Question

这应该非常简单，但我无法解决。我从str_match_all函数得出我认为是矩阵的东西。尽管出现，但事实并非如此。通过对sapply [1,2]中的索引进行硬编码，我能够从“矩阵”的第一行提取时间戳记值。我想对矩阵中的最后一个条目执行相同的操作，并认为我可以轻松地提取矩阵中的行数来执行此操作，例如[nrow（sm），2]，但不能！参见下文：

sm <- str_match_all(regex_text, regex_list[row, "regex_pattern"] )
      print(sm)
#This gives me this (which is good):
#      [[1]]
#      [,1]                                       [,2]               [,3]                      
#      [1,] "09/08/2014 13:01CONTENT_ACCESS.preparing" "09/08/2014 13:01" "CONTENT_ACCESS.preparing"
#      [2,] "09/08/2014 13:06CONTENT_ACCESS.preparing" "09/08/2014 13:06" "CONTENT_ACCESS.preparing"
#      [3,] "09/08/2014 13:08CONTENT_ACCESS.preparing" "09/08/2014 13:08" "CONTENT_ACCESS.preparing"

#Get the first timestamp
      start_t_stamp <- sapply(sm, function(x) x[1,2])
      print(start_t_stamp)
# Also good, I get [1] "09/08/2014 13:01" 

#Get the last timestamp. How do extract the 'number of rows' in sm?      
#This returns NULL 
      print(nrow(sm)) 
#transform to matrix??? 
      t_sm <- t(sm)
#This then prints "[1,] Character,9" 
      print(t_sm)
#Therfore this prints 1      
      print(nrow(t_sm))

预先感谢...

R nrow函数返回NULL或1

0 个答案: