如何对数据框属性逻辑测试导致NA-行

Question

我有以下形式的数据帧：

>df 
stationid station  gear sample  lat lon  date depth 
1  25679   CORBOX150 UE4 53.9015 7.8617 15.07.1987 19 
2  25681 UE9 Kern CORCRB050 UE9 54.0167 7.3982 15.07.1987 33 
3  NA       54.0167 7.3982 15.07.1987 33 

上stationid逻辑测试给我，旁边的正确的第一线，一个恼人的线路全程的NAs：

> df[df$stationid=="25679",] 
stationid station  gear sample  lat lon  date depth 
1  25679   CORBOX150 UE4 53.9015 7.8617 15.07.1987 19 
NA  NA <NA>  <NA> <NA>  NA  NA  <NA> NA 

这是为什么？

df第3行的某处，我猜想事情会搞砸。

继承人的数据：

df<-structure(list(stationid = c(25679L, 25681L, NA), station = structure(c(2L, 
3L, 1L), .Label = c("", " ", "UE9 Kern"), class = "factor"), 
gear = structure(c(2L, 3L, 1L), .Label = c("", "CORBOX150", 
"CORCRB050"), class = "factor"), sample = structure(c(2L, 
3L, 1L), .Label = c("", "UE4", "UE9"), class = "factor"), 
lat = c(53.9015, 54.0167, 54.0167), lon = c(7.8617, 7.3982, 
7.3982), date = structure(c(1L, 1L, 1L), .Label = "15.07.1987", class = "factor"), 
depth = c(19L, 33L, 33L)), .Names = c("stationid", "station", 
"gear", "sample", "lat", "lon", "date", "depth"), class = "data.frame", row.names = c(NA, 
-3L)) 

Answer 1

与NA任何比较导致的结果NA（见http://cran.r-project.org/doc/manuals/R-intro.html#Missing-values）...您可以使用

df[df$stationid==25679 & !is.na(df$stationid),] 

或（如建议在上面的注释）

df[which(df$stationid==25679),] 

或

subset(df,stationid==25679) 

（subset has t他有时不想要的副作用，但是在这种情况下，它正是你想要的做的想要的）

Answer 2

另一种解决方案是df[df$stationid==25679 & !is.na(df$stationid),]。时间更长，但更明确。