如何创建一个先前调查过的区域的子集，涵盖所有调查团队和土地类型？

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英文:

How do I create a subset of previously surveyed areas which covers all survey teams and land class types?

问题

我目前只使用虚拟数据（附加），但情况是：

  • 我将有大量不同的原始调查区域（在虚拟数据中为100个）
  • 每个原始调查区域都将由一个调查团队进行调查（在我的虚拟数据中，我包括了5个不同的调查团队：a、b、c、d、e）
  • 每个调查区域还被分配了一个土地类别（在我的虚拟数据中，我包括了9个土地类别：1 - 9）

我想编写一个脚本，它将为我确定重新调查的某个数量（举例来说，假设为25%）的调查区域。这些被确定重新调查的区域必须：

  • 尽可能均匀地覆盖所有调查团队（即每个团队5个）
    并且作为其中的子集
  • 尽可能均匀地覆盖所有土地类别

在R中是否可能实现这一点？或者是其他系统？我也可以访问AGOL和ArcPRO。

虚拟数据：（代码部分不翻译）

英文:

I am currently only working with dummy data (attached) but the situation is:

  • I will have a high number of different original survey areas (in dummy data, 100)
  • Each original survey area will have been surveyed by a survey team (in my dummy data, I have included 5 different survey teams: a, b, c, d, e)
  • Each survey area has also been allocated a land class type (in my dummy data, I have included 9 land classes: 1 - 9)

I want to write a script which will identify a certain number (for examples sake, let's say 25%) of survey areas for me to resurvey for quality assurance. These identified areas for resurvey must:

  • Evenly (as much as possible) cover all survey teams (i.e. 5 per team)
    AND as a subset of that
  • Evenly (as much as possible) cover all land classes

Is this possible within R? Or an alternate system? I have access to AGOL and ArcPRO too.

Dummy data:

  1. Completed survey area |	Survey team |Land Class
  2. 1	a	1
  3. 2	b	2
  4. 3	c	3
  5. 4	d	4
  6. 5	e	5
  7. 6	a	6
  8. 7	b	7
  9. 8	c	8
  10. 9	d	9
  11. 10	e	1
  12. 11	a	2
  13. 12	b	3
  14. 13	c	4
  15. 14	d	5
  16. 15	e	6
  17. 16	a	7
  18. 17	b	8
  19. 18	c	9
  20. 19	d	1
  21. 20	e	2
  22. 21	a	3
  23. 22	b	4
  24. 23	c	5
  25. 24	d	6
  26. 25	e	7
  27. 26	a	8
  28. 27	b	9
  29. 28	c	1
  30. 29	d	2
  31. 30	e	3
  32. 31	a	4
  33. 32	b	5
  34. 33	c	6
  35. 34	d	7
  36. 35	e	8
  37. 36	a	9
  38. 37	b	1
  39. 38	c	2
  40. 39	d	3
  41. 40	e	4
  42. 41	a	5
  43. 42	b	6
  44. 43	c	7
  45. 44	d	8
  46. 45	e	9
  47. 46	a	1
  48. 47	b	2
  49. 48	c	3
  50. 49	d	4
  51. 50	e	5
  52. 51	a	6
  53. 52	b	7
  54. 53	c	8
  55. 54	d	9
  56. 55	e	1
  57. 56	a	2
  58. 57	b	3
  59. 58	c	4
  60. 59	d	5
  61. 60	e	6
  62. 61	a	7
  63. 62	b	8
  64. 63	c	9
  65. 64	d	1
  66. 65	e	2
  67. 66	a	3
  68. 67	b	4
  69. 68	c	5
  70. 69	d	6
  71. 70	e	7
  72. 71	a	8
  73. 72	b	9
  74. 73	c	1
  75. 74	d	2
  76. 75	e	3
  77. 76	a	4
  78. 77	b	5
  79. 78	c	6
  80. 79	d	7
  81. 80	e	8
  82. 81	a	9
  83. 82	b	1
  84. 83	c	2
  85. 84	d	3
  86. 85	e	4
  87. 86	a	5
  88. 87	b	6
  89. 88	c	7
  90. 89	d	8
  91. 90	e	9
  92. 91	a	1
  93. 92	b	2
  94. 93	c	3
  95. 94	d	4
  96. 95	e	5
  97. 96	a	6
  98. 97	b	7
  99. 98	c	8
  100. 99	d	9
  101. 100	e	1

I am yet to try anything as not sure where to begin.

答案1

得分: 0

这是你想要的吗？我创建了一个更大的示例数据集（n = 10,000），因为你提供的示例数据集可能太小，无法得到你想要的结果。请注意，由于"Survey team"值相对于"Land Class"的分布，略多于25%的调查站点被返回。观察到一点：我建议使用语法上有效的列名，例如避免空格和特殊字符，这样你在声明列名时就不必每次都使用反引号。

英文:

Is this what your after? I created a larger example dataset (n = 10,000) as the example dataset you gave was I believe too small to get the result you're wanting. Note also that due to the distribution of the "Survey team" values relative to "Land Class", slightly more than 25% of survey sites are returned:

  1. library(tidyr)
  2. library(dplyr)
  3. # set.seed() to make results below reproducible
  4. set.seed(1)
  6. # Example df based on values given in your example df
  7. df <- data.frame(`Completed survey area` = 1:10000,
  8.                  `Survey team` = rep(letters[1:5], 2000),
  9.                  `Land Class` = c(rep(1:9, 1111), 1),
  10.                  check.names = FALSE)
  12. # Return ~25% sample of dataset with even distribution of teams and land classes
  13. resurvey <- df %>%
  14.   group_by(`Survey team`, `Land Class`) %>%
  15.   sample_frac(size=.25) # Sample size as fraction of total dataset e.g. 25%
  17. resurvey
  18. # # A tibble: 2,520 × 3
  19. # # Groups:   Survey team, Land Class [45]
  20. # `Completed survey area` `Survey team` `Land Class`
  21. #                   <int>         <chr>        <dbl>
  22. # 1                  3016             a            1
  23. # 2                  7471             a            1
  24. # 3                  5761             a            1
  25. # 4                  7246             a            1
  26. # 5                  9631             a            1
  27. # 6                  1891             a            1
  28. # 7                   586             a            1
  29. # 8                  9406             a            1
  30. # 9                  8371             a            1
  31. # 10                 2251             a            1
  32. # # … with 2,510 more rows# ℹ Use `print(n = ...)` to see more rows
  34. # Check sample distribution
  35. check <- resurvey %>% group_by(`Survey team`,`Land Class`) %>% tally()
  36. print(check, n = nrow(check))
  37. # # A tibble: 45 × 3
  38. # # Groups:   Survey team [5]
  39. # `Survey team`     `Land Class`   n
  40. # <chr>                  <dbl> <int>
  41. # 1 a                        1    56
  42. # 2 a                        2    56
  43. # 3 a                        3    56
  44. # 4 a                        4    56
  45. # 5 a                        5    56
  46. # 6 a                        6    56
  47. # 7 a                        7    56
  48. # 8 a                        8    56
  49. # 9 a                        9    56
  50. # 10 b                       1    56
  51. # 11 b                       2    56
  52. # 12 b                       3    56
  53. # 13 b                       4    56
  54. # 14 b                       5    56
  55. # 15 b                       6    56
  56. # 16 b                       7    56
  57. # 17 b                       8    56
  58. # 18 b                       9    56
  59. # 19 c                       1    56
  60. # 20 c                       2    56
  61. # 21 c                       3    56
  62. # 22 c                       4    56
  63. # 23 c                       5    56
  64. # 24 c                       6    56
  65. # 25 c                       7    56
  66. # 26 c                       8    56
  67. # 27 c                       9    56
  68. # 28 d                       1    56
  69. # 29 d                       2    56
  70. # 30 d                       3    56
  71. # 31 d                       4    56
  72. # 32 d                       5    56
  73. # 33 d                       6    56
  74. # 34 d                       7    56
  75. # 35 d                       8    56
  76. # 36 d                       9    56
  77. # 37 e                       1    56
  78. # 38 e                       2    56
  79. # 39 e                       3    56
  80. # 40 e                       4    56
  81. # 41 e                       5    56
  82. # 42 e                       6    56
  83. # 43 e                       7    56
  84. # 44 e                       8    56
  85. # 45 e                       9    56

One observation: I would recommend using syntactically valid column names - e.g. avoid spaces and special characters - that way you don't have to use back ticks every time you declare a column name.

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  • 本文由 发表于 2023年4月19日 19:25:06
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