创建一个新变量仅当满足条件时。

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英文:

How to create a new variable only if a condition if satisfied?

问题

我想创建一个新变量，从两个不同数据集中减去两个变量。但我需要R仅计算相同值的差异... 在我的情况下，"相同的事物" 是第三个变量 COD_PROV，它指示城市。

我需要R计算数据集1的 Real Wage 变量和数据集2的 Real Wage 变量之间的差异，但仅当这些实际工资的 COD_PROV 相同时。

这是数据集1的示例：

  1. COD_PROV     Real wage 
  2. 1            1962.18
  3. 6            1742.85
  4. 5            1541.81
  5. 96           1612.2
  6. 4            1574
  7. 3            1823.53
  8. 103          1584.49
  9. 2            1666.21
  10. 7            1747.81
  11. 10           2066.42
  12. 8            1498.01
  13. 11           1871.34
  14. 9            1770.41
  15. 15           2240.03
  16. 16           1729.17
  17. 17           1773.38
  18. 13           1832.57

数据集2具有相同的框架，但缺少一些 COD_PROV 值：

  1. COD_PROV     Real wage 
  2. 1            4962.18
  3. 6            1542.85
  4. 5            3541.81
  5. 4            1564
  6. 3            1223.53
  7. 2            1446.21
  8. 7            1557.81
  9. 10           2226.42
  10. 8            1458.01
  11. 11           1843.34
  12. 16           1439.17
  13. 17           1883.38
  14. 13           1992.57

我尝试了这个：

  1. new <- mutate(dataset1, `Wage Difference` = dataset1$`Real wage` - dataset2$`Real wage`)

但R当然会回复：

  1. Error in mutate():
  2. ℹ In argument: Wage difference = ... - dataset1$Real wage.
  3. Caused by error:
  4. ! Wage difference must be size 105 or 1, not 106.
  5. Run rlang::last_error() to see where the error occurred.

我认为原因是数据集2的观测数量比数据集1少（特别是缺少一些 COD_PROV 值）... 我如何只对相同的 COD_PROV 值应用差异？

英文:

I want to create a new variable subtracting two variables from two different datasets. But I need that R make the difference only for the values that are referred to the same thing... The "same thing" in my case is the a third variable COD_PROV which indicates the city.

I need that R make the difference between the Real Wage variable of dataset 1 and the Real Wage variable of dataset 2, but only if the COD_PROV of these real wages is the same.

This is an example of the dataset 1

  1. COD_PROV     Real wage 
  2. 1            1962,18
  3. 6            1742,85
  4. 5            1541,81
  5. 96           1612,2
  6. 4            1574
  7. 3            1823,53
  8. 103          1584,49
  9. 2            1666,21
  10. 7            1747,81
  11. 10           2066,42
  12. 8            1498,01
  13. 11           1871,34
  14. 9            1770,41
  15. 15           2240,03
  16. 16           1729,17
  17. 17           1773,38
  18. 13           1832,57

Datset 2 has the same framework, but some values of COD_PROV are missing

  1. COD_PROV     Real wage 
  2. 1            4962,18
  3. 6            1542,85
  4. 5            3541,81
  5. 4            1564
  6. 3            1223,53
  7. 2            1446,21
  8. 7            1557,81
  9. 10           2226,42
  10. 8            1458,01
  11. 11           1843,34
  12. 16           1439,17
  13. 17           1883,38
  14. 13           1992,57

I've tried this

  1. new &lt;- mutate( dataset1, `Wage Difference ` = dataset1$`Real wage` - dataset2$`Real wage` )

but R of course replies

  1. Error in mutate():
  2. ℹ In argument: Wage difference = ... - dataset1$Real wage.
  3. Caused by error:
  4. ! Wage difference must be size 105 or 1, not 106.
  5. Run rlang::last_error() to see where the error occurred.

I suppose that the reason is that dataset 2 has less observations than dataset1 ( in particular some values of COD_PROV are missing)... How can I apply the difference only for the same values of COD_PROV ?

答案1

得分: 1

我认为你应该将它们合并在一起，然后计算差异。

  1. library(dplyr)
  2. dataset1 %>%
  3.   left_join(dataset2, by = "COD_PROV", suffix = c("", ".y")) %>%
  4.   mutate(diff = `Real wage` - `Real wage.y`)
  5. #    COD_PROV Real wage Real wage.y  diff
  6. # 1         1   1962.18     4962.18 -3000
  7. # 2         6   1742.85     1542.85   200
  8. # 3         5   1541.81     3541.81 -2000
  9. # 4        96   1612.20          NA    NA
  10. # 5         4   1574.00     1564.00    10
  11. # 6         3   1823.53     1223.53   600
  12. # 7       103   1584.49          NA    NA
  13. # 8         2   1666.21     1446.21   220
  14. # 9         7   1747.81     1557.81   190
  15. # 10       10   2066.42     2226.42  -160
  16. # 11        8   1498.01     1458.01    40
  17. # 12       11   1871.34     1843.34    28
  18. # 13        9   1770.41          NA    NA
  19. # 14       15   2240.03          NA    NA
  20. # 15       16   1729.17     1439.17   290
  21. # 16       17   1773.38     1883.38  -110
  22. # 17       13   1832.57     1992.57  -160

有关"join/merge"的更多信息，请参见https://stackoverflow.com/q/1299871/3358272，https://stackoverflow.com/q/5706437/3358272。

数据：

  1. dataset1 <- structure(list(COD_PROV = c(1L, 6L, 5L, 96L, 4L, 3L, 103L, 2L, 7L, 10L, 8L, 11L, 9L, 15L, 16L, 17L, 13L), "Real wage" = c(1962.18, 1742.85, 1541.81, 1612.2, 1574, 1823.53, 1584.49, 1666.21, 1747.81, 2066.42, 1498.01, 1871.34, 1770.41, 2240.03, 1729.17, 1773.38, 1832.57)), row.names = c(NA, -17L), class = "data.frame")
  2. dataset2 <- structure(list(COD_PROV = c(1L, 6L, 5L, 4L, 3L, 2L, 7L, 10L, 8L, 11L, 16L, 17L, 13L), "Real wage" = c(4962.18, 1542.85, 3541.81, 1564, 1223.53, 1446.21, 1557.81, 2226.42, 1458.01, 1843.34, 1439.17, 1883.38, 1992.57)), row.names = c(NA, -13L), class = "data.frame")
英文:

I think you should join/merge these together and then calculate the difference.

  1. library(dplyr)
  2. dataset1 %&gt;%
  3.   left_join(dataset2, by = &quot;COD_PROV&quot;, suffix = c(&quot;&quot;, &quot;.y&quot;)) %&gt;%
  4.   mutate(diff = `Real wage` - `Real wage.y`)
  5. #    COD_PROV Real wage Real wage.y  diff
  6. # 1         1   1962.18     4962.18 -3000
  7. # 2         6   1742.85     1542.85   200
  8. # 3         5   1541.81     3541.81 -2000
  9. # 4        96   1612.20          NA    NA
  10. # 5         4   1574.00     1564.00    10
  11. # 6         3   1823.53     1223.53   600
  12. # 7       103   1584.49          NA    NA
  13. # 8         2   1666.21     1446.21   220
  14. # 9         7   1747.81     1557.81   190
  15. # 10       10   2066.42     2226.42  -160
  16. # 11        8   1498.01     1458.01    40
  17. # 12       11   1871.34     1843.34    28
  18. # 13        9   1770.41          NA    NA
  19. # 14       15   2240.03          NA    NA
  20. # 15       16   1729.17     1439.17   290
  21. # 16       17   1773.38     1883.38  -110
  22. # 17       13   1832.57     1992.57  -160

For more information on what join/merge means, see https://stackoverflow.com/q/1299871/3358272, https://stackoverflow.com/q/5706437/3358272.

Data

  1. dataset1 &lt;- structure(list(COD_PROV = c(1L, 6L, 5L, 96L, 4L, 3L, 103L, 2L, 7L, 10L, 8L, 11L, 9L, 15L, 16L, 17L, 13L), &quot;Real wage&quot; = c(1962.18, 1742.85, 1541.81, 1612.2, 1574, 1823.53, 1584.49, 1666.21, 1747.81, 2066.42, 1498.01, 1871.34, 1770.41, 2240.03, 1729.17, 1773.38, 1832.57)), row.names = c(NA, -17L), class = &quot;data.frame&quot;)
  2. dataset2 &lt;- structure(list(COD_PROV = c(1L, 6L, 5L, 4L, 3L, 2L, 7L, 10L, 8L, 11L, 16L, 17L, 13L), &quot;Real wage&quot; = c(4962.18, 1542.85, 3541.81, 1564, 1223.53, 1446.21, 1557.81, 2226.42, 1458.01, 1843.34, 1439.17, 1883.38, 1992.57)), row.names = c(NA, -13L), class = &quot;data.frame&quot;)

答案2

得分: 0

我们可以使用 {powerjoin}:

  1. powerjoin::power_left_join(dataset1, dataset2, by = "COD_PROV", conflict = `-`)
  2. #>    COD_PROV  实际工资
  3. #> 1         1   -3000
  4. #> 2         6    200
  5. #> 3         5   -2000
  6. #> 4        96   <NA>
  7. #> 5         4     10
  8. #> 6         3    600
  9. #> 7       103   <NA>
  10. #> 8         2    220
  11. #> 9         7    190
  12. #> 10       10   -160
  13. #> 11        8     40
  14. #> 12       11     28
  15. #> 13        9   <NA>
  16. #> 14       15   <NA>
  17. #> 15       16    290
  18. #> 16       17   -110
  19. #> 17       13   -160

创建于2023年3月17日，使用 reprex v2.0.2

英文:

We can use {powerjoin}:

  1. powerjoin::power_left_join(dataset1, dataset2, by = &quot;COD_PROV&quot;, conflict = `-`)
  2. #&gt;    COD_PROV Real wage
  3. #&gt; 1         1     -3000
  4. #&gt; 2         6       200
  5. #&gt; 3         5     -2000
  6. #&gt; 4        96        NA
  7. #&gt; 5         4        10
  8. #&gt; 6         3       600
  9. #&gt; 7       103        NA
  10. #&gt; 8         2       220
  11. #&gt; 9         7       190
  12. #&gt; 10       10      -160
  13. #&gt; 11        8        40
  14. #&gt; 12       11        28
  15. #&gt; 13        9        NA
  16. #&gt; 14       15        NA
  17. #&gt; 15       16       290
  18. #&gt; 16       17      -110
  19. #&gt; 17       13      -160

<sup>Created on 2023-03-17 with reprex v2.0.2</sup>

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  • 本文由 发表于 2023年2月27日 01:27:23
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  • conditional-statements
  • dataset
  • dplyr
  • r
  • variables
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