我有一个data.frame，如下所示
soc_sec group_count total_creds group_start group_end
(chr) (int) (dbl) (date) (date)
1 AA2105480 5 14.0 2005-01-09 2005-05-16
2 AA2105480 7 17.0 2004-08-26 2004-12-10
3 AB4378973 1 0.0 2004-01-21 2004-05-07
4 AB4990257 2 1.0 2014-09-01 2014-12-14
5 AB7777777 5 12.0 2004-01-21 2005-03-22
6 AB7777777 6 15.0 2004-08-26 2004-12-10
7 AB7777777 5 15.0 2005-01-09 2005-05-12
8 AC4285291 2 3.0 2014-09-01 2014-12-14
9 AC4285291 1 3.0 2015-01-12 2015-04-15
10 AC6039874 9 17.5 2010-01-06 2010-05-06
11 AC6039874 7 16.0 2011-01-05 2011-04-29
12 AC6039874 8 12.5 2010-08-31 2010-12-21
13 AC6039874 7 13.5 2011-08-31 2011-12-21
14 AC6547645 7 18.0 2005-01-09 2005-05-12
15 AC6547645 6 17.0 2004-08-26 2004-12-10
16 AC6547645 1 2.0 2005-04-20 2005-06-01
17 AD1418577 7 13.0 2013-01-09 2013-05-17
18 AD1418577 8 16.0 2013-08-28 2013-12-13
19 AD1418577 6 15.0 2014-01-08 2014-05-05
20 AD1418577 7 13.0 2015-08-26 2015-12-15
我尝试做的是创建一个列，以后我可以根据group_start和{{1}之间的天数顺序将其用作xts对象的逐日索引}。我知道我能够使用group_end计算一列的向量我甚至可以在v <- seq(df$group_start[1], df$group_end[1], by="days")之后对行进行必要的重复：
dplyr::bind_rows(df,v)我无法做的是对data.frame中的每一行进行矢量化。
我尝试过的不起作用的事情
df$len <- apply(df, 1, function(x){
length(seq(as.Date(x["group_start"]), as.Date(x["group_end"]), by="days"))
})
df <- df[rep(seq_len(nrow(df)), df$len),]
错误：create_date_vector <- function(x){
flog.debug("id: %s", x["soc_sec"])
seq(as.Date(x["group_start"]), as.Date(x["group_end"]), by = "days")
}
date_vec <- c()
date_vec <- c(date_vec, apply(df, 1, create_date_vector))
Error in seq.int(0, to0 - from, by) : wrong sign in 'by' argument错误：date_vec <- c()
for(i in 1:nrow(df)){
date_vec <- c(date_vec, seq(from=as.Date(df$group_start[as.integer(i)]), to=as.Date(df$group_end[as.integer(i)])), by="days")
}
非常感谢任何帮助。谢谢。
dput
Error in seq.Date(from = as.Date(ags_df$group_start[as.integer(i)]), to = as.Date(ags_df$group_end[as.integer(i)])) :
exactly two of 'to', 'by' and 'length.out' / 'along.with' must be specified答案 0 :(得分：3)
我不知道在你找到一个可行的解决方案一个多月后这有多大用处，但我还是把你的代码缩小到更紧凑的东西。
library(dplyr)
df <- structure(list(soc_sec = c("AA2105480", "AA2105480", "AB4378973",
"AB4990257", "AB7777777", "AB7777777", "AB7777777", "AC4285291",
"AC4285291", "AC6039874", "AC6039874", "AC6039874", "AC6039874",
"AC6547645", "AC6547645", "AC6547645", "AD1418577", "AD1418577",
"AD1418577", "AD1418577"), group_count = c(5L, 7L, 1L, 2L, 5L,
6L, 5L, 2L, 1L, 9L, 7L, 8L, 7L, 7L, 6L, 1L, 7L, 8L, 6L, 7L),
total_creds = c(14, 17, 0, 1, 12, 15, 15, 3, 3, 17.5, 16,
12.5, 13.5, 18, 17, 2, 13, 16, 15, 13), group_start = structure(c(12792,
12656, 12438, 16314, 12438, 12656, 12792, 16314, 16447, 14615,
14979, 14852, 15217, 12792, 12656, 12893, 15714, 15945, 16078,
16673), class = "Date"), group_end = structure(c(12919, 12762,
12545, 16418, 12864, 12762, 12915, 16418, 16540, 14735, 15093,
14964, 15329, 12915, 12762, 12935, 15842, 16052, 16195, 16784
), class = "Date")), .Names = c("soc_sec", "group_count",
"total_creds", "group_start", "group_end"), class = c("tbl_df",
"data.frame"), row.names = c(NA, -20L))
# Essentially the same as the calc_day_nums() and apply() part of
# your solution. It returns an object of class difftime, but that
# doesn't seem to cause any problems
diffs <- abs(with(df, group_start-group_end))+1
# This will repeat row[i] diffs[i] number of times
df.rep <- df[rep(1:nrow(df), times=diffs), ]
reps <- rep(diffs, times=diffs)
# Creating the time sequences. Many ways to skin this cat, I suspect.
# This is but one
dates.l <- apply(
df[colnames(df) %in% c("group_start", "group_end")], 1,
function(x) {
seq(min(as.Date(x)), max(as.Date(x)), by="days")
})
# Converting the list into one long vector. Essentially the same as
# unlist(), except it retains the Date class.
days <- do.call(c, dates.l)
# Combining the elements by column
df.long <- cbind(df.rep, reps, days)
str(df.long)
# dplyr isn't exactly my forte. This is just to convert the output
# into the same tbl format as the input
library(tibble)
df.long <- as_tibble(df.long)
答案 1 :(得分：1)
所以，我设法弄明白了，我想我应该把解决方案放在这里以防万一。这需要多个步骤，所以如果有人能想到更好的方法，请告诉我。
首先，我创建了一个列来计算两个日期之间的天数。我需要这个，以便我知道每行要重复多少次
calc_day_nums <- function(x){
if(as.numeric(as.Date(x["group_start"])) < as.numeric(as.Date(x["group_end"]))){
len <- length(seq(as.Date(x["group_start"]), as.Date(x["group_end"]), by="days"))
} else if (as.numeric(as.Date(x["group_start"])) > as.numeric(as.Date(x["group_end"]))){
len <- length(seq(as.Date(x["group_end"]), as.Date(x["group_start"]), by="days"))
} else {
len <- 1 #basically these are records whose start and end are the same
}
return(len)
}
df$reps <- apply(df, 1, calc_day_nums)
然后，我自己创建了一个载体
date_vec <- function(i, x, y){
if(as.Date(x[i]) != as.Date(y[i])){
as.Date(as.Date(x[i]):as.Date(y[i]), origin="1970-01-01")
} else{
as.Date(x[i])
}
}
vec <- lapply(seq_along(df$group_start), date_vec, x=df$group_start, y=df$group_end)
vec <- unlist(vec)
vec <- as.Date(vec)
之后，我对data.frame
进行了正确的行重复次数df <- df[rep(seq_len(nrow(df)), df$reps),]
最后，我将向量绑定到data.frame。此时我也可以将vec定义为xts索引xt <- xts(x = df, order.by = vec)，但我想将其添加到data.frame
df <- bind_cols(df, data.frame(days=vec))