Question

我想从标题

中寻求提取信息的帮助

我在一个文件（下面的示例）中有一个包含数百行和1000列（相等）的表，就像这个一样，我想创建一个循环来从标题部分（新列）中提取日期并重新排序行中的值。

R2n_19970919__105056604_2_BF.MER_A123_DAY_00.nc <- c(0.09,0.09,0.08,0.08,0.06,0.07,0.09,0.08,0.08,"NA")
R2n_19970920__105056604_2_BF.MER_A123_DAY_00.nc <- c(0.08,0.08,0.08,0.07,"NA",0.05,0.08,0.08,0.08,"NA")
R2n_19970921__105056604_2_BF.MER_A123_DAY_00.nc <- c(0.07,"NA",0.08,"NA","NA",0.07,0.06,"NA",0.08,"NA")
data <- data.frame(R2n_19970919__105056604_2_BF.MER_A123_DAY_00.nc,R2n_19970920__105056604_2_BF.MER_A123_DAY_00.nc,R2n_19970921__105056604_2_BF.MER_A123_DAY_00.nc)

如何做到最好？非常感谢帮助。

这是我的预期结果：

R2n_ 19970919 __ 105056604_2_BF.MER_A123_DAY_00.nc = 1997/09/19。

Date        R2n.nc        
1997/09/19  0.09        
1997/09/19  0.09
1997/09/19  0.08
1997/09/19  0.08
1997/09/19  0.06
1997/09/19  0.07
1997/09/19  0.09
1997/09/19  0.08
1997/09/19  0.08
1997/09/19  NA
1999/09/20  0.08        
1999/09/20  0.08
1999/09/20  0.08
1999/09/20  0.07
1999/09/20  NA
1999/09/20  0.05
1999/09/20  0.08
1999/09/20  0.08
1999/09/20  0.08
1999/09/20  NA
2001/09/21  ...
.
.
.

Answer 1

这是一个解决方案，使用@RomanLuštrik建议的提示：

library(stringr) # str_sub() function
library(reshape2) # melt() function

# Modify columns names (if date information is always at the same position)
names(data) = paste0(str_sub(names(data), 5,8), "-", str_sub(names(data), 9,10), "-",str_sub(names(data), 11, 12))
data$id = seq(1,nrow(data))

# Melt the data
data_melt = melt(data, id = "id")

> data_melt
   id   variable value
1   1 1997-09-19  0.09
2   2 1997-09-19  0.09
3   3 1997-09-19  0.08
4   4 1997-09-19  0.08
5   5 1997-09-19  0.06
...

Answer 2

library(anytime)

df <- stack(data)
df$ind <- anydate(substr(df$ind, 5, 12))

head(df)
##    values        ind
## 1    0.09 1997-09-19
## 2    0.09 1997-09-19
## 3    0.08 1997-09-19
## 4    0.08 1997-09-19
## 5    0.06 1997-09-19
## 6    0.07 1997-09-19

虽然我可能会这样做：

library(anytime)
library(dplyr)

tbl_df(data) %>% 
  stack() %>% 
  mutate(ind=anydate(substr(ind, 5, 12)))
## # A tibble: 30 × 2
##    values        ind
##     <chr>     <date>
## 1    0.09 1997-09-19
## 2    0.09 1997-09-19
## 3    0.08 1997-09-19
## 4    0.08 1997-09-19
## 5    0.06 1997-09-19
## 6    0.07 1997-09-19
## 7    0.09 1997-09-19
## 8    0.08 1997-09-19
## 9    0.08 1997-09-19
## 10     NA 1997-09-19
## # ... with 20 more rows

代替。

R使用循环从标题创建新列并重新排序表

2 个答案: