Question

如果我有一个看起来像这样的表（用read.table读入）：

http://i.stack.imgur.com/gaef6.png

（0是占位符，所以我有一致的行数）

有没有办法可以添加具有相同相应名称的值（a的两个值，l的所有4个值）并将其作为数据框输出？此外，行不一致（即一些列有4个，有些有2个）

结果应如下所示：

http://i.stack.imgur.com/HrCt5.png

我可以使用a，b等将数据框组合为行名称和列值相加后的列，但我无法弄清楚如何根据相应的名称对它们求和。

这就是我目前正在接近这个：

read.table()
定义空data.frame
使用循环来提取和添加相应名称的列值？＆lt; -need help with this
cbind()我刚刚生成的数据框和列
继续循环结束

- 使用row.names（）和colnames（）来更改行名和列名

到目前为止我的代码：

setwd(wd)
read.table(dat.txt,sep="\t")->x
read.table(total.txt, sep="\t")->total
met<-c("a","b","c","d","e","f","g","h","i","j","k","l","m","n","o","p","q","r")
y<-data.frame(),
total[2,]->name
for g in 1:ncol(x){
  #column will be the column with the combined values according to name
  cbind(y,column)
}
row.names(y)<-met
colnames(y)<-name
write.table(y,file="data.txt",sep="\t")

非常感谢任何帮助。

Answer 1

你可以分解为两部分：

（1）汇总每一列（2）结合成一张漂亮的桌子

使用* apply可以很好地完成。我把它分解为for循环以便于阅读

  ## sample df 
  df <- structure(list(thing1.met = c("a", "a", "b", "b", "c", "c", "d", "d", "e", "e", "f", "f", "g", "g", "h", "h", "i", "i", "j", "j", "k", "k", "l", "l", "m", "n", "o", "p", "q", "r"), thing1 = c(-0.57, 0.42, -1.12, -0.5, -0.94, -1.87, -2.22, -0.33, 1.45, 0.46, -1.96, -0.35, -1.01, 0.72, 0.04, -0.21, 0.81, 1.28, -0.52, -1.19, 0.03, -1.71, 0.53, -1.96, 1.58, -0.1, -0.88, 0.92, 0.02, -0.91), thing2.met = c("a", "a", "a", "b", "b", "c", "c", "c", "d", "e", "e", "f", "f", "g", "g", "h", "h", "i", "i", "j", "j", "k", "l", "l", "m", "n", "o", "p", "q", "r"), thing2 = c(-0.06, -0.7, 0.16, 1.96, 0.78, -0.65, -0.17, 0.89, 0.68, -0.93, -1.44, -0.16, -0.52, -0.19, 1.15, -0.77, 0.69, -0.48, 1.75, 1.62, -0.68, -1.06, -1.2, 1.42, -0.2, 1.33, 2.24, 0.35, 2, -1.21), thing3.met = c("a", "a", "a", "b", "c", "c", "d", "d", "e", "e", "f", "f", "g", "g", "g", "g", "g", "i", "k", "l", "l", "l", "m", "o", "o", "o", "p", "p", "p", "q"),     thing3 = c(1.27, 4.45, -2.42, -9.53, 3.33, 5.58, -2.94, 2.54,     12.44, 12.41, 7.6, 0.63, 5.67, -3.79, 12.28, 1.77, -0.4,     -0.04, 0.95, 4.93, 1.77, 0.37, -2.79, 2.36, 12.76, -5.4,     -4.73, -1.8, 0.52, -4.97)), .Names = c("thing1.met", "thing1", "thing2.met", "thing2", "thing3.met", "thing3"), row.names = c(NA, -30L), class = "data.frame")

  # create blank data frame
  results <- data.frame(row.names=met)

  # grab every other column
  cols <- seq(2, ncol(df), 2) 

  # aggregate over all m, over every pair of columns
  for (m in met) { 
    for (c in cols) {
      results[m, c/2] <- sum(df[,c][df[,c-1]==m])
    }
  }

  # clean up column names
  names(results) <- names(df)[cols]

  # final output
  results

示例：

    > results
       thing1 thing2 thing3
    a  -0.15  -0.60   3.30
    b  -1.62   2.74  -9.53
    c  -2.81   0.07   8.91
    d  -2.55   0.68  -0.40
    e   1.91  -2.37  24.85
    f  -2.31  -0.68   8.23
    g  -0.29   0.96  15.53
    h  -0.17  -0.08   0.00
    i   2.09   1.27  -0.04
    j  -1.71   0.94   0.00
    k  -1.68  -1.06   0.95
    l  -1.43   0.22   7.07
    m   1.58  -0.20  -2.79
    n  -0.10   1.33   0.00
    o  -0.88   2.24   9.72
    p   0.92   0.35  -6.01
    q   0.02   2.00  -4.97
    r  -0.91  -1.21   0.00

将具有相同名称的项目组合到数据框中

1 个答案: