R: adding rows to a table from a for loop

Question

I have a 1 column table with postcodes in it: I would like to loop through each postcode using the postcode_lookup() function in the postcodeioR library. My current attempts are the following:

x <- data.frame()

for(i in 1:3){
  x[i, ] <- postcode_lookup(table$Var1[i])
}

So i instantiated a new table and tried to add the result of postcode_lookup to a new row every time. But I get nothing. What i get is data frame with 3 obs. and 0 variables. the data should look like this: imagine 31 columns and multiple rows: table

slava-kohut · Accepted Answer · 2019-11-01 17:19:34Z

1

You need to explicitly specify the number of columns when creating a data frame:

df <- as.data.frame(matrix(NA, 0, 1))

set.seed(123)
val <- runif(20)

for (i in 1:3){
  df[i, ] <- val[[i]]
}

In this case, a matrix with 0 rows and 1 column is converted to a data frame. This is a convenient way to create an empty data frame with the required number of columns.

In your case, you have a data frame with 0 columns. Hence, nothing gets populated.

answered Nov 1, 2019 at 17:19

slava-kohut

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3 Comments

uhakdt Over a year ago

That worked very well - thank you!! - One more question if you dont mind - I get the following error when i go beyong 299 times in the for loop: Error in x[[jj]][iseq] <- vjj : replacement has length zero: This is my code: df <- as.data.frame(matrix(NA, 2560, 31)) for (i in 1:length(table$Var1)){ df[i, ] <- postcode_lookup(table$Var1[i]) }

uhakdt Over a year ago

Your right there is an empty vector - problem is that the for loop stops - is it possible for the for loop to keep going and try the next one instead of terminating?

slava-kohut Over a year ago

then you need to use two sets of indices in your loop and adjust them accordingly. while loop will be more appropriate

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