df <- data.frame(
cola = c('a','b','c','d','e','e','1',NA,'c','d'),
colb = c("A",NA,"C","D",'a','b','c','d','c','d'),stringsAsFactors = FALSE)
#equal 2 dataframe
df2<-df
df['cola'] <- lapply(df['cola'], function(x) droplevels(factor(x,levels=c('a','b','c','d','e','f','1'),ordered = FALSE)))
df2['cola'] <- lapply(df2['cola'], function(x) factor(x,ordered = FALSE))
#should be eqaul
dplyr::all_equal(df,df2)
#check levels
levels(df$cola)
levels(df2$cola)
上述脚本的输出为:
> dplyr::all_equal(df,df2)
[1] "Factor levels not equal for column `cola`"
> levels(df$cola)
[1] "a" "b" "c" "d" "e" "1"
> levels(df2$cola)
[1] "1" "a" "b" "c" "d" "e"
至于ordered = FALSE
,"a" "b" "c" "d" "e" "1"
应该等于"1" "a" "b" "c" "d" "e"
为什么all_equal
告诉我Factor levels not equal
?
如何比较这两个因素水平是否相等?
如果使用原始all.equal
,原因会变得更加清晰。
all.equal(df, df2)
# [1] "Component “cola”: Attributes: < Component “levels”: 6 string mismatches >"
您的级别只是彼此不匹配。这对于 data.frame 或两个字符串向量的压缩列是简洁的:
all.equal(letters[c(3, 1, 2)], letters[c(2, 3, 1)])
# [1] "3 string mismatches"
您可以改用sort
。
sort(levels(df$cola)) == sort(levels(df2$cola))
# [1] TRUE TRUE TRUE TRUE TRUE TRUE
要选中所有内容,请使用 all
。
all(sort(levels(df$cola)) == sort(levels(df2$cola)))
# [1] TRUE
你可以把它包装成一个函数。
checkEqualLevels <- function(l, x, y) {
if (all(sort(levels(x[[l]])) == sort(levels(y[[l]]))))
cat(paste0("Factor levels are equal for column ", "'", l, "'"))
else
cat(paste0("Factor levels not equal for column ", "'", l, "'"))
}
checkEqualLevels("cola", df, df2)
# Factor levels are equal for column 'cola'