如果我两个月回报的滞后时间小于-0.7,我想将下一个x <- x %>%
group_by(seriesid) %>%
mutate(first = { twomonthreturn <= (-0.7) } %>% { . * !duplicated(.) } ) %>%
mutate(first = ifelse(first==1,datem,NA)) #%>%
mutate(closeret = ifelse(datem<=sum(first,na.rm = TRUE),closeret,NA))
观察值设置为NA。
我尝试过:
date seriesid price closeret twomonthreturn
2018-07-25 50005 3.100 NA NA
2018-08-14 50005 2.500-0.19354839 NA
2018-09-28 50005 2.350 -0.06000000 -0.24193548
2018-10-27 50005 0.800 -0.65957447 -0.68000000
2018-11-27 50005 0.500 -0.37500000 -0.7872340
2018-12-31 50005 0.300 -0.40000000 -0.62500000
2019-01-26 50005 0.360 0.20000000 -0.28000000
2019-02-23 50005 0.300 -0.16666667 0.00000000
2017-01-21 50006 7.000 NA NA
2017-03-28 50006 9.750 NA NA
2017-04-14 50006 8.875 -0.08974359 NA
2017-05-20 50006 9.000 0.01408451 -0.07692308
2017-06-22 50006 9.000 0.00000000 0.01408451
其中datem =年(日期)* 12 +月(日期)
date seriesid price closeret twomonthreturn
2018-07-25 50005 3.100 NA NA
2018-08-14 50005 2.500-0.19354839 NA
2018-09-28 50005 2.350 -0.06000000 -0.24193548
2018-10-27 50005 0.800 -0.65957447 -0.68000000
2018-11-27 50005 0.500 -0.37500000 -0.7872340
2018-12-31 50005 0.300 NA -0.62500000
2019-01-26 50005 0.360 NA -0.28000000
2019-02-23 50005 0.300 NA 0.00000000
2017-01-21 50006 7.000 NA NA
2017-03-28 50006 9.750 NA NA
2017-04-14 50006 8.875 -0.08974359 NA
2017-05-20 50006 9.000 0.01408451 -0.07692308
2017-06-22 50006 9.000 0.00000000 0.01408451
我正在寻找每个组的解决方案,我的解决方案适用于第一个组,但是如果没有第一个解决方案(因为它是NA),那么此解决方案将不起作用。
service
答案 0 :(得分:0)
希望这将起作用:
将import urllib.request as urllib2
# change of IP address
page = urllib2.urlopen("http://example.com/").read()
和replace
一起用作cumsum(lag(replace_na(twomonthreturn, Inf), , Inf) <= -.7) > 0
中 - 0.7
之后组中所有值的索引,因此它将全部替换twomonthreturn
和closered
中的此类值。
NA
library(tidyverse)
dat %>%
arrange(seriesid, date) %>%
group_by(seriesid) %>%
mutate(
closeret = replace(
closeret,
cumsum(lag(replace_na(twomonthreturn, Inf), , Inf) <= -.7) > 0,
NA_real_
)
) %>%
ungroup()
# A tibble: 13 x 5
date seriesid price closeret twomonthreturn
<date> <dbl> <dbl> <dbl> <dbl>
1 2018-07-25 50005 3.1 NA NA
2 2018-08-14 50005 2.5 -0.194 NA
3 2018-09-28 50005 2.35 -0.06 -0.242
4 2018-10-27 50005 0.8 -0.660 -0.68
5 2018-11-27 50005 0.5 -0.375 -0.787
6 2018-12-31 50005 0.3 NA -0.625
7 2019-01-26 50005 0.36 NA -0.28
8 2019-02-23 50005 0.3 NA 0
9 2017-01-21 50006 7 NA NA
10 2017-03-28 50006 9.75 NA NA
11 2017-04-14 50006 8.88 -0.0897 NA
12 2017-05-20 50006 9 0.0141 -0.0769
13 2017-06-22 50006 9 0 0.0141