[英]Using ifelse on factor in R
I am restructuring a dataset of species names. 我正在重组一个物种名称的数据集。 It has a column with latin names and column with trivial names when those are available.
它有一个带有拉丁名称的列和具有普通名称的列(如果可用)。 I would like to make a 3rd column which gives the trivial name when available, otherwise the latin name.
我想制作第3列,在可用时给出琐碎的名称,否则为拉丁名称。 Both trivial names and latin names are in factor-class.
琐碎的名字和拉丁名字都属于因子级。 I have tried with an if-loop:
我试过if-loop:
if(art2$trivname==""){
art2$artname=trivname
}else{
art2$artname=latname
}
It gives me the correct trivnames, but only gives NA when supplying latin names. 它给了我正确的trivnames,但只在提供拉丁名字时给出了NA。
And when I use ifelse I only get numbers. 当我使用ifelse时,我只会得到数字。
As always, all help appreciated :) 一如既往,所有帮助赞赏:)
Example: 例:
art <- data.frame(trivname = c("cat", "", "deer"), latname = c("cattus", "canis", "cervus"))
art$artname <- with(art, ifelse(trivname == "", as.character(latname), as.character(trivname)))
print(art)
# trivname latname artname
# 1 cat cattus cat
# 2 canis canis
# 3 deer cervus deer
(I think options(stringsAsFactors = FALSE) as default would be easier for most people, but there you go...) (我认为默认选项(stringsAsFactors = FALSE)对大多数人来说会更容易,但是你去......)
Getting only numbers suggests that you just need to add as.character
to your assignments, and the if-else
would probably work you also seem to not be referring to the data frame in the assignment? 只获得数字表明你只需要在你的作业中添加
as.character
,而if-else
可能也可能你也似乎没有引用作业中的数据框?
if(as.character(art2$trivname)==""){
art2$artname=as.character(art2$trivname)
}else{
art2$artname=as.character(art2$latname)
}
Option 2: Using ifelse
: 选项2:使用
ifelse
:
art2$artname= ifelse(as.character(art2$trivname) == "", as.character(art2$latname),as.character(art2$trivname))
It is probably easier (and more "R-thonic" because it avoids the loop) just to assign artname
to trivial
across the board, then overwrite the blank ones with latname
... 它可能更容易(并且更多“R-thonic”,因为它避免了循环)只是将
artname
分配给artname
的trivial
,然后用latname
覆盖空白的...
art2 = art
art2$artname = as.character(art$trivname)
changeme = which(art2$artname=="")
art2$artname[changeme] = as.character(art$latname[changeme])
If art2 is the dataframe, and artname the new column, another possible solution: 如果art2是数据框 , artname是新列,则另一种可能的解决方案:
art2$artname <- as.character(art2$trivname)
art2[art$artname == "",'artname'] <- as.character(art2[art2$artname == "", 'latname'])
And if you want factors in the new column: 如果您想要新列中的因素:
art2$artname <- as.factor(art2$artname)
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