r - 使用连接值拆分数据框行

Question

我有一个如下所示的 data.frame:

df <- data.frame(col1=c("a","b","c","d"), col2=c("1","1;2;3","5","3;2;5;5;3"), col3=c("0","1;1;0","0","0;0;1;1;0"))

#   col1      col2      col3
# 1    a         1         0
# 2    b     1;2;3     1;1;0
# 3    c         5         0
# 4    d 3;2;5;5;3 0;0;1;1;0

换言之，某些行的列中的值由“;”连接。在读取 data.frame 之前，我不知道哪些列将包含连接值，但我确实知道对于所有具有该值的行，它们将是相同的。我还知道，对于包含具有串联值的列的行，所有这些列中的串联值的数量是相同的(第 2 行在 col2 和 col3 中都有 3 个值，第 4 行在这些列中有 5 个值)

我想创建一个新的 data.frame，其中这些连接的值被拆分为单独的行。对于这些行，没有连接值的列中的值应按连接值的数量进行复制。

生成的 data.frame 将是:

df <- data.frame(col1=c("a","b","b","b","c","d","d","d","d","d"), col2=c("1","1","2","3","5","3","2","5","5","3"), col3=c("0","1","1","0","0","0","0","1","1","0"))

#    col1 col2 col3
# 1     a    1    0
# 2     b    1    1
# 3     b    2    1
# 4     b    3    0
# 5     c    5    0
# 6     d    3    0
# 7     d    2    0
# 8     d    5    1
# 9     d    5    1
# 10    d    3    0

Answer 1

这里有一个选择

df <- data.frame(col1=c("a","b","c","d"), col2=c("1","1;2;3","5","3;2;5;5;3"), col3=c("0","1;1;0","0","0;0;1;1;0"))

df2 <- data.frame(col1=c("a","b","b","b","c","d","d","d","d","d"), col2=c("1","1","2","3","5","3","2","5","5","3"), col3=c("0","1","1","0","0","0","0","1","1","0"))


## reshape `col1` to make it look like the others
v <- Vectorize(gsub)
df$col1 <- v('\\b\\d\\b', df$col1, df$col2)

#        col1      col2      col3
# 1         a         1         0
# 2     b;b;b     1;2;3     1;1;0
# 3         c         5         0
# 4 d;d;d;d;d 3;2;5;5;3 0;0;1;1;0


## split on white space or `;` and coerce back into a data frame
data.frame(do.call('cbind', lapply(df, function(x)
  unlist(strsplit(as.character(x), '[\\s;]')))))

#    col1 col2 col3
# 1     a    1    0
# 2     b    1    1
# 3     b    2    1
# 4     b    3    0
# 5     c    5    0
# 6     d    3    0
# 7     d    2    0
# 8     d    5    1
# 9     d    5    1
# 10    d    3    0