user3375672 - 1 year ago 61

R Question

Given two data frames

`s`

`q`

`set.seed(8)`

s <- data.frame(id=sample(c('Z','X'), 5, T),

t0=sample(1:10, 5, T),

t1 = sample(11:30, 5, T))

q <- data.frame(id=sample(c('Z','X'), 5, T),

t0=sample(1:10, 5, T),

t1 = sample(11:30, 5, T))

> s

id t0 t1

1 Z 8 20

2 Z 3 12

3 X 10 19

4 X 8 21

5 Z 7 13

> q

id t0 t1

1 X 3 30

2 Z 5 12

3 Z 7 23

4 Z 3 21

5 X 7 27

The midpoint for the observations between the variables t0 and t1 is (e.g. for

`s`

`s$t0+(s$t1-s$t0)/2`

To find the index of the (first) observation in

`s`

`q`

`i <- which.min(abs((s$t0+(s$t1-s$t0)/2 - (q$t0[1]+(q$t1[1]-q$t0[1])/2)))`

s[i,]

gives:

`id t0 t1`

3 X 10 19

But I cannot figure out how to find the same index in the original data

`s`

`which.min(....) & s$id == q$id[1]`

Again: I need a index to be used in the original 5-row data set.

Answer Source

Set the `which.min`

argument to infinity when your condition is not obeyed:

```
val <- abs((s$t0+(s$t1-s$t0)/2 - (q$t0[1]+(q$t1[1]-q$t0[1])/2))
val[s$id != q$id[1]] <- Inf
i <- which.min(val)
```

By the way, you can simplify the expression in the first character as:

```
val <- abs((s$t0+s$t1)/2-(q$t0[1]+q$t1[1])/2)
```

or even

```
val <- abs(s$t0+s$t1-q$t0[1]-q$t1[1])/2
```