为这是绑定行所需要的。

我们可以使用

lappy

将列表转换为单行数据帧，然后使用

bind\u rows

将多个数据帧绑定在一起

library(dplyr)
bind_rows(lapply(employeesList, function(x) rbind.data.frame(c(t(x)))))

#   X.Al. X.Jones. X.Barb. X.Smith. X.Carol. X.Adams.
#1    Al    Jones    <NA>     <NA>     <NA>     <NA>
#2    Al    Jones    Barb    Smith     <NA>     <NA>
#3    Al    Jones    Barb    Smith    Carol    Adams
#4    Al    Jones    <NA>     <NA>     <NA>     <NA>

库（dplyr）
绑定行（lappy（employeesList，function（x）rbind.data.frame（c（t（x）щщ）'））
#X.Al。琼斯先生。倒钩。史密斯先生。卡罗尔。亚当斯。
#1艾尔·琼斯
#2艾尔·琼斯·巴伯·史密斯
#3艾尔·琼斯·巴伯·史密斯·卡罗尔·亚当斯
#4艾尔·琼斯

---

根据我们的偏好，我们可以稍后使用

setNames

重命名列。

可能是使用data.table

library('data.table')

rbindlist( l = lapply( employeesList, function(x) {
                  dcast( data    = melt( setDT( x ), measure.vars = c( 'first', 'second'))[, V1 := seq_along(value), by = variable][],
                         formula = " . ~ variable + V1")[, -1]
                }), 
           fill = TRUE, 
           use.names = TRUE )

#    first_1 second_1 first_2 second_2 first_3 second_3
# 1:      Al    Jones      NA       NA      NA       NA
# 2:      Al    Jones    Barb    Smith      NA       NA
# 3:      Al    Jones    Barb    Smith   Carol    Adams
# 4:      Al    Jones      NA       NA      NA       NA

---

这里有一个选项带有

聚集/扩散

library(tidyverse)
employeesList %>% 
   map_df(~ .x %>% 
               mutate_all(as.character) %>% # convert columns to character class
               mutate(n = row_number(), n = replace(n, n==1, "")),
             .id = 'grp') %>% 
  group_by(grp) %>%
  gather(key, val, first:second) %>% # gather to long format
  arrange(grp, n) %>% 
  unite(keyn, key, n, sep="") %>% # unite columns to create new column
  ungroup %>% 
  mutate(keyn = factor(keyn, levels = unique(keyn))) %>% # for column order
  spread(keyn, val) %>% # spread to wide format
  select(-grp)
# A tibble: 4 x 6
#  first second first2 second2 first3 second3
#  <chr> <chr>  <chr>  <chr>   <chr>  <chr>  
#1 Al    Jones  NA     NA      NA     NA     
#2 Al    Jones  Barb   Smith   NA     NA     
#3 Al    Jones  Barb   Smith   Carol  Adams  
#4 Al    Jones  NA     NA      NA     NA     

库（tidyverse）
员工名单%>%
map_df（~.x%>%
mutate_all（as.character）%>%#将列转换为字符类
mutate（n=row_number（），n=replace（n，n==1，“”），
.id='grp'）%>%
分组依据（grp）%>%
聚集（键、值、第一个：第二个）%>%#聚集为长格式
排列（grp，n）%>%
联合（键，键，n，sep=“”）%>%#联合列以创建新列
解组%>%
对列顺序进行变异（keyn=因子（keyn，levels=唯一（keyn））%>%#
排列（键，值）%>%#排列为宽格式
选择（-grp）
#一个tibble:4x6
#第一秒第一2秒第二2秒第一3秒3
#              
#1艾尔·琼斯·纳纳
#2艾尔·琼斯·巴伯·史密斯·纳纳
#3艾尔·琼斯·巴伯·史密斯·卡罗尔·亚当斯
#4艾尔·琼斯·纳纳

我在提交示例数据时犯了一个错误。在两个不同的方面，它不够笼统。 列名的变化可能不一致，数据的变化可能比指示的要大得多。 然后我在r-help问了我的问题。在那里，这个问题得到了多方面的回答。 以下是其他人提出的解决方案以及我的时间研究

# input data (list of data frames and data frames may have multiple rows)
employees4List = list(data.frame(first1 = "Al", second1 =
                                   "Jones"),
                      data.frame(first2 = c("Al2", "Barb"),
                                 second2 = c("Jones", "Smith")),
                      data.frame(first3 = c("Al3", "Barbara",
                                            "Carol"),
                                 second3 = c("Jones", "Smith",
                                             "Adams")),
                      data.frame(first4 = ("Al"), second4 =
                                   "Jones2"))
employees4List

# intermediate step (list of data frames with each just one row)
df1 = data.frame(First1 = "Al", Second1 = "Jones",
                 First2 = NA, Second2 = NA,
                 First3 = NA, Second3 = NA,
                 First4 = NA, Second4 = NA)
df2 = data.frame(First1 = "Al2", Second1 = "Jones",
                 First2 = "Barb", Second2 = "Smith",
                 First3 = NA, Second3 = NA,
                 First4 = NA, Second4 = NA)
df3 = data.frame(First1 = "Al3", Second1 = "Jones",
                 First2 = "Barbara", Second2 = "Smith",
                 First3 = "Carol", Second3 = "Adams",
                 First4 = NA, Second4 = NA)
df4 = data.frame(First1 = "Al", Second1 = "Jones2",
                 First2 = NA, Second2 = NA,
                 First3 = NA, Second3 = NA,
                 First4 = NA, Second4 = NA)
listFinal = list(df1, df2, df3, df4)
listFinal

# Expected final step, except that all columns should be character
# Just one data frame
dplyr::bind_rows(listFinal)
sapply(dplyr::bind_rows(listFinal), class)

# Solution 1 using base R by Sarah Goslee

dfbycol <- function(x) {
  x <- lapply(x, function(y)as.vector(t(as.matrix(y))))
  x <- lapply(x, function(y){length(y) <- max(sapply(x, length)); y})
  x <- do.call(rbind, x)
  x <- data.frame(x, stringsAsFactors=FALSE)
  colnames(x) <- paste0(c("first", "last"), rep(seq(1, ncol(x)/2), each=2))
  x
}

dfbycol(listFinal)

##########
# Solution 2 by Jeff Newmiller (Base R)

myrename2 <- function( DF, m ) {
  # if a pair of columns is not present, raise an error
  stopifnot( 2 == length( DF ) )
  n <- nrow( DF )
  # use memory layout of elements of matrix
  # t() automatically converts to matrix (nrow=2)
  # matrix(,nrow=1) re-interprets the column-major output of t()
  # as a single row matrix
  result <- as.data.frame( matrix( t( DF ), nrow = 1 )
                           , stringsAsFactors = FALSE
  )
  if ( n < m ) {
    result[ , seq( 2 * n + 1, 2 * m ) ] <- NA
  }
  setNames( result
            , sprintf( "%s%d"
                       , c( "First", "Second" )
                       , rep( seq.int( m ), each = 2 )
            )
  )
}

m <- max( unlist( lapply( employees4List, nrow ) ) )
listFinal2 <- lapply( employees4List, myrename2, m = m )
listFinal2

result2 <- do.call( rbind, listFinal2 )
result2

##########
# Solution 3 by Jeff Newmiller (uses dplyr)
myrename3 <- function( DF ) {
  # if a pair of columns is not present, raise an error
  stopifnot( 2 == length( DF ) )
  n <- nrow( DF )
  # use memory layout of elements of matrix
  # t() automatically converts to matrix (nrow=2)
  # matrix(,nrow=1) re-interprets the column-major output of t()
  # as a single row matrix
  setNames( as.data.frame( matrix( t( DF ), nrow = 1 )
                           , stringsAsFactors = FALSE
  )
  , sprintf( "%s%d"
             , c( "First", "Second" )
             , rep( seq.int( n ), each = 2 )
  )
  )
}

listFinal3 <- lapply( employees4List, myrename3 )
listFinal3
result3 <- dplyr::bind_rows( listFinal3 )
result3

# Solution 4 by Jeff Newmiller (uses dplyr and tidyr)

library(dplyr)
library(tidyr)
myrename4 <- function( DF ) {
  # if a pair of columns is not present, raise an error
  stopifnot( 2 == length( DF ) )
  names( DF ) <- c( "a", "b" )
  m <- nrow( DF )
  (  DF
    %>% mutate_all( as.character )
    %>% mutate( rw = LETTERS[ seq.int( n() ) ] )
    %>% gather( col, val, -rw )
    %>% tidyr::unite( "labels", rw, col, sep="" )
    %>% spread( labels, val )
    %>% setNames( sprintf( "%s%d"
                           , c( "First", "Second" )
                           , rep( seq.int( m ), each = 2 )
    )
    )
  )
}

listFinal4 <- lapply( employees4List, myrename3)
listFinal4
result4 <- dplyr::bind_rows(listFinal4)
result4

#####
# Timing
# Create a large dataset
firsts = c("Al", "Barb", "Carol")
seconds = c("Washington", "Adams", "Jefferson" )
numReplications = 10000
set.seed(2018)

# Create data frames
sim_list1 = replicate(n = numReplications,
                      expr = {data.frame(first = base::sample(x = firsts, size = 1, replace = TRUE),
                                         second = base::sample(x = seconds, size = 1, replace = TRUE))},
                      simplify = F)

sim_list2 = replicate(n = numReplications,
                      expr = {data.frame(first = base::sample(x = firsts, size = 2, replace = TRUE),
                                         second = base::sample(x = seconds, size = 2, replace = TRUE))},
                      simplify = F)

sim_list3 = replicate(n = numReplications,
                      expr = {data.frame(first = base::sample(x = firsts, size = 3, replace = TRUE),
                                         second = base::sample(x = seconds, size = 3, replace = TRUE))},
                      simplify = F)

# Create list
employeesList = c(sim_list1, sim_list2, sim_list3)

# Method 1

system.time(res1 <- dfbycol(employeesList))
# > system.time(dfbycol(employeesList))
# user  system elapsed 
# 757.87    0.18  758.62 
# res1
rm(res1)

#####
# Method 2

system.time(m <- max( unlist( lapply( employeesList, nrow ) ) ))
#    user  system elapsed 
#    0.22    0.00    0.22

system.time(listFinal2 <- lapply( employeesList, myrename2, m = m ) )
listFinal2
# user  system elapsed 
# 16.16    0.01   16.18 

system.time(result2 <- do.call( rbind, listFinal2 ) )
# result2
# user  system elapsed 
# 3.96    0.00    3.96
rm(listFinal2)
rm(result2)

#####
# Method 3

system.time(listFinal3 <- lapply( employeesList, myrename3))
# user  system elapsed 
# 7.33    0.00    7.33
listFinal3
system.time(result3 <- dplyr::bind_rows( listFinal3 ))
# user  system elapsed 
# 0.17    0.00    0.17
rm(listFinal3)
rm(result3)

#####
# Method 4

system.time(listFinal4 <- lapply( employeesList, myrename4) )
# user  system elapsed 
# 400.05    0.04  400.24 
listFinal4
system.time(result4 <- dplyr::bind_rows( listFinal4 ) )
#    user  system elapsed 
#   0.17    0.00    0.17 
# result4

#输入数据（数据帧列表和数据帧可能有多行）
employees4List=list（data.frame（first1=“Al”，second1=
“琼斯”），
数据帧（first2=c（“Al2”，“倒钩”），
second2=c（“琼斯”、“史密斯”），
data.frame（first3=c（“Al3”，“Barbara”，
“卡罗尔”），
second3=c（“琼斯”、“史密斯”，
"亚当斯"),，
data.frame（first4=（“Al”），second4=
“琼斯2号”）
雇员4名单
#中间步骤（每个数据帧只有一行的数据帧列表）
df1=data.frame（First1=“Al”，Second1=“Jones”，
第一个2=NA，第二个2=NA，
第一个3=NA，第二个3=NA，
第一个4=NA，第二个4=NA）
df2=data.frame（First1=“Al2”，Second1=“Jones”，
First2=“倒钩”，Second2=“史密斯”，
第一个3=NA，第二个3=NA，
第一个4=NA，第二个4=NA）
df3=data.frame（First1=“Al3”，Second1=“Jones”，
First2=“芭芭拉”，Second2=“史密斯”，
First3=“卡罗尔”，Second3=“亚当斯”，
第一个4=NA，第二个4=NA）
df4=data.frame（First1=“Al”，Second1=“Jones2”，
第一个2=NA，第二个2=NA，
第一个3=NA，第二个3=NA，
第一个4=NA，第二个4=NA）
listFinal=列表（df1、df2、df3、df4）
决赛
#预期为最后一步，但所有列都应为字符
#只有一个数据帧
dplyr:：绑定_行（listFinal）
sapply（dplyr:：bind_行（listFinal），类）
#解决方案1使用Sarah Goslee的base R
我决定对这三种方法进行时间测试。这种方法显然是赢家。使用此链接，我创建了数据帧列表。数据帧有一行、两行或三行。这三种类型各有10000个数据帧。方法1 dplyr#用户系统运行时间#44.94 0.00 45.03方法2数据表#用户系统运行时间#225.36 4.68 229.61方法3 tidyverse#用户系统运行时间#513.72 0.23 519.52
# input data (list of data frames and data frames may have multiple rows)
employees4List = list(data.frame(first1 = "Al", second1 =
                                   "Jones"),
                      data.frame(first2 = c("Al2", "Barb"),
                                 second2 = c("Jones", "Smith")),
                      data.frame(first3 = c("Al3", "Barbara",
                                            "Carol"),
                                 second3 = c("Jones", "Smith",
                                             "Adams")),
                      data.frame(first4 = ("Al"), second4 =
                                   "Jones2"))
employees4List

# intermediate step (list of data frames with each just one row)
df1 = data.frame(First1 = "Al", Second1 = "Jones",
                 First2 = NA, Second2 = NA,
                 First3 = NA, Second3 = NA,
                 First4 = NA, Second4 = NA)
df2 = data.frame(First1 = "Al2", Second1 = "Jones",
                 First2 = "Barb", Second2 = "Smith",
                 First3 = NA, Second3 = NA,
                 First4 = NA, Second4 = NA)
df3 = data.frame(First1 = "Al3", Second1 = "Jones",
                 First2 = "Barbara", Second2 = "Smith",
                 First3 = "Carol", Second3 = "Adams",
                 First4 = NA, Second4 = NA)
df4 = data.frame(First1 = "Al", Second1 = "Jones2",
                 First2 = NA, Second2 = NA,
                 First3 = NA, Second3 = NA,
                 First4 = NA, Second4 = NA)
listFinal = list(df1, df2, df3, df4)
listFinal

# Expected final step, except that all columns should be character
# Just one data frame
dplyr::bind_rows(listFinal)
sapply(dplyr::bind_rows(listFinal), class)

# Solution 1 using base R by Sarah Goslee

dfbycol <- function(x) {
  x <- lapply(x, function(y)as.vector(t(as.matrix(y))))
  x <- lapply(x, function(y){length(y) <- max(sapply(x, length)); y})
  x <- do.call(rbind, x)
  x <- data.frame(x, stringsAsFactors=FALSE)
  colnames(x) <- paste0(c("first", "last"), rep(seq(1, ncol(x)/2), each=2))
  x
}

dfbycol(listFinal)

##########
# Solution 2 by Jeff Newmiller (Base R)

myrename2 <- function( DF, m ) {
  # if a pair of columns is not present, raise an error
  stopifnot( 2 == length( DF ) )
  n <- nrow( DF )
  # use memory layout of elements of matrix
  # t() automatically converts to matrix (nrow=2)
  # matrix(,nrow=1) re-interprets the column-major output of t()
  # as a single row matrix
  result <- as.data.frame( matrix( t( DF ), nrow = 1 )
                           , stringsAsFactors = FALSE
  )
  if ( n < m ) {
    result[ , seq( 2 * n + 1, 2 * m ) ] <- NA
  }
  setNames( result
            , sprintf( "%s%d"
                       , c( "First", "Second" )
                       , rep( seq.int( m ), each = 2 )
            )
  )
}

m <- max( unlist( lapply( employees4List, nrow ) ) )
listFinal2 <- lapply( employees4List, myrename2, m = m )
listFinal2

result2 <- do.call( rbind, listFinal2 )
result2

##########
# Solution 3 by Jeff Newmiller (uses dplyr)
myrename3 <- function( DF ) {
  # if a pair of columns is not present, raise an error
  stopifnot( 2 == length( DF ) )
  n <- nrow( DF )
  # use memory layout of elements of matrix
  # t() automatically converts to matrix (nrow=2)
  # matrix(,nrow=1) re-interprets the column-major output of t()
  # as a single row matrix
  setNames( as.data.frame( matrix( t( DF ), nrow = 1 )
                           , stringsAsFactors = FALSE
  )
  , sprintf( "%s%d"
             , c( "First", "Second" )
             , rep( seq.int( n ), each = 2 )
  )
  )
}

listFinal3 <- lapply( employees4List, myrename3 )
listFinal3
result3 <- dplyr::bind_rows( listFinal3 )
result3

# Solution 4 by Jeff Newmiller (uses dplyr and tidyr)

library(dplyr)
library(tidyr)
myrename4 <- function( DF ) {
  # if a pair of columns is not present, raise an error
  stopifnot( 2 == length( DF ) )
  names( DF ) <- c( "a", "b" )
  m <- nrow( DF )
  (  DF
    %>% mutate_all( as.character )
    %>% mutate( rw = LETTERS[ seq.int( n() ) ] )
    %>% gather( col, val, -rw )
    %>% tidyr::unite( "labels", rw, col, sep="" )
    %>% spread( labels, val )
    %>% setNames( sprintf( "%s%d"
                           , c( "First", "Second" )
                           , rep( seq.int( m ), each = 2 )
    )
    )
  )
}

listFinal4 <- lapply( employees4List, myrename3)
listFinal4
result4 <- dplyr::bind_rows(listFinal4)
result4

#####
# Timing
# Create a large dataset
firsts = c("Al", "Barb", "Carol")
seconds = c("Washington", "Adams", "Jefferson" )
numReplications = 10000
set.seed(2018)

# Create data frames
sim_list1 = replicate(n = numReplications,
                      expr = {data.frame(first = base::sample(x = firsts, size = 1, replace = TRUE),
                                         second = base::sample(x = seconds, size = 1, replace = TRUE))},
                      simplify = F)

sim_list2 = replicate(n = numReplications,
                      expr = {data.frame(first = base::sample(x = firsts, size = 2, replace = TRUE),
                                         second = base::sample(x = seconds, size = 2, replace = TRUE))},
                      simplify = F)

sim_list3 = replicate(n = numReplications,
                      expr = {data.frame(first = base::sample(x = firsts, size = 3, replace = TRUE),
                                         second = base::sample(x = seconds, size = 3, replace = TRUE))},
                      simplify = F)

# Create list
employeesList = c(sim_list1, sim_list2, sim_list3)

# Method 1

system.time(res1 <- dfbycol(employeesList))
# > system.time(dfbycol(employeesList))
# user  system elapsed 
# 757.87    0.18  758.62 
# res1
rm(res1)

#####
# Method 2

system.time(m <- max( unlist( lapply( employeesList, nrow ) ) ))
#    user  system elapsed 
#    0.22    0.00    0.22

system.time(listFinal2 <- lapply( employeesList, myrename2, m = m ) )
listFinal2
# user  system elapsed 
# 16.16    0.01   16.18 

system.time(result2 <- do.call( rbind, listFinal2 ) )
# result2
# user  system elapsed 
# 3.96    0.00    3.96
rm(listFinal2)
rm(result2)

#####
# Method 3

system.time(listFinal3 <- lapply( employeesList, myrename3))
# user  system elapsed 
# 7.33    0.00    7.33
listFinal3
system.time(result3 <- dplyr::bind_rows( listFinal3 ))
# user  system elapsed 
# 0.17    0.00    0.17
rm(listFinal3)
rm(result3)

#####
# Method 4

system.time(listFinal4 <- lapply( employeesList, myrename4) )
# user  system elapsed 
# 400.05    0.04  400.24 
listFinal4
system.time(result4 <- dplyr::bind_rows( listFinal4 ) )
#    user  system elapsed 
#   0.17    0.00    0.17 
# result4