如何删除gridExtra：：tableGrob中的行名？

o75abkj4 于 2023-03-27 发布在其他

关注(0)|答案(1)|浏览(94)

我想用ggplot绘制一个表，但是，我似乎无法设法丢失row.names。
我试过：

row.names(cov_table_a)<-NULL # outside the plot
row.names = FALSE            # inside the annotation_custom

这是我的数据：

cov_table_a<-
structure(list(Year = structure(1:16, .Label = c("1970", "1971", 
"1972", "1973", "1974", "1975", "1977", "1979", "1980", "1981", 
"1982", "1983", "1984", "1985", "1986", "1987", "1988", "1990", 
"1991", "1992", "2000", "2001", "2003", "2004", "2007", "2008", 
"2009", "2010", "2011", "2012", "2013", "2014"), class = "factor"), 
`Percentage \nof missing data` = structure(c(4L, 2L, 4L, 6L, 
1L, 12L, 5L, 21L, 21L, 11L, 10L, 13L, 9L, 17L, 18L, 14L), .Label = c("0%", 
"1%", "100%", "17%", "24%", "31%", "35%", "5%", "58%", "59%", 
"60%", "70%", "71%", "72%", "8%", "86%", "90%", "92%", "95%", 
"98%", "99%"), class = "factor")), .Names = c("Year", "Percentage \nof missing data"
), row.names = c(NA, -16L), class = "data.frame")

这就是我尝试绘制它的方式：

library(ggplot2)
library(gridExtra)

# First I create an empty graph with absolutely nothing :
qplot(1:10, 1:10, geom = "blank") + theme_bw() + theme(line = element_blank(), text = element_blank()) +
# Then I add my table :
annotation_custom(grob = tableGrob(cov_table_a), xmin=-5, xmax=10,ymin=1, ymax=10)

r

来源：https://stackoverflow.com/questions/40788778/how-can-i-remove-rownames-in-gridextratablegrob

1条答案

按热度按时间

bttbmeg01#

注意：* 如果要在空画布上绘制表格，则不需要使用ggplot作为容器：

grid.table(cov_table_a, rows = NULL)

或

grid.draw(tableGrob(cov_table_a, rows = NULL))

也可以使用。
但是也可以使用ggplot：

qplot(1:10, 1:10, geom = "blank") + 
  theme_void() +
  annotation_custom(
    grob = tableGrob(cov_table_a, rows = NULL), 
    xmin = -5, xmax = 10, ymin = 1, ymax = 10
  )

赞(0）回复(0）举报 2023-03-27

我来回答

如何删除gridExtra：：tableGrob中的行名？

1条答案

相关问题

热门标签

最新问答