使用查找表创建新变量

Question

我想使用查找表创建一个新变量。数据框如下所示：

  id    sex     age length
   1    Female  1   45
   2    Female  2   54
   3    Female  3   56
   4    Female  4   60
   5    Female  5   60
   6    Female  6   61
   7    Female  7   63
   8    Male    1   55
   9    Male    2   54
   10   Male    3   58
   11   Male    4   61
   12   Male    5   65
   13   Male    6   63
   14   Male    7   65
   15   Male    8   67
   16   Male    9   68
   17   Male    10  69

查找表如下所示

sex    age  length
Female  1   50
Female  2   53
Female  3   56
Female  4   58
Female  5   60
Female  6   61
Female  7   63
Male    1   50
Male    2   54
Male    3   57
Male    4   60
Male    5   62
Male    6   63
Male    7   65
Male    8   66
Male    9   67
Male    10  69

我想新建一个变量growth.rate，有两个级别：“Normal”和“Low”，所以最终的数据框是这样的，

id   sex   age  length  growth.rate
1   Female  1   45  Low
2   Female  2   54  Normal
3   Female  3   56  Low
4   Female  4   60  Normal
5   Female  5   60  Low
6   Female  6   61  Low
7   Female  7   63  Low
8   Male    1   55  Normal
9   Male    2   54  Low
10  Male    3   58  Normal
11  Male    4   61  Normal
12  Male    5   65  Normal
13  Male    6   63  Low
14  Male    7   65  Low
15  Male    8   67  Normal
16  Male    9   68  Normal
17  Male    10  69  Low

在此示例中，id 1 的growth.rate 为“Low”，因为她的长度低于查找表中针对 1 岁女性的值。

相反，id 2 的 growth.rate 为“正常”，因为她的长度高于查找表中 2 岁女性的值。

我试图调整这个解决方案但没有成功Getting contextstack overflow error - too many nested ifelse statements within for loop?

非常感谢任何帮助

Answer 1

在base R中，我们可以使用merge将sex和age这两个数据框连接起来，并通过ifelse检查条件来创建一个新列。

transform(merge(df, lookup, all.x = TRUE, by = c("sex", "age")), 
          growth.rate = ifelse(length.x > length.y, "Normal", "Low"))

#      sex age id length.x length.y growth.rate
#1  Female   1  1       45       50         Low
#2  Female   2  2       54       53      Normal
#3  Female   3  3       56       56         Low
#4  Female   4  4       60       58      Normal
#5  Female   5  5       60       60         Low
#6  Female   6  6       61       61         Low
#7  Female   7  7       63       63         Low
#8    Male   1  8       55       50      Normal
#9    Male  10 17       69       69         Low
#10   Male   2  9       54       54         Low
#11   Male   3 10       58       57      Normal
#12   Male   4 11       61       60      Normal
#13   Male   5 12       65       62      Normal
#14   Male   6 13       63       63         Low
#15   Male   7 14       65       65         Low
#16   Male   8 15       67       66      Normal
#17   Male   9 16       68       67      Normal

您可以删除不需要的列。

Answer 2

如果我们在第一个和基于“性别”、“年龄”的查找数据集之间进行left_join，我们会得到两个“长度”列，在这些列之间进行比较并使用ifelse 或@ 创建一个新列987654323@

library(dplyr)
left_join(df1, lookup, by = c('sex', 'age')) %>%
    transmute(id, sex, age, 
      growth.rate = case_when(length.x <= length.y ~ "Low", 
        TRUE ~ "Normal"), length = length.x)
#   id    sex age growth.rate length
#1   1 Female   1         Low     45
#2   2 Female   2      Normal     54
#3   3 Female   3         Low     56
#4   4 Female   4      Normal     60
#5   5 Female   5         Low     60
#6   6 Female   6         Low     61
#7   7 Female   7         Low     63
#8   8   Male   1      Normal     55
#9   9   Male   2         Low     54
#10 10   Male   3      Normal     58
#11 11   Male   4      Normal     61
#12 12   Male   5      Normal     65
#13 13   Male   6         Low     63
#14 14   Male   7         Low     65
#15 15   Male   8      Normal     67
#16 16   Male   9      Normal     68
#17 17   Male  10         Low     69

---

在data.table，这可以更紧凑

library(data.table)
setDT(df1)[lookup, growth.rate := fcase(length <= i.length, "Low", 
           "Normal"), on = .(sex, age)]

或带索引

setDT(df1)[lookup, growth.rate := 
       c("Normal", "Low")[1 + (length <= i.length)], on = .(sex, age)]

数据

df1 <- structure(list(id = 1:17, sex = c("Female", "Female", "Female", 
"Female", "Female", "Female", "Female", "Male", "Male", "Male", 
"Male", "Male", "Male", "Male", "Male", "Male", "Male"), age = c(1L, 
2L, 3L, 4L, 5L, 6L, 7L, 1L, 2L, 3L, 4L, 5L, 6L, 7L, 8L, 9L, 10L
), length = c(45L, 54L, 56L, 60L, 60L, 61L, 63L, 55L, 54L, 58L, 
61L, 65L, 63L, 65L, 67L, 68L, 69L)), class = "data.frame", row.names = c(NA, 
-17L))

lookup <- structure(list(sex = c("Female", "Female", "Female", "Female", 
"Female", "Female", "Female", "Male", "Male", "Male", "Male", 
"Male", "Male", "Male", "Male", "Male", "Male"), age = c(1L, 
2L, 3L, 4L, 5L, 6L, 7L, 1L, 2L, 3L, 4L, 5L, 6L, 7L, 8L, 9L, 10L
), length = c(50L, 53L, 56L, 58L, 60L, 61L, 63L, 50L, 54L, 57L, 
60L, 62L, 63L, 65L, 66L, 67L, 69L)), class = "data.frame", row.names = c(NA, 
-17L))