I feel that my basic problem is how to regress multiple series on a single series. Although my series are not equal in time, even when I use equal time length series for stock and benchmark(I can provide data I made equal manually if needed) I get an error. I want to estimate a market model (that is, regressing a stock's returns on benchmark's return for each day, for all stocks) and to make a data frame of beta values from the regression in long format. So, for the sample provided, there will be 4 beta values(2 for ABC and 2 for XYZ) in the beta value data frame. This is a sample of two stock prices
我觉得我的基本问题是如何在一个系列中回归多个系列。虽然我的系列在时间上是不相等的,即使我使用相同的时间长度系列库存和基准(我可以提供我在需要时手动相等的数据)我得到一个错误。我想估计一个市场模型(即,对于所有股票,回归每股基准回报的股票回报率),并以长格式从回归中得出β值的数据框。因此,对于提供的样本,β值数据框中将有4个β值(ABC为2,XYZ为2)。这是两个股票价格的样本
idf <- structure(list(Firm = c("ABC", "ABC", "ABC", "ABC", "ABC", "ABC", "ABC",
"ABC", "ABC", "ABC", "ABC", "ABC", "ABC", "ABC", "ABC", "XYZ", "XYZ", "XYZ",
"XYZ", "XYZ", "XYZ", "XYZ", "XYZ", "XYZ", "XYZ", "XYZ", "XYZ", "XYZ", "XYZ",
"XYZ"), Date = structure(c(NA, 1451642400, 1451646000, 1451649600, 1451653200,
1451656800, 1451660400, 1451664000, 1451898000, 1451901600, 1451905200,
1451908800, 1451912400, 1451916000, 1451919600, NA, 1451642400, 1451646000,
1451649600, 1451653200, 1451656800, 1451660400, 1451664000, 1451898000,
1451901600, 1451905200, 1451908800, 1451912400, 1451916000, 1451919600),
tzone = "UTC", class = c("POSIXct", "POSIXt")), Price = c(1270.9, 1277,
1273.25, 1273.85, 1273.75, 1272, 1265.35, 1265.35, 1253.1, 1248.1, 1242,
1248.15, 1241.1, 1246.5, 1242.5, 225.75, 225.7, 225.5, 225.45, 228.6, 227.7,
227.8, 227.8, 226, 225.1, 222.35, 222.25, 221.1, 221.2, 220.7), rt = c(NA,
0.00478826614451489, -0.0029408902678254, 0.000471124032113579,
-7.8505259892836e-05, -0.00137484063686699, -0.00524170116535849, 0,
-0.00972828256347036, -0.00399808624683118, -0.00489941143098349,
0.00493947152112462, -0.00566437187907365, 0.00434154082813709,
-0.00321414499282824, NA, -0.000221508473604359, -0.000886524880757023,
-0.000221754075639957, 0.0138753464936165, -0.00394477829101625,
0.000439077943387822, 0, -0.00793305174079517, -0.00399025135959974,
-0.0122920309559813, -0.000449842562691316, -0.00518778653061336,
0.000452181784779349, -0.00226295638548901), day = structure(c(NA, 16801,
16801, 16801, 16801, 16801, 16801, 16801, 16804, 16804, 16804, 16804, 16804,
16804, 16804, NA, 16801, 16801, 16801, 16801, 16801, 16801, 16801, 16804,
16804, 16804, 16804, 16804, 16804, 16804), class = "Date")),
.Names = c("Firm", "Date", "Price", "rt", "day"),
row.names = c(NA, -30L), class = c("tbl_df", "data.frame"))
and a sample of benchmark index
以及基准指数的样本
imdf <- structure(list(Date = structure(c(NA, 1451642400, 1451646000, 1451649600,
1451653200, 1451656800, 1451660400, 1451664000, 1451898000, 1451901600,
1451905200, 1451908800, 1451912400, 1451916000, 1451919600, 1451923200),
class = c("POSIXct", "POSIXt"), tzone = "UTC"), Price = c(3443.1, 3450.5,
3453.85, 3447.9, 3456.9, 3468.45, 3472.1, 3472.1, 3484.75, 3466.45, 3417.5,
3416.05, 3401, 3425.75, 3425.25, 3425.25), rt = c(NA, 0.0021469197059254,
0.000970402793278424, -0.00172420081111291, 0.00260688364525663,
0.00333557458000655, 0.00105178994070698, 0, 0.0036367074012027,
-0.00526529010184795, -0.0142217259100423, -0.000424376794422088,
-0.00441540679648789, 0.00725091981911596, -0.000145964093093198,
0), day = structure(c(NA, 16801, 16801, 16801, 16801, 16801, 16801, 16801,
16804, 16804, 16804, 16804, 16804, 16804, 16804, 16804), class = "Date")),
.Names = c("Date", "Price", "rt", "day"), row.names = c(NA, -16L),
class = c("tbl_df", "tbl", "data.frame"))
Following is my dplyr
based code for estimating current beta for stocks ABC and XYZ for two days by regressing their intraday return on benchmark return.
以下是基于dplyr的代码,通过回归基准回报的日内回报,估算两天股票ABC和XYZ的当前beta。
require(dplyr)
q3 <- idf %>%
group_by(Firm, day) %>%
summarise(Beta = PerformanceAnalytics::CAPM.beta(idf$rt, imdf$rt)) %>%
na.omit()
But it gives the following error:
但它给出了以下错误:
Error: The data cannot be converted into a time series. If you are trying to pass in names from a data object with one column, you should use the form 'data[rows, columns, drop = FALSE]'. Rownames should have standard date formats, such as '1985-03-15'.
For
对于
q3 <- idf %>%
group_by(Firm, day) %>%
summarise(Beta = coef(summary(lm(idf$rt~imdf$rt)))[2,1]) %>%
na.omit()
I get the following error:
我收到以下错误:
Error: error in evaluating the argument 'object' in selecting a method for function 'summary': Error in model.frame.default(formula = idf$rt ~ imdf$rt, drop.unused.levels = TRUE) : variable lengths differ (found for 'imdf$rt')`
I also tried the following based on aggregate function:
我还尝试了基于聚合函数的以下内容:
beta.est= function(x) coef(summary(lm(x~imdf$rt)))[2,1]
betas = aggregate(cbind(rt) ~ Firm + day , idf, FUN = beta.est)
But it also gives same error:
但它也给出了同样的错误:
Error in summary(lm(x ~ imdf$rt)) : error in evaluating the argument 'object' in selecting a method for function 'summary': Error in model.frame.default(formula = x ~ imdf$rt, drop.unused.levels = TRUE) : variable lengths differ (found for 'imdf$rt')
Then I tried a 'for loop':
然后我尝试了'for循环':
betas=list()
for(i in idf$Firm ){
for(j in idf$day){
betas$day= coef(summary(lm(idf[i,4]|j~imdf$rt|j)))[2,1]}}
Again, I get this:
我再说一遍:
Error in summary(lm(idf[i, 4] | j ~ imdf$rt | j)) : error in evaluating the argument 'object' in selecting a method for function 'summary': Error in model.frame.default(formula = idf[i, 4] | j ~ imdf$rt | j, drop.unused.levels = TRUE) : variable lengths differ (found for 'imdf$rt | j')
When I use equal time length series for stock and benchmark I get the following error:
当我对股票和基准使用相等的时间长度系列时,我得到以下错误:
Error in summary(lm(idf[i, 4] | j ~ imdf$rt | j)) : error in evaluating the argument 'object' in selecting a method for function 'summary': Error in model.frame.default(formula = idf[i, 4] | j ~ imdf$rt | j, drop.unused.levels = TRUE) : variable lengths differ (found for 'imdf$rt | j')
Although the above codes are for estimating current beta values, it would be nice if I can also regress lag of benchmark index on the stocks. Kindly help me.
虽然上面的代码用于估算当前的β值,但如果我还可以回归股票基准指数的滞后,那将是很好的。请帮助我。
1 个解决方案
#1
5
I hope you don't mind that I use some additional packages. I my opinion a combination of dplyr
,tidyr
,purrr
and broom
really simplifys working with multiple linear regressions:
我希望你不介意我使用一些额外的包。我认为dplyr,tidyr,purrr和扫帚的组合真的可以简化多线性回归:
To ensure we don't confuse the data column names with the benchmark column names we simply add the benchmarks name:
为确保我们不会将数据列名称与基准列名称混淆,我们只需添加基准名称:
names(imdf) <- paste("bm1", names(imdf), sep ="_")
You can also add a lagged Date variable and/or join another benchmark
您还可以添加滞后日期变量和/或加入另一个基准
set.seed(123)
imdf$bm1_Date_lag <- imdf$bm1_Date - 3600 # shift Date by 1 hour
imdf$bm2_rt <- rnorm(nrow(imdf), sd = 0.005) # add a dummy benchmark
First you have to group your data by Firm
and day
, join it with the benchmark data on Date
and nest it, this creates for each group a data.frame
called data
containing all the other variables.
首先,您必须按Firm和day对数据进行分组,将其与Date上的基准数据相结合并嵌套,这将为每个组创建一个称为包含所有其他变量的数据的data.frame。
idf_grouped <- idf %>% group_by(Firm, day) %>%
left_join(imdf, by = c("Date" = "bm1_Date")) %>%
nest()
Remove the groups without day
values so that you are able to fit a linear model on each group and add it as a new column with map
from purrr
. You can do this for multiple models in one step.
删除没有日值的组,以便您能够在每个组上拟合线性模型,并将其添加为带有purrr映射的新列。您可以一步完成多个模型的操作。
idf_grouped <- idf_grouped %>% filter(!is.na(day)) %>%
mutate(model = map(data,~lm(rt~bm1_rt, data = .)),
model2 = map(data,~lm(rt~bm2_rt, data = .)))
To extract the coefficients of linear models I prefer tidy
from broom
because it returns a data.frame
which is - in the context of dplyr
- easier to handle. unnest()
transforms your data back into a single data.frame
. Because you aren't interested in the intercept we filter for bm1_rt
为了提取线性模型的系数,我更喜欢从扫帚中整理,因为它返回一个data.frame,它在dplyr的上下文中更容易处理。 unnest()将您的数据转换回单个data.frame。因为你对截距不感兴趣,我们过滤bm1_rt
idf_grouped %>%
unnest(bm1 = model %>% map(tidy, quick = TRUE),
bm2 = model2 %>% map(tidy, quick = TRUE),.sep = "_") %>%
filter(bm1_term == "bm1_rt")
## Firm day bm1_term bm1_estimate bm2_term bm2_estimate
## <chr> <date> <fctr> <dbl> <fctr> <dbl>
## 1 ABC 2016-01-01 bm1_rt 0.08879514 bm2_rt -0.2781275
## 2 ABC 2016-01-04 bm1_rt 0.25888489 bm2_rt 0.3845765
## 3 XYZ 2016-01-01 bm1_rt 0.66791986 bm2_rt -0.2891714
## 4 XYZ 2016-01-04 bm1_rt 0.45735812 bm2_rt -0.5014824
To compute these betas on lagged benchmarks just join by "Date" = "bm1_Date_lag"
during the first step.
要在滞后基准测试中计算这些测试版,只需在第一步中加入“Date”=“bm1_Date_lag”。
Edit: In order to implement an industry return you must have a mapping which industry each firm belongs to. For illustration purpose I simply added another firm "DEF"
as a copy of "XYZ"
that I map to the same industry as "ABC"
编辑:为了实现行业回报,您必须拥有每家公司所属行业的映射。为了说明的目的,我简单地添加了另一个公司“DEF”作为“XYZ”的副本,我将其映射到与“ABC”相同的行业
idf_2 <- idf %>% filter(Firm == "XYZ") %>% mutate(Firm = "DEF") %>% bind_rows(idf)
firm_map <- data.frame(Firm = c("ABC", "DEF", "XYZ"), Industry = c(1,1,2), stringsAsFactors = FALSE)
simply join this into idf_2
只需将其加入idf_2即可
idf_map <- idf_2 %>% left_join(firm_map, by = "Firm")
and compute eg. each industry's average return
并计算例如。每个行业的平均回报
idf_map %>% left_join(idf_map %>% group_by(Industry, Date) %>%
summarise(ind_rt = mean(rt, na.rm = TRUE)), by = c("Industry", "Date"))
Now ind_rt
can be used as an explanatory variable in the regression.
现在ind_rt可以用作回归中的解释变量。
#1
5
I hope you don't mind that I use some additional packages. I my opinion a combination of dplyr
,tidyr
,purrr
and broom
really simplifys working with multiple linear regressions:
我希望你不介意我使用一些额外的包。我认为dplyr,tidyr,purrr和扫帚的组合真的可以简化多线性回归:
To ensure we don't confuse the data column names with the benchmark column names we simply add the benchmarks name:
为确保我们不会将数据列名称与基准列名称混淆,我们只需添加基准名称:
names(imdf) <- paste("bm1", names(imdf), sep ="_")
You can also add a lagged Date variable and/or join another benchmark
您还可以添加滞后日期变量和/或加入另一个基准
set.seed(123)
imdf$bm1_Date_lag <- imdf$bm1_Date - 3600 # shift Date by 1 hour
imdf$bm2_rt <- rnorm(nrow(imdf), sd = 0.005) # add a dummy benchmark
First you have to group your data by Firm
and day
, join it with the benchmark data on Date
and nest it, this creates for each group a data.frame
called data
containing all the other variables.
首先,您必须按Firm和day对数据进行分组,将其与Date上的基准数据相结合并嵌套,这将为每个组创建一个称为包含所有其他变量的数据的data.frame。
idf_grouped <- idf %>% group_by(Firm, day) %>%
left_join(imdf, by = c("Date" = "bm1_Date")) %>%
nest()
Remove the groups without day
values so that you are able to fit a linear model on each group and add it as a new column with map
from purrr
. You can do this for multiple models in one step.
删除没有日值的组,以便您能够在每个组上拟合线性模型,并将其添加为带有purrr映射的新列。您可以一步完成多个模型的操作。
idf_grouped <- idf_grouped %>% filter(!is.na(day)) %>%
mutate(model = map(data,~lm(rt~bm1_rt, data = .)),
model2 = map(data,~lm(rt~bm2_rt, data = .)))
To extract the coefficients of linear models I prefer tidy
from broom
because it returns a data.frame
which is - in the context of dplyr
- easier to handle. unnest()
transforms your data back into a single data.frame
. Because you aren't interested in the intercept we filter for bm1_rt
为了提取线性模型的系数,我更喜欢从扫帚中整理,因为它返回一个data.frame,它在dplyr的上下文中更容易处理。 unnest()将您的数据转换回单个data.frame。因为你对截距不感兴趣,我们过滤bm1_rt
idf_grouped %>%
unnest(bm1 = model %>% map(tidy, quick = TRUE),
bm2 = model2 %>% map(tidy, quick = TRUE),.sep = "_") %>%
filter(bm1_term == "bm1_rt")
## Firm day bm1_term bm1_estimate bm2_term bm2_estimate
## <chr> <date> <fctr> <dbl> <fctr> <dbl>
## 1 ABC 2016-01-01 bm1_rt 0.08879514 bm2_rt -0.2781275
## 2 ABC 2016-01-04 bm1_rt 0.25888489 bm2_rt 0.3845765
## 3 XYZ 2016-01-01 bm1_rt 0.66791986 bm2_rt -0.2891714
## 4 XYZ 2016-01-04 bm1_rt 0.45735812 bm2_rt -0.5014824
To compute these betas on lagged benchmarks just join by "Date" = "bm1_Date_lag"
during the first step.
要在滞后基准测试中计算这些测试版,只需在第一步中加入“Date”=“bm1_Date_lag”。
Edit: In order to implement an industry return you must have a mapping which industry each firm belongs to. For illustration purpose I simply added another firm "DEF"
as a copy of "XYZ"
that I map to the same industry as "ABC"
编辑:为了实现行业回报,您必须拥有每家公司所属行业的映射。为了说明的目的,我简单地添加了另一个公司“DEF”作为“XYZ”的副本,我将其映射到与“ABC”相同的行业
idf_2 <- idf %>% filter(Firm == "XYZ") %>% mutate(Firm = "DEF") %>% bind_rows(idf)
firm_map <- data.frame(Firm = c("ABC", "DEF", "XYZ"), Industry = c(1,1,2), stringsAsFactors = FALSE)
simply join this into idf_2
只需将其加入idf_2即可
idf_map <- idf_2 %>% left_join(firm_map, by = "Firm")
and compute eg. each industry's average return
并计算例如。每个行业的平均回报
idf_map %>% left_join(idf_map %>% group_by(Industry, Date) %>%
summarise(ind_rt = mean(rt, na.rm = TRUE)), by = c("Industry", "Date"))
Now ind_rt
can be used as an explanatory variable in the regression.
现在ind_rt可以用作回归中的解释变量。