Question

我将此数据集保存为名为s2的矩阵

       [,1]
0         4
0.5       1
1         6
10       61
15       28
2         8
20       25
23        1
25        4
3         3
30       44

我想通过它们的行名对它们进行分组，例如

         [,1]
0-10      22
10-20     89
20-30     30
30-40     48

我想知道是否有比s[1,]=s[1,]+s2[2,]+..更快的方式（因为我的数据集比这大得多），然后删除所有行？我尝试使用aggregate，但我读到一些有关wordStem()的信息，但我什么都没找到。谢谢

Answer 1

假设矩阵为m，您可以这样做：

library(tidyverse)

# specify your breaks for the grouping
brks = c(0,10,20,30,40)

data.frame(m) %>%                              # create a dataframe from your matrix
  rownames_to_column() %>%                     # add rownames as a column
  mutate(rowname = as.numeric(rowname)) %>%    # make that column numeric (in order to group)
  group_by(group = cut(rowname, breaks = brks, right = F)) %>%  # use your breaks to group 
  summarise(m = sum(m)) %>%                    # get the sum of values for each group
  data.frame() %>%                             # create a dataframe from tibble (in order to have rownames)
  column_to_rownames("group")                  # add rownames from your group column

#          m
# [0,10)  22
# [10,20) 89
# [20,30) 30
# [30,40) 44

将多行合并为一

1 个答案: