我有以下数据:
> dat
ID Gene Value1 Value2
1 NM_013468 Ankrd1 Inf Inf
2 NM_023785 Ppbp Inf Inf
3 NM_178666 Themis NaN Inf
4 NM_001161790 Mefv Inf Inf
5 NM_001161791 Mefv Inf Inf
6 NM_019453 Mefv Inf Inf
7 NM_008337 Ifng Inf Inf
8 NM_022430 Ms4a8a Inf Inf
9 PBANKA_090410 Rab6 NaN Inf
10 NM_011328 Sct Inf Inf
11 NM_198411 Inf2 1.152414 1.445595
12 NM_177363 Tarm1 NaN Inf
13 NM_001136068 Klrc1 NaN Inf
14 NM_019418 Tnfsf14 Inf Inf
15 NM_010652 Klrc1 NaN Inf
我想要做的是包含Inf
和Nan
的行
仅返回NM_198411 Inf2 1.152414 1.445595
但为什么这段代码失败了?
dat <- structure(list(ID = structure(c(7L, 11L, 13L, 2L, 3L, 9L, 4L,
10L, 15L, 6L, 14L, 12L, 1L, 8L, 5L), .Label = c("NM_001136068 ",
"NM_001161790 ", "NM_001161791 ", "NM_008337 ", "NM_010652 ",
"NM_011328 ", "NM_013468 ", "NM_019418 ", "NM_019453 ", "NM_022430 ",
"NM_023785 ", "NM_177363 ", "NM_178666 ", "NM_198411 ", "PBANKA_090410 "
), class = "factor"), Gene = structure(c(1L, 7L, 11L, 5L, 5L,
5L, 2L, 6L, 8L, 9L, 3L, 10L, 4L, 12L, 4L), .Label = c("Ankrd1 ",
"Ifng ", "Inf2 ", "Klrc1 ", "Mefv ", "Ms4a8a ", "Ppbp ", "Rab6 ",
"Sct ", "Tarm1 ", "Themis ", "Tnfsf14 "), class = "factor"),
Value1 = c(Inf, Inf, NaN, Inf, Inf, Inf, Inf, Inf, NaN, Inf,
1.152414042, NaN, NaN, Inf, NaN), Value2 = c(Inf, Inf, Inf,
Inf, Inf, Inf, Inf, Inf, Inf, Inf, 1.445594931, Inf, Inf,
Inf, Inf)), .Names = c("ID", "Gene", "Value1", "Value2"), class = "data.frame", row.names = c(NA,
-15L))
dat[apply(dat,1,function(x) any(x >=1 | x != Inf | x != NaN) ),]
答案 0 :(得分:11)
您无法使用常规比较运算符检查NaN
。您可以Inf
执行此操作,但您还必须检查否定的情况。最好使用其中一个功能:https://stat.ethz.ch/R-manual/R-devel/library/base/html/is.finite.html
编辑:tonytonov指出,is.finite(NaN)
是FALSE
,这足以在这种情况下使用。你只需要
dat[is.finite(dat$Value1) & is.finite(dat$Value2), ]