3

我有一个数据框（示例如下），如下：

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，但由于某种原因出现错误。

输出应如下所示（转换 df 的第一行）：

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

这样 df 的每一行在输出中变成 2 行。

我尝试过的不起作用（实际上我不确定该怎么做）：

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

我得到：

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

有任何想法吗？

4

3 回答 3

3 
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
于 2019-06-20T11:44:52.887 回答
2

这是另一个使用extractfromtidyr

library(dplyr)
library(tidyr)

df %>%
  mutate(row = row_number()) %>%
  gather(key, value, -row) %>%
  extract(key, c("key", "Time"), regex = "(Stage|Clinical.*)(\\d+y)") %>%
  spread(key, value) %>%
  select(-row)

#  Time  ClinicalActivity Stage
#   <chr> <chr>            <chr>
# 1 1y    TRUE             3.1  
# 2 2y    FALSE            NA   
# 3 1y    TRUE             1    
# 4 2y    TRUE             2    
# 5 1y    TRUE             4    
# 6 2y    TRUE             3.2  
# 7 1y    TRUE             2    
# 8 2y    TRUE             2    
# 9 1y    FALSE            NA   
#10 2y    FALSE            NA
于 2019-06-20T11:59:09.957 回答
0

我们可以很容易地做到这一点，data.table其中可以使用参数melt的多个列measure

library(data.table)
melt(setDT(df), measure = patterns("^Stage", "Clinical"), 
         value.name = c("Stage", "Clinical"),
         variable.name = "Time")[, Time := paste0(Time, "y")][]
#    Time Stage Clinical
# 1:   1y   3.1     TRUE
# 2:   1y     1     TRUE
# 3:   1y     4     TRUE
# 4:   1y     2     TRUE
# 5:   1y    NA    FALSE
# 6:   2y    NA    FALSE
# 7:   2y     2     TRUE
# 8:   2y   3.2     TRUE
# 9:   2y     2     TRUE
#10:   2y    NA    FALSE
于 2019-06-20T14:17:16.150 回答