Question

我想编写一些代码可以获取给定的数据框，检查是否缺少任何列，如果是，则添加缺少的列填充0或NA。这就是我所拥有的：

> df
   x1 x2 x4
1   0  1  3
2   3  1  3
3   1  2  1

> nameslist <- c("x1","x2","x3","x4")
> miss.names <- !nameslist %in% colnames(df)
> holder <- rbind(nameslist,miss.names)
> miss.cols <- subset(holder[1,], holder[2,] == "TRUE")

除此之外，我无法弄清楚如何在没有硬编码的情况下添加丢失的列（“x3”）。理想情况下，我希望新的完整数据框的列也与nameslist一样。

有什么想法吗？我当前的代码可以忽略，没问题。

Answer 1

这是一种直截了当的方法

df <- data.frame(a=1:4, e=4:1)
nms <- c("a", "b", "d", "e")   # Vector of columns you want in this data.frame

Missing <- setdiff(nms, names(df))  # Find names of missing columns
df[Missing] <- 0                    # Add them, filled with '0's
df <- df[nms]                       # Put columns in desired order
#   a b d e
# 1 1 0 0 4
# 2 2 0 0 3
# 3 3 0 0 2
# 4 4 0 0 1

Answer 2

library(stringr)
df <- data.frame(X1=1:4,X2=1:4,X5=1:4)
>df
  X1 X2 X5
1  1  1  1
2  2  2  2
3  3  3  3
4  4  4  4
current <- as.numeric(str_extract(names(df),"[0-9]"))
missing <-seq(min(current),max(current))

df[paste("X",missing[!missing %in% current],sep="")]<-0

>df[,order(colnames(df))]
  X1 X2 X3 X4 X5
1  1  1  0  0  1
2  2  2  0  0  2
3  3  3  0  0  3
4  4  4  0  0  4

Answer 3

谢谢你们，感谢你们，我设法用一组数据帧（文件）和另一个列名列表（ncolunas）来做到这一点。

 for (i in serieI)    {
     if ((identical(colnames(Files[[i]]),ncolunas)) == FALSE) {

         nms   = ncolunas
          df =   Files[[i]]
          aux = colnames(df)
          aux1 = row.names(df)

          Missing = setdiff(nms, colnames(df))  

          serie = seq(1,length(Missing)) #creating indices 1-5 for loop
          for (j in serie)  {    #loop to add colums with zeros
              df = cbind(df,c(0))
          }
          colnames(df) = c(aux,Missing)   #updates columns names

          df = df[,order(colnames(df))]  #put colums into order
          df = t(as.matrix(df))          #hanges into matrix
           row.names(df) = aux1          #update lines' names
           Files[[i]] = df              #updates object from list
     } 

 }

R：查找缺少的列，如果缺少则添加到数据框

3 个答案: