用 R 和 dplyr 显示拟合值

Showing fitted values with R and dplyr

我有数据框DF。我正在使用 Rdplyr 来分析它。

DF 包含:

>   glimpse(DF)
Observations: 1244160
Variables:
$ Channel  (int) 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0...
$ Row      (int) 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15, 16, 17,...
$ Col      (int) 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1...
$ mean     (dbl) 776.0667, 786.6000, 833.4667, 752.3333, 831.6667, 772.9333...

我适合它:

Fit <-  DF %>%
    group_by(Channel) %>% 
    do(fit = lm(mean ~ Col + poly(Row, 2), data = .))

如何使用数据在 DF 中获取另一列(给定 ChannelRowCol) 由 Fit?

拟合

如果您可以提供除 Channel、Col 和 Row 之外的更多字段,我们会给出更好的指导。现在,我已经为给定的 Channel & Col 准备了解决方案。您可以随时添加 Row 并使用您拥有的其他字段定义 lm。

我认为以下内容对您有用,

library(dplyr)
df = data.frame(Channel=c(rep(0,50),rep(1,50),rep(2,100)), 
                Row = 1:200, 
                Col = c(rep(1,50),rep(2,100),rep(3,50)), 
                mean = rnorm(200))
glimpse(df)
Fit <-  df %>%
  group_by(Channel,Col) %>% 
  do(fit = lm(mean ~ poly(Row, 2), data = .))
Fit    
Source: local data frame [4 x 3]
Groups: <by row>    
  Channel Col     fit
1       0   1 <S3:lm>
2       1   2 <S3:lm>
3       2   2 <S3:lm>
4       2   3 <S3:lm>
Fit$fit
[[1]]    
Call:
lm(formula = mean ~ poly(Row, 2), data = .)    
Coefficients:
  (Intercept)  poly(Row, 2)1  poly(Row, 2)2  
       0.1403         0.2171        -0.6281
[[2]]    
Call:
lm(formula = mean ~ poly(Row, 2), data = .)
Coefficients:
  (Intercept)  poly(Row, 2)1  poly(Row, 2)2  
     -0.07416       -0.39332        0.57889
[[3]]
Call:
lm(formula = mean ~ poly(Row, 2), data = .)
Coefficients:
  (Intercept)  poly(Row, 2)1  poly(Row, 2)2  
       0.1349        -0.3405         1.5679
[[4]]
Call:
lm(formula = mean ~ poly(Row, 2), data = .)
Coefficients:
  (Intercept)  poly(Row, 2)1  poly(Row, 2)2  
       0.0379         1.2867        -1.1028

我根据这个问题找到了一个简单的解决方案dplyr::do() requires named function?

Fit <-  DF %>%
    group_by(Channel) %>% 
    do({
        fit = lm(mean ~ Col + poly(Row, 2), data = .)
        pred <- predict(fit)
        data.frame(., pred)
    })