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我想将来自斜体数据框的名称绘制为 ylab。R plot的解决方案:在这种情况下,在标题中使用斜体和变量不起作用。使用substituteor expressionR 从字面上读取我的代码
substitute
expression
例如substitute (paste 'Species = ', italic (rownames (genR) [i]))
substitute (paste 'Species = ', italic (rownames (genR) [i]))
将在测试中出现:
物种 =行名 (genR)[1]
我想要
物种 = Musa paradisiaca
有什么建议吗?
非常感谢!
你可以试试bquote
bquote
nm1 <- 'Musa paradisiaca' plot(1,1, main=bquote('Species = '~italic(.(nm1))))
或使用substitute
plot(1,1, main=eval(substitute(expr=expression(paste('Species = ', italic(x))), env=list(x=as.name(nm1)))))