Kate N
Reputation: 453
In R how to fill in numbers that do not appear in the sequence?
I have a data set that list the percentiles for a set of scores like this:
> percentiles
Score Percentile
1 231 0
2 385 1
3 403 2
4 413 3
5 418 4
6 424 5
7 429 6
8 434 7
9 437 8
10 441 9
11 443 10
I would like the "Score" column to run from 100 to 500. That is, I would like Scores 100 to 231 to be associated with a Percentile of 0, Scores 232 to 385 to be associated with a Percentile of 1, etc. Is there a simple way to fill in the values that do not appear in the sequence of "Score" values so it looks like the below data set?
> percentiles
Score Percentile
1 100 0
2 101 0
3 102 0
4 103 0
5 104 0
6 105 0
7 106 0
8 107 0
9 108 0
10 109 0
--------------------
130 229 0
131 230 0
132 231 0
133 232 1
134 233 1
135 234 1
136 235 1
137 236 1
138 237 1
139 238 1
140 239 1
Upvotes: 4
Views: 332
Answers (6)
hello_friend
Reputation: 5788
A bit hacky Base R:
# Create a dataframe with all score values in the range:
score_range <- merge(data.frame(Score = c(100:500)), percentiles, by = "Score", all.x = TRUE)
# Reverse the order of the dataframe:
score_range <- score_range[rev(order(score_range$Score)),]
# Change the first NA to the maximum score:
score_range$Percentile[which(is.na(score_range$Percentile))][1] <- max(score_range$Percentile, na.rm = TRUE)
# Replace all NAs with the value before them:
score_range$Percentile <- na.omit(score_range$Percentile)[cumsum(!is.na(score_range$Percentile))]
Data:
percentiles <- structure(list(Score = c(231L, 385L, 403L, 413L, 418L, 424L,
429L, 434L, 437L, 441L, 443L),
Percentile = 0:10), class = "data.frame",
row.names = c("1",
"2", "3", "4", "5", "6", "7", "8", "9", "10", "11"))
Upvotes: 0
ThomasIsCoding
Reputation: 101753
Here is a base R solution, where cut() and match() are the key points to make it, i.e.,
df <- data.frame(Score = (x <- 100:500),
percentile = percentiles$Percentile[match(s <-cut(x,c(0,percentiles$Score)),levels(s))])
such that
> df
Score percentile
1 100 0
2 101 0
3 102 0
4 103 0
5 104 0
6 105 0
7 106 0
8 107 0
9 108 0
10 109 0
11 110 0
12 111 0
13 112 0
14 113 0
15 114 0
16 115 0
17 116 0
18 117 0
19 118 0
20 119 0
21 120 0
22 121 0
23 122 0
24 123 0
25 124 0
26 125 0
27 126 0
28 127 0
29 128 0
30 129 0
31 130 0
32 131 0
33 132 0
34 133 0
35 134 0
36 135 0
37 136 0
38 137 0
39 138 0
40 139 0
41 140 0
42 141 0
43 142 0
44 143 0
45 144 0
46 145 0
47 146 0
48 147 0
49 148 0
50 149 0
51 150 0
52 151 0
53 152 0
54 153 0
55 154 0
56 155 0
57 156 0
58 157 0
59 158 0
60 159 0
61 160 0
62 161 0
63 162 0
64 163 0
65 164 0
66 165 0
67 166 0
68 167 0
69 168 0
70 169 0
71 170 0
72 171 0
73 172 0
74 173 0
75 174 0
76 175 0
77 176 0
78 177 0
79 178 0
80 179 0
81 180 0
82 181 0
83 182 0
84 183 0
85 184 0
86 185 0
87 186 0
88 187 0
89 188 0
90 189 0
91 190 0
92 191 0
93 192 0
94 193 0
95 194 0
96 195 0
97 196 0
98 197 0
99 198 0
100 199 0
101 200 0
102 201 0
103 202 0
104 203 0
105 204 0
106 205 0
107 206 0
108 207 0
109 208 0
110 209 0
111 210 0
112 211 0
113 212 0
114 213 0
115 214 0
116 215 0
117 216 0
118 217 0
119 218 0
120 219 0
121 220 0
122 221 0
123 222 0
124 223 0
125 224 0
126 225 0
127 226 0
128 227 0
129 228 0
130 229 0
131 230 0
132 231 0
133 232 1
134 233 1
135 234 1
136 235 1
137 236 1
138 237 1
139 238 1
140 239 1
141 240 1
142 241 1
143 242 1
144 243 1
145 244 1
146 245 1
147 246 1
148 247 1
149 248 1
150 249 1
151 250 1
152 251 1
153 252 1
154 253 1
155 254 1
156 255 1
157 256 1
158 257 1
159 258 1
160 259 1
161 260 1
162 261 1
163 262 1
164 263 1
165 264 1
166 265 1
167 266 1
168 267 1
169 268 1
170 269 1
171 270 1
172 271 1
173 272 1
174 273 1
175 274 1
176 275 1
177 276 1
178 277 1
179 278 1
180 279 1
181 280 1
182 281 1
183 282 1
184 283 1
185 284 1
186 285 1
187 286 1
188 287 1
189 288 1
190 289 1
191 290 1
192 291 1
193 292 1
194 293 1
195 294 1
196 295 1
197 296 1
198 297 1
199 298 1
200 299 1
201 300 1
202 301 1
203 302 1
204 303 1
205 304 1
206 305 1
207 306 1
208 307 1
209 308 1
210 309 1
211 310 1
212 311 1
213 312 1
214 313 1
215 314 1
216 315 1
217 316 1
218 317 1
219 318 1
220 319 1
221 320 1
222 321 1
223 322 1
224 323 1
225 324 1
226 325 1
227 326 1
228 327 1
229 328 1
230 329 1
231 330 1
232 331 1
233 332 1
234 333 1
235 334 1
236 335 1
237 336 1
238 337 1
239 338 1
240 339 1
241 340 1
242 341 1
243 342 1
244 343 1
245 344 1
246 345 1
247 346 1
248 347 1
249 348 1
250 349 1
251 350 1
252 351 1
253 352 1
254 353 1
255 354 1
256 355 1
257 356 1
258 357 1
259 358 1
260 359 1
261 360 1
262 361 1
263 362 1
264 363 1
265 364 1
266 365 1
267 366 1
268 367 1
269 368 1
270 369 1
271 370 1
272 371 1
273 372 1
274 373 1
275 374 1
276 375 1
277 376 1
278 377 1
279 378 1
280 379 1
281 380 1
282 381 1
283 382 1
284 383 1
285 384 1
286 385 1
287 386 2
288 387 2
289 388 2
290 389 2
291 390 2
292 391 2
293 392 2
294 393 2
295 394 2
296 395 2
297 396 2
298 397 2
299 398 2
300 399 2
301 400 2
302 401 2
303 402 2
304 403 2
305 404 3
306 405 3
307 406 3
308 407 3
309 408 3
310 409 3
311 410 3
312 411 3
313 412 3
314 413 3
315 414 4
316 415 4
317 416 4
318 417 4
319 418 4
320 419 5
321 420 5
322 421 5
323 422 5
324 423 5
325 424 5
326 425 6
327 426 6
328 427 6
329 428 6
330 429 6
331 430 7
332 431 7
333 432 7
334 433 7
335 434 7
336 435 8
337 436 8
338 437 8
339 438 9
340 439 9
341 440 9
342 441 9
343 442 10
344 443 10
345 444 NA
346 445 NA
347 446 NA
348 447 NA
349 448 NA
350 449 NA
351 450 NA
352 451 NA
353 452 NA
354 453 NA
355 454 NA
356 455 NA
357 456 NA
358 457 NA
359 458 NA
360 459 NA
361 460 NA
362 461 NA
363 462 NA
364 463 NA
365 464 NA
366 465 NA
367 466 NA
368 467 NA
369 468 NA
370 469 NA
371 470 NA
372 471 NA
373 472 NA
374 473 NA
375 474 NA
376 475 NA
377 476 NA
378 477 NA
379 478 NA
380 479 NA
381 480 NA
382 481 NA
383 482 NA
384 483 NA
385 484 NA
386 485 NA
387 486 NA
388 487 NA
389 488 NA
390 489 NA
391 490 NA
392 491 NA
393 492 NA
394 493 NA
395 494 NA
396 495 NA
397 496 NA
398 497 NA
399 498 NA
400 499 NA
401 500 NA
Upvotes: 1
user2474226
Reputation: 1502
In base R you could use the findInterval function to break up your sequence 100:500 into buckets determined by the Score, then index into the Percentile column:
x <- 100:500
ind <- findInterval(x, percentiles$Score, left.open = TRUE)
output <- data.frame(Score = x, Percentile = percentiles$Percentile[ind + 1])
Values of x above 443 will receive a percentile of NA.
Upvotes: 1
camille
Reputation: 16842
You might think about this differently: instead of a data frame to fill, as a set of breaks for binning your scores. Use the scores as the breaks with -Inf tacked on to have the lower bound. If you need something different to happen for the scores above the highest break, add Inf to the end of the breaks, but you'll need to come up with an additional label.
library(dplyr)
dat <- data.frame(Score = 100:500) %>%
mutate(Percentile = cut(Score, breaks = c(-Inf, percentiles$Score),
labels = percentiles$Percentile,
right = T, include.lowest = F))
Taking a look at a few of the breaking points:
slice(dat, c(129:135, 342:346))
#> Score Percentile
#> 1 228 0
#> 2 229 0
#> 3 230 0
#> 4 231 0
#> 5 232 1
#> 6 233 1
#> 7 234 1
#> 8 441 9
#> 9 442 10
#> 10 443 10
#> 11 444 <NA>
#> 12 445 <NA>
Upvotes: 2
IceCreamToucan
Reputation: 28695
If you convert percentiles to a data.table, you could do a rolling join with a new table of all scores 100:500. The rolling join with roll = -Inf gives a fill-backward behavior by itself, but still the 444:500 values are NA so a forward nafill is added at the end.
library(data.table)
setDT(percentiles)
percentiles[data.table(Score = 100:500), on = .(Score), roll = -Inf
][, Percentile := nafill(Percentile, 'locf')]
# Score Percentile
# 1: 100 0
# 2: 101 0
# 3: 102 0
# 4: 103 0
# 5: 104 0
# ---
# 397: 496 10
# 398: 497 10
# 399: 498 10
# 400: 499 10
# 401: 500 10
Upvotes: 4
akrun
Reputation: 887231
We could use complete
library(dplyr)
library(tidyr)
out <- complete(percentiles, Score = 100:500) %>%
fill(Percentile, .direction = "updown")
out %>%
slice(c(1:10, 130:140)) %>%
as.data.frame
# Score Percentile
#1 100 0
#2 101 0
#3 102 0
#4 103 0
#5 104 0
#6 105 0
#7 106 0
#8 107 0
#9 108 0
#10 109 0
#11 229 0
#12 230 0
#13 231 0
#14 232 1
#15 233 1
#16 234 1
#17 235 1
#18 236 1
#19 237 1
#20 238 1
#21 239 1
data
percentiles <- structure(list(Score = c(231L, 385L, 403L, 413L, 418L, 424L,
429L, 434L, 437L, 441L, 443L), Percentile = 0:10), class = "data.frame",
row.names = c("1",
"2", "3", "4", "5", "6", "7", "8", "9", "10", "11"))
Upvotes: 1
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