Dmi*_*nko 6 r dataframe r-factor
数据框 AEbySOC 包含两列 - 具有字符级别的因子 SOC 和整数计数 Count:
> str(AEbySOC)
'data.frame': 19 obs. of 2 variables:
$ SOC : Factor w/ 19 levels "","Blood and lymphatic system disorders",..: 1 2 3 4 5 6 7 8 9 10 ...
$ Count: int 25 50 7 3 1 49 49 2 1 9 ...
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SOC 的级别之一是空字符串:
> l = levels(AEbySOC$SOC)
> l[1]
[1] ""
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我想用非空字符串替换此级别的值,例如“未指定”。这不起作用:
> library(plyr)
> revalue(AEbySOC$SOC, c(""="Not specified"))
Error: attempt to use zero-length variable name
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这也不行:
> AEbySOC$SOC[AEbySOC$SOC==""] = "Not specified"
Warning message:
In `[<-.factor`(`*tmp*`, AEbySOC$SOC == "", value = c(NA, 2L, 3L, :
invalid factor level, NA generated
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实现这一点的正确方法是什么?我感谢任何输入/评论。
levels(AEbySOC$SOC)[1] <- "Not specified"
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创建了一个玩具示例:
df<- data.frame(a= c("", "a", "b"))
df
# a
#1
#2 a
#3 b
levels(df$a)
#[1] "" "a" "b"
levels(df$a)[1] <- "Not specified"
levels(df$a)
#[1] "Not specified" "a" "b"
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编辑
根据 OP 的评论,如果我们需要根据值找到它,那么在这种情况下,我们可以尝试
levels(AEbySOC$SOC)[levels(AEbySOC$SOC) == ""] <- "Not specified"
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