Question

我有一个看起来像这样的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

简而言之，某些行的列的值由＆＃34;;＆＃34;连接。在读取data.frame之前，我不知道哪些列将包含连接值，但我知道它们对于具有该值的所有行都是相同的。我还知道，对于具有连接值的列的行，连接值的数量在所有这些列中是相同的（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，不应具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

这是一个选项

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

Answer 2

这是我写的＆＃34; splitstackshape＆＃34;包裹。您可以使用cSplit，如下所示：

library(splitstackshape)
cSplit(df, c("col2", "col3"), ";", "long")
#     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 3

不像rawr的答案那么复杂，但也许更容易看到发生了什么

df1 <- 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"),
                  stringsAsFactors=FALSE)

df1_rows   <- nrow(df1)
col1_split <- strsplit(df1$col1,";") 
col2_split <- strsplit(df1$col2,";") 
col3_split <- strsplit(df1$col3,";") 

df2 <- data.frame(col1=character(), 
                  col2=character(), 
                  col3=character(),
                  stringsAsFactors=FALSE) 

for (n in 1:df1_rows){ df2 <- rbind(df2, 
       data.frame(col1=col1_split[[n]],
                  col2=col2_split[[n]],
                  col3=col3_split[[n]], 
                  stringsAsFactors=FALSE))}

给出了

> df2 
   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

使用连接值拆分数据框行

3 个答案: