使用数据集df:
df
confint row Index
0.3407,0.4104 1 1
0.2849,0.4413 2 2
0.2137,0.2674 3 3
0.1910,0.4575 4 1
0.4039,0.4905 5 2
0.403,0.4822 6 3
0.0301,0.0646 7 1
0.0377,0.0747 8 2
0.0835,0.0918 9 3
0.0437,0.0829 10 1
0.0417,0.0711 11 2
0.0718,0.0798 12 3
0.0112,0.0417 13 1
0.019,0.0237 14 2
0.0213,0.0293 15 3
0.0121,0.0393 16 1
0.0126,0.0246 17 2
0.0318,0.0428 18 3
0.0298,0.0631 19 1
0.018,0.0202 20 2
0.1031,0.1207 21 3
这应该是一个相当容易的数据集,可以将它从长(7)(行)x 3(列)的数据帧转换成长格式。结果应该有3个以Index
命名的列和7行(21/3 = 7)。代码如下:
df <- spread(df,Index, confint, convert = FALSE)
但是,通过使用Spread(),我收到了以下错误:
错误:行(1、4、7、10、13、16、19),(2、5、8、11、14、17、20),(3、6、9、12, 15、18、21)
任何帮助将不胜感激!
答案 0 :(得分:3)
我们需要创建一个序列列,然后创建spread
library(tidyverse)
df %>%
group_by(Index) %>%
mutate(ind = row_number()) %>%
spread(Index, confint, convert = FALSE)
注意:这将是原始数据集中的问题,而不是帖子中显示的示例数据中的问题