可重复的例子:
# test data with NaN and NA values
df <- data.frame(
v1 = c(1, 2, 3),
v2 = c(NaN, 4, 5),
v3 = c(6, NA, 7)
)
# extract NA values: works
df[is.na(df)]
# extract NaN values: fails!
is.nan(df)
# The long way to do that...
df[sapply(df, is.nan)]
第二种解决方案正常。但是,为什么第一个不适用于is.na
?