计算 90 个百分位数并用 R 中的组中位数替换它
[英]calculation of 90 percentile and replacement of it by median by groups in R
问题描述
Here part of data.
这里是部分数据。
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.我有 2 组变量代码+项目。 Here two groups:这里有两组：
123 234
222 333
Also i have action column.我也有行动专栏。 It can have only two values(category) zero(0) or one(1).它只能有两个值（类别）零（0）或一（1）。
i need calculate 90 percentile by zero category of action of return column, which go before one category of action.我需要通过返回列的零类动作来计算90 percentile ，它在一类动作之前。 Then i need calculate the median by zero category of action of return column, which go before one category of action.然后我需要通过返回列的零类动作来计算中median ，它在一类动作之前。 (After one we don't touch zeros) （一后我们不碰零）
These statistics must be calculate by 14 zeros preceding the one category of action这些统计数据必须由一类动作之前的 14 个零计算
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.然后我必须找到上面计算的超过 90 个百分位数的值，然后这些值必须由计算的中位数代替。
After one category of action go zero category again for return column.一类动作后，返回列再次归零类。 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).为此，我也必须找到上面计算的更多 90 个百分位数的值，然后必须用上面计算的中位数替换该值（计算 14 个零时）。
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请注意， calculation是由一类动作之前的 14 个零完成的，但replacing为中位数是针对所有zero category动作完成的，并为每个组code+item
The result can be in output column.结果可以在输出列中。
to be more clear here desired output.在这里更清楚所需的输出。
for 123+234 group the 90 perc=39,5 median=12对于123+234组，90 perc=39,5 中位数=12
for 222+333 the 90 perc=39,5 median=12对于222+333 333，90 perc=39,5 中位数=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. ** 我标记了值被中位数替换的行。
2 个解决方案
解决方案1
1 2018-08-20 11:41:43
With tidyverse :使用tidyverse ：
mydat%>%
group_by(code,item)%>%
mutate(output=ifelse(return>quantile(return,.9) & action==0,median(return),return))
解决方案2
0 已采纳 2018-08-21 14:02:42
If I understand correctly, the OP wants to如果我理解正确，OP 想要
- compute the 90% quantile and the median for the last 14 "zero action rows" (ie, rows with
action == 0) before the first row withaction == 1, separately for eachcode,itemgroup.计算90％分位点和中值的最后14“零点行动的行”（即，与行action == 0）的第一行与前action == 1，分别对每个code，item组。 (This implies that there is only one streak ofaction == 1values per group.（这意味着每组只有一个连续action == 1值。 - Copy the
returnvalues to theoutputfor all rows withaction == 0.将action == 0所有行的return值复制到output中。 - 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.如果output值大于 90% 分位数，则替换为每组中第一个action == 1行之前所有零操作行中的中位数。
This can be solved by updating in a non-equi join with some preparations这可以通过在非对等连接中更新并进行一些准备来解决
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:由于mydat有 94 行，我们将只显示更新的行：
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
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