从 R 中的整个数据帧中删除空格

如果我理解正确，那么您希望删除整个数据帧中的所有空格，我猜您使用的代码有助于删除列名中的空格.我觉得你应该试试这个:

 apply(myData,2,function(x)gsub('\\s+', '',x))

希望这能奏效.

但是，这将返回一个矩阵，如果要将其更改为数据帧，请执行以下操作:

as.data.frame(apply(myData,2,function(x)gsub('\\s+', '',x)))

EDIT In 2020:

将lapply和trimws函数与both=TRUE一起使用可以删除前导和尾随空格，但不能删除其中的空格.因为OP没有提供输入数据，所以我添加了一个虚拟示例来生成结果.

DATA:

df <- data.frame(val = c(" abc"," kl m","dfsd "),val1 = c("klm ","gdfs","123"),num=1:3,num1=2:4,stringsAsFactors = FALSE)

#situation: 1(使用基数R)，当我们想要删除字符串值中的空格only at the leading and trailing ends NOT inside时，我们可以使用trimws

cols_to_be_rectified <- names(df)[vapply(df, is.character, logical(1))]
df[,cols_to_be_rectified] <- lapply(df[,cols_to_be_rectified], trimws)

# situation: 2(使用Base R)，当我们想要删除字符列中数据帧中每个位置的空格时(inside of a string as well as at the leading and trailing ends).

(This was the initial solution proposed using apply, please note a solution using apply seems to work but would be very slow, also the with the question its apparently not very clear if OP really wanted to remove leading/trailing blank or every blank in the data)

cols_to_be_rectified <- names(df)[vapply(df, is.character, logical(1))]
df[,cols_to_be_rectified] <- lapply(df[,cols_to_be_rectified], function(x)gsub('\\s+','',x))

## situation: 1(使用data.table，仅删除前导空格和尾随空格)

library(data.table)
setDT(df)
cols_to_be_rectified <- names(df)[vapply(df, is.character, logical(1))]
df[,c(cols_to_be_rectified) := lapply(.SD, trimws), .SDcols = cols_to_be_rectified]

Output从situation1:

    val val1 num num1
1:  abc  klm   1    2
2: kl m gdfs   2    3
3: dfsd  123   3    4

## situation: 2(使用data.table，删除每个内部空白以及前导/尾随空白)

cols_to_be_rectified <- names(df)[vapply(df, is.character, logical(1))]
df[,c(cols_to_be_rectified) := lapply(.SD, function(x)gsub('\\s+', '', x)), .SDcols = cols_to_be_rectified]

Output从situation2:

    val val1 num num1
1:  abc  klm   1    2
2:  klm gdfs   2    3
3: dfsd  123   3    4

请注意第2行中两种情况的输出之间的差异:您可以看到，对于trimws，我们可以删除前导空格和尾随空格，但是对于正则表达式解决方案，我们可以删除每个空格.

我希望这有帮助，谢谢