在R中,使用一个函数引用另一个数据框,向数据框添加一列。

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英文:

Add a column to data frame using a function referencing another data frame in R

问题

对于每个人,对于每个t值,我需要取该t值的天花板值,找到df2中该人和天花板t值的行,然后取exp(c1b1+c2b2+c3*b3)的行,其中b1=1,b2=1,b3=1。

所以对于Ben:

ceiling(1.1)=2,在df2中对于Ben和t=2,我们得到exp(1b1+0b2+0*b3)

ceiling (2.3) = 3,在df2中对于Ben和t=3,我们得到exp(1b1+0b2+1*b3)

然后exp(1b1+0b2+0b3) + exp(1b1+0b2+1b3)

对于Lucy:

ceiling(1.2) = 2,在df2中对于Lucy和t=2,我们得到exp(0b1+0b2+1*b3)

ceiling(2.5) = 3,在df2中对于Lucy和t=3,我们得到exp(0b1+0b2+2*b3)

ceiling(2.7) = 3,在df2中对于Lucy和t=3,我们得到exp(0b1+0b2+2*b3)

然后exp(0b1+0b2+1b3) + exp(0b1+0b2+2b3) + exp(0b1+0b2+2*b3)

英文:
df1= 

  Name   t
1  Ben 1.1
2  Ben 2.3
3 Lucy 1.2
4 Lucy 2.5
5 Lucy 2.7
df2 =

  Name t c1 c2 c3
1  Ben 1  0  0  0
2  Ben 2  1  0  0
3  Ben 3  1  0  1
4 Lucy 1  1  1  0
5 Lucy 2  0  0  1
6 Lucy 3  0  0  2

For each person, for each t value, I need to take the ceiling of that t value, find the row with that person and ceiling t value in df2 and take exp(c1b1+c2*b2+c3*b3) of that row where b1=1, b2=1, b3=1.

So for Ben:

ceiling(1.1)=2, in df2 for Ben and t=2 we get exp(1b1+0b2+0*b3)

ceiling (2.3) = 3, in df2 for Ben and t=3 we get exp(1b1+0b2+1*b3)

and then exp(1b1+0b2+0b3) + exp(1b1+0b2+1b3)

For Lucy:

ceiling(1.2) = 2, in df2 for Lucy and t=2 we get exp(0b1+0b2+1*b3)

ceiling(2.5) = 3, in df2 for Lucy and t=3 we get exp(0b1+0b2+2*b3)

ceiling(2.7) = 3, in df2 for Lucy and t=3 we get exp(0b1+0b2+2*b3)

and then exp(0b1+0b2+1b3) + exp(0b1+0b2+2b3) + exp(0b1+0b2+2*b3)

答案1

得分: 2

使用dplyr包,我们可以通过left_join,然后使用mutate获取每一行的值,最后使用summarize来获取总和:

library(dplyr)

# 将b1、b2和b3设置为1
b1 <- b2 <- b3 <- 1

df1 %>%
  mutate(ceiling_t = ceiling(t)) %>%
  left_join(df2, by = c(Name = "Name", ceiling_t = "t")) %>%
  mutate(result = exp(c1 * b1 + c2 * b2 + c3 * b3)) %>%
  group_by(Name) %>%
  summarize(result = sum(result))
#> # A tibble: 2 x 2
#>   Name  result
#>   <chr>  <dbl>
#> 1 Ben     10.1
#> 2 Lucy    17.5

创建于2023-02-15,使用reprex v2.0.2

英文:

Using the dplyr package, we could do this via a left_join, then mutate to get each row's value and finally summarize to get the sum:

library(dplyr)

# Set b1, b2 and b3 to 1
b1 &lt;- b2 &lt;- b3 &lt;- 1

df1 %&gt;% 
  mutate(ceiling_t = ceiling(t)) %&gt;%
  left_join(df2, by = c(Name = &quot;Name&quot;, ceiling_t = &quot;t&quot;)) %&gt;%
  mutate(result = exp(c1 * b1 + c2 * b2 + c3 * b3)) %&gt;%
  group_by(Name) %&gt;%
  summarize(result = sum(result))
#&gt; # A tibble: 2 x 2
#&gt;   Name  result
#&gt;   &lt;chr&gt;  &lt;dbl&gt;
#&gt; 1 Ben     10.1
#&gt; 2 Lucy    17.5

<sup>Created on 2023-02-15 with reprex v2.0.2</sup>

答案2

得分: 2

在基本的R中:

a <- merge(transform(df1, t = ceiling(t)), df2)
rowsum(exp(as.matrix(a[-(1:2)])%*%c(b1 = 1,b2 = 1,b3 = 1)), a$Name)
         [,1]
Ben  10.10734
Lucy 17.49639
英文:

in base R:

a &lt;- merge(transform(df1, t = ceiling(t)), df2)
rowsum(exp(as.matrix(a[-(1:2)])%*%c(b1 = 1,b2 = 1,b3 = 1)), a$Name)
         [,1]
Ben  10.10734
Lucy 17.49639

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  • 本文由 发表于 2023年2月16日 06:20:29
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