Question

我导入的数据看起来像这样;

ID col1 col2 col3 col4
1  a    e    i    r
             j    s
             k    t
2  b    f    l    u
             m    v
             n    w
             o    x
3  c    g    p    y
4  d    h    q    z

并希望它被转换，以便每行有一个唯一的ID，IE：

ID col1 col2 col3 col4 col5 col6 col7 col8 col9 col10
1  a    e    i    r    j    s    k    t
2  b    f    l    u    m    v    n    w    o    x
3  c    g    p    y
4  d    h    q    z

易于消化的数据：

df <- data.frame(ID = c(1, NA, NA, 2, NA, NA, NA, 3, 4),
                 col1 = c('a', NA, NA, 'b', NA, NA, NA, 'c', 'd'),
                 col2 = c('e', NA, NA, 'f', NA, NA, NA, 'g', 'h'),
                 col3 = letters[9:17],
                 col4 = letters[18:26])

Answer 1

需要注意的是，对于这样的情况，长形式几乎总是更有用，有两种选择：

color.setRGB( Math.random(), Math.random(), Math.random() );

或将所有内容折叠为字符串并分开：

library(tidyverse)

df <- data.frame(ID = c(1, NA, NA, 2, NA, NA, NA, 3, 4),
                 col1 = c('a', NA, NA, 'b', NA, NA, NA, 'c', 'd'),
                 col2 = c('e', NA, NA, 'f', NA, NA, NA, 'g', 'h'),
                 col3 = letters[9:17],
                 col4 = letters[18:26])

df %>% fill(ID) %>% 
    gather(var, val, -ID) %>% 
    drop_na(val) %>% 
    group_by(ID) %>% 
    mutate(var = sprintf('col%02d', row_number())) %>% 
    spread(var, val)

#> # A tibble: 4 × 11
#> # Groups: ID [4]
#>      ID col01 col02 col03 col04 col05 col06 col07 col08 col09 col10
#> * <dbl> <chr> <chr> <chr> <chr> <chr> <chr> <chr> <chr> <chr> <chr>
#> 1     1     a     e     i     j     k     r     s     t  <NA>  <NA>
#> 2     2     b     f     l     m     n     o     u     v     w     x
#> 3     3     c     g     p     y  <NA>  <NA>  <NA>  <NA>  <NA>  <NA>
#> 4     4     d     h     q     z  <NA>  <NA>  <NA>  <NA>  <NA>  <NA>

Answer 2

tidyverse解决方案：

df %>%
  mutate(ID = zoo::na.locf(ID)) %>%
  mutate(row = row_number()) %>%
  tidyr::gather(col, val, col1:col4) %>%
  filter(!is.na(val)) %>%
  arrange(ID, row, col) %>%
  select(-row) %>%
  group_by(ID) %>%
  mutate(col = row_number()) %>%
  mutate(col = paste0('col', stringr::str_pad(col, side = 'left', pad = '0', width = 2))) %>%
  tidyr::spread(col, val)

Answer 3

以下是使用dplyr和tidyr的组合以及一些基础的解决方案：

library(dplyr)
library(tidyr)

df <- fill(df, ID, .direction = 'down')
numCols <- max(sapply(split(df, df$ID), function(x) sum(!is.na(x[, -1]))))

df %>%
  group_by(ID) %>%
  do(summarise(., l = paste(unlist(.[, -1])[!is.na(unlist(.[, -1]))], collapse = ' '))) %>%
  separate(l, into = paste0('col', 1:numCols), sep = ' ')

输出如下：

     ID  col1  col2  col3  col4  col5  col6  col7  col8  col9 col10
* <dbl> <chr> <chr> <chr> <chr> <chr> <chr> <chr> <chr> <chr> <chr>
1     1     a     e     i     j     k     r     s     t  <NA>  <NA>
2     2     b     f     l     m     n     o     u     v     w     x
3     3     c     g     p     y  <NA>  <NA>  <NA>  <NA>  <NA>  <NA>
4     4     d     h     q     z  <NA>  <NA>  <NA>  <NA>  <NA>  <NA>

Answer 4

基本上有时候一半都不好：

tmp <- na.omit(data.frame(id=cummax(replace(df$ID, is.na(df$ID), 0)), col=unlist(df[-1]) ))
reshape(transform(tmp, time=ave(id,id,FUN=seq_along)), direction="wide", idvar="id", sep="")

#      id col1 col2 col3 col4 col5 col6 col7 col8 col9 col10
#col11  1    a    e    i    j    k    r    s    t <NA>  <NA>
#col14  2    b    f    l    m    n    o    u    v    w     x
#col18  3    c    g    p    y <NA> <NA> <NA> <NA> <NA>  <NA>
#col19  4    d    h    q    z <NA> <NA> <NA> <NA> <NA>  <NA>

转换R中的数据

4 个答案: