在R中有条件地修改多个列

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英文:

Conditionally mutate multiple columns in R

问题

我有一个包含j个级别的因子列的数据框，以及j个长度为k的向量。我想要在前一个数据框中根据因子的条件填充k列，使用后一个向量中的值。

简化的例子（三个级别，三个向量，两个值）：

  1. df1 <- data.frame("Factor" = rep(c("A", "B", "C"), times = 5))
  2. vecA <- c(1, 2)
  3. vecB <- c(2, 1)
  4. vecC <- c(3, 3)

这里是使用嵌套的ifelse语句的解决方案：

  1. library(tidyverse)
  2. df1 %>%
  3.   mutate(V1 = ifelse(Factor == "A", vecA[1], 
  4.                      ifelse(Factor == "B", vecB[1], vecC[1])),
  5.          V2 = ifelse(Factor == "A", vecA[2], 
  6.                      ifelse(Factor == "B", vecB[2], vecC[2])))

我想要避免嵌套的ifelse语句。理想情况下，我也想避免单独对每一列进行突变。

英文:

I have a dataframe with a factor column with j levels, as well as j vectors of length k. I would like to populate k columns in the former dataframe with values from the latter vectors, conditional on the factor.

Simplified example (three levels, three vectors, two values):

  1. df1 &lt;- data.frame(&quot;Factor&quot; = rep(c(&quot;A&quot;, &quot;B&quot;, &quot;C&quot;), times = 5))
  2. vecA &lt;- c(1, 2)
  3. vecB &lt;- c(2, 1)
  4. vecC &lt;- c(3, 3)

Here is a solution using nested ifelse statements:

  1. library(tidyverse)
  2. df1 %&gt;%
  3.   mutate(V1 = ifelse(Factor == &quot;A&quot;, vecA[1], 
  4.                      ifelse(Factor == &quot;B&quot;, vecB[1], vecC[1])),
  5.          V2 = ifelse(Factor == &quot;A&quot;, vecA[2], 
  6.                      ifelse(Factor == &quot;B&quot;, vecB[2], vecC[2])))

I would like to avoid the nested ifelse statements. Ideally, I would also like to avoid mutating each column separately.

答案1

得分: 1

以下是一个想法。在全局环境中，获取所有以“vec”开头的对象，使用mget()完成。这将创建一个列表。对于列表中的每个元素，使用下划线“_”连接数字。然后，在以下连接过程中排列向量的名称。在连接之后，使用cSplit()拆分列中的值。我希望这个方法对你的实际情况适用。

  1. library(tidyverse)
  2. library(splitstackshape)
  4. # 创建一个字符向量。
  5. mychr <- map_chr(.x = mget(ls(pattern = "vec")),
  6.                  .f = function(x) {paste0(x, collapse = "_")})
  8. # 移除名称中的“vec”。
  9. names(mychr) <- sub(x = names(mychr), pattern = "vec", replacement = "")
  11. #   A     B     C 
  12. # "1_2" "2_1" "3_3"
  14. # stack()创建一个数据框。在left_join()中使用它。
  15. # 然后，拆分列中的值为两列。你可能有多个列，所以我决定在这里使用cSplit()。
  17. left_join(df1, stack(mychr), by = c("Factor" = "ind")) %>%
  18. cSplit(splitCols = "values", sep = "_", direction = "wide", type.convert = FALSE)
  20. #   Factor values_1 values_2
  21. # 1:      A        1        2
  22. # 2:      B        2        1
  23. # 3:      C        3        3
  24. # 4:      A        1        2
  25. # 5:      B        2        1
  26. # 6:      C        3        3
  27. # 7:      A        1        2
  28. # 8:      B        2        1
  29. # 9:      C        3        3
  30. #10:      A        1        2
  31. #11:      B        2        1
  32. #12:      C        3        3
  33. #13:      A        1        2
  34. #14:      B        2        1
  35. #15:      C        3        3
英文:

Here is one idea. In the global environment, get all objects that begin with "vec", which is done by mget(). This creates a list. For each element in the list, paste the numbers with "_" in between. Then, arrange names in the vector for the following join process. After join, split the column, values with cSplit(). I hope this approach will be applicable to your real situation.

  1. library(tidyverse)
  2. library(splitstackshape)
  4. # Create a character vector.
  5. mychr &lt;- map_chr(.x = mget(ls(pattern = &quot;vec&quot;)),
  6.                  .f = function(x) {paste0(x, collapse = &quot;_&quot;)})
  8. # Remove &quot;vec&quot; in names.
  9. names(mychr) &lt;- sub(x = names(mychr), pattern = &quot;vec&quot;, replacement = &quot;&quot;)
  11. #   A     B     C 
  12. #&quot;1_2&quot; &quot;2_1&quot; &quot;3_3&quot;
  14. # stack() creates a data frame. Use it in left_join().
  15. # Then, split the column, values into two columns. You probably have more than
  16. # two. So I decided to use cSplit() here.
  18. left_join(df1, stack(mychr), by = c(&quot;Factor&quot; = &quot;ind&quot;)) %&gt;%
  19. cSplit(splitCols = &quot;values&quot;, sep = &quot;_&quot;, direction = &quot;wide&quot;, type.convert = FALSE)
  21. #    Factor values_1 values_2
  22. # 1:      A        1        2
  23. # 2:      B        2        1
  24. # 3:      C        3        3
  25. # 4:      A        1        2
  26. # 5:      B        2        1
  27. # 6:      C        3        3
  28. # 7:      A        1        2
  29. # 8:      B        2        1
  30. # 9:      C        3        3
  31. #10:      A        1        2
  32. #11:      B        2        1
  33. #12:      C        3        3
  34. #13:      A        1        2
  35. #14:      B        2        1
  36. #15:      C        3        3

答案2

得分: 1

以下是翻译好的代码部分：

使用 base R 选项：

  1. df1[c('V1', 'V2')] <- do.call(Map, c(f = c, mget(ls(pattern='^vec[A-C]$'))))
  2. df1
  3. #    Factor V1 V2
  4. #1       A  1  2
  5. #2       B  2  1
  6. #3       C  3  3
  7. #4       A  1  2
  8. #5       B  2  1
  9. #6       C  3  3
  10. #7       A  1  2
  11. #8       B  2  1
  12. #9       C  3  3
  13. #10      A  1  2
  14. #11      B  2  1
  15. #12      C  3  3
  16. #13      A  1  2
  17. #14      B  2  1
  18. #15      C  3  3

或者使用 purrr 中的 transpose：

  1. library(dplyr)
  2. library(purrr)
  3. mget(ls(pattern='^vec[A-C]$')) %>%
  4.      transpose %>%
  5.      setNames(c('V1', 'V2')) %>%
  6.      cbind(df1, .)
英文:

Here is a base R option

  1. df1[c(&#39;V1&#39;, &#39;V2&#39;)] &lt;- do.call(Map, c(f = c, mget(ls(pattern=&quot;^vec[A-C]$&quot;))))
  2. df1
  3. #    Factor V1 V2
  4. #1       A  1  2
  5. #2       B  2  1
  6. #3       C  3  3
  7. #4       A  1  2
  8. #5       B  2  1
  9. #6       C  3  3
  10. #7       A  1  2
  11. #8       B  2  1
  12. #9       C  3  3
  13. #10      A  1  2
  14. #11      B  2  1
  15. #12      C  3  3
  16. #13      A  1  2
  17. #14      B  2  1
  18. #15      C  3  3

Or with transpose from purrr

  1. library(dplyr)
  2. library(purrr)
  3. mget(ls(pattern=&quot;^vec[A-C]$&quot;)) %&gt;% 
  4.      transpose %&gt;% 
  5.      setNames(c(&#39;V1&#39;, &#39;V2&#39;)) %&gt;% 
  6.      cbind(df1, .)

答案3

得分: 0

这是一种方法：

  1. # 修改向量
  2. l <- list('A' = vecA, 'B' = vecB, 'C' = vecC)
  4. # 创建带映射的数据框
  5. df2 = data.frame(t(sapply(df1$Factor, function(x) l[[x]])))
  6. colnames(df2) <- c('V1', 'V2')
  8. new_df = cbind(df1, df2)
  10.    Factor V1 V2
  11. 1       A  1  2
  12. 2       B  2  1
  13. 3       C  3  3
  14. 4       A  1  2
  15. 5       B  2  1
  16. 6       C  3  3
  17. 7       A  1  2
  18. 8       B  2  1
  19. 9       C  3  3
  20. 10      A  1  2
  21. 11      B  2  1
  22. 12      C  3  3
  23. 13      A  1  2
  24. 14      B  2  1
  25. 15      C  3  3
英文:

Here's a way to do:

  1. # modify the vectors
  2. l &lt;- list(&#39;A&#39; = vecA, &#39;B&#39; = vecB, &#39;C&#39; = vecC)
  4. # create df with mapping
  5. df2 = data.frame(t(sapply(df1$Factor, function(x) l[[x]])))
  6. colnames(df2) &lt;- c(&#39;V1&#39;, &#39;V2&#39;)
  8. new_df = cbind(df1, df2)
  10.    Factor V1 V2
  11. 1       A  1  2
  12. 2       B  2  1
  13. 3       C  3  3
  14. 4       A  1  2
  15. 5       B  2  1
  16. 6       C  3  3
  17. 7       A  1  2
  18. 8       B  2  1
  19. 9       C  3  3
  20. 10      A  1  2
  21. 11      B  2  1
  22. 12      C  3  3
  23. 13      A  1  2
  24. 14      B  2  1
  25. 15      C  3  3

huangapple
  • 本文由 发表于 2020年1月6日 21:01:57
  • 转载请务必保留本文链接：https://go.coder-hub.com/59612602.html
  • dplyr
  • if-statement
  • r
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