合并列并在没有值的地方添加NA

Question

我想从不同的数据帧合并19个不同长度的列并进行比较.这是一个例子:

df1: 
PA0001
PA0002
PA0003
PA0004
PA0005

df2: 
PA0001
PA0003
PA0006
PA0007


df3: 
PA0001
PA0007

等等...

输出是这样的:

PA0001  | PA0001 | PA0001
PA0002  |  NA    | NA
PA0003  | PA0003 | NA                   
PA0004  | NA     | NA
PA0005  | NA     | NA
NA      | PA0006 | NA
NA      | PA0007 | PA0007

我使用compare或merge功能,但我没有一个好结果.我试图使用这个问题的功能:链接

但我得到了这个错误:

Error in attributes(.Data) <- c(attributes(.Data), attrib) : 
  'names' attribute [5254] must be the same length as the vector [2]

在这里你是一个例子:

test1 <- data.frame(c("PA0001","PA0002","PA0003","PA0004","PA0005","PA0006"))
test2 <- data.frame(c("PA0001","PA0002","PA0004","PA0005","PA0007"))
test3 <- data.frame(c("PA0001","PA0004","PA0005","PA0007", "PA0008"))

非常感谢你.

Answer 1

如果我们需要输出作为OP的预期,将在数据集list,rbind的list元素,同时创造了"GRP"列rbindlist,然后dcast从"长"到"宽",同时通过创建公式中的一个顺序列match荷兰国际集团的"带有unique'id'元素的id'

library(data.table)
dcast(rbindlist(list(test1, test2, test3), idcol = 'grp'),
         match(id, unique(id)) ~ paste0("col", grp))[, id := NULL][]
#     col1   col2   col3
#1: PA0001 PA0001 PA0001
#2: PA0002     NA     NA
#3: PA0003 PA0003     NA
#4: PA0004     NA     NA
#5: PA0005     NA     NA
#6:     NA PA0006     NA
#7:     NA PA0007 PA0007

或者@jogo分割代码以使其更清晰,在第一步中,通过指定参数创建'grp'列的rbind所有数据集listidcol

t_all <- rbindlist(list(test1, test2, test3), idcol='grp');

然后dcast,以'宽'格式并将'id'列分配给NULL

dcast(t_all, id ~ grp, value.var='id')[, id := NULL][]

数据

test1 <- data.frame(id = c("PA0001","PA0002","PA0003","PA0004","PA0005"))
test2 <- data.frame(id = c("PA0001","PA0003","PA0006","PA0007"))
test3 <- data.frame(id = c("PA0001", "PA0007"))

Answer 2

您可以尝试tidyverse解决方案

library(tidyverse)
d1 <- read.table(text="PA0001
           PA0002
           PA0003
           PA0004
           PA0005")
d2 <- read.table(text="PA0001
           PA0003
           PA0006
           PA0007")
d3 <- read.table(text="PA0001
           PA0007")

list(d1, d2, d3) %>% 
  bind_rows(.id = "df") %>% 
  mutate(n = TRUE) %>% 
  spread(df, n, fill = FALSE)
      V1     1     2     3
1 PA0001  TRUE  TRUE  TRUE
2 PA0002  TRUE FALSE FALSE
3 PA0003  TRUE  TRUE FALSE
4 PA0004  TRUE FALSE FALSE
5 PA0005  TRUE FALSE FALSE
6 PA0006 FALSE  TRUE FALSE
7 PA0007 FALSE  TRUE  TRUE

我们的想法是包括所有data.frames在列表中,然后将它们绑定横行,添加逻辑TRUE和使用tidyr的spread功能,以获得结果.当然你也可以使用以下方法获得你的wxpected输出:

list(d1, d2, d3) %>% 
  bind_rows(.id="df") %>% 
  mutate(n=V1) %>% 
  spread(df, n) %>% 
  select(-1)
       1      2      3
1 PA0001 PA0001 PA0001
2 PA0002   <NA>   <NA>
3 PA0003 PA0003   <NA>
4 PA0004   <NA>   <NA>
5 PA0005   <NA>   <NA>
6   <NA> PA0006   <NA>
7   <NA> PA0007 PA000

在baseR你可以尝试:

Reduce(function(x, y) merge(x, y, by="V1", all.x = TRUE, all.y = TRUE),
              lapply(list(d1, d2, d3), function(x) cbind(x,V2=x$V1)))[,-1]
    V2.x   V2.y     V2
1 PA0001 PA0001 PA0001
2 PA0002   <NA>   <NA>
3 PA0003 PA0003   <NA>
4 PA0004   <NA>   <NA>
5 PA0005   <NA>   <NA>
6   <NA> PA0006   <NA>
7   <NA> PA0007 PA0007