R function `paste` is inverting the order of its arguments

Question

I am simply trying to print the output of a statistical process I'm running in R, and I find that the paste function appears to be inverting its inputs. Here's a MWE:

df = data.frame(p_values=c(0.01, 0.001, 0.1, 0.01))
min_index = which.min(df$p_values)
print(paste(str(min_index), 
            " with p-value ",
            str(df$p_values[min_index])))

I'm expecting output more-or-less like this:

2 with p-value 0.001

Instead, I'm getting the highly unintuitive result

 int 2
 num 0.001
[1] "  with p-value  "

In addition to the unexpected order of the printing, I'm getting the int and num and [1], as well as not all being on one line.

What's going on here? I had thought paste was nearly a drop-in replacement for Python's concat operator +, but this result has me scratching my head.

Answer 1

structure not string

str() in R is not the same as the Python str() function, which coerces an object to a string. In R, the str() function exists to:

Compactly display the internal structure of an R object, a diagnostic function and an alternative to summary (and to some extent, dput).

This means when you do str(min_index), R tells you that the structure of min_index is that it's an integer vector of length one, and the value of its element is 2.

The equivalent R function to Python's str() would be toString(), or perhaps as.character(). However, in general R is more forgiving about types than Python. paste() and all other string concatenation or printing commands I can think of will coerce numbers to strings for you, so you can just do:

paste(min_index, "with p-value", df$p_values[min_index])
# [1] "2  with p-value 0.001"

Note that I deleted your spaces, as by default paste() adds one, though that can be changed by supplying a different sep argument or using paste0().

Why the output order seems inverted

Accounting for this, you might expect your output to be "int 2 with p-value num 0.001". However, it is:

 int 2
 num 0.001
[1] "  with p-value  "

This is because str() prints its output and returns nothing:

x  <- str(1) # "num 1" is printed
print(x) # NULL

Your command can basically be interpreted as:

str(min_index) # int 2
str(df$p_values[min_index]) # num 0.001
paste(NULL, "with p-value", NULL) # [1] " with p-value "

This why it prints in the order you see.

A note on displaying output

As Onyambu says in the comments, if you want to display the output without the [1] you can use cat():

cat(
  min_index, "with p-value", df$p_values[min_index],
  fill = TRUE
)
# 2 with p-value 0.001

Note that you need fill = TRUE to format this correctly, including a new line at the end. Depending on the purpose of your output, you may also want to look at message().

Answer 2

Why you are using str()? it returns the structure of the object. if you use:

print(paste(min_index[1],"with p-value", df$p_values[min_index]))

it would work i think

Answer 3

As another option you can consider glue

library(glue)
glue({min_index}, " with p-value ", {df$p_values[min_index]})
2 with p-value 0.001