[英]Calculate for date difference in days R
如何計算ID是否連續出現少於5天? 還計算相同ID記錄之間的日差。 我真的無法得到這個問題的邏輯,我不知道我可以從什么開始。
(下面給出的樣本數據只是一個樣本,我的實際數據量很大。因此,需要進行優化。)
樣本數據 :
sample<- data.frame(
id=c("A","B","C","D","A","C","D","A","C","D","A","D","A","C"),
date=c("1/3/2013","1/3/2013", "1/3/2013","1/3/2013","2/3/2013","2/3/2013",
"2/3/2013","3/3/2013","3/3/2013",
"3/3/2013",
"4/3/2013",
"4/3/2013",
"5/3/2013",
"5/3/2013"
)
)
預期產出:
output<- data.frame(
id=c("A","A","A","A","A","B","C","C","C","C","D","D","D","D","D","D","D"),
date=c("1/3/2013",
"2/3/2013",
"3/3/2013",
"4/3/2013",
"5/3/2013",
"1/3/2013",
"1/3/2013",
"2/3/2013",
"3/3/2013",
"5/3/2013",
"1/3/2013",
"2/3/2013",
"3/3/2013",
"4/3/2013",
"5/3/2013",
"6/3/2013",
"7/3/2013" ),
num=c(0,1,2,3,4,0,0,1,2,4,0,1,2,3,4,5,6)
)
計算邏輯:
計算日期差異。 例如,1/3到2/3是1天的差異所以2/3行,列idu:1。 2/3到3/3是1天的差異所以加1行3/3,列idu:2。 3/3到5/3是2天差異所以加2到idu。 第5/3行,列idu:4。 (基於相同的ID)
Date | idu
1/3 | 0
2/3 | 1
3/3 | 2
5/3 | 4
提前致謝。
sample<- data.frame(
id=c("A","B","C","D","A","C","D","A","C","D","A","D","A","C"),
date=c("1/3/2013","1/3/2013", "1/3/2013","1/3/2013","2/3/2013","2/3/2013",
"2/3/2013","3/3/2013","3/3/2013",
"3/3/2013",
"4/3/2013",
"4/3/2013",
"5/3/2013",
"5/3/2013"), stringsAsFactors = F)
library(lubridate)
sample$date <- dmy(sample$date)
sample1 <- sample[order(sample$id, sample$date), ]
sample1$idu <- unlist(sapply(rle(sample1$id)$lengths, seq_len)) -1
id date idu
1 A 2013-03-01 0
5 A 2013-03-02 1
8 A 2013-03-03 2
11 A 2013-03-04 3
13 A 2013-03-05 4
2 B 2013-03-01 0
3 C 2013-03-01 0
6 C 2013-03-02 1
9 C 2013-03-03 2
14 C 2013-03-05 3
4 D 2013-03-01 0
7 D 2013-03-02 1
10 D 2013-03-03 2
12 D 2013-03-04 3
為了添加時間延遲列,可以使用多個選項。 我只是這樣做
sample1$diff <- c(0, int_diff(sample1$date)/days(1))
# Remainder cannot be expressed as fraction of a period.
# Performing %/%.
> sample1
id date idu diff
1 A 2013-03-01 0 0
5 A 2013-03-02 1 1
8 A 2013-03-03 2 1
11 A 2013-03-04 3 1
13 A 2013-03-05 4 1
2 B 2013-03-01 0 -4
3 C 2013-03-01 0 0
6 C 2013-03-02 1 1
9 C 2013-03-03 2 1
14 C 2013-03-05 3 2
4 D 2013-03-01 0 -4
7 D 2013-03-02 1 1
10 D 2013-03-03 2 1
12 D 2013-03-04 3 1
並根據需要進行進一步的更改。 用0替換所有負值。
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