Ryan Rothman - 8 months ago 33

R Question

I have some data that I am looking at in R. One particular column, titled "Height", contains a few rows of NA.

I am looking to subset my data-frame so that all Heights above a certain value are excluded from my analysis.

`df2 <- subset ( df1 , Height < 40 )`

However whenever I do this, R automatically removes all rows that contain NA values for Height. I do not want this. I have tried including arguments for na.rm

`f1 <- function ( x , na.rm = FALSE ) {`

df2 <- subset ( x , Height < 40 )

}

f1 ( df1 , na.rm = FALSE )

but this does not seem to do anything; the rows with NA still end up disappearing from my data-frame. Is there a way of subsetting my data as such, without losing the NA rows?

Answer

If we decide to use `subset`

function, then we need to watch out:

```
For ordinary vectors, the result is simply ‘x[subset & !is.na(subset)]’.
```

So only non-NA values will be retained.

If you want to keep `NA`

cases, use logical or condition to tell R not to drop `NA`

cases:

```
subset(df1, Height < 40 | is.na(Height))
# or `df1[df1$Height < 40 | is.na(df1$Height), ]`
```

Don't use directly (to be explained soon):

```
df2 <- df1[df1$Height < 40, ]
```

**Example**

```
df1 <- data.frame(Height = c(NA, 2, 4, NA, 50, 60), y = 1:6)
subset(df1, Height < 40 | is.na(Height))
# Height y
#1 NA 1
#2 2 2
#3 4 3
#4 NA 4
df1[df1$Height < 40, ]
# Height y
#1 NA NA
#2 2 2
#3 4 3
#4 NA NA
```

The reason that the latter fails, is that indexing by `NA`

gives `NA`

. Consider this simple example with a vector:

```
x <- 1:4
ind <- c(NA, TRUE, NA, FALSE)
x[ind]
# [1] NA 2 NA
```

We need to somehow replace those `NA`

with `TRUE`

. The most straightforward way is to add another "or" condition `is.na(ind)`

:

```
x[ind | is.na(ind)]
# [1] 1 2 3
```

This is exactly what will happen in your situation. If your `Height`

contains `NA`

, then logical operation `Height < 40`

ends up a mix of `TRUE`

/ `FALSE`

/ `NA`

, so we need replace `NA`

by `TRUE`

as above.

Source (Stackoverflow)