R loop change variable names

Question

Looked over several posts on this topic, but still couldn't figure out. Thought I'd just ask:

I wrote a for-loop:

for (i in 0:5) {
     est16_y2016$pov50_[i] <- est16_y2016$pop[i]*est16_y2016$ITPR_0.5
               }

to achieve the same results as the following code:

 est16_y2016$pov50_0 <- est16_y2016$pop0 * est16_y2016$ITPR_0.5
 est16_y2016$pov50_1 <- est16_y2016$pop1 * est16_y2016$ITPR_0.5
 est16_y2016$pov50_2 <- est16_y2016$pop2 * est16_y2016$ITPR_0.5 
 est16_y2016$pov50_3 <- est16_y2016$pop3 * est16_y2016$ITPR_0.5 
 est16_y2016$pov50_4 <- est16_y2016$pop4 * est16_y2016$ITPR_0.5 
 est16_y2016$pov50_5 <- est16_y2016$pop5 * est16_y2016$ITPR_0.5

But the loop doesn't work. No error message, no new variables generated either. Help! Thanks.

Seems like an XY problem. What are you trying to do? Please provide a little more context. — John Coleman
– John Coleman, Commented Aug 24, 2018 at 18:31
@JohnColeman I'm trying to generate some new variables (you probably can tell from my code), and I want to use for-loop to make the code more compact. — Eric Wang
– Eric Wang, Commented Aug 24, 2018 at 18:37
@EricWang Sorry, this should help for (i in 0:5) est16_y2016[,paste0('pov50_',i)] <- est16_y2016[,paste0('pop',i)]*est16_y2016$ITPR_0.5 } — A. Suliman
– A. Suliman, Commented Aug 24, 2018 at 18:49

Rui Barradas · Accepted Answer · 2018-08-24 19:51:59Z

3

Though the construct data$column_name is very convenient when in an interactive R session, when programming it may cause some problems. @A.Suliman's comment presents a way to solve those problems, here is another one.

for(i in 0:5){
    target <- paste("pov50", i, sep = "_")
    pop <- paste0("pop",i)
    est16_y2016[[target]] <- est16_y2016[[pop]]*est16_y2016[["ITPR_0.5"]]
}

answered Aug 24, 2018 at 19:51

Rui Barradas

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Comments

GordonShumway · Accepted Answer · 2018-08-24 20:25:16Z

0

It's kind of hard to answer your question without a reproducible example but I will give it a shot. est16_y2016$pop[i] will give you the ith element of est16_y2016$pop (which probably doesn't even exist. What you want instead is est16_y2016[paste0("pop",i)] so you code should look like this:

for (i in 0:5) 
    {
     est16_y2016[[paste0("pov50_",i)]] <- est16_y2016[[paste0("pop",i)]]*est16_y2016$ITPR_0.5
     }

(edited)

edited Aug 24, 2018 at 20:25

answered Aug 24, 2018 at 18:46

GordonShumway

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6 Comments

Rui Barradas Over a year ago

Do the same with paste0("pov50_",i). And this looks like a df, use double [[ as in est16_y2016[[paste0("pop",i)]]. You should edit the answer.

r2evans Over a year ago

GordonShumway, your assignment is wrong. The OP wants columns named pov50_0, pov50_1, etc. The suggestion to use paste0("pop",i) is for the lhs, not rhs.

Eric Wang Over a year ago

Please see @A. Suliman 's answer above. Not all questions require a reproducible example.

Rui Barradas Over a year ago

@EricWang Yes they do, see the amount of comments, with a reproducible example it would all have been more clear to start with.

Eric Wang Over a year ago

@RuiBarradas they do? We've already got a solution. Matter of fact, the very first comment by A.Suliman is a solution. It's funny that so many people who don't know the answer just keep commenting.

|

FewKey · Accepted Answer · 2019-03-19 19:09:33Z

-1

It is easy to create a new variable name with paste(), and the problem is how to use the corresponding variable instead of the name of a variable.

  for (i in 0:5){ 
    # Create new variable names
    pov.name = paste0("est16_y2016$pov50_",i)
    pop.name = paste0("est16_y2016$pop",i)

    assign(pov.name,eval(parse(text = pop.name))*est16_y2016$ITPR_0.5)
    }

In this code,

eval(parse(text = pop.name) uses the string "pop.name" as a variable name

assign(pov.name,value1)creates a variable named pov.name and assigns value1 to pov.name

This way, you can get six new variables without using a dataframe.

edited Mar 19, 2019 at 19:09

answered Aug 24, 2018 at 21:52

FewKey

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