How can I generate two consecutive vectors of numbers in a loop?

Question

How can I generate two consecutive vectors of numbers in a loop?

I have the number of trees in a hectare (400) and I want to number them. I know that the species "a" is 80% of the hectare, while species "b" is 20% of the hectare.

In a loop similar to the one below (this is a simplification of a much more complex code), the first round of the loop should create a numeric vector from 1 to 320, in the second round it should create a numeric vector from 321 to 400.

percentage <- c(.8, .2) 
init_density <- 400

for (s in 1:length(percentage)) {
        #number the trees
        temp <- vector(1:init_density, length= init_density*percentage[s])
}

Or is there a way to create the two vectors automatically given the first two variables?

Answer 1

Maybe this:

#Data
percentage <- c(.5, 0.3,0.2)
n <- 400
vals <- cumsum(n*percentage)
vals2 <- c(1,vals[-length(vals)]+1)
#Store
List <- list()
#Loop
for(i in 1:length(vals))
{
  List[[i]] <- seq(vals2[i],vals[i],by = 1)
}
#Print
List

Output:

List
[[1]]
  [1]   1   2   3   4   5   6   7   8   9  10  11  12  13  14  15  16  17  18  19
 [20]  20  21  22  23  24  25  26  27  28  29  30  31  32  33  34  35  36  37  38
 [39]  39  40  41  42  43  44  45  46  47  48  49  50  51  52  53  54  55  56  57
 [58]  58  59  60  61  62  63  64  65  66  67  68  69  70  71  72  73  74  75  76
 [77]  77  78  79  80  81  82  83  84  85  86  87  88  89  90  91  92  93  94  95
 [96]  96  97  98  99 100 101 102 103 104 105 106 107 108 109 110 111 112 113 114
[115] 115 116 117 118 119 120 121 122 123 124 125 126 127 128 129 130 131 132 133
[134] 134 135 136 137 138 139 140 141 142 143 144 145 146 147 148 149 150 151 152
[153] 153 154 155 156 157 158 159 160 161 162 163 164 165 166 167 168 169 170 171
[172] 172 173 174 175 176 177 178 179 180 181 182 183 184 185 186 187 188 189 190
[191] 191 192 193 194 195 196 197 198 199 200

[[2]]
  [1] 201 202 203 204 205 206 207 208 209 210 211 212 213 214 215 216 217 218 219
 [20] 220 221 222 223 224 225 226 227 228 229 230 231 232 233 234 235 236 237 238
 [39] 239 240 241 242 243 244 245 246 247 248 249 250 251 252 253 254 255 256 257
 [58] 258 259 260 261 262 263 264 265 266 267 268 269 270 271 272 273 274 275 276
 [77] 277 278 279 280 281 282 283 284 285 286 287 288 289 290 291 292 293 294 295
 [96] 296 297 298 299 300 301 302 303 304 305 306 307 308 309 310 311 312 313 314
[115] 315 316 317 318 319 320

[[3]]
 [1] 321 322 323 324 325 326 327 328 329 330 331 332 333 334 335 336 337 338 339 340
[21] 341 342 343 344 345 346 347 348 349 350 351 352 353 354 355 356 357 358 359 360
[41] 361 362 363 364 365 366 367 368 369 370 371 372 373 374 375 376 377 378 379 380
[61] 381 382 383 384 385 386 387 388 389 390 391 392 393 394 395 396 397 398 399 400

Answer 2

Instead of the seq function I would go for the cut function and use the rle s, because looping seq might yield incomplete numbering when probabilities are low (see demonstration below).

FUN <- function(p, n) {
  l <- cut(1:n, breaks=c(0, cumsum(sort(p*n))))
  lapply(rle(as.character(l))$lengths, seq_len)
}

FUN(p=c(.8, .15, .05), n=400)
# [[1]]
# [1]  1  2  3  4  5  6  7  8  9 10 11 12 13 14 15 16 17 18 19 20
# 
# [[2]]
# [1]  1  2  3  4  5  6  7  8  9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31
# [32] 32 33 34 35 36 37 38 39 40 41 42 43 44 45 46 47 48 49 50 51 52 53 54 55 56 57 58 59 60
# 
# [[3]]
# [1]   1   2   3   4   5   6   7   8   9  10  11  12  13  14  15  16  17  18  19  20  21  22  23
# [24]  24  25  26  27  28  29  30  31  32  33  34  35  36  37  38  39  40  41  42  43  44  45  46
# [47]  47  48  49  50  51  52  53  54  55  56  57  58  59  60  61  62  63  64  65  66  67  68  69
# [70]  70  71  72  73  74  75  76  77  78  79  80  81  82  83  84  85  86  87  88  89  90  91  92
# [93]  93  94  95  96  97  98  99 100 101 102 103 104 105 106 107 108 109 110 111 112 113 114 115
# [116] 116 117 118 119 120 121 122 123 124 125 126 127 128 129 130 131 132 133 134 135 136 137 138
# [139] 139 140 141 142 143 144 145 146 147 148 149 150 151 152 153 154 155 156 157 158 159 160 161
# [162] 162 163 164 165 166 167 168 169 170 171 172 173 174 175 176 177 178 179 180 181 182 183 184
# [185] 185 186 187 188 189 190 191 192 193 194 195 196 197 198 199 200 201 202 203 204 205 206 207
# [208] 208 209 210 211 212 213 214 215 216 217 218 219 220 221 222 223 224 225 226 227 228 229 230
# [231] 231 232 233 234 235 236 237 238 239 240 241 242 243 244 245 246 247 248 249 250 251 252 253
# [254] 254 255 256 257 258 259 260 261 262 263 264 265 266 267 268 269 270 271 272 273 274 275 276
# [277] 277 278 279 280 281 282 283 284 285 286 287 288 289 290 291 292 293 294 295 296 297 298 299
# [300] 300 301 302 303 304 305 306 307 308 309 310 311 312 313 314 315 316 317 318 319 320

Demonstrating the issue with looping seq

n <- 400
length(unlist(lapply(c(.8, .199, .001), function(x, n) seq_len(n*x), n)))
# [1] 399

or respectively,

out <- NULL
probs <- c(.8, .199, .001)
for (i in seq(probs)) {
  out[[i]] <- seq_len(400*probs[i])
}
length(unlist(out))
# [1] 399

whereas:

length(unlist(FUN(p=c(.8, .199, .001), n=400)))
# [1] 400

Answer 3

Take a look at the seq functions. You have briefly,

## Default S3 method:
seq(from = 1, to = 1, by = ((to - from)/(length.out - 1)),
    length.out = NULL, along.with = NULL, ...)

seq.int(from, to, by, length.out, along.with, ...)

seq_along(along.with)
seq_len(length.out)

For your loop, you could easily use an implicit "foreach",

percentage <- c(.8, .2)
init_density <- 400

for (s in percentage) {
   temp <- seq.int(from=1, to= init_density*s)
}

(your original code would not/might not work as expected, as you are requesting the numbers from 1 to n, but the length of the output should be less than n)

If you wanted a generic approach to return an interval, based on a fraction of a whole, we could do this instead:

percentages <- c(4,1)
percentages <- percentages / sum(percentages) # ensure it sums to 1
n <- 400

create.interval <- function(i) {
  start <- if (i == 1) 0 else (sum(percentages[1:(i-1)])*n)
  end <- sum(percentages[1:i])*n
  seq.int(from=start+1, to=end)
}

Answer 4

A simple function:

multiple_vector <- function(vector,s){
  rule <- max(vector)*c(0,s,1)
  split(vector,cut(vector,rule))
}

Your example:

percentage <- .8 
init_density <- 400
vec <- 1:init_density

multiple_vector(vec,percentage)

EDIT: for multiple species:

# New species
spec1 <- 1:300
spec2 <- 1:700
spec3 <- 1:10
species <- list(spec1,spec2,spec3)

If you have an unique percentage vector for each specie

results_species <- lapply(species,multiple_vector,percentage)

If you have a percentage vector per specie, the easiest way is to use a mapper. Use purrr .

## A percentage per species
percentage1 <- c(.1,.9)
percentage2 <- c(.5,.6,.8)
percentage3 <- c(.2)
percentages <- list(percentage1,percentage2,percentage3)

## species and percentages should have the same length
purrr::map2(species,percentages,multiple_vector)