[英]Removing column names in pandoc table - r markdown
Is it possible to remove (not to show) column names in pandoc tables?是否可以删除(不显示)pandoc 表中的列名?
If I use pander
(or pandoc.table
) function it prints column names automatically.如果我使用
pander
(或pandoc.table
)功能,它会自动打印列名。
> pander(iris[1:4, ])
-------------------------------------------------------------------
Sepal.Length Sepal.Width Petal.Length Petal.Width Species
-------------- ------------- -------------- ------------- ---------
5.1 3.5 1.4 0.2 setosa
4.9 3 1.4 0.2 setosa
4.7 3.2 1.3 0.2 setosa
4.6 3.1 1.5 0.2 setosa
-------------------------------------------------------------------
Expected output should be:预期输出应该是:
-------------------------------------------------------------------
-------------- ------------- -------------- ------------- ---------
5.1 3.5 1.4 0.2 setosa
4.9 3 1.4 0.2 setosa
4.7 3.2 1.3 0.2 setosa
4.6 3.1 1.5 0.2 setosa
-------------------------------------------------------------------
I'd fix this outside of pander
by simply removing the column headers:我想解决这个问题之外
pander
通过简单地移除列标题:
> df <- iris[1:4, ]
> names(df) <- NULL
> pander(df)
--- --- --- --- ------
5.1 3.5 1.4 0.2 setosa
4.9 3 1.4 0.2 setosa
4.7 3.2 1.3 0.2 setosa
4.6 3.1 1.5 0.2 setosa
--- --- --- --- ------
Would this suffice?这样就够了吗?
pandoc.table({temp <- iris; names(temp) <- rep(" ", ncol(temp)); temp[1:4,]})
to yield.屈服。
----------------------
--- --- --- --- ------
5.1 3.5 1.4 0.2 setosa
4.9 3 1.4 0.2 setosa
4.7 3.2 1.3 0.2 setosa
4.6 3.1 1.5 0.2 setosa
----------------------
I'd replace Benjamin's   with NULL but otherwise agree:我会用 NULL 替换本杰明的   ,但否则同意:
temp <- iris[1:4,]; names(temp) <- rep(NULL, ncol(temp)); temp[1:4,]
pandoc.table(temp)
setting column names to NULL
like so names(dt) <- NULL
actually gives an error.像这样将列名设置为
NULL
names(dt) <- NULL
实际上会产生错误。 In my case就我而言
Error in alloc.col(newx) : Internal error: length of names (0) is not length of dt (2)
using a data.table
with 2 columns.使用具有 2 列的
data.table
。
Instead, I'd go with相反,我会去
names(dt) <- rep(" ", ncol(dt)) pander(dt)
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