r - 在 R 中整理数据

Question

我想从下图中的左表转到右表，但似乎无法找到背后的编码逻辑来使用 R 达到结果。

表

非常感谢您的帮助 ！

Answer 1

我创建了一个最小的例子，应该做你想做的事。这里的主要问题是表达你的问题，因为我认为有比我更好的答案来将滞后值与模态匹配。

library(dplyr)

# --- v0 is your data simplified
v0 <- c("cA", "t1", "t2", "cB", "t3") 

# --- indic tels us what are the groups
indic <- v0 %>%  stringr::str_detect(string = ., pattern = "c") %>% 
        cumsum()

# --- here you can try the code line by line (without the %>% (pipe) operator to understand the code
dfr <- tibble(v0, indic)
dfr %>% 
        group_by(indic) %>% 
        mutate(v1 = v0[which(stringr::str_detect(v0, "c") )] ) %>% 
        ungroup() %>% 
        filter(! stringr::str_detect(v0, "c")) %>% 
        select(v1, v0)

#> # A tibble: 3 x 2
#>   v1    v0   
#>   <chr> <chr>
#> 1 cA    t1   
#> 2 cA    t2   
#> 3 cB    t3

# you could also use a loop

Answer 2

使用 base 的示例R：

data <- c(
  "cinema A", "17:45", "20:00", "cinema B", "13:00", "15:45", "16:00", 
  "cinema C", "08:20"
)

time_rows <- grep("cinema", data, invert = TRUE)
data.frame(
  time = data[time_rows],
  cinema = grep("cinema", data, value = TRUE)[cumsum(grepl("cinema", data))][time_rows]
)

Answer 3

如评论中所述，请为以后的帖子提供一些示例数据。在这种情况下，我根据您所附的图片为您制作了它。

有很多方法可以解决这个问题。这是一个三步法。

library(tidyverse)
library(stringr)

# Create the data
df <- tibble(
  X1 = c("cinema A", 17.45, 20.00, "cinema B", 13.00, 15.45, 16.00, "cinema C", 8.20))

df
#> # A tibble: 9 x 1
#>   X1      
#>   <chr>   
#> 1 cinema A
#> 2 17.45   
#> 3 20      
#> 4 cinema B
#> 5 13      
#> 6 15.45   
#> 7 16      
#> 8 cinema C
#> 9 8.2

# Step 1: detect where the cinema values are and copy them to a new column
df$cinema <- ifelse(str_detect(df$X1, "cinema"), df$X1, NA)
df
#> # A tibble: 9 x 2
#>   X1       cinema  
#>   <chr>    <chr>   
#> 1 cinema A cinema A
#> 2 17.45    <NA>    
#> 3 20       <NA>    
#> 4 cinema B cinema B
#> 5 13       <NA>    
#> 6 15.45    <NA>    
#> 7 16       <NA>    
#> 8 cinema C cinema C
#> 9 8.2      <NA>

# Step 2: replace NA values in the new column with the values above
df <- fill(df, cinema)
df
#> # A tibble: 9 x 2
#>   X1       cinema  
#>   <chr>    <chr>   
#> 1 cinema A cinema A
#> 2 17.45    cinema A
#> 3 20       cinema A
#> 4 cinema B cinema B
#> 5 13       cinema B
#> 6 15.45    cinema B
#> 7 16       cinema B
#> 8 cinema C cinema C
#> 9 8.2      cinema C

# Step 3: remove the rows where X1 contains cinema information
df <- filter(df, !str_detect(df$X1, "cinema"))
df
#> # A tibble: 6 x 2
#>   X1    cinema  
#>   <chr> <chr>   
#> 1 17.45 cinema A
#> 2 20    cinema A
#> 3 13    cinema B
#> 4 15.45 cinema B
#> 5 16    cinema B
#> 6 8.2   cinema C

^{由reprex 包（v0.3.0）于 2019-11-26 创建}

Answer 4

这是一个解决方案base R。

假设输入作为数据框给出，即：

df <- data.frame(X = c("cinema A", 17.45, 20.00, "cinema B", 13.00, 15.45, 16.00, "cinema C", 8.20))
> df
         X
1 cinema A
2    17.45
3       20
4 cinema B
5       13
6    15.45
7       16
8 cinema C
9      8.2

以下代码可以帮助您获得右侧的表格：

lst <- split(df,findInterval(seq(nrow(df)),grep("cinema",df$X)-1,left.open = T))
res <- Reduce(rbind,lapply(lst, function(v) data.frame(ViewingTime = v[-1,],CinemaName = v[1,])))

输出res如下所示：

> res
  ViewingTime CinemaName
1       17.45   cinema A
2          20   cinema A
3          13   cinema B
4       15.45   cinema B
5          16   cinema B
6         8.2   cinema C