按名称重命名多个列

 names(x) <- gsub("q", "A", gsub("e", "B", names(x) ) )

rename(df,newname=oldname.x,newname=oldname.y)

您可以获取名称集，将其保存为列表，然后对字符串进行批量重命名。一个很好的例子是当您在数据集上执行长到宽的转换时：

names(labWide)
      Lab1    Lab10    Lab11    Lab12    Lab13    Lab14    Lab15    Lab16
1 35.75366 22.79493 30.32075 34.25637 30.66477 32.04059 24.46663 22.53063

nameVec <- names(labWide)
nameVec <- gsub("Lab","LabLat",nameVec)

names(labWide) <- nameVec
"LabLat1"  "LabLat10" "LabLat11" "LabLat12" "LabLat13" "LabLat14""LabLat15"    "LabLat16" " 

附注：如果您希望将一个字符串连接到所有列名中，可以使用以下简单代码。

colnames(df) <- paste("renamed_",colnames(df),sep="")

有很多类似答案，因此我编写了该函数，以便您可以复制/粘贴。

rename <- function(x, old_names, new_names) {
    stopifnot(length(old_names) == length(new_names))
    # pull out the names that are actually in x
    old_nms <- old_names[old_names %in% names(x)]
    new_nms <- new_names[old_names %in% names(x)]

    # call out the column names that don't exist
    not_nms <- setdiff(old_names, old_nms)
    if(length(not_nms) > 0) {
        msg <- paste(paste(not_nms, collapse = ", "), 
            "are not columns in the dataframe, so won't be renamed.")
        warning(msg)
    }

    # rename
    names(x)[names(x) %in% old_nms] <- new_nms
    x
}

 x = data.frame(q = 1, w = 2, e = 3)
 rename(x, c("q", "e"), c("Q", "E"))

   Q w E
 1 1 2 3

names(data) <- data[row,]

假设data是您的数据框，row是包含新值的行号。

然后，您可以使用以下代码删除包含名称的行：

data <- data[-row,]

names(x) <- setNames(c(newnames, names(x)), c(oldnames, names(x)))[names(x)]

names(x) <- (\(.) setNames(c(newnames, .), c(oldnames, .))[.])(names(x)) #Variant

x
#  A w B
#1 1 2 3

使用 transform。

names(x) <- do.call(transform, c(list(as.list(setNames(names(x), names(x)))),
                                 as.list(setNames(newnames, oldnames))))

数据

x = data.frame(q=1,w=2,e=3)
oldnames = c("q","e")
newnames = c("A","B")

上面有很多使用专业软件包的好答案。这是一种仅使用基本R的简单方法。

df.rename.cols <- function(df, col2.list) {
  tlist <- transpose(col2.list)
    
  names(df)[which(names(df) %in% tlist[[1]])] <- tlist[[2]]

  df
} 

这里是一个例子：

df1 <- data.frame(A = c(1, 2), B = c(3, 4), C = c(5, 6), D = c(7, 8))
col.list <- list(c("A", "NewA"), c("C", "NewC"))
df.rename.cols(df1, col.list)

  NewA B NewC D
1    1 3    5 7
2    2 4    6 8

我最近在@agile bean的答案基础上构建了一个函数（使用rename_with，以前是rename_at），如果数据框中存在列名，则更改列名，这样当适用时，可以使异构数据框的列名相匹配。

循环肯定可以改进，但我想分享给后人。

创建示例数据框：

x= structure(list(observation_date = structure(c(18526L, 18784L, 
17601L), class = c("IDate", "Date")), year = c(2020L, 2021L, 
2018L)), sf_column = "geometry", agr = structure(c(id = NA_integer_, 
common_name = NA_integer_, scientific_name = NA_integer_, observation_count = NA_integer_, 
country = NA_integer_, country_code = NA_integer_, state = NA_integer_, 
state_code = NA_integer_, county = NA_integer_, county_code = NA_integer_, 
observation_date = NA_integer_, time_observations_started = NA_integer_, 
observer_id = NA_integer_, sampling_event_identifier = NA_integer_, 
protocol_type = NA_integer_, protocol_code = NA_integer_, duration_minutes = NA_integer_, 
effort_distance_km = NA_integer_, effort_area_ha = NA_integer_, 
number_observers = NA_integer_, all_species_reported = NA_integer_, 
group_identifier = NA_integer_, year = NA_integer_, checklist_id = NA_integer_, 
yday = NA_integer_), class = "factor", .Label = c("constant", 
"aggregate", "identity")), row.names = c("3", "3.1", "3.2"), class = "data.frame")

函数

match_col_names <- function(x){

  col_names <- list(date = c("observation_date", "date"),
                    C =    c("observation_count", "count","routetotal"),
                    yday  = c("dayofyear"),
                    latitude  = c("lat"),
                    longitude = c("lon","long")
                    )

  for(i in seq_along(col_names)){
    newname=names(col_names)[i]
    oldnames=col_names[[i]]

  toreplace = names(x)[which(names(x) %in% oldnames)]
  x <- x %>%
    rename_with(~newname, toreplace)
}

return(x)

}

应用函数

x <- match_col_names(x)