将多行合并到一个数据框的单个单元格

debugcn 发表于 Dev

普拉布

我有一个这样的数据框

  day V1 V9 V13 V14 V15 V16
1    1  2  3   0   0   0   0
2    1  5  3   L   L   0   0
3    1  5  1   S   L   L   0
4    1  4  3   S   0   0   0
5    1  5  2   S   L   L   0
6    1  5  1   S   M   L   0
7    1  4  1   L   L   0   0
8    1  4  1   M   S   0   0
9    1  1  3   0   0   0   0
10   1  4  1   S   L   0   0
11   2  1  3   0   0   0   0
12   2  2  1   0   0   0   0
13   2  3  3   0   0   0   0
14   2  1  3   0   0   0   0
15   2  1  1   0   0   0   0
16   2  4  3   S   0   0   0
17   2  4  3   S   0   0   0
18   2  3  3   0   0   0   0
19   2  3  1   S   0   0   0
20   2  4  1   L   M   0   0

我需要将其转换为如下所示

 day V1 V9 V13 V14 V15 V16
1    1  2  3   0   0   0   0
2       5  3   L   L   0   0
3       5  1   S   L   L   0
4       4  3   S   0   0   0
5       5  2   S   L   L   0
6       5  1   S   M   L   0
7       4  1   L   L   0   0
8       4  1   M   S   0   0
9       1  3   0   0   0   0
10      4  1   S   L   0   0
11   2  1  3   0   0   0   0
12      2  1   0   0   0   0
13      3  3   0   0   0   0
14      1  3   0   0   0   0
15      1  1   0   0   0   0
16      4  3   S   0   0   0
17      4  3   S   0   0   0
18      3  3   0   0   0   0
19      3  1   S   0   0   0
20      4  1   L   M   0   0

这类似于对特定列的多行使用excel merge命令。我无法弄清楚，也许我可以将日值保留在中间以更好地表示。有人可以帮忙吗？

罗纳克·沙

您可以使用空值duplicated替换day列中的重复值。

df$day[duplicated(df$day)] <- ''
df

#   day V1 V9 V13 V14 V15 V16
#1    1  2  3   0   0   0   0
#2       5  3   L   L   0   0
#3       5  1   S   L   L   0
#4       4  3   S   0   0   0
#5       5  2   S   L   L   0
#6       5  1   S   M   L   0
#7       4  1   L   L   0   0
#8       4  1   M   S   0   0
#9       1  3   0   0   0   0
#10      4  1   S   L   0   0
#11   2  1  3   0   0   0   0
#12      2  1   0   0   0   0
#13      3  3   0   0   0   0
#14      1  3   0   0   0   0
#15      1  1   0   0   0   0
#16      4  3   S   0   0   0
#17      4  3   S   0   0   0
#18      3  3   0   0   0   0
#19      3  1   S   0   0   0
#20      4  1   L   M   0   0

这会将day列转换为字符，可能更安全地替换为字符NA。

数据

df <- structure(list(day = c(1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 
2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L), V1 = c(2L, 5L, 5L, 4L, 
5L, 5L, 4L, 4L, 1L, 4L, 1L, 2L, 3L, 1L, 1L, 4L, 4L, 3L, 3L, 4L
), V9 = c(3L, 3L, 1L, 3L, 2L, 1L, 1L, 1L, 3L, 1L, 3L, 1L, 3L, 
3L, 1L, 3L, 3L, 3L, 1L, 1L), V13 = c("0", "L", "S", "S", "S", 
"S", "L", "M", "0", "S", "0", "0", "0", "0", "0", "S", "S", "0", 
"S", "L"), V14 = c("0", "L", "L", "0", "L", "M", "L", "S", "0", 
"L", "0", "0", "0", "0", "0", "0", "0", "0", "0", "M"), V15 = c("0", 
"0", "L", "0", "L", "L", "0", "0", "0", "0", "0", "0", "0", "0", 
"0", "0", "0", "0", "0", "0"), V16 = c(0L, 0L, 0L, 0L, 0L, 0L, 
0L, 0L, 0L, 0L, 0L, 0L, 0L, 0L, 0L, 0L, 0L, 0L, 0L, 0L)), 
class = "data.frame", row.names = c(NA, -20L))

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