问题
This is a continued question from the post Remove the first row from each group if the second row meets a condition
Below is a sample dataset:
df <- data.frame(id=c("9","9","9","5","5","4","4","4","4","4","20","20"),
Date=c("11/29/2018","11/29/2018","11/29/2018","2/13/2019","2/13/2019",
"6/15/2018","6/20/2018","8/17/2018","8/20/2018","8/23/2018","12/25/2018","12/25/2018"),
Buyer= c("John","John","John","Maria","Maria","Sandy","Sandy","Sandy","Sandy","Sandy","Paul","Paul"),
Amount= c("959","1158","596","922","922","1849","4193","4256","65","100","313","99"), stringsAsFactors = F) %>%
group_by(Buyer,id) %>% mutate(diffs = c(NA, diff(as.Date(Date, format = "%m/%d/%Y"))))
which would look like:
| id | Date | Buyer | diff | Amount |
|----|:----------:|------:|------|--------|
| 9 | 11/29/2018 | John | NA | 959 |
| 9 | 11/29/2018 | John | 0 | 1158 |
| 9 | 11/29/2018 | John | 0 | 596 |
| 5 | 2/13/2019 | Maria | 76 | 922 |
| 5 | 2/13/2019 | Maria | 0 | 922 |
| 4 | 6/15/2018 | Sandy | -243 | 1849 |
| 4 | 6/20/2018 | Sandy | 5 | 4193 |
| 4 | 8/17/2018 | Sandy | 58 | 4256 |
| 4 | 8/20/2018 | Sandy | 3 | 65 |
| 4 | 8/23/2018 | Sandy | 3 | 100 |
| 20 | 12/25/2018 | Paul | 124 | 313 |
| 20 | 12/25/2018 | Paul | 0 | 99 |
I need to retain those records where based on each buyer and id, the sum of amount between consecutive rows >5000 if the difference between two consecutive rows <=5. So, for example, Buyer 'Sandy' with id '4' has two transactions of 1849 and 4193 on '6/15/2018' and '6/20/2018' within a gap of 5 days, and since the sum of these two amounts>5000, the output would have these records. Whereas, for the same Buyer 'Sandy' with id '4' has another transactions of 4256, 65 and 100 on '8/17/2018', '8/20/2018' and '8/23/2018' within a gap of 3 days each, but the output will not have these records as the sum of this amount <5000. The final output would look like:
| id | Date | Buyer | diff | Amount |
|----|:---------:|------:|------|--------|
| 4 | 6/15/2018 | Sandy | -243 | 1849 |
| 4 | 6/20/2018 | Sandy | 5 | 4193 |
回答1:
df <- data.frame(id=c("9","9","9","5","5","4","4","4","4","4","20","20"),
Date=c("11/29/2018","11/29/2018","11/29/2018","2/13/2019","2/13/2019",
"6/15/2018","6/20/2018","8/17/2018","8/20/2018","8/23/2018","12/25/2018","12/25/2018"),
Buyer= c("John","John","John","Maria","Maria","Sandy","Sandy","Sandy","Sandy","Sandy","Paul","Paul"),
Amount= c("959","1158","596","922","922","1849","4193","4256","65","100","313","99"), stringsAsFactors = F) %>%
group_by(Buyer,id) %>% mutate(diffs = c(NA, diff(as.Date(Date, format = "%m/%d/%Y"))))
Changing Date from character to Date and Amount from character to numeric:
df$Date<-as.Date(df$Date, '%m/%d/%y')
df$Amount<-as.numeric(df$Amount)
Now here I group the dataset by id, arrange it with Date, and create a rank within each id (so for example Sandy is going to have rank from 1 through 5 for 5 different days in which she has shopped), then I define a new variable called ConsecutiveSum which adds the Value of each row to it's previous row's Value (lag gives you the previous row). The ifelse statement forces consecutive sum to output a 0 if the previous row's Value doesn't exists. The next step is just enforcing your conditions:
df %>%
group_by(id) %>%
arrange(Date) %>%
mutate(rank=dense_rank(Date)) %>%
mutate(ConsecutiveSum = ifelse(is.na(lag(Amount)),0,Amount + lag(Amount , default = 0)))%>%
filter(diffs<=5 & ConsecutiveSum>=5000 | ConsecutiveSum==0 & lead(ConsecutiveSum)>=5000)
# id Date Buyer Amount diffs rank ConsecutiveSum
# <chr> <chr> <chr> <dbl> <dbl> <int> <dbl>
# 1 4 6/15/2018 Sandy 1849 NA 1 0
# 2 4 6/20/2018 Sandy 4193 5 2 6042
回答2:
I would use a combination of techniques available in tidyverse:
First create a grouping variable (new_id) and use the original id and new_id in combination to add together based on a grouping. Then we can filter by the criteria of the sum of the Amount > 5000. We can take this and filter then join or semi_join to filter based on the criteria.
ids is a dataset that finds the total Amount based on id and new_id and filters for when Dollars > 5000. This gives you the id and new_id that meets your criteria
df <- data.frame(id=c("9","9","9","5","5","4","4","4","4","4","20","20"),
Date=c("11/29/2018","11/29/2018","11/29/2018","2/13/2019","2/13/2019",
"6/15/2018","6/20/2018","8/17/2018","8/20/2018","8/23/2018","12/25/2018","12/25/2018"),
Buyer= c("John","John","John","Maria","Maria","Sandy","Sandy","Sandy","Sandy","Sandy","Paul","Paul"),
Amount= c(959,1158,596,922,922,1849,4193,4256,65,100,313,99), stringsAsFactors = F) %>%
group_by(Buyer,id) %>% mutate(diffs = c(NA, diff(as.Date(Date, format = "%m/%d/%Y"))))
library(tidyverse)
df1 <- df %>% mutate(Date = as.Date(Date , format = "%m/%d/%Y"),
tf1 = (id != lag(id, default = 0)),
tf2 = (is.na(diffs) | diffs > 5))
df1$new_id <- cumsum(df1$tf1 + df1$tf2 > 0)
>df1
id Date Buyer Amount diffs days_post tf1 tf2 new_id
<chr> <date> <chr> <dbl> <dbl> <date> <lgl> <lgl> <int>
1 9 2018-11-29 John 959 NA 2018-12-04 TRUE TRUE 1
2 9 2018-11-29 John 1158 0 2018-12-04 FALSE FALSE 1
3 9 2018-11-29 John 596 0 2018-12-04 FALSE FALSE 1
4 5 2019-02-13 Maria 922 NA 2019-02-18 TRUE TRUE 2
5 5 2019-02-13 Maria 922 0 2019-02-18 FALSE FALSE 2
6 4 2018-06-15 Sandy 1849 NA 2018-06-20 TRUE TRUE 3
7 4 2018-06-20 Sandy 4193 5 2018-06-25 FALSE FALSE 3
8 4 2018-08-17 Sandy 4256 58 2018-08-22 FALSE TRUE 4
9 4 2018-08-20 Sandy 65 3 2018-08-25 FALSE FALSE 4
10 4 2018-08-23 Sandy 100 3 2018-08-28 FALSE FALSE 4
11 20 2018-12-25 Paul 313 NA 2018-12-30 TRUE TRUE 5
12 20 2018-12-25 Paul 99 0 2018-12-30 FALSE FALSE 5
ids <- df1 %>%
group_by(id, new_id) %>%
summarise(dollar = sum(Amount)) %>%
ungroup() %>% filter(dollar > 5000)
id new_id dollar
<chr> <int> <dbl>
1 4 3 6042
df1 %>% semi_join(ids)
来源：https://stackoverflow.com/questions/57928394/calculate-sum-of-a-column-if-the-difference-between-consecutive-rows-meets-a-con