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R 使用查找表创建新变量_R_Lookup_Sapply - Fatal编程技术网

R 使用查找表创建新变量

R 使用查找表创建新变量,r,lookup,sapply,R,Lookup,Sapply,我想使用查找表创建一个新变量。数据帧如下所示: 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

我想使用查找表创建一个新变量。数据帧如下所示:

  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
我想创建一个新变量
growth.rate
,它有两个级别:“Normal”和“Low”,因此最终的数据框如下所示

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
在本例中,id 1的growth.rate为“Low”,因为她的长度低于查找表中1岁女性的值

相反,id 2的生长率为“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


非常感谢任何帮助

如果我们在第一个数据集和基于“性别”、“年龄”的查找数据集之间进行
左联接
,我们将得到两个“长度”列,在这些列之间进行比较,并在时使用
ifelse
case\u创建一个新列

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
或者使用索引

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

c(“正常”、“低”)[1+(长度在基数R中,我们可以使用
merge
通过
sex
age
将两个数据帧连接起来,并通过使用
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。

我收到以下错误“请在…中提供偶数个参数,包括逻辑条件、结果值对(按该顺序);收到3个输入。”@Chris我无法用我在帖子中显示的数据重现错误谢谢。它在我使用索引时有效。但在我使用fcase@Chris它位于data.table的devel版本中。对不起,我应该提到它
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))
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