在dplyr中，如何删除和重命名不存在的列，操作所有名称，并使用字符串命名新变量？

How can I simplify or perform the following operations using dplyr:

如何使用dplyr简化或执行以下操作：

Run a function on all data.frame names, like mutate_each(funs()) for values, e.g.

在所有data.frame名称上运行一个函数，例如mutate_each（funs（））的值，例如：
```
names(iris) <- make.names(names(iris))
```
Delete columns that do NOT exist (i.e. delete nothing), e.g.

删除不存在的列（即不删除任何内容），例如
```
iris %>% select(-matches("Width")) # ok
iris %>% select(-matches("X"))     # returns empty data.frame, why?
```

Add a new column by name (string), e.g.

按名称（字符串）添加新列，例如

iris %>% mutate_("newcol" = 0) # ok

x <- "newcol"
iris %>% mutate_(x = 0) # adds a column with name "x" instead of "newcol"

Rename a data.frame colname that does not exist

重命名不存在的data.frame colname

names(iris)[names(iris)=="X"] <- "Y"

iris %>% rename(sl=Sepal.Length) # ok
iris %>% rename(Y=X)             # error, instead of no change

2 个解决方案

#1

I would use setNames for this:
我会使用setNames：

iris %>% setNames(make.names(names(.)))

Include everything() as an argument for select:
包括everything（）作为select的参数：

iris %>% select(-matches("Width"), everything())
iris %>% select(-matches("X"), everything())

To my understanding there's no other shortcut than explicitly naming the string like you already do:
根据我的理解，除了显式命名字符串之外别无其他快捷方式：

iris %>% mutate_("newcol" = 0)

#2

1 through 3 are answered above. I came here because I had the same problem as number 4. Here is my solution:

上面回答了1到3。我来到这里是因为我遇到了与4号相同的问题。这是我的解决方案：

df <- iris

Set a name key with the columns to be renamed and the new values:

使用要重命名的列和新值设置名称键：

name_key <- c(
  sl = "Sepal.Length",
  sw = "Sepal.Width",
  Y = "X"
)

Set values not in data frame to NA. This works for my purpose better. You could probably just remove it from name_key.

将不在数据框中的值设置为NA。这更符合我的目的。你可能只是从name_key中删除它。

for (var in names(name_key)) {
  if (!(name_key[[var]] %in% names(df))) {
    name_key[var] <- NA
  }
}

Get a vector of column names in the data frame.

获取数据框中的列名称向量。

cols <- names(name_key[!is.na(name_key)])

Rename columns

重命名列

for (nm in names(name_key)) {
  names(df)[names(df) == name_key[[nm]]] <- nm
}

Select columns

选择列

df2 <- df %>%
  select(cols)

I'm almost positive this can be done more elegantly, but this is what I have so far. Hope this helps, if you haven't solved it already!

我几乎肯定这可以做得更优雅，但这是我到目前为止所做的。希望这有帮助，如果你还没有解决它！

#1

I would use setNames for this:
我会使用setNames：

iris %>% setNames(make.names(names(.)))

Include everything() as an argument for select:
包括everything（）作为select的参数：

iris %>% select(-matches("Width"), everything())
iris %>% select(-matches("X"), everything())

To my understanding there's no other shortcut than explicitly naming the string like you already do:
根据我的理解，除了显式命名字符串之外别无其他快捷方式：

iris %>% mutate_("newcol" = 0)

#2