使用df，我正在创建一个新数据框（final.df），其中startdate和enddate之间的每个日期都有一行df } datadframe。
df <- data.frame(claimid = c("123A",
"125B",
"151C",
"124A",
"325C"),
startdate = as.Date(c("2018-01-01",
"2017-05-20",
"2017-12-15",
"2017-11-05",
"2018-02-06")),
enddate = as.Date(c("2018-01-06",
"2017-06-21",
"2018-01-02",
"2017-11-15",
"2018-02-18")))
下面的嵌套函数是我目前用于创建final.df的函数，但是当循环数十万个声明时，这种创建final.df的方法需要数小时才能运行。我正在寻找可以更有效地创建final.df的替代方案。
claim_level <- function(a) {
specific_row <- df[a, ]
dates <- seq(specific_row$startdate, specific_row$enddate, by="days")
day_level <- function(b) {
day <- dates[b]
data.frame(claimid = specific_row$claimid, date = day)
}
do.call("rbind", lapply(c(1:length(dates)), function(b) day_level(b)))
}
final.df <- do.call("rbind", lapply(c(1:nrow(df)), function(a) claim_level(a)))
print(subset(final.df, claimid == "123A"))
#claimid date
#123A 2018-01-01
#123A 2018-01-02
#123A 2018-01-03
#123A 2018-01-04
#123A 2018-01-05
#123A 2018-01-06
答案 0 :(得分：4)
您可以使用gather中的tidyr转换为长格式，然后使用pad中的padr在开始日期和结束日期之间创建新的日期行。 group = "claimid"参数允许您指定分组变量：
library(dplyr)
library(tidyr)
library(padr)
df %>%
gather(var, date, -claimid) %>%
pad(group = "claimid") %>%
select(-var)
或data.table提高效率：
library(data.table)
setDT(df)[,.(date = seq(startdate, enddate, "days")), claimid]
<强>结果：
claimid date
1 123A 2018-01-01
2 123A 2018-01-02
3 123A 2018-01-03
4 123A 2018-01-04
5 123A 2018-01-05
6 123A 2018-01-06
7 124A 2017-11-05
8 124A 2017-11-06
9 124A 2017-11-07
10 124A 2017-11-08
11 124A 2017-11-09
12 124A 2017-11-10
13 124A 2017-11-11
14 124A 2017-11-12
15 124A 2017-11-13
16 124A 2017-11-14
17 124A 2017-11-15
18 125B 2017-05-20
19 125B 2017-05-21
20 125B 2017-05-22
...
<强>基准：
初始化函数：
library(tidyverse)
library(padr)
library(data.table)
# OP's function
claim_level <- function(a) {
specific_row <- df[a, ]
dates <- seq(specific_row$startdate, specific_row$enddate, by="days")
day_level <- function(b) {
day <- dates[b]
data.frame(claimid = specific_row$claimid, date = day)
}
do.call("rbind", lapply(c(1:length(dates)), function(b) day_level(b)))
}
OP_f = function(){
do.call("rbind", lapply(c(1:nrow(df)), function(a) claim_level(a)))
}
# useR's tidyverse + padr
f1 = function(){
df %>%
gather(var, date, -claimid) %>%
pad(interval = "day", group = "claimid") %>%
select(-var)
}
# useR's data.table
DT = df
setDT(DT)
f2 = function(){
DT[,.(date = seq(startdate, enddate, "days")), claimid]
}
# Moody_Mudskipper's Base R
f3 = function(){
do.call(rbind,
Map(function(claimid, startdate, enddate)
data.frame(claimid, date=as.Date(startdate:enddate, origin = "1970-01-01")),
df$claimid, df$startdate, df$enddate))
}
# Moody_Mudskipper's tidyverse
f4 = function(){
df %>%
group_by(claimid) %>%
mutate(date = list(as.Date(startdate:enddate, origin = "1970-01-01"))) %>%
select(1, 4) %>%
unnest %>%
ungroup
}
# MKR's tidyr expand
f5 = function(){
df %>%
group_by(claimid) %>%
expand(date = seq(startdate, enddate, by="day"))
}
检查是否相同：
> identical(OP_f() %>% arrange(claimid), data.frame(f1()))
[1] TRUE
> identical(OP_f(), data.frame(f2()))
[1] TRUE
> identical(OP_f(), data.frame(f3()))
[1] TRUE
> identical(OP_f(), data.frame(f4()))
[1] TRUE
> identical(OP_f() %>% arrange(claimid), data.frame(f5()))
[1] TRUE
基准测试结果：
library(microbenchmark)
microbenchmark(OP_f(), f1(), f2(), f3(), f4(), f5())
Unit: milliseconds
expr min lq mean median uq max neval
OP_f() 26.421534 27.697194 30.342682 28.981143 31.537396 58.071238 100
f1() 36.133364 38.179196 40.749812 39.870931 41.367655 58.428888 100
f2() 1.005843 1.261449 1.450633 1.383232 1.559689 4.058900 100
f3() 2.373679 2.534148 2.786888 2.633035 2.797452 6.941421 100
f4() 22.659097 23.341435 25.275457 24.111411 26.499893 40.840061 100
f5() 46.445622 48.148606 52.565480 51.185478 52.845829 176.912276 100
data.table是速度方面的赢家，@ Moody_Mudskipper的Base R解决方案是第二好的。虽然padr::pad和tidyr::expand似乎是最方便的，但它们也是最慢的（甚至比OP的原始程序还慢）。
答案 1 :(得分：2)
在基地people = int(input("Will you be travelling by yourself (1), or as a group of two (2)?: "))
while people:
if people == 1:
print("\nAh, a holiday for one! How adventurous.")
break
elif people == 2:
print("\nOoh, bringing a friend! Sounds like fun!")
break
else:
print("\nPlease enter either 1 or 2 to determine the number of travellers.")
people = int(input("Will you be travelling by yourself (1), or as a group of two (2)?: "))
：
R仅使用do.call(rbind,
Map(function(claimid, startdate, enddate)
data.frame(claimid, date=as.Date(startdate:enddate, origin = "1970-01-01")),
df$claimid, df$startdate, df$enddate))
# claimid date
# 1 123A 2018-01-01
# 2 123A 2018-01-02
# 3 123A 2018-01-03
# 4 123A 2018-01-04
# 5 123A 2018-01-05
# 6 123A 2018-01-06
#...
：
tidyverse答案 2 :(得分：2)
一种选择是使用tidyr::expand功能在startdate到enddate之间展开行。
library(tidyverse)
df %>% group_by(claimid) %>%
expand(date = seq(startdate, enddate, by="day")) %>%
as.data.frame()
# claimid date
# 1 123A 2018-01-01
# 2 123A 2018-01-02
# 3 123A 2018-01-03
# 4 123A 2018-01-04
# 5 123A 2018-01-05
# 6 123A 2018-01-06
# 7 124A 2017-11-05
# 8 124A 2017-11-06
# 9 124A 2017-11-07
# 10 124A 2017-11-08
# 11 124A 2017-11-09
# 12 124A 2017-11-10
#
# 70 more rows