Question

我正在研究清理一些数据的解决方案，但我不能100％确定最佳解决方案是什么。我找到了一个有效的解决方案，但想知道是否有更简单的方法（特别是在尝试扩展它时）。我想要做的是分离数据框中的所有元素（用分号分隔），将每个组件应用到数据框的元素，然后将结果合并到一个新的数据框中。示例如下：

test <- data.frame(class=c("a1", "a2","a3","a4"),
        person=c("p1;p3;p4","p2;p4","p4;p5;p6","p1;p5"),
        stringsAsFactors = F)

test1 <- c()
test2 <- c()

for (i in 1:nrow(test)){
    test1 <-append(test1, strsplit(test[i,2],";")[[1]])
    test2 <- append(test2, rep(test[i,1],length(strsplit(test[i,2],";")[[1]])))
}

Answer 1

如果我完全理解你要做的事情，那么这个略显冗长的oneliner会做到这一点：

do.call("rbind",apply(test, 1, function(x) expand.grid(x[1], unlist(strsplit(x[2], split=";")))))

   Var1 Var2
1    a1   p1
2    a1   p3
3    a1   p4
4    a2   p2
5    a2   p4
6    a3   p4
7    a3   p5
8    a3   p6
9    a4   p1
10   a4   p5

我在原始数据框的每一行使用expand.grid，包括第一个变量和第二个变量;作为输入。由于apply是一个列表，我使用do.call和rbind将其重新放回数据框。

Answer 2

我们可以使用来自[0, 1, 2, 5, 6] [0, 1, 2, 5, 7] [0, 1, 2, 5, 8] [0, 1, 2, 5, 9] [0, 1, 2, 6, 7] [0, 1, 2, 6, 8] [0, 1, 2, 6, 9] [0, 1, 2, 7, 8] [0, 1, 2, 7, 9] [0, 1, 2, 8, 9] [0, 1, 3, 5, 6] [0, 1, 3, 5, 7] [0, 1, 3, 5, 8] [0, 1, 3, 5, 9] [0, 1, 3, 6, 7] [0, 1, 3, 6, 8] [0, 1, 3, 6, 9] [0, 1, 3, 7, 8] [0, 1, 3, 7, 9] [0, 1, 3, 8, 9] [0, 1, 4, 5, 6] [0, 1, 4, 5, 7] [0, 1, 4, 5, 8] [0, 1, 4, 5, 9] [0, 1, 4, 6, 7] [0, 1, 4, 6, 8] [0, 1, 4, 6, 9] [0, 1, 4, 7, 8] [0, 1, 4, 7, 9] [0, 1, 4, 8, 9] [0, 2, 3, 5, 6] [0, 2, 3, 5, 7] [0, 2, 3, 5, 8] [0, 2, 3, 5, 9] [0, 2, 3, 6, 7] [0, 2, 3, 6, 8] [0, 2, 3, 6, 9] [0, 2, 3, 7, 8] [0, 2, 3, 7, 9] [0, 2, 3, 8, 9] [0, 2, 4, 5, 6] [0, 2, 4, 5, 7] [0, 2, 4, 5, 8] [0, 2, 4, 5, 9] [0, 2, 4, 6, 7] [0, 2, 4, 6, 8] [0, 2, 4, 6, 9] [0, 2, 4, 7, 8] [0, 2, 4, 7, 9] [0, 2, 4, 8, 9] [0, 3, 4, 5, 6] [0, 3, 4, 5, 7] [0, 3, 4, 5, 8] [0, 3, 4, 5, 9] [0, 3, 4, 6, 7] [0, 3, 4, 6, 8] [0, 3, 4, 6, 9] [0, 3, 4, 7, 8] [0, 3, 4, 7, 9] [0, 3, 4, 8, 9] [1, 2, 3, 5, 6] [1, 2, 3, 5, 7] [1, 2, 3, 5, 8] [1, 2, 3, 5, 9] [1, 2, 3, 6, 7] [1, 2, 3, 6, 8] [1, 2, 3, 6, 9] [1, 2, 3, 7, 8] [1, 2, 3, 7, 9] [1, 2, 3, 8, 9] [1, 2, 4, 5, 6] [1, 2, 4, 5, 7] [1, 2, 4, 5, 8] [1, 2, 4, 5, 9] [1, 2, 4, 6, 7] [1, 2, 4, 6, 8] [1, 2, 4, 6, 9] [1, 2, 4, 7, 8] [1, 2, 4, 7, 9] [1, 2, 4, 8, 9] [1, 3, 4, 5, 6] [1, 3, 4, 5, 7] [1, 3, 4, 5, 8] [1, 3, 4, 5, 9] [1, 3, 4, 6, 7] [1, 3, 4, 6, 8] [1, 3, 4, 6, 9] [1, 3, 4, 7, 8] [1, 3, 4, 7, 9] [1, 3, 4, 8, 9] [2, 3, 4, 5, 6] [2, 3, 4, 5, 7] [2, 3, 4, 5, 8] [2, 3, 4, 5, 9] [2, 3, 4, 6, 7] [2, 3, 4, 6, 8] [2, 3, 4, 6, 9] [2, 3, 4, 7, 8] [2, 3, 4, 7, 9] [2, 3, 4, 8, 9]的{{1}}（此处我提到了一个包并使用了一个而不包含另一个包）来分割“＃1}}＆＃39;分隔符cSplit的列，并将splitstackshape指定为;，以重新定义为“{1}}长期＆＃39;分割后的格式。

direction

或来自long的另一个单行（不使用任何套餐）

library(splitstackshape)
cSplit(test, 'person', ';', 'long')
#    class person
# 1:    a1     p1
# 2:    a1     p3
# 3:    a1     p4
# 4:    a2     p2
# 5:    a2     p4
# 6:    a3     p4
# 7:    a3     p5
# 8:    a3     p6
# 9:    a4     p1
#10:    a4     p5

Answer 3

使用tidyr：

library(dplyr)
library(tidyr)

         # separate person into multiple columns
test %>% separate(person, into = paste0('person', 1:5), fill = 'right') %>% 
    # gather from wide to long
    gather(key = id, value = person, -class, na.rm = TRUE) %>% 
    # clean up extra column
    select(-id)

#    class person
# 1     a1     p1
# 2     a2     p2
# 3     a3     p4
# 4     a4     p1
# 5     a1     p3
# 6     a2     p4
# 7     a3     p5
# 8     a4     p5
# 9     a1     p4
# 11    a3     p6

循环跨数据框架以创建新的数据框架

3 个答案: