复制每一行并将一列更改为二进制值

Question

df <- data.frame(n = c(3, 2, 2), 
                 survive = c(2, 1, 2), 
                 a = c(1,1,0), 
                 b = c(0,0,1))

如何扩展上面 data.frame 的最后两列，以便每行出现在列“n”中指定的次数。而第二列“survive”根据“survive”的值变为二进制值0/1

换句话说:

n  survive a  b
3  2       1  0
2  1       1  0
2  2       0  1

对此

survive a  b
1       1  0
1       1  0
0       1  0
1       1  0
0       1  0
1       0  1
1       0  1

Answer 1

几种替代解决方案:

1) 使用基础 R:

rn <- rep(1:nrow(df), df$n)
df2 <- df[rn,]
df2$survive <- as.integer(df2$survive >= ave(rn, rn, FUN = seq_along))

这使:

> df2[,-1]
   survive a b
1:       1 1 0
2:       1 1 0
3:       0 1 0
4:       1 1 0
5:       0 1 0
6:       1 0 1
7:       1 0 1

2) 使用 data.table 包:

library(data.table)
df2 <- setDT(df)[, rid := .I
                 ][, .(survive = c(rep(1, survive), rep(0, n - survive)), a, b)
                   , by = rid
                   ][, rid := NULL][]

这使:

> df2
   survive a b
1:       1 1 0
2:       1 1 0
3:       0 1 0
4:       1 1 0
5:       0 1 0
6:       1 0 1
7:       1 0 1

或者更短一点:

df2 <- setDT(df)[, .(survive = c(rep(1, survive), rep(0, n - survive)), a, b), by = 1:nrow(df)
                 ][, nrow := NULL]

3) 使用 dplyr 包:

library(dplyr)
df %>% 
  mutate(rid = row_number()) %>% 
  .[rep(1:nrow(df), df$n),] %>% 
  group_by(rid) %>% 
  mutate(survive = c(rep(1, unique(survive)), rep(0, unique(n) - unique(survive))) ) %>% 
  ungroup() %>% 
  select(-n, -rid)

这使:

# A tibble: 7 × 3
  survive     a     b
    <dbl> <dbl> <dbl>
1       1     1     0
2       1     1     0
3       0     1     0
4       1     1     0
5       0     1     0
6       1     0     1
7       1     0     1

使用数据:

df <- data.frame(n = c(3, 2, 2), 
                 survive = c(2, 1, 2), 
                 a = c(1,1,0), 
                 b = c(0,0,1))