识别 R 中序列中大致等值数值的序列

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英文:

identify sequences of approximately equivalent values in a series using R

问题

我有一系列的数值，其中包括彼此接近的数值字符串，例如下面的序列。请注意，我已经在V1中标记了V2中具有不同值的值的范围，V2中标记为1的所有值彼此之间的点数变化在20点以内。V2中标记为2的所有值都在彼此之间的点数变化在20点以内，以此类推。请注意，这些值并不相同（它们都是不同的）。但相反，它们围绕着一个共同的值聚集。

我手动识别了这些群集。如何自动化处理它？

  1.             V1 V2
  2.     1  399.710  1
  3.     2  403.075  1
  4.     3  405.766  1
  5.     4  407.112  1
  6.     5  408.458  1
  7.     6  409.131  1
  8.     7  410.477  1
  9.     8  411.150  1
  10.     9  412.495  1
  11.     10 332.419  2
  12.     11 330.400  2
  13.     12 329.054  2
  14.     13 327.708  2
  15.     14 326.363  2
  16.     15 325.017  2
  17.     16 322.998  2
  18.     17 319.633  2
  19.     18 314.923  2
  20.     19 288.680  3
  21.     20 285.315  3
  22.     21 283.969  3
  23.     22 281.950  3
  24.     23 279.932  3
  25.     24 276.567  3
  26.     25 273.875  3
  27.     26 272.530  3
  28.     27 271.857  3
  29.     28 272.530  3
  30.     29 273.875  3
  31.     30 274.548  3
  32.     31 275.894  3
  33.     32 275.894  3
  34.     33 276.567  3
  35.     34 277.240  3
  36.     35 278.586  3
  37.     36 279.932  3
  38.     37 281.950  3
  39.     38 284.642  3
  40.     39 288.007  3
  41.     40 291.371  3
  42.     41 294.063  4
  43.     42 295.409  4
  44.     43 296.754  4
  45.     44 297.427  4
  46.     45 298.100  4
  47.     46 299.446  4
  48.     47 300.792  4
  49.     48 303.484  4
  50.     49 306.848  4
  51.     50 327.708  5
  52.     51 309.540  6
  53.     52 310.213  6
  54.     53 309.540  6
  55.     54 306.848  6
  56.     55 304.156  6
  57.     56 302.811  6
  58.     57 302.811  6
  59.     58 304.156  6
  60.     59 305.502  6
  61.     60 306.175  6
  62.     61 306.175  6
  63.     62 304.829  6

我还没有尝试任何方法，不知道如何处理这个问题。

英文:

I have a series of values that includes strings of values that are close to each other, for example the sequences below. Note that roughly around the places I have categorized the values in V1 with distinct values in V2, the range of the values changes. That is, all the values called 1 in V2 are within 20 points of each other. All the values marked 2 in V2 are within 20 points of each other. All the values marked 3 are within 20 points of each other, etc. Notice that the values are not identical (they are all different). But instead, they cluster around a common value.

I identified these clusters manually. How could I automate it?

  1.         V1 V2
  2. 1  399.710  1
  3. 2  403.075  1
  4. 3  405.766  1
  5. 4  407.112  1
  6. 5  408.458  1
  7. 6  409.131  1
  8. 7  410.477  1
  9. 8  411.150  1
  10. 9  412.495  1
  11. 10 332.419  2
  12. 11 330.400  2
  13. 12 329.054  2
  14. 13 327.708  2
  15. 14 326.363  2
  16. 15 325.017  2
  17. 16 322.998  2
  18. 17 319.633  2
  19. 18 314.923  2
  20. 19 288.680  3
  21. 20 285.315  3
  22. 21 283.969  3
  23. 22 281.950  3
  24. 23 279.932  3
  25. 24 276.567  3
  26. 25 273.875  3
  27. 26 272.530  3
  28. 27 271.857  3
  29. 28 272.530  3
  30. 29 273.875  3
  31. 30 274.548  3
  32. 31 275.894  3
  33. 32 275.894  3
  34. 33 276.567  3
  35. 34 277.240  3
  36. 35 278.586  3
  37. 36 279.932  3
  38. 37 281.950  3
  39. 38 284.642  3
  40. 39 288.007  3
  41. 40 291.371  3
  42. 41 294.063  4
  43. 42 295.409  4
  44. 43 296.754  4
  45. 44 297.427  4
  46. 45 298.100  4
  47. 46 299.446  4
  48. 47 300.792  4
  49. 48 303.484  4
  50. 49 306.848  4
  51. 50 327.708  5
  52. 51 309.540  6
  53. 52 310.213  6
  54. 53 309.540  6
  55. 54 306.848  6
  56. 55 304.156  6
  57. 56 302.811  6
  58. 57 302.811  6
  59. 58 304.156  6
  60. 59 305.502  6
  61. 60 306.175  6
  62. 61 306.175  6
  63. 62 304.829  6

I haven't tried anything yet, I don't know how to do this.

答案1

得分: 1

使用dist和hclust以及cutree来检测聚类，但在断点处具有唯一级别。

  1. hc <- hclust(dist(x))
  2. cl <- cutree(hc, k=6)
  3. data.frame(x, seq=cumsum(c(0, diff(cl)) != 0) + 1)
  4. #          x seq
  5. # 1  399.710   1
  6. # 2  403.075   1
  7. # 3  405.766   1
  8. # 4  407.112   1
  9. # 5  408.458   1
  10. # 6  409.131   1
  11. # 7  410.477   1
  12. # 8  411.150   1
  13. # 9  412.495   1
  14. # 10 332.419   2
  15. # 11 330.400   2
  16. # 12 329.054   2
  17. # 13 327.708   2
  18. # 14 326.363   2
  19. # 15 325.017   2
  20. # 16 322.998   2
  21. # 17 319.633   3
  22. # 18 314.923   3
  23. # 19 288.680   4
  24. # 20 285.315   4
  25. # 21 283.969   4
  26. # 22 281.950   4
  27. # 23 279.932   4
  28. # 24 276.567   5
  29. # 25 273.875   5
  30. # 26 272.530   5
  31. # 27 271.857   5
  32. # 28 272.530   5
  33. # 29 273.875   5
  34. # 30 274.548   5
  35. # 31 275.894   5
  36. # 32 275.894   5
  37. # 33 276.567   5
  38. # 34 277.240   5
  39. # 35 278.586   6
  40. # 36 279.932   6
  41. # 37 281.950   6
  42. # 38 284.642   6
  43. # 39 288.007   6
  44. # 40 291.371   6
  45. # 41 294.063   7
  46. # 42 295.409   7
  47. # 43 296.754   7
  48. # 44 297.427   7
  49. # 45 298.100   7
  50. # 46 299.446   7
  51. # 47 300.792   7
  52. # 48 303.484   7
  53. # 49 306.848   7
  54. # 50 327.708   8
  55. # 51 309.540   9
  56. # 52 310.213   9
  57. # 53 309.540   9
  58. # 54 306.848   9
  59. # 55 304.156   9
  60. # 56 302.811   9
  61. # 57 302.811   9
  62. # 58 304.156   9
  63. # 59 305.502   9
  64. # 60 306.175   9
  65. # 61 306.175   9
  66. # 62 304.829   9

然而，树状图表明有4个簇，而不是6个，但这是主观的。

  1. plot(hc)
  2. abline(h=30, lty=2, col=2)
  3. abline(h=18.5, lty=2, col=3)
  4. abline(h=14, lty=2, col=4)
  5. legend('topright', lty=2, col=2:4, legend=paste(c(4, 5, 7), 'cluster'), cex=.8)

数据：

  1. x <- c(399.71, 403.075, 405.766, 407.112, 408.458, 409.131, 410.477, 
  2.     411.15, 412.495, 332.419, 330.4, 329.054, 327.708, 326.363, 325.017, 
  3.     322.998, 319.633, 314.923, 288.68, 285.315, 283.969, 281.95, 
  4.     279.932, 276.567, 273.875, 272.53, 271.857, 272.53, 273.875, 
  5.     274.548, 275.894, 275.894, 276.567, 277.24, 278.586, 279.932, 
  6.     281.95, 284.642, 288.007, 291.371, 294.063, 295.409, 296.754, 
  7.     297.427, 298.1, 299.446, 300.792, 303.484, 306.848, 327.708, 
  8.     309.54, 310.213, 309.54, 306.848, 304.156, 302.811, 302.811, 
  9.     304.156, 305.502, 306.175, 306.175, 304.829)
英文:

Using dist and hclust with cutree to detect clusters, but with unique levels at the breaks.

  1. hc &lt;- hclust(dist(x))
  2. cl &lt;- cutree(hc, k=6)
  3. data.frame(x, seq=cumsum(c(0, diff(cl)) != 0) + 1)
  4. #          x seq
  5. # 1  399.710   1
  6. # 2  403.075   1
  7. # 3  405.766   1
  8. # 4  407.112   1
  9. # 5  408.458   1
  10. # 6  409.131   1
  11. # 7  410.477   1
  12. # 8  411.150   1
  13. # 9  412.495   1
  14. # 10 332.419   2
  15. # 11 330.400   2
  16. # 12 329.054   2
  17. # 13 327.708   2
  18. # 14 326.363   2
  19. # 15 325.017   2
  20. # 16 322.998   2
  21. # 17 319.633   3
  22. # 18 314.923   3
  23. # 19 288.680   4
  24. # 20 285.315   4
  25. # 21 283.969   4
  26. # 22 281.950   4
  27. # 23 279.932   4
  28. # 24 276.567   5
  29. # 25 273.875   5
  30. # 26 272.530   5
  31. # 27 271.857   5
  32. # 28 272.530   5
  33. # 29 273.875   5
  34. # 30 274.548   5
  35. # 31 275.894   5
  36. # 32 275.894   5
  37. # 33 276.567   5
  38. # 34 277.240   5
  39. # 35 278.586   6
  40. # 36 279.932   6
  41. # 37 281.950   6
  42. # 38 284.642   6
  43. # 39 288.007   6
  44. # 40 291.371   6
  45. # 41 294.063   7
  46. # 42 295.409   7
  47. # 43 296.754   7
  48. # 44 297.427   7
  49. # 45 298.100   7
  50. # 46 299.446   7
  51. # 47 300.792   7
  52. # 48 303.484   7
  53. # 49 306.848   7
  54. # 50 327.708   8
  55. # 51 309.540   9
  56. # 52 310.213   9
  57. # 53 309.540   9
  58. # 54 306.848   9
  59. # 55 304.156   9
  60. # 56 302.811   9
  61. # 57 302.811   9
  62. # 58 304.156   9
  63. # 59 305.502   9
  64. # 60 306.175   9
  65. # 61 306.175   9
  66. # 62 304.829   9

However, the dendrogram suggests rather k=4 clusters instead of 6, but it is arbitrary.

  1. plot(hc)
  2. abline(h=30, lty=2, col=2)
  3. abline(h=18.5, lty=2, col=3)
  4. abline(h=14, lty=2, col=4)
  5. legend(&#39;topright&#39;, lty=2, col=2:4, legend=paste(c(4, 5, 7), &#39;cluster&#39;), cex=.8)

识别 R 中序列中大致等值数值的序列

Data:

  1. x &lt;- c(399.71, 403.075, 405.766, 407.112, 408.458, 409.131, 410.477, 
  2. 411.15, 412.495, 332.419, 330.4, 329.054, 327.708, 326.363, 325.017, 
  3. 322.998, 319.633, 314.923, 288.68, 285.315, 283.969, 281.95, 
  4. 279.932, 276.567, 273.875, 272.53, 271.857, 272.53, 273.875, 
  5. 274.548, 275.894, 275.894, 276.567, 277.24, 278.586, 279.932, 
  6. 281.95, 284.642, 288.007, 291.371, 294.063, 295.409, 296.754, 
  7. 297.427, 298.1, 299.446, 300.792, 303.484, 306.848, 327.708, 
  8. 309.54, 310.213, 309.54, 306.848, 304.156, 302.811, 302.811, 
  9. 304.156, 305.502, 306.175, 306.175, 304.829)

答案2

得分: 0

这个解决方案遍历每个数值，检查到该点为止组内所有值的范围，并且如果范围大于阈值，则开始一个新的组。

  1. maxrange <- 18
  3. grp_start <- 1
  4. grp_num <- 1
  5. V3 <- numeric(length(dat$V1))
  6. for (i in seq_along(dat$V1)) {
  7.   grp <- dat$V1[grp_start:i]
  8.   if (max(grp) - min(grp) > maxrange) {
  9.     grp_num <- grp_num + 1 
  10.     grp_start <- i
  11.   }
  12.   V3[[i]] <- grp_num
  13. }
  15. cbind(dat, V3)
英文:

This solution iterates over every value, checks the range of all values in the group up to that point, and starts a new group if the range is greater than a threshold.

  1. maxrange &lt;- 18
  3. grp_start &lt;- 1
  4. grp_num &lt;- 1
  5. V3 &lt;- numeric(length(dat$V1))
  6. for (i in seq_along(dat$V1)) {
  7.   grp &lt;- dat$V1[grp_start:i]
  8.   if (max(grp) - min(grp) &gt; maxrange) {
  9.     grp_num &lt;- grp_num + 1 
  10.     grp_start &lt;- i
  11.   }
  12.   V3[[i]] &lt;- grp_num
  13. }
  15. cbind(dat, V3)
  1.         V1 V2 V3
  2. 1  399.710  1  1
  3. 2  403.075  1  1
  4. 3  405.766  1  1
  5. 4  407.112  1  1
  6. 5  408.458  1  1
  7. 6  409.131  1  1
  8. 7  410.477  1  1
  9. 8  411.150  1  1
  10. 9  412.495  1  1
  11. 10 332.419  2  2
  12. 11 330.400  2  2
  13. 12 329.054  2  2
  14. 13 327.708  2  2
  15. 14 326.363  2  2
  16. 15 325.017  2  2
  17. 16 322.998  2  2
  18. 17 319.633  2  2
  19. 18 314.923  2  2
  20. 19 288.680  3  3
  21. 20 285.315  3  3
  22. 21 283.969  3  3
  23. 22 281.950  3  3
  24. 23 279.932  3  3
  25. 24 276.567  3  3
  26. 25 273.875  3  3
  27. 26 272.530  3  3
  28. 27 271.857  3  3
  29. 28 272.530  3  3
  30. 29 273.875  3  3
  31. 30 274.548  3  3
  32. 31 275.894  3  3
  33. 32 275.894  3  3
  34. 33 276.567  3  3
  35. 34 277.240  3  3
  36. 35 278.586  3  3
  37. 36 279.932  3  3
  38. 37 281.950  3  3
  39. 38 284.642  3  3
  40. 39 288.007  3  3
  41. 40 291.371  3  4
  42. 41 294.063  4  4
  43. 42 295.409  4  4
  44. 43 296.754  4  4
  45. 44 297.427  4  4
  46. 45 298.100  4  4
  47. 46 299.446  4  4
  48. 47 300.792  4  4
  49. 48 303.484  4  4
  50. 49 306.848  4  4
  51. 50 327.708  5  5
  52. 51 309.540  6  6
  53. 52 310.213  6  6
  54. 53 309.540  6  6
  55. 54 306.848  6  6
  56. 55 304.156  6  6
  57. 56 302.811  6  6
  58. 57 302.811  6  6
  59. 58 304.156  6  6
  60. 59 305.502  6  6
  61. 60 306.175  6  6
  62. 61 306.175  6  6
  63. 62 304.829  6  6

A threshold of 18 reproduces your groups, except that group 4 starts one row earlier. You could use a higher threshold, but then group 6 would start later than you have it.

huangapple
  • 本文由 发表于 2023年2月19日 03:04:47
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