4列宽数据帧到3列长数据帧 [英] Wide data frame with 4 columns to long data frame with 3 columns

问题描述

我有一个数据框（下面的示例），如下所示：

I have a data frame (sample below), as follows:

df = structure(list(Stage1yBefore = c("3.1", "1", "4", "2", "NA"), 
Stage2yBefore = c("NA", "2", "3.2", "2", "NA"), ClinicalActivity1yBefore = 
c(TRUE, 
TRUE, TRUE, TRUE, FALSE), ClinicalActivity2yBefore = c(FALSE, 
TRUE, TRUE, TRUE, FALSE)), class = c("tbl_df", "tbl", "data.frame"
), row.names = c(NA, -5L))

我想使用dplyr将其转换为长格式，但由于某种原因会出现错误。

I would like to convert it to a long format using dplyr, but for some reason get an error.

输出应如下所示（将df的第一行）：

The output should look like (converting the first row of df):

Output = data_frame(TimeFrame = c("1y", "2y"), Stage = c(3, NA), Clinical = 
c(T, F))

df的行在输出中变成2行。

So that each row of df becomes 2 rows in the output.

我尝试的方法不起作用（而且我实际上不确定到底该怎么做）：

What I tried doesnt work (and I'm actually not sure exactly how to do this):

Output = gather(df, TimeFrame, Stage, Clinical, Stage1yBefore:ClinicalActivity2yBefore)

我得到：

Error in .f(.x[[i]],...): Object 'Clinical' not found.

有什么想法吗？

推荐答案

library(dplyr)
library(stringr)
library(tidyr)
df %>% rownames_to_column() %>% 
       gather(TimeFrame, Stage, Stage1yBefore:ClinicalActivity2yBefore) %>% 
       #From TimeFrame extract a digit followed by y, also Stage or Clinical 
       mutate(Time=str_extract(TimeFrame,'\\dy'), Key=str_extract(TimeFrame,'Stage|Clinical')) %>% 
       dplyr::select(-TimeFrame) %>% 
       spread(Key,Stage)

# A tibble: 10 x 4
  rowname Time  Clinical Stage
  <chr>   <chr> <chr>    <chr>
  1 1       1y    TRUE     3.1  
  2 1       2y    FALSE    NA   
  3 2       1y    TRUE     1    
  4 2       2y    TRUE     2    
  5 3       1y    TRUE     4    
  6 3       2y    TRUE     3.2  
  7 4       1y    TRUE     2    
  8 4       2y    TRUE     2    
  9 5       1y    FALSE    NA   
 10 5       2y    FALSE    NA

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