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Below you can recreate my data in R. I would like to generate a sequence of numbers based on two individual columns. In this example of real data my column names are :

df= or10x1BC

"Tank" "Core" "BCl"  "BCu"  "Mid"  "TL"   "SL"

I wish to use the value in each row from BCu and BCl to generate a sequence by 0.001. For example seq(BCu[1], BCl[1], 0.001) will generate a sequence based on the first row in each, I wish to have this work for each row down the list.

Ultimately this sequence will be used in my function to make an average of the sequence, i.e. mean(function(seq(Bcu[i], BCl[j], 0.001)) and be added to a new column or10x1BC["meanBVF"] = mean(function(seq(Bcu[i], BCl[j], 0.001)).

See data below:

structure(list(Tank = structure(c(1L, 1L, 1L, 1L, 1L, 1L, 1L, 
1L, 1L), .Label = "1", class = "factor"), Core = structure(c(1L, 
1L, 1L, 2L, 2L, 2L, 3L, 3L, 3L), .Label = c("A", "B", "C"), class = "factor"), 
    BCl = structure(c(8L, 5L, 2L, 6L, 3L, 1L, 9L, 7L, 4L), .Label = c("17", 
    "18", "22", "22.3", "23", "26", "27.3", "28", "29"), class = "factor"), 
    BCu = structure(c(8L, 5L, 2L, 6L, 3L, 1L, 9L, 7L, 4L), .Label = c("12.5", 
    "13.5", "17", "17.8", "18", "22", "22.3", "23", "27.3"), class = "factor"), 
    Mid = structure(c(8L, 5L, 2L, 6L, 3L, 1L, 9L, 7L, 4L), .Label = c("14.75", 
    "15.75", "19.5", "20.05", "20.5", "24", "24.8", "25.5", "28.15"
    ), class = "factor"), TL = structure(c(2L, 2L, 2L, 1L, 1L, 
    1L, 3L, 3L, 3L), .Label = c("26", "28", "29"), class = "factor"), 
    SL = structure(c(4L, 4L, 3L, 2L, 4L, 3L, 1L, 4L, 3L), .Label = c("1.7", 
    "4", "4.5", "5"), class = "factor")), .Names = c("Tank", 
"Core", "BCl", "BCu", "Mid", "TL", "SL"), row.names = c(NA, -9L
), class = "data.frame")
MockCommunity1
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  • It sounds like you have two questions: the sequence generation and the functions based of values of `CORE`. I'd suggest you remove the `functiona` `functionb` `functionc` info from this question so it is focused on a single problem. You can ask a new question about the functions. – Gregor Thomas Nov 16 '15 at 22:29
  • Also, the mean of an evenly-spaced sequence between two numbers will be the same as the mean of the two numbers. So if you're just generating sequences to take means, you can skip the sequence step and take the means directly. – Gregor Thomas Nov 16 '15 at 22:31
  • Editing now. functiona is actually a step function. So by generating a sequence for my range (BCl to BCu) I can calculate a weighted average for a range that falls oddly across steps. Will update the question with more detailed background information. – MockCommunity1 Nov 16 '15 at 22:35
  • Also, in your `dput`, all your data are factors. You should conver all the numbers (at least BCl, BCu, Mid, etc.) to numeric using `as.numeric(as.character(x))` – Gregor Thomas Nov 16 '15 at 22:41

1 Answers1

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mapply is like apply, or lapply, but with multiple arguments:

First, as I mentioned in the comment, we need to convert your data to numeric. I did it like this, to convert all but the second column:

df[, -2] = lapply(df[, -2], as.character)
df[, -2] = lapply(df[, -2], as.numeric)

We can then use mapply like this to generate the sequences:

seqs = mapply(FUN = function(a, b) {
      seq(from = a, to = b, by = .001)
  }, a = df$BCu, b = df$BCl)

It seems messy to put that in the data frame, but you can if you'd like:

df$seqs = seqs

If it were me, I'd probably leave it as a list of vectors outside of the data frame.

Gregor Thomas
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  • This produces accurate sequences. The result of mapply is not able to be directly used in a function later though, i.e. this simple function `mean(seqs)` where the identical sequence returns a result `mean(seq(23, 28, 0.001))`. – MockCommunity1 Nov 17 '15 at 00:28
  • That's because `mean` is vectorized. You'll need to deliberately apply mean to each sequence, e.g., `lapply(seqs, mean)`. If you end up with a list of sequences of weights, you could use `mapply` again: `mapply(weighted.mean, x = seqs, w = list_of_weight_vectors)`. – Gregor Thomas Nov 17 '15 at 00:45