根据R中的行值合并Dataframe

Posted 2021-05-04

我有5个数据帧，每个数据帧有不同的行数。我需要根据每个数据帧的col1中的值将它们合并在一起。守

df1 <- read.table(text="
   col1    col2
    A        5
    B        3
    C        6
    E        7", header=TRUE, stringsAsFactors=FALSE)

df2 <- read.table(text="
   col1    col2
    A        5
    B        6
    C       7
    M       8
    Z       9", header=T, stringsAsFactors=FALSE)

但我需要它来产生：

   newdf
     col1    col2(#from df1)   col3(#from df2)
      A          5                  5
      B          3                  6
      C          6                  7  
      E          7                  0
      M          0                  8
      Z          0                  9

我曾尝试合并一些by='col1'，但没有运气。有小费吗？

我尝试过的：

posidf<-merge(df1,df2,df3,df4,df5,all.x=TRUE)
#wont execute
posidf<-merge(df1,df2,df3,df4,df5,by="col1",all.x=TRUE)
#wont execute
posidf<-merge(df1,df2,df3,df4,df5,by="col1")
Error in fix.by(by.x, x) : 
'by' must specify one or more columns as numbers, names or logical

答案

我怀疑你正在寻找类似这样的例子：

merge(df1, df2, by = "col1", all.x=TRUE, all.y=TRUE)

编辑：

col1 <- c('A', 'B', 'C', 'E')
col2 <- c(5, 3, 6 ,7)
df1 <- data.frame(col1, col2)

col1 <- c('A', 'B', 'C', 'M', 'Z')
col2 <- c(5, 6, 7 ,8, 9)
df2 <- data.frame(col1, col2)

col1 <- c('A', 'B', 'C')
col2 <- c(10, 29, 7)
df3 <- data.frame(col1, col2)

col1 <- c('A', 'S', 'T')
col2 <- c(7 ,8, 9)
df4 <- data.frame(col1, col2)

col1 <- c('B', 'C')
col2 <- c(7 ,8)
df5 <- data.frame(col1, col2)

frame_list <- list(df1, df2, df3, df4, df5)
frame_names <- list('df1', 'df2', 'df3', 'df4', 'df5')
counter <- 0

df <- data.frame(Date=as.Date(character()),
                 File=character(), 
                 User=character(), 
                 stringsAsFactors=FALSE) 

df <- data.frame(matrix(ncol = 2, nrow = 0))
colnames(df) <- c("col1","col2")

for (d in frame_list) {
  counter <- counter + 1
  colnames(d) <- c("col1", paste0('col2_',frame_names[counter]))
  df <- merge(df, d, by = "col1", all.x=TRUE, all.y=TRUE)
}

df$col2 <- NULL
df[is.na(df)] <- 0

另一答案

假设您的合并看起来像这样，并且您的示例代表您的数据，则没有问题。

newdf <- merge(x = df1, y = df2,'col1')

##  col1 col2.x col2.y
##1    A      5      5
##2    B      3      6
##3    C      6      7

如果这不是您期望的输出，您可能需要查看不同类型的合并：内部联接（上方），外部联接，左外侧和右外侧。

外联接

merge(x = df1, y = df2, by = 'col1', all = TRUE)

##  col1 col2.x col2.y
##1    A      5      5
##2    B      3      6
##3    C      6      7
##4    E      7     NA
##5    M     NA      8
##6    Z     NA      9

左外

merge(x = df1, y = df2, by = 'col1', all.x = TRUE)

##  col1 col2.x col2.y
##1    A      5      5
##2    B      3      6
##3    C      6      7
##4    E      7     NA

右外

merge(x = df1, y = df2, by = 'col1', all.y = TRUE)

##  col1 col2.x col2.y
##1    A      5      5
##2    B      3      6
##3    C      6      7
##4    M     NA      8
##5    Z     NA      9

另一答案

我想你想要的是一个full_join。使用dplyr：

library(dplyr)

df1 <- data.frame(
  col1 = c("A", "B", "C", "E"),
  col2 = c(5, 3, 6, 7)
)

df2 <- data.frame(
  col1 = c("A", "B", "C", "M", "Z"),
  col2 = c(5, 6, 7, 8, 9)
)


df_merged <- full_join(x = df1, y = df2, by = "col1") 

如果你想要一个0而不是NA，请用它们替换它们

df_merged[is.na(df_merged)] <- 0

  col1 col2.x col2.y
1    A      5      5
2    B      3      6
3    C      6      7
4    E      7      0
5    M      0      8
6    Z      0      9

编辑多个数据框

将它们存储在列表中并使用reduce和相应的连接，这里是full_join

set.seed(123)
df_list <- replicate(5, data.frame(col1 = LETTERS[sample(1:26, 5)], col2 = sample(1:9, 5)), simplify = F)
reduce(df_list, full_join, by = "col1")

   col1 col2.x col2.y col2.x.x col2.y.y col2
1     E      9     NA       NA        1    4
2     F      5     NA       NA       NA   NA
3     N      2     NA       NA        8   NA
4     X      7      4       NA       NA    6
5     P      8     NA       NA        5   NA
6     D     NA      8       NA       NA   NA
7     Q     NA      5       NA       NA   NA
8     J     NA      1       NA       NA   NA
9     U     NA      2       NA        9    8
10    V     NA     NA        1       NA   NA
11    M     NA     NA        7       NA   NA
12    B     NA     NA        8       NA   NA
13    H     NA     NA        9       NA   NA
14    I     NA     NA        4       NA   NA
15    K     NA     NA       NA        6   NA
16    W     NA     NA       NA       NA    9
17    O     NA     NA       NA       NA    3

但正如@ zx8754在评论中所建议的那样：从这里复制Simultaneously merge multiple data.frames in a list