Question

下面是示例数据，该数据是包含不同数据帧的列表。我想根据以下两个条件从中获取一个数据帧。

第一：

对于列表开始第1列中的每个数据框，请保留rbind() ing列，这些列的名称与前一列完全相同。遇到不同的列名称时，请将其和所有列放到最后一个。
例如：如果将列1命名为Banana，则将列2命名为Banana，但是将列3命名为Orange，然后将列4再次命名为Banana。然后，第1列和第2列将rbind()，第3列和第4列将被删除。
另一个示例：如果将列1命名为Banana，则将列2命名为Orange，但是将列3命名为Banana，则只有列1才能作为列2的开始名称是不同的，即使与第1列相同，我也不在乎第3列的名称。

第二：

在通过上述条件运行数据帧列表之后，我想合并列表中的所有数据帧以获得一个数据帧，我认为可以使用以下代码来实现。
在这里，lst2是第一个条件的输出。

do.call(rowr::cbind.fill, c(lst2, list(fill = 0)))

以上代码功劳@akrun。任何建议都会有所帮助。

样本数据

list(A = structure(list(`A-DIODE` = c(1.2, 0.4), `A-DIODE` = c(1.3, 
0.6)), row.names = c(NA, -2L), class = "data.frame"), B = structure(list(
    `B-DIODE` = c(1.4, 0.8), `B-ACC1` = c(1.5, 1), `B-ACC2` = c(1.6, 
    1.2), `B-ANA0` = c(1.7, 1.4), `B-ANA1` = c(1.8, 1.6), `B-BRICKID` = c(1.9, 
    1.8), `B-CC0` = c(2L, 2L), `B-CC1` = c(2.1, 2.2), `B-DIGDN` = c(2.2, 
    2.4), `B-DIGDP` = c(2.3, 2.6), `B-DN1` = c(2.4, 2.8), `B-DN2` = c(2.5, 
    3), `B-DP1` = c(2.6, 3.2), `B-DP2` = c(2.7, 3.4), `B-SCL` = c(2.8, 
    3.6), `B-SDA` = c(2.9, 3.8), `B-USB0DN` = 3:4, `B-USB0DP` = c(3.1, 
    4.2), `B-USB1DN` = c(3.2, 4.4), `B-USB1DP` = c(3.3, 4.6), 
    `B-ACC1` = c(3.4, 4.8), `B-ACC2` = c(3.5, 5), `B-ANA0` = c(3.6, 
    5.2), `B-ANA1` = c(3.7, 5.4), `B-BRICKID` = c(3.8, 5.6), 
    `B-CC0` = c(3.9, 5.8), `B-CC1` = c(4L, 6L), `B-DIGDN` = c(4.1, 
    6.2), `B-DIGDP` = c(4.2, 6.4), `B-DN1` = c(4.3, 6.6), `B-DN2` = c(4.4, 
    6.8), `B-DP1` = c(4.5, 7), `B-DP2` = c(4.6, 7.2), `B-SCL` = c(4.7, 
    7.4), `B-SDA` = c(4.8, 7.6), `B-USB0DN` = c(4.9, 7.8), `B-USB0DP` = c(5L, 
    8L), `B-USB1DN` = c(5.1, 8.2), `B-USB1DP` = c(5.2, 8.4), 
    `B-NA` = c(5.3, 8.6), `B-ACC2PWRLKG_0v4` = c(5.4, 8.8), `B-ACC2PWRLKG_0v4` = c(5.5, 
    9), `B-P_IN_Leak` = c(5.6, 9.2)), row.names = c(NA, -2L), class = "data.frame"))

更新1

在@ØysteinS回答之后，我意识到也应该有第三个条件：

第三：

如果列表中的一个数据框中只有一列，则仅将该列添加到父数据框中。

Answer 1

这应该可以完成工作：

data <- list(A = structure(list(`A-DIODE` = c(1.2, 0.4), `A-DIODE` = c(1.3, 
0.6)), row.names = c(NA, -2L), class = "data.frame"), B = structure(list(
    `B-DIODE` = c(1.4, 0.8), `B-ACC1` = c(1.5, 1), `B-ACC2` = c(1.6, 
    1.2), `B-ANA0` = c(1.7, 1.4), `B-ANA1` = c(1.8, 1.6), `B-BRICKID` = c(1.9, 
    1.8), `B-CC0` = c(2L, 2L), `B-CC1` = c(2.1, 2.2), `B-DIGDN` = c(2.2, 
    2.4), `B-DIGDP` = c(2.3, 2.6), `B-DN1` = c(2.4, 2.8), `B-DN2` = c(2.5, 
    3), `B-DP1` = c(2.6, 3.2), `B-DP2` = c(2.7, 3.4), `B-SCL` = c(2.8, 
    3.6), `B-SDA` = c(2.9, 3.8), `B-USB0DN` = 3:4, `B-USB0DP` = c(3.1, 
    4.2), `B-USB1DN` = c(3.2, 4.4), `B-USB1DP` = c(3.3, 4.6), 
    `B-ACC1` = c(3.4, 4.8), `B-ACC2` = c(3.5, 5), `B-ANA0` = c(3.6, 
    5.2), `B-ANA1` = c(3.7, 5.4), `B-BRICKID` = c(3.8, 5.6), 
    `B-CC0` = c(3.9, 5.8), `B-CC1` = c(4L, 6L), `B-DIGDN` = c(4.1, 
    6.2), `B-DIGDP` = c(4.2, 6.4), `B-DN1` = c(4.3, 6.6), `B-DN2` = c(4.4, 
    6.8), `B-DP1` = c(4.5, 7), `B-DP2` = c(4.6, 7.2), `B-SCL` = c(4.7, 
    7.4), `B-SDA` = c(4.8, 7.6), `B-USB0DN` = c(4.9, 7.8), `B-USB0DP` = c(5L, 
    8L), `B-USB1DN` = c(5.1, 8.2), `B-USB1DP` = c(5.2, 8.4), 
    `B-NA` = c(5.3, 8.6), `B-ACC2PWRLKG_0v4` = c(5.4, 8.8), `B-ACC2PWRLKG_0v4` = c(5.5, 
    9), `B-P_IN_Leak` = c(5.6, 9.2)), row.names = c(NA, -2L), class = "data.frame"))

# Use lapply to apply the same function to each data frame in the list.
combined_frames <- lapply(data, function(df){
  first_name <- names(df)[[1]]
  result <- df[, 1, drop = FALSE]
  # Keep adding if name is the same as the first
  if (ncol(df) != 1) {
   for(i in seq(2, length(names(df)), by = 1)){
     if(names(df)[[i]] == names(df)[[1]]){
       result <- rbind(result, df[, i, drop = FALSE])
     } else { 
       # Otherwise, break out of loop
       break
     }
   }
  }
  return(result)
})

# Yes, your suggested code seems to work as expected for the last task
do.call(rowr::cbind.fill, c(combined_frames, list(fill = 0)))
#>   A.DIODE B.DIODE
#> 1     1.2     1.4
#> 2     0.4     0.8
#> 3     1.3     0.0
#> 4     0.6     0.0

Answer 2

一个简单的选择是遍历list，获取列名的运行长度ID，仅提取等于1的列名，unlist，转换为data.frame首先是列名，然后是cbind.fill，将data.frame的list绑定在一起

library(data.table)
lst1 <- lapply(data, function(x) 
       setNames(data.frame(unlist(x[rleid(names(x)) == 1])), names(x)[1]))
do.call(rowr::cbind.fill, c(lst1, list(fill = 0)))
#    A.DIODE B.DIODE
#1     1.2     1.4
#2     0.4     0.8
#3     1.3     0.0
#4     0.6     0.0

如何根据条件在数据帧列表中行合并和删除列

2 个答案: