有没有办法将数据帧的列与完整列名向量相匹配?

你可以用 setdiff 获取缺少的列名,然后添加新列。

column_names <- c("col1","col2","col3","col4","col5","col6","col7")

df <- data.frame(col1 = c(20,30,10,20,45),col2=c(1,10,2,45,2),col4=c(9,10,22,11,12),col5=c(87,7,7,7,14),col6=c(89,90,99,12,1))serdi(colnames(df),column_names)

missing_cols <- setdiff(column_names,colnames(df))
df[missing_cols] <- NA

df <- df[column_names]

df

  col1 col2 col3 col4 col5 col6 col7
1   20    1   NA    9   87   89   NA
2   30   10   NA   10    7   90   NA
3   10    2   NA   22    7   99   NA
4   20   45   NA   11    7   12   NA
5   45    2   NA   12   14    1.  NA

cbind(df1, lapply(
  setdiff(column_names, names(df1)),
  \(x) setNames(data.frame(NA), x)
))[column_names]

#   col1 col2 col3 col4 col5 col6 col7
# 1   20    1    9   NA   87   89   NA
# 2   30   10   10   NA    7   90   NA
# 3   10    2   22   NA    7   99   NA
# 4   20   45   11   NA    7   12   NA
# 5   45    2   12   NA   14    1   NA

我提出了这个矢量化的解决方案:

## original number of columns in `df1`
nc <- length(df1)
## augment `df1` with an extra column of NA
df1$na <- NA
## map all columns not in `df1` to this extra NA column
df2 <- df1[match(column_names, names(df1)[1:nc], nomatch = nc + 1)]
## fix names to desired ones
names(df2) <- column_names
## see the result
df2
#  col1 col2 col3 col4 col5 col6 col7
#1   20    1    9   NA   87   89   NA
#2   30   10   10   NA    7   90   NA
#3   10    2   22   NA    7   99   NA
#4   20   45   11   NA    7   12   NA
#5   45    2   12   NA   14    1   NA

在这里,我处理数据。框架作为命名列表并使用 [] 用于所有索引操作。同样的想法也适用于命名列表和命名向量。