问题
Here part of data.
mydat=structure(list(code = c(123L, 123L, 123L, 123L, 123L, 123L, 123L,
123L, 123L, 123L, 123L, 123L, 123L, 123L, 123L, 123L, 123L, 123L,
123L, 123L, 123L, 123L, 123L, 123L, 123L, 123L, 123L, 123L, 123L,
123L, 123L, 123L, 123L, 123L, 123L, 123L, 123L, 123L, 123L, 123L,
123L, 123L, 123L, 123L, 123L, 123L, 123L, 222L, 222L, 222L, 222L,
222L, 222L, 222L, 222L, 222L, 222L, 222L, 222L, 222L, 222L, 222L,
222L, 222L, 222L, 222L, 222L, 222L, 222L, 222L, 222L, 222L, 222L,
222L, 222L, 222L, 222L, 222L, 222L, 222L, 222L, 222L, 222L, 222L,
222L, 222L, 222L, 222L, 222L, 222L, 222L, 222L, 222L, 222L),
item = c(234L, 234L, 234L, 234L, 234L, 234L, 234L, 234L,
234L, 234L, 234L, 234L, 234L, 234L, 234L, 234L, 234L, 234L,
234L, 234L, 234L, 234L, 234L, 234L, 234L, 234L, 234L, 234L,
234L, 234L, 234L, 234L, 234L, 234L, 234L, 234L, 234L, 234L,
234L, 234L, 234L, 234L, 234L, 234L, 234L, 234L, 234L, 333L,
333L, 333L, 333L, 333L, 333L, 333L, 333L, 333L, 333L, 333L,
333L, 333L, 333L, 333L, 333L, 333L, 333L, 333L, 333L, 333L,
333L, 333L, 333L, 333L, 333L, 333L, 333L, 333L, 333L, 333L,
333L, 333L, 333L, 333L, 333L, 333L, 333L, 333L, 333L, 333L,
333L, 333L, 333L, 333L, 333L, 333L), return = c(25L, 25L,
21L, 37L, 23L, 27L, 19L, 7L, 16L, 12L, 33L, 24L, 6L, 14L,
4L, 25L, 90L, 27L, 3L, 16L, 7L, 1L, 13L, 11L, 36L, 5L, 6L,
14L, 11L, 41L, 11L, 6L, 4L, 11L, 3L, 6L, 21L, 41L, 28L, 30L,
92L, 4L, 1L, 83L, 3L, 16L, 4L, 25L, 25L, 21L, 37L, 23L, 27L,
19L, 7L, 16L, 12L, 33L, 24L, 6L, 14L, 4L, 25L, 90L, 27L,
3L, 16L, 7L, 1L, 13L, 11L, 36L, 5L, 6L, 14L, 11L, 41L, 11L,
6L, 4L, 11L, 3L, 6L, 21L, 41L, 28L, 30L, 92L, 4L, 1L, 83L,
3L, 16L, 4L), action = c(0L, 0L, 0L, 0L, 0L, 0L, 0L, 0L,
0L, 0L, 0L, 0L, 0L, 0L, 0L, 0L, 0L, 0L, 0L, 0L, 0L, 0L, 0L,
0L, 0L, 0L, 0L, 0L, 0L, 0L, 1L, 1L, 1L, 1L, 0L, 0L, 0L, 0L,
0L, 0L, 0L, 0L, 0L, 0L, 0L, 0L, 0L, 0L, 0L, 0L, 0L, 0L, 0L,
0L, 0L, 0L, 0L, 0L, 0L, 0L, 0L, 0L, 0L, 0L, 0L, 0L, 0L, 0L,
0L, 0L, 0L, 0L, 0L, 0L, 0L, 0L, 0L, 1L, 1L, 1L, 1L, 0L, 0L,
0L, 0L, 0L, 0L, 0L, 0L, 0L, 0L, 0L, 0L, 0L)), .Names = c("code",
"item", "return", "action"), class = "data.frame", row.names = c(NA,
-94L))
I have 2 group vars code+item. Here two groups:
123 234
222 333
Also i have action column. It can have only two values(category) zero(0) or one(1).
i need calculate 90 percentile by zero category of action of return column, which go before one category of action.
Then i need calculate the median by zero category of action of return column, which go before one category of action. (After one we don't touch zeros)
These statistics must be calculate by 14 zeros preceding the one category of action
Then i have to find values that more 90 percentile which was calculated above, then such values must be replce by the median which was calculated.
After one category of action go zero category again for return column. For it i too have to find value that more 90 percentile which was calculeted above, then such value must be replaced by the median which was calculeted above(when calculate for 14 zeros).
Note the calculation is done by 14 zeros preceding the one category of action
but replacing by median is done for all zero category of action
and performing for each groups code+item
The result can be in output column.
to be more clear here desired output.
for 123+234 group
the 90 perc=39,5
median=12
for 222+333
the 90 perc=39,5
median=12
code item return action output
1 123 234 25 0 25
2 123 234 25 0 25
3 123 234 21 0 21
4 123 234 37 0 16
5 123 234 23 0 23
6 123 234 27 0 27
7 123 234 19 0 19
8 123 234 7 0 7
9 123 234 16 0 16
10 123 234 12 0 12
11 123 234 33 0 33
12 123 234 24 0 24
13 123 234 6 0 6
14 123 234 14 0 14
15 123 234 4 0 4
16 123 234 25 0 25
17 123 234 90 0 **12**
18 123 234 27 0 27
19 123 234 3 0 3
20 123 234 16 0 16
21 123 234 7 0 7
22 123 234 1 0 1
23 123 234 13 0 13
24 123 234 11 0 11
25 123 234 36 0 36
26 123 234 5 0 5
27 123 234 6 0 6
28 123 234 14 0 14
29 123 234 11 0 11
30 123 234 41 0 16
31 123 234 11 1 Na
32 123 234 6 1 Na
33 123 234 4 1 Na
34 123 234 11 1 Na
35 123 234 3 0 3
36 123 234 6 0 6
37 123 234 21 0 21
38 123 234 41 0 **12**
39 123 234 28 0 28
40 123 234 30 0 30
41 123 234 92 0 **12**
42 123 234 4 0 4
43 123 234 1 0 1
44 123 234 83 0 **12**
45 123 234 3 0 3
46 123 234 16 0 16
47 123 234 4 0 4
48 222 333 25 0 25
49 222 333 25 0 25
50 222 333 21 0 21
51 222 333 37 0 16
52 222 333 23 0 23
53 222 333 27 0 27
54 222 333 19 0 19
55 222 333 7 0 7
56 222 333 16 0 16
57 222 333 12 0 12
58 222 333 33 0 33
59 222 333 24 0 24
60 222 333 6 0 6
61 222 333 14 0 14
62 222 333 4 0 4
63 222 333 25 0 25
64 222 333 90 0 **12**
65 222 333 27 0 27
66 222 333 3 0 3
67 222 333 16 0 16
68 222 333 7 0 7
69 222 333 1 0 1
70 222 333 13 0 13
71 222 333 11 0 11
72 222 333 36 0 36
73 222 333 5 0 5
74 222 333 6 0 6
75 222 333 14 0 14
76 222 333 11 0 11
77 222 333 41 0 16
78 222 333 11 1 Na
79 222 333 6 1 Na
80 222 333 4 1 Na
81 222 333 11 1 Na
82 222 333 3 0 3
83 222 333 6 0 6
84 222 333 21 0 21
85 222 333 41 0 **12**
86 222 333 28 0 28
87 222 333 30 0 30
88 222 333 92 0 **12**
89 222 333 4 0 4
90 222 333 1 0 1
91 222 333 83 0 **12**
92 222 333 3 0 3
93 222 333 16 0 16
94 222 333 4 0 4
** i marked rows where value was replced by median.
回答1:
If I understand correctly, the OP wants to
- compute the 90% quantile and the median for the last 14 "zero action rows" (i.e., rows with
action == 0) before the first row withaction == 1, separately for eachcode,itemgroup. (This implies that there is only one streak ofaction == 1values per group. - Copy the
returnvalues to theoutputfor all rows withaction == 0. - Replace the
outputvalue if it is larger than the 90% quantile by the median in all zero action rows before the firstaction == 1row in each group.
This can be solved by updating in a non-equi join with some preparations
library(data.table)
# mark the zero acton rows before the the action period
setDT(mydat)[, zero_before := cummax(action), by = .(code, item)]
# compute median and 90% quantile for that last 14 rows before each action period
agg <- mydat[zero_before == 0,
quantile(tail(return, 14L), c(0.5, 0.9)) %>%
as.list() %>%
set_names(c("med", "q90")) %>%
c(.(zero_before = 0)), by = .(code, item)]
agg
code item med q90 zero_before
1: 123 234 12 39.5 0
2: 222 333 12 39.5 0
# append output column
mydat[action == 0, output := as.double(return)][
# replace output values greater q90 in an update non-equi join
agg, on = .(code, item, action, return > q90), output := as.double(med)][
# remove helper column
, zero_before := NULL]
As mydat has 94 rows, we will show only the updated rows:
mydat[return != output]
code item return action output 1: 123 234 90 0 12 2: 123 234 41 0 12 3: 123 234 41 0 12 4: 123 234 92 0 12 5: 123 234 83 0 12 6: 222 333 90 0 12 7: 222 333 41 0 12 8: 222 333 41 0 12 9: 222 333 92 0 12 10: 222 333 83 0 12
回答2:
With tidyverse:
mydat%>%
group_by(code,item)%>%
mutate(output=ifelse(return>quantile(return,.9) & action==0,median(return),return))
来源：https://stackoverflow.com/questions/51929941/calculation-of-90-percentile-and-replacement-of-it-by-median-by-groups-in-r