从动态变量中选择一列

Question

How can I select the second column of a dynamically named variable?

I create variables of the form "population.USA", "population.Mexico", "population.Canada". Each variable has a column for the year, and another column for the population value. I would like to select the second column from each of these variables during a loop.

I use this syntax:

sprintf("population.%s", country)[, 2]

R returns the error: Error in sprintf("population.%s", country)[, 2] : incorrect number of dimensions

Answer 1

Based on your sequence of questions over the last few minutes, I have two general recommendations for you as you get familiar with R:

1. Don't use sprintf .
2. Don't use assign .

Now, obviously, those functions are both useful at times. But you've learned about them too early, before you've mastered some basic stuff about R's data structures. Try to write code without those crutches (for the time being!), as they're just causing you problems.

Rather than creating separate individual variables for each nation's population, place them in a list.

population <- vector("list",3)
names(population) <- c('USA','Mexico','Russia')

Then you can access each using the string representation of the name of each country:

population[['USA']] <- 10000

Or,

region <- 'USA'
population[[region]]

In this example, I've assigned a single value to a list element, lists will hold any other data type, including matrices or data frames. It will be a lot less typing than using sprintf and assign , and a lot safer and more efficient as well.

Answer 2

See ?get . Here is an example:

> country <- "FOO"
> assign(sprintf("population.%s", country), data.frame(runif(5), runif(5)))
> 
> get(sprintf("population.%s", country))[,2]
[1] 0.2241105 0.5640709 0.5945869 0.1830719 0.1895938

It is critically important to look at the object returned by a function if you get an error. It is immediately clear why your example fails if you just look at what it returns:

> sprintf("population.%s", country)
[1] "population.FOO"

At that point it would be immediately clear, if you didn't already know or have thought to read ?sprintf , that sprintf() returns a string not the object of that name. Armed with that knowledge you would have narrowed down the problem to how to recall an object from the computed name?