Why does apply(x, 1, paste0(collapse=“”) leave white space between positive values?

Question

When I apply across columns in this example I get a white space for positive valued numbers but not for negative values? Why is this? Shouldn't paste0 remove whitespace between elements? The context behind this problem is that I am trying to form endpoints for the googlemaps directions api.

library(dplyr)
stop_latlon <- data.frame(lat = paste0("via:", rnorm(10)), lon = rnorm(10))

stop_latlon %>% 
  apply(1, function(x) paste0(x, collapse = "%7")) 

edit: I think that it has something to do with running apply on a dataframe with different data types (lat is a character and lon is a numeric)

Answer 1

Why does the white space show up?

paste0 doesn't add white space - nor does it remove it. You can test this by just calling paste0 on your vector.

apply runs on matrices and arrays, not data frames. When you pass a data frame to apply, it is coerced to a matrix. The main thing about a matrix, of course, is that all elements must be the same type. Since strings or factors can't generally be coerced to numerics, your numeric is coerced to a string or factor to match the first column. If you examine as.matrix.data.frame, you'll see that format is used for this conversion, and ?format shows a default trim = FALSE that says

trim; if FALSE, logical, numeric and complex values are right-justified to a common width: if TRUE the leading blanks for justification are suppressed.

So there's your problem!

---

What's the solution?

paste and paste0 are vectorized, so there's no reason to apply them one row at a time. You can just paste the columns together directly:

with(stop_latlon, paste0(lat, "%7", lon))

In a more complicated case where apply really would be necessary, the solution would be to handle your own matrix conversion rather than relying on apply to do it with defaults. If you made all the columns strings before passing the data to apply, (or if you used a character matrix instead of a data frame), the conversion would be straightforward (or unnecessary).

Answer 2

Since you are already using dplyr, a dplyr solution is to use

stop_latlon %>% rowwise() %>% 
  summarise(latlon = paste0(lat, "%7", lon))

# A tibble: 10 x 1
                                       latlon
                                        <chr>
1     via:1.222988975822%7-0.0916195541513781
2     via:0.159343465931011%72.13195314768885
3    via:-1.20468509249113%70.207717129395512
4  via:-0.134019685121819%7-0.912028913867691
5    via:-0.279895116522155%71.93812564387851
6     via:1.34379237820276%70.500525410068601
7   via:0.808272181619927%7-0.942578996972991
8    via:-1.17359899808855%70.126116638988962
9      via:1.1859602145711%7-1.00865269561505
10    via:1.77635906904826%70.685722866041471

Using a tibble instead of a data.frame by default will not convert your vector to a factor, which I think is desirable in this instance.

Additionally, regarding your question about paste0, it does not remove whitespace between words, it just doesn't add them when concatenating. str_trim in the stringr package will trim whitespace for you.

stop_latlon <- data.frame(lat = paste0("via:", rnorm(10)),
                          lon = rnorm(10), stringsAsFactors = FALSE)


library(stringr)
stop_latlon %>% 
  apply(1, function(x) paste0(str_trim(x), collapse = "%7")) 

Will also provide the desired result.