根据特定ID找出月份中的日期与R中的求解之间的差异

Question

Below is my sample data set which has many unique ID 's and I have to find out the difference between the first occurrence and the second occurrence in the First_Diff column and the difference between the second occurrence and the third in Second_Diff column and so on. They are many occurrences in the table and trying to give a sample.

Input:

Date    ID
1/3/2006    209
1/3/2006    489
1/3/2006    502
1/3/2006    439
1/3/2006    534
1/3/2006    474
1/3/2006    566
1/3/2006    591
1/4/2006    209
1/4/2007    489
1/5/2007    502
1/7/2006    439
1/3/2008    534
1/3/2007    474
1/3/2008    566
1/7/2009    439
1/3/2009    534
1/3/2009    474
1/3/2010    566

Output:

ID  First_Diff  Second_Diff Third_DIff
209 1   0   0
489 13  0   0
502 14  0   0
439 3   0   0
534 24  0   0
474 12  0   0
566 24  12  0
591 0   0   0

Can anyone please help me in this. As this very complicated for me and did not able to solve this for my further findings.

Answer 1

This could help with your problem

# Create fake data frame
dates = c(rep(c("2015-01-09", "2015-02-09", "2015-03-08"), 3))
id  = c(rep(c("A", "B", "C"), each = 3))

df =data.frame(date = as.Date(dates), id = as.factor(id)) 

You can calculate the differences using lag() function

# Calculate difference between times
df = df %>% 
  group_by(id) %>%
  mutate(datediff = difftime(date, lag(date)))

And then transform to wide format using spread

df_wide = df %>% spread(date, datediff) 

names(df_wide) <-  c("id", "first_diff", "second_diff", "third_diff")

The output is:

# A tibble: 3 x 4
# Groups:   id [3]
      id first_diff second_diff third_diff
* <fctr>     <time>      <time>     <time>
1      A    NA days     31 days    27 days
2      B    NA days     31 days    27 days
3      C    NA days     31 days    27 days

Answer 2

First use ave to create a seq column that is 0 for the first occurrence of any ID , 1 for the second and so on.

Then use tapply to create a matrix mat with each ID being a row and each seq being a column and the content being the number of months since the Epoch. as.yearmon internally converts each date to a year + frac where jan = 0, feb = 1/12, etc., tapply will convert that to a number and multiplying by 12 gives the number of months.

Finally create the result by using the rownames of mat as the ID column and for the remaining columns difference the columns of mat appropriate. We have used NA rather than 0 to indicate a value that cannot be calculated since otherwise we would not be able to distinguish two occurrences that are 0 months apart and a value that cannot be calculated.

library(zoo)

DF2 <- transform(DF, seq = ave(ID, ID, FUN = seq_along) - 1)
mat <- with(DF2, 12 * tapply(Date, list(ID, seq), as.yearmon, format = "%d/%m/%Y"))
cbind(ID = rownames(mat), as.data.frame(mat[, -1] - mat[, -ncol(mat)]))

giving:

     ID  1  2
209 209  1 NA
439 439  4 36
474 474 12 24
489 489 13 NA
502 502 14 NA
534 534 24 12
566 566 24 24
591 591 NA NA

Variation

We could also write the above code in a magrittr pipeline:

library(zoo)
library(magrittr)

DF %>%
   transform(seq = ave(ID, ID, FUN = seq_along) - 1) %>%
   with(12 * tapply(Date, list(ID, seq), as.yearmon, format = "%d/%m/%Y")) %>%
   { cbind(ID = rownames(.), as.data.frame(.[, -1] - mat[, -ncol(.)])) }

Note

The input in reproducible form:

Lines <- "Date    ID
1/3/2006    209
1/3/2006    489
1/3/2006    502
1/3/2006    439
1/3/2006    534
1/3/2006    474
1/3/2006    566
1/3/2006    591
1/4/2006    209
1/4/2007    489
1/5/2007    502
1/7/2006    439
1/3/2008    534
1/3/2007    474
1/3/2008    566
1/7/2009    439
1/3/2009    534
1/3/2009    474
1/3/2010    566"
DF <- read.table(text = Lines, header = TRUE)