Question

如果我想将名称从2列更改为结尾，为什么我的命令不起作用？

fredTable <- structure(list(Symbol = structure(c(3L, 1L, 4L, 2L, 5L), .Label = c("CASACBM027SBOG", 
"FRPACBW027SBOG", "TLAACBM027SBOG", "TOTBKCR", "USNIM"), class = "factor"), 
    Name = structure(1:5, .Label = c("bankAssets", "bankCash", 
    "bankCredWk", "bankFFRRPWk", "bankIntMargQtr"), class = "factor"), 
    Category = structure(c(1L, 1L, 1L, 1L, 1L), .Label = "Banks", class = "factor"), 
    Country = structure(c(1L, 1L, 1L, 1L, 1L), .Label = "USA", class = "factor"), 
    Lead = structure(c(1L, 1L, 3L, 3L, 2L), .Label = c("Monthly", 
    "Quarterly", "Weekly"), class = "factor"), Freq = structure(c(2L, 
    1L, 3L, 3L, 4L), .Label = c("1947-01-01", "1973-01-01", "1973-01-03", 
    "1984-01-01"), class = "factor"), Start = structure(c(1L, 
    1L, 1L, 1L, 1L), .Label = "Current", class = "factor"), End = c(TRUE, 
    TRUE, TRUE, TRUE, FALSE), SeasAdj = c(FALSE, FALSE, FALSE, 
    FALSE, TRUE), Percent = structure(c(1L, 1L, 1L, 1L, 1L), .Label = "Fed", class = "factor"), 
    Source = structure(c(1L, 1L, 1L, 1L, 1L), .Label = "Res", class = "factor"), 
    Series = structure(c(1L, 1L, 1L, 1L, 2L), .Label = c("Level", 
    "Ratio"), class = "factor")), .Names = c("Symbol", "Name", 
"Category", "Country", "Lead", "Freq", "Start", "End", "SeasAdj", 
"Percent", "Source", "Series"), row.names = c("1", "2", "3", 
"4", "5"), class = "data.frame")

然后，为了将第二列名称更改为结尾，我使用以下顺序但不起作用

names(fredTable[,-1]) = paste("case", 1:ncol(fredTable[,-1]), sep = "")

或

names(fredTable)[,-1] = paste("case", 1:ncol(fredTable)[,-1], sep = "")

一般来说，如何更改特定列的列名称 2到结束，2到7等等，并将其设置为名称s /他喜欢

Answer 1

通过在函数外部进行子集化替换特定的列名，而不是像第一次尝试中的names函数一样：

> names(fredTable)[-1] <- paste("case", 1:ncol(fredTable[,-1]), sep = "")

<强>解释

如果我们将新名称保存在向量newnames中，我们可以使用替换函数调查底层的内容。

#These are the names that will replace the old names
newnames <- paste("case", 1:ncol(fredTable[,-1]), sep = "")

我们应该始终用以下格式替换特定的列名：

#The right way to replace the second name only
names(df)[2] <- "newvalue"

#The wrong way
names(df[2]) <- "newvalue"

问题是您正在尝试创建新的列名向量，然后将输出分配给数据框。这两项操作在正确更换时同时完成。

正确的方式[内部]

我们可以使用：

扩展函数调用

#We enter this:
names(fredTable)[-1] <- newnames

#This is carried out on the inside
`names<-`(fredTable, `[<-`(names(fredTable), -1, newnames))

错误的方式[内部]

更换错误方式的内部是这样的：

#Wrong way
names(fredTable[-1]) <- newnames

#Wrong way Internal
`names<-`(fredTable[-1], newnames)

请注意，没有`[<-`分配。子集化数据框fredTable[-1]在全局环境中不存在，因此不会发生`names<-`的分配。

更改数据框中某些列的名称

1 个答案: