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This is my dataframe in R: The days_id column is the output that I want How can I achieve this in R considering that I only have the first 2 columns?

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Laura
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1 Answers1

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Here is my answer: based on the first column by using group_by and group_indices to get the answer(third column).

Here is my example based ggplot2::economic dataset

  1. group_by dataset based on date column and save the group_indices into a tmp vector

  2. create a new column from tmp using mutate

Here is the code and result

> economics |>group_by(date) |> group_indices() ->tmp
> tmp

[1] 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 [19] 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 [37] 37 38 39 40 41 42 43 44 45 46 47 48 49 50 51 52 53 54 [55] 55 56 57 58 59 60 61 62 63 64 65 66 67 68 69 70 71 72 [73] 73 74 75 76 77 78 79 80 81 82 83 84 85 86 87 88 89 90 [91] 91 92 93 94 95 96 97 98 99 100 101 102 103 104 105 106 107 108 [109] 109 110 111 112 113 114 115 116 117 118 119 120 121 122 123 124 125 126 [127] 127 128 129 130 131 132 133 134 135 136 137 138 139 140 141 142 143 144 [145] 145 146 147 148 149 150 151 152 153 154 155 156 157 158 159 160 161 162 [163] 163 164 165 166 167 168 169 170 171 172 173 174 175 176 177 178 179 180 [181] 181 182 183 184 185 186 187 188 189 190 191 192 193 194 195 196 197 198 [199] 199 200 201 202 203 204 205 206 207 208 209 210 211 212 213 214 215 216 [217] 217 218 219 220 221 222 223 224 225 226 227 228 229 230 231 232 233 234 [235] 235 236 237 238 239 240 241 242 243 244 245 246 247 248 249 250 251 252 [253] 253 254 255 256 257 258 259 260 261 262 263 264 265 266 267 268 269 270 [271] 271 272 273 274 275 276 277 278 279 280 281 282 283 284 285 286 287 288 [289] 289 290 291 292 293 294 295 296 297 298 299 300 301 302 303 304 305 306 [307] 307 308 309 310 311 312 313 314 315 316 317 318 319 320 321 322 323 324 [325] 325 326 327 328 329 330 331 332 333 334 335 336 337 338 339 340 341 342 [343] 343 344 345 346 347 348 349 350 351 352 353 354 355 356 357 358 359 360 [361] 361 362 363 364 365 366 367 368 369 370 371 372 373 374 375 376 377 378 [379] 379 380 381 382 383 384 385 386 387 388 389 390 391 392 393 394 395 396 [397] 397 398 399 400 401 402 403 404 405 406 407 408 409 410 411 412 413 414 [415] 415 416 417 418 419 420 421 422 423 424 425 426 427 428 429 430 431 432 [433] 433 434 435 436 437 438 439 440 441 442 443 444 445 446 447 448 449 450 [451] 451 452 453 454 455 456 457 458 459 460 461 462 463 464 465 466 467 468 [469] 469 470 471 472 473 474 475 476 477 478 479 480 481 482 483 484 485 486 [487] 487 488 489 490 491 492 493 494 495 496 497 498 499 500 501 502 503 504 [505] 505 506 507 508 509 510 511 512 513 514 515 516 517 518 519 520 521 522 [523] 523 524 525 526 527 528 529 530 531 532 533 534 535 536 537 538 539 540 [541] 541 542 543 544 545 546 547 548 549 550 551 552 553 554 555 556 557 558 [559] 559 560 561 562 563 564 565 566 567 568 569 570 571 572 573 574

> economics |> mutate(date_id = tmp)

# A tibble: 574 x 7
   date         pce    pop psavert uempmed unemploy date_id
   <date>     <dbl>  <dbl>   <dbl>   <dbl>    <dbl>   <int>
 1 1967-07-01  507. 198712    12.6     4.5     2944       1
 2 1967-08-01  510. 198911    12.6     4.7     2945       2
 3 1967-09-01  516. 199113    11.9     4.6     2958       3
 4 1967-10-01  512. 199311    12.9     4.9     3143       4
 5 1967-11-01  517. 199498    12.8     4.7     3066       5
 6 1967-12-01  525. 199657    11.8     4.8     3018       6
 7 1968-01-01  531. 199808    11.7     5.1     2878       7
 8 1968-02-01  534. 199920    12.3     4.5     3001       8
 9 1968-03-01  544. 200056    11.7     4.1     2877       9
10 1968-04-01  544  200208    12.3     4.6     2709      10
# ... with 564 more rows
# i Use `print(n = ...)` to see more rows

Best Regards,

WangYong

Yong Wang
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