Question

如何创建带有A列* B列的数据框。

例如，在“年”列（2018年至2025年）中，每年从1:52开始的“周”列中。

基本上，我想要一种更好的方法来获得此结果：

a =data.table( c(2018) , c(1:52)) 
x <- c("year", "week")
colnames(a) <- x


b =data.table(c(2019) , c(1:52)) 
x <- c("year", "week")
colnames(b) <- x

c =data.table(c(2020) , c(1:52)) 
x <- c("year", "week")
colnames(c) <- x

d = rbind(a, b, c)

编辑：谢谢！

Answer 1

d <- expand.grid(year = c(2018:2020), week = c(1:52))

Answer 2

使用tidyr包中的crossing。像这样：

library(tidyr)
library(data.table)
crossing(
  data.table(year=2018:2020),
  data.table(week=1:52))

有关更多详细信息，请参见https://stackoverflow.com/a/49630818/1358308

Answer 3

使用基础R

data.frame(year = rep(2018:2020, 52), week = rep(1:52, length(year)))

Answer 4

由于您似乎使用data.table，因此这里还有一个选择。

library(data.table)
CJ('year' = 2018:2020, 'week' = 1:52)
#     year week
#  1: 2018    1
#  2: 2018    2
#  3: 2018    3
#  4: 2018    4
#  5: 2018    5
# ---        
#152: 2020   48
#153: 2020   49
#154: 2020   50
#155: 2020   51
#156: 2020   52

Answer 5

您只需要致电$collection = User::join('locations as l', 'users.location_id', '=', 'l.id') ->select('users.*', DB::raw('(6371 * acos(cos(radians(' . $coordinates['latitude'] . ')) * cos(radians(`lat`)) * cos(radians(`lng`) - radians(' . $coordinates['longitude'] . ')) + sin(radians(' . $coordinates['latitude'] . ')) * sin(radians(`lat`)))) as distances')) ->having('distances', '<', 32.688888) ->orderBy('distances', 'ASC') ->get();

data.frame

Answer 6

基本上

year = rep(c(2018:2025),each = 52)
week = rep(c(1:52), length(c(2018:2025)))
d = as.data.frame(cbind(year, week))

使用A列* B列创建R数据框

6 个答案: