英文:
R: Set Time Point as Zero and then Ascending and Descending Integers Around
问题
我尝试将特定时间点标记为零点，然后将前面的行设置为降序整数（-1、-2等），后面的行设置为升序整数（+1、+2等），但我完全不知道如何做到这一点。这是一个示例数据集：
Time = c("2023-07-28 22:14:00", "2023-07-28 22:15:00", "2023-07-28 22:16:00", "2023-07-28 22:17:00", "2023-07-28 22:18:00")
num = c(1000, 1200, 1234, 980, 1300)
df <- data.frame(Time = rep(Time, num), num = sequence(num))
df[paste0('rand', seq_along(Time))] <- rnorm(length(Time) * sum(num))
head(df)
目标是将2023-07-28 22:16:00设为0，然后在时间点之前设置降序整数，在时间点之后设置升序整数。谢谢您提前的帮助。
英文:
I'm trying to make mark a specific time point as point zero and then the previous rows set as descending integers (-1, -2, etc) and the following rows as ascending integers (+1, +2, etc), but I'm completely stumped on how to do this. Here is a sample dataset:
Time = c("2023-07-28 22:14:00", "2023-07-28 22:15:00", "2023-07-28 22:16:00", "2023-07-28 22:17:00", "2023-07-28 22:18:00")
num = c(1000,1200,1234,980,1300)
df <- data.frame(Time = rep(Time, num), num = sequence(num))
df[paste0('rand', seq_along(Time))] <- rnorm(length(Time) * sum(num))
head(df)
The goal would be to make 2023-07-28 22:16:00 = 0 and then descending integers before and ascending integers after time point. Thanks in advance
答案1
得分: 1
以下是翻译好的部分：
如果您的目标是根据给定的“零点”创建一个升序或降序的新列，您可以尝试以下方法。
Time = c("2023-07-28 22:14:00", "2023-07-28 22:15:00", "2023-07-28 22:16:00", "2023-07-28 22:17:00", "2023-07-28 22:18:00")
num = c(1000, 1200, 1234, 980, 1300)
df <- data.frame(Time = rep(Time, num), num = sequence(num))
df[paste0('rand', seq_along(Time))] <- rnorm(length(Time) * sum(num))
head(df)
zero = "2023-07-28 22:16:00"
id_col = 'Time'
add_index = function(df, zero, id_col){
df = df[order(df[, id_col]),]
df$new_col = c(-rev(sequence(sum(df[, id_col] < zero))), rep(0, sum(df[, id_col] == zero)), sequence(sum(df[, id_col] > zero)))
return(df)
}
df %>%
add_index(zero, id_col) %>%
head()
还有以下输出结果：
Time num rand1 rand2 rand3 rand4 rand5 new_col
1 2023-07-28 22:14:00 1 0.2995052 1.64549561 -0.79780340 -1.07715548 0.4070298 -2200
2 2023-07-28 22:14:00 2 -1.6490923 -0.35256450 0.06296558 0.06870601 -0.3095747 -2199
3 2023-07-28 22:14:00 3 -0.5708513 -1.65916717 0.83986290 -2.09933862 0.3588580 -2198
4 2023-07-28 22:14:00 4 -0.3760261 0.89072105 0.56148014 0.08404197 -0.6248979 -2197
5 2023-07-28 22:14:00 5 -1.2998215 0.62532664 0.14171299 -0.25804508 0.5117224 -2196
6 2023-07-28 22:14:00 6 -0.4678034 0.03267064 0.76075794 -0.32648332 -3.3211990 -2195
希望这有助于您理解代码和输出。
英文:
EDIT: using the updated code in the question...
If your goal is to create a new column with numbers ascending or descending depending on a given "zero-point" you could try something like this.
Time = c("2023-07-28 22:14:00", "2023-07-28 22:15:00", "2023-07-28 22:16:00", "2023-07-28 22:17:00", "2023-07-28 22:18:00")
num = c(1000,1200,1234,980,1300)
df <- data.frame(Time = rep(Time, num), num = sequence(num))
df[paste0('rand', seq_along(Time))] <- rnorm(length(Time) * sum(num))
head(df)
zero = "2023-07-28 22:16:00"
id_col = 'Time'
add_index = function(df,zero,id_col){
df = df[order(df[,id_col]),]
df$new_col = c(-rev(sequence(sum(df[,id_col] < zero))),rep(0,sum(df[,id_col] == zero)),sequence(sum(df[,id_col] > zero)))
return(df)
}
df %>%
add_index(zero,id_col) %>%
head()
There are probably other ways of doing this, but here is the output of this approach:
Time num rand1 rand2 rand3 rand4 rand5 new_col
1 2023-07-28 22:14:00 1 0.2995052 1.64549561 -0.79780340 -1.07715548 0.4070298 -2200
2 2023-07-28 22:14:00 2 -1.6490923 -0.35256450 0.06296558 0.06870601 -0.3095747 -2199
3 2023-07-28 22:14:00 3 -0.5708513 -1.65916717 0.83986290 -2.09933862 0.3588580 -2198
4 2023-07-28 22:14:00 4 -0.3760261 0.89072105 0.56148014 0.08404197 -0.6248979 -2197
5 2023-07-28 22:14:00 5 -1.2998215 0.62532664 0.14171299 -0.25804508 0.5117224 -2196
6 2023-07-28 22:14:00 6 -0.4678034 0.03267064 0.76075794 -0.32648332 -3.3211990 -2195
答案2
得分: 1
Sure, here's the translated code with the requested part:
df %>%
mutate(Time = ymd_hms(Time),
dif = Time - ymd_hms('2023-07-28 22:16:00'),
rank = dense_rank(dif) - row_number()[dif == 0]) %>%
select(-dif)
Please let me know if you need any further assistance.
英文:
df %>%
mutate(Time = ymd_hms(Time),
dif = Time - ymd_hms('2023-07-28 22:16:00'),
rank = dense_rank(dif) - row_number()[dif== 0])
Time num dif rank
1 2023-07-28 22:14:00 1000 -120 secs -2
2 2023-07-28 22:15:00 1200 -60 secs -1
3 2023-07-28 22:16:00 1234 0 secs 0
4 2023-07-28 22:17:00 980 60 secs 1
5 2023-07-28 22:18:00 1300 120 secs 2
You can then remove the dif column
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