Error in as.environment(pos) : invalid 'pos' argument

[zman3]

I've been working with R for the past week to transfer my VBA code. I worked on a loop with my father to assign a data frame column with values from another data frame column, but I am getting an error back: Error in as.environment(pos) : invalid 'pos' argument. Since my knowledge and experience of R is limited, I do not know if I should be using a different function or how to fix my line of code.

The Q_new$Vol..mL variable contains 517 rows of numbers such as 69, 68, 68, 67, 66, 65, 64, 64, 63, 62, 61,60, 79, 78, 77 .... The Infil_Data.Time.1$m contains 517 rows of no data (a blank cell). What I want to do is take the first number of every 12th row from the Q_new$Vol..mL variable and duplicate it down 12 times in the Infil_Data.Time.1$m variable. So it should look like 69,69,69,69,69,69,69,69,69,69,69,69,79,79,79,79,79,79,79,79,79,79,79,79....

I used the reprex as R Studio Community recommend. Let me know if what I have below is not properly formatted. Or additional information is needed.

jeff <- while (!is.null(Q_new$Vol..mL)) {
    for (x in 1:11) {
        R1 <- R1 + 1 %>% assign(Infil_Data.Time.1$m, Q_new[R2, 
            1]) %>% (next)(x)
    }
    R2 = R2 + 1
}
#> Error in eval(expr, envir, enclos): object 'Q_new' not found

[andresrcs]

The error message in your title it doesn't seems related to the code you are showing, also a reprex is a self contained chunck of code (including sample data) that can be directly copied and executed elsewhere
You may want to read this

FAQ: What's a reproducible example (`reprex`) and how do I create one?

Why reprex? Getting unstuck is hard. Your first step here is usually to create a reprex, or reproducible example. The goal of a reprex is to package your code, and information about your problem so that others can run it and feel your pain. Then, hopefully, folks can more easily provide a solution. What's in a Reproducible Example? Parts of a reproducible example: background information - Describe what you are trying to do. What have you already done? complete set up - include any library() calls and data to reproduce your issue. data for a reprex: Here's a discussion on setting up data for a reprex make it run - include the minimal code required to reproduce your error on the data…

Have in mind that VBA and R have very different sintaxis, so translation is not straight forward.

[zman3]

Do you need to see my data frames then?

[andresrcs]

You can use datapasta to include a small but representative part of your data in a reprex, see this blog post by mara

reprex-ing with {datapasta} 🍝

[zman3]

Thank you. I really appreciate the help so that my future questions will not lack in quality. So here is part of the Q_new data frame:

tibble::tribble(~Site.ID, ~Vol..mL., "H1", 63L, "H2", 82L, "H3", 
    69L)
#> # A tibble: 3 x 2
#>   Site.ID Vol..mL.
#>   <chr>      <int>
#> 1 H1            63
#> 2 H2            82
#> 3 H3            69

Data frame Infil_Data.Time.1 looks like:

tibble::tribble(~m, ~Time, ~Site.ID, ~Vol..mL., NA, 0L, "H1", 
    63, NA, 30L, "H1", 62, NA, 60L, "H1", 60, NA, 90L, "H1", 
    59, NA, 120L, "H1", 58, NA, 150L, "H1", 56, NA, 180L, "H1", 
    54, NA, 210L, "H1", 52.5, NA, 240L, "H1", 50, NA, 270L, "H1", 
    48.5, NA, 300L, "H1", 46.5, NA, 0L, "H2", 82, NA, 30L, "H2", 
    77, NA, 60L, "H2", 73, NA, 90L, "H2", 68, NA, 120L, "H2", 
    65, NA, 150L, "H2", 61, NA, 180L, "H2", 56, NA, 210L, "H2", 
    52, NA, 240L, "H2", 47.5, NA, 270L, "H2", 42.5, NA, 300L, 
    "H2", 37.5)
#> # A tibble: 22 x 4
#>    m      Time Site.ID Vol..mL.
#>    <lgl> <int> <chr>      <dbl>
#>  1 NA        0 H1          63  
#>  2 NA       30 H1          62  
#>  3 NA       60 H1          60  
#>  4 NA       90 H1          59  
#>  5 NA      120 H1          58  
#>  6 NA      150 H1          56  
#>  7 NA      180 H1          54  
#>  8 NA      210 H1          52.5
#>  9 NA      240 H1          50  
#> 10 NA      270 H1          48.5
#> # ... with 12 more rows

And lastly, I want Infil_Data.Time to look like:

tibble::tribble(~m, ~Time, ~Site.ID, ~Vol..mL., 63L, 0L, "H1", 
    63, 63L, 30L, "H1", 62, 63L, 60L, "H1", 60, 63L, 90L, "H1", 
    59, 63L, 120L, "H1", 58, 63L, 150L, "H1", 56, 63L, 180L, 
    "H1", 54, 63L, 210L, "H1", 52.5, 63L, 240L, "H1", 50, 63L, 
    270L, "H1", 48.5, 63L, 300L, "H1", 46.5, 82L, 0L, "H2", 82, 
    82L, 30L, "H2", 77, 82L, 60L, "H2", 73, 82L, 90L, "H2", 68, 
    82L, 120L, "H2", 65, 82L, 150L, "H2", 61, 82L, 180L, "H2", 
    56, 82L, 210L, "H2", 52, 82L, 240L, "H2", 47.5, 82L, 270L, 
    "H2", 42.5, 82L, 300L, "H2", 37.5)
#> # A tibble: 22 x 4
#>        m  Time Site.ID Vol..mL.
#>    <int> <int> <chr>      <dbl>
#>  1    63     0 H1          63  
#>  2    63    30 H1          62  
#>  3    63    60 H1          60  
#>  4    63    90 H1          59  
#>  5    63   120 H1          58  
#>  6    63   150 H1          56  
#>  7    63   180 H1          54  
#>  8    63   210 H1          52.5
#>  9    63   240 H1          50  
#> 10    63   270 H1          48.5
#> # ... with 12 more rows

The code I wrote should give back the third data frame that I provided.

R1 <- 0
R2 <- 1
jeff <- while (!is.null(Q_new$Vol..mL)) {
    for (x in 1:11) {
        R1 <- R1 + 1 %>% assign(Infil_Data.Time.1$m, Q_new[R2, 
            1]) %>% (next)(x)
    }
    R2 = R2 + 1
}
#> Error in eval(expr, envir, enclos): object 'Q_new' not found

[cderv]

Look at assign() documentation

assign function - RDocumentation

assign(x, value, pos = -1, envir = as.environment(pos),
       inherits = FALSE, immediate = TRUE)

As you use the function in a pipe, first x is the LHS of %>%, value is Infil_Data.Time.1$m, and pos is Q_new[R2, 1]. it is why you have the error, because it is not a correct pos argument. pos must be

The pos argument can specify the environment in which to assign the object in any of several ways: as -1 (the default), as a positive integer (the position in the search list); as the character string name of an element in the search list; or as an environment (including using sys.frame to access the currently active function calls).

What do you want to do exactly with assign call ?
I would not use that in a regular pipe with %>%, and inside a for-loop.
If you can explain what you want to achieve, we can help further.

[zman3]

What I am trying to achieve is to take the first Vol..mL value that corresponds to Time = 0 in Data.Time.1 and repeat it 12 times in a new variable called "m". Then take the next Vol..mL value that corresponds to Time = 0 and repeat it 12 times. I want to continue this through the whole data frame until there are no more Time = 0 cells.

So if you look at the data frame Infil_Data.Time.1, the "m" column contains nothing. Just a column with 517 rows of nothing. Now look at the Q_new data frame. The Vol..mL. values correspond to when Time = 0 (see Infil_Data.Time.1). I would like the value 63 to be repeated 12 times in column "m". And then value 82 to be repeated 12 times in column "m". I would like the final data frame to look like the last table I provided in the post above.

Does that make sense or do you need more information?

Cheers

-Z

[zman3]

I used the assign function because I thought that I could assign each row from column "m" a value. I also tried messing around with the match function, but didn't come out with my desired results. Plus, I am not sure if I was using it correctly. Any help would be most appreciated.

Cheers,

-Z

[andresrcs]

For future reference this is the way you should present a reprex, and I believe this is what you want to do but using data wrangling instead of loops

library(dplyr)
Q_new <- tibble::tribble(~Site.ID, ~Vol..mL., "H1", 63L, "H2", 82L, "H3", 
                         +                 69L)
Infil_Data.Time.1 <- tibble::tribble(~m, ~Time, ~Site.ID, ~Vol..mL., NA, 0L, "H1", 
                                     63, NA, 30L, "H1", 62, NA, 60L, "H1", 60, NA, 90L, "H1", 
                                     59, NA, 120L, "H1", 58, NA, 150L, "H1", 56, NA, 180L, "H1", 
                                     54, NA, 210L, "H1", 52.5, NA, 240L, "H1", 50, NA, 270L, "H1", 
                                     48.5, NA, 300L, "H1", 46.5, NA, 0L, "H2", 82, NA, 30L, "H2", 
                                     77, NA, 60L, "H2", 73, NA, 90L, "H2", 68, NA, 120L, "H2", 
                                     65, NA, 150L, "H2", 61, NA, 180L, "H2", 56, NA, 210L, "H2", 
                                     52, NA, 240L, "H2", 47.5, NA, 270L, "H2", 42.5, NA, 300L, 
                                     "H2", 37.5)
Infil_Data.Time <- Infil_Data.Time.1 %>% 
    left_join(Q_new, by = 'Site.ID') %>% 
    mutate(m = Vol..mL..y) %>% 
    select(m, Time, Site.ID, Vol..mL. = Vol..mL..x)

Infil_Data.Time
#> # A tibble: 22 x 4
#>        m  Time Site.ID Vol..mL.
#>    <int> <int> <chr>      <dbl>
#>  1    63     0 H1          63  
#>  2    63    30 H1          62  
#>  3    63    60 H1          60  
#>  4    63    90 H1          59  
#>  5    63   120 H1          58  
#>  6    63   150 H1          56  
#>  7    63   180 H1          54  
#>  8    63   210 H1          52.5
#>  9    63   240 H1          50  
#> 10    63   270 H1          48.5
#> # ... with 12 more rows

^{Created on 2019-01-04 by the reprex package (v0.2.1)}

[zman3]

Thank you andrescs. I will make sure to present a reprex like that. I entered in what you provided. However, I am getting another error.

Infil_Data.Time.1 <- Infil_Data.Time.1 %>% left_join(Q_new, by = "site.ID") %>%
mutate(m = Vol..mL..y) %>% select(m, Time, Site.ID, Vol..mL. = Vol..mL..x)
#> Error in Infil_Data.Time.1 %>% left_join(Q_new, by = "site.ID") %>% mutate(m = Vol..mL..y) %>% : could not find function "%>%"

[andresrcs]

You have to load dplyr first library(dplyr)

[zman3]

I did install the package and I reinstalled it twice more. I found the issue. Your code suggestion was correct. I used a lowercase "s" for 'Site.ID' instead of an uppercase "S". Thank you for your help again. That is exactly what I wanted.

Could you explain why my code on the original post does not work?

[andresrcs]

I'm sorry I really don't understand the logic behind your code, I just realized what you are trying to do when I saw the example dataframes

If you are going to be ussing r for analizing your data you need to have in mind that unlike VBA there is not need to manually iterate over your data, R is specificaly tailored for wrangling data in a more efficient way.

I think that a good resourse for you to read is this book https://r4ds.had.co.nz/ by Hadley Wickham

[zman3]

I'll keep that in mind for future problems. Thanks again!

[andresrcs]

If your question's been answered, would you mind choosing a solution? It helps other people see which questions still need help, or find solutions if they have similar problems. Here’s how to do it:

FAQ: How do I mark a solution?

If your question has been answered, don't forget to mark the solution! How do I mark a solution? Find the reply you want to mark as the solution and look for the row of small gray icons at the bottom of that reply. Click the one that looks like a box with a checkmark in it: [image] Hovering over the mark solution button shows the label, "Select if this reply solves the problem". If you don't see the mark solution button, try clicking the three dots button ( ••• ) to expand the full set of options. When a solution is chosen, the icon turns green and the hover label changes to: "Unselect if this reply no longer solves the problem". Success! [solution_reply_author] …

[cderv]

@zman3

Could you explain why my code on the original post does not work?

You code in the original post does not work for several reason

• while (!is.null(Q_new$Vol..mL)) will do an infinite loop, because in your data the column Vol.mL in Q_new is not null, and won't be null ever. I assumed you want to iterate on this colum
• x %>% f(y, z) works like f(x, y, z). So you can't to a pipe chain like you've done.
• assign takes a character as first argument. You can pipe it with the previous result you have
• (next)(x) is not a correct call. next is not needed, unless you want to go to next x before the end of the for loop.

The solution given by @andresrcs is way better than a for loop with index. However, you can indeed use for loops and also vectorisation to go through this

Infil_Data.Time.1 <- tibble::tribble(~m, ~Time, ~Site.ID, ~Vol..mL, NA, 0L, "H1", 
                                     63, NA, 30L, "H1", 62, NA, 60L, "H1", 60, NA, 90L, "H1", 
                                     59, NA, 120L, "H1", 58, NA, 150L, "H1", 56, NA, 180L, "H1", 
                                     54, NA, 210L, "H1", 52.5, NA, 240L, "H1", 50, NA, 270L, "H1", 
                                     48.5, NA, 300L, "H1", 46.5, NA, 0L, "H2", 82, NA, 30L, "H2", 
                                     77, NA, 60L, "H2", 73, NA, 90L, "H2", 68, NA, 120L, "H2", 
                                     65, NA, 150L, "H2", 61, NA, 180L, "H2", 56, NA, 210L, "H2", 
                                     52, NA, 240L, "H2", 47.5, NA, 270L, "H2", 42.5, NA, 300L, 
                                     "H2", 37.5)

R1 <- 0
# iterate on the data for Times == 0
for (val in Infil_Data.Time.1[Infil_Data.Time.1$Time == 0, "Vol..mL", drop = TRUE]) {
  # with vectorisation, fill value 12 by 12 according to vall
  Infil_Data.Time.1$m[(1:11) + (R1 * 11)] <- val
  # increment the indice for next twelve value
  R1 = R1 + 1
}
print(Infil_Data.Time.1, n = 22)
#> # A tibble: 22 x 4
#>        m  Time Site.ID Vol..mL
#>    <dbl> <int> <chr>     <dbl>
#>  1    63     0 H1         63  
#>  2    63    30 H1         62  
#>  3    63    60 H1         60  
#>  4    63    90 H1         59  
#>  5    63   120 H1         58  
#>  6    63   150 H1         56  
#>  7    63   180 H1         54  
#>  8    63   210 H1         52.5
#>  9    63   240 H1         50  
#> 10    63   270 H1         48.5
#> 11    63   300 H1         46.5
#> 12    82     0 H2         82  
#> 13    82    30 H2         77  
#> 14    82    60 H2         73  
#> 15    82    90 H2         68  
#> 16    82   120 H2         65  
#> 17    82   150 H2         61  
#> 18    82   180 H2         56  
#> 19    82   210 H2         52  
#> 20    82   240 H2         47.5
#> 21    82   270 H2         42.5
#> 22    82   300 H2         37.5

You see that

• you don't need to use %>% in all situation
• you can assign to a column in a data.frame simply using <-
• you can use vectorisation to a vector directly. Here, one val is assign to 12 values in one assignment

Hope it is clearer. Look into @andresrcs advices for training about data wrangling. it will help you a lot, and code will be more readable and clearer.