Question

我在 R 中有一个包含300000行的数据框，如下所示：

11 2990000 3000000 0.00000000
11 3000000 3010000 2.30247191
11 3010000 3020000 0.32213483
11 3020000 3030000 0.91696629
11 3030000 3040000 1.81595506
11 3040000 3050000 0.27269663
11 3050000 3060000 2.21988764
11 3060000 3070000 3.44640449
11 3070000 3080000 2.02134831
11 3080000 3090000 1.22123596 #10th row
11 3090000 3100000 3.47089888
11 3100000 3110000 3.08921348
11 3110000 3120000 3.11786517
11 3120000 3130000 1.44325843
11 3130000 3140000 0.00000000
11 3140000 3150000 0.00000000
11 3150000 3160000 2.55146067
11 3160000 3170000 0.63460674
11 3170000 3180000 1.08415730
11 3180000 3190000 2.73101124 #20th row

我想制作一个新的数据框，将每10行的列4求和，并分别输出所考虑的第一行和第十行的列2和列3 。输出应为：

11 2990000 3090000 14.5391
11 3090000 3190000 18.12247

Answer 1

您可以尝试tidyverse

library(tidyverse)
d %>% 
  group_by(gr=gl(n()/10,10)) %>% 
  summarise(Sum=sum(V4)) 
# A tibble: 2 x 2
  gr      Sum
  <fct> <dbl>
1 1      14.5
2 2      18.1

或尝试mutate

d %>% 
  group_by(gr=gl(n()/10,10)) %>% 
  mutate(Sum=sum(V4)) %>%
  slice(10)
# A tibble: 2 x 6
# Groups:   gr [2]
     V1      V2      V3    V4 gr      Sum
  <int>   <int>   <int> <dbl> <fct> <dbl>
1    11 3080000 3090000  1.22 1      14.5
2    11 3180000 3190000  2.73 2      18.1

在基础R中，您可以使用ave

ave(d$V4, factor(rep(1:(nrow(d)/10), each=10)), FUN=sum)[seq(10, nrow(d), 10)]

Answer 2

这是data.table的解决方案：

library("data.table")
D <- fread(
"11 2990000 3000000 0.00000000
11 3000000 3010000 2.30247191
11 3010000 3020000 0.32213483
11 3020000 3030000 0.91696629
11 3030000 3040000 1.81595506
11 3040000 3050000 0.27269663
11 3050000 3060000 2.21988764
11 3060000 3070000 3.44640449
11 3070000 3080000 2.02134831
11 3080000 3090000 1.22123596
11 3090000 3100000 3.47089888
11 3100000 3110000 3.08921348
11 3110000 3120000 3.11786517
11 3120000 3130000 1.44325843
11 3130000 3140000 0.00000000
11 3140000 3150000 0.00000000
11 3150000 3160000 2.55146067
11 3160000 3170000 0.63460674
11 3170000 3180000 1.08415730
11 3180000 3190000 2.73101124")

D[, .(V2=V2[1], V3=V3[.N], V4=sum(V4)), by=gl(D[, .N]/10, 10)]
# > D[, .(V2=V2[1], V3=V3[.N], V4=sum(V4)), by=gl(D[, .N]/10, 10)]
#    gl      V2      V3       V4
# 1:  1 2990000 3090000 14.53910
# 2:  2 3090000 3190000 18.12247

Answer 3

将以下方面的现有答案汇总在一起：

Select every nth row from dataframe
Sum every nth points

v = D$V4
n = 10

#using @jogo's example data
cbind(
  D[ seq(1, nrow(D), n), 1:3 ],
  tapply(v, (seq_along(v)-1) %/% n, sum)
  )

#    V1      V2      V3       V2
# 1: 11 2990000 3000000 14.53910
# 2: 11 3090000 3100000 18.12247

数据帧R中每10行后的列值总和

3 个答案: