问题

在绑定所有小表时，数据框的名称具有扩展名“.x”、“ .y”、“ .xX”、“ .yY”等，但我希望在数据框中保持相同的名称。

例如，在我的数据框中，我希望保持相同的列名“names”。

data <- data.frame(part=c("DD","MM","TT"),
                   Team1 = c("kamal", "vimal", "shamal"),
                   names = c(45,23,35),
                   Team2 = c("kamal", "vimal", "shamal"),
                   names = c(46,24,36),
                   Team3 = c("kamal", "vimal", "shamal"),
                   names = c(47,25,37),
                   Team2 = c("kamal", "vimal", "shamal"),
                   names = c(48,26,38),
                   age = c(45,23,35),
                   weight = c(50,34,42))

所需输出应该如下所示。

data <- data.frame(part=c("DD","MM","TT"),
Team1 = c("kamal", "vimal", "shamal"),
"names.x" = c(45,23,35),
Team2 = c("kamal", "vimal", "shamal"),
"names.y" = c(46,24,36),
Team3 = c("kamal", "vimal", "shamal"),
"names.x.x" = c(47,25,37),
Team2 = c("kamal", "vimal", "shamal"),
"names.y.y" = c(48,26,38),
age = c(45,23,35),
weight = c(50,34,42))

答案1

得分: 1

使用`sub`函数只保留“names”，然后使用`setNames`函数更改列名。

setNames(data, sub("(names).*", "\1", colnames(data)))

part Team1 names Team2 names Team3 names Team2.1 names age weight
1 DD kamal 45 kamal 46 kamal 47 kamal 48 45 50
2 MM vimal 23 vimal 24 vimal 25 vimal 26 23 34
3 TT shamal 35 shamal 36 shamal 37 shamal 38 35 42

<hr>

或者移除点号后的内容。

setNames(data, sub("\..*", "", colnames(data)))

另一种方法是使用`colnames(data) <-`语法，它将立即替换`data`中的列名（请谨慎使用）。

colnames(data) <- sub("\..*", "", colnames(data))

### 输出

part Team1 names Team2 names Team3 names Team2 names age weight
1 DD kamal 45 kamal 46 kamal 47 kamal 48 45 50
2 MM vimal 23 vimal 24 vimal 25 vimal 26 23 34
3 TT shamal 35 shamal 36 shamal 37 shamal 38 35 42

<hr>

### 更新:

要将`names`与顺序数字组合，可以使用以下方法:

colnames(data)[grepl("names", colnames(data))] <- paste0("names", seq_len(sum(grepl("names", colnames(data)))))

part Team1 names1 Team2 names2 Team3 names3 Team2.1 names4 age weight
1 DD kamal 45 kamal 46 kamal 47 kamal 48 45 50
2 MM vimal 23 vimal 24 vimal 25 vimal 26 23 34
3 TT shamal 35 shamal 36 shamal 37 shamal 38 35 42

<details>
<summary>英文:</summary>

You can use `sub` to only keep &quot;names&quot;, then `setNames` to change the column names.

setNames(data, sub("(names).*", "\1", colnames(data)))

part Team1 names Team2 names Team3 names Team2.1 names age weight
1 DD kamal 45 kamal 46 kamal 47 kamal 48 45 50
2 MM vimal 23 vimal 24 vimal 25 vimal 26 23 34
3 TT shamal 35 shamal 36 shamal 37 shamal 38 35 42

<hr>

Or remove anything after a dot `.`.

setNames(data, sub("\..*", "", colnames(data)))

Another way of doing it is to use the `colnames(data) <-` syntax, which will immediately replace the column names in `data` (use with caution).

colnames(data) <- sub("\..*", "", colnames(data))

### Output 

part Team1 names Team2 names Team3 names Team2 names age weight
1 DD kamal 45 kamal 46 kamal 47 kamal 48 45 50
2 MM vimal 23 vimal 24 vimal 25 vimal 26 23 34
3 TT shamal 35 shamal 36 shamal 37 shamal 38 35 42

<hr>

### Update:

To combine `names` with sequential numbers, do:

colnames(data)[grepl("names", colnames(data))] <- paste0("names", seq_len(sum(grepl("names", colnames(data)))))

part Team1 names1 Team2 names2 Team3 names3 Team2.1 names4 age weight
1 DD kamal 45 kamal 46 kamal 47 kamal 48 45 50
2 MM vimal 23 vimal 24 vimal 25 vimal 26 23 34
3 TT shamal 35 shamal 36 shamal 37 shamal 38 35 42

</details>