计算数据框列中的单元格之间的差异

时间:2019-03-13 16:56:53

标签: r dataframe finance quantitative-finance

我有一个来自from here的STOXX投资领域:

 head(df)

        Date   SX5P   SX5E  SXXP  SXXE  SXXF  SXXA   DK5F  DKXF
1 1986-12-31 775.00 900.82 82.76 98.58 98.06 69.06 645.26 65.56
2 1987-01-01 775.00 900.82 82.76 98.58 98.06 69.06 645.26 65.56
3 1987-01-02 770.89 891.78 82.57 97.80 97.43 69.37 647.62 65.81
4 1987-01-05 771.89 898.33 82.82 98.60 98.19 69.16 649.94 65.82
5 1987-01-06 775.92 902.32 83.28 99.19 98.83 69.50 652.49 66.06
6 1987-01-07 781.21 899.15 83.78 98.96 98.62 70.59 651.97 66.20

了解操作分配的原则。我必须确定每个月底的 分配,以便每个股票对总投资组合造成相同的风险。

然后我跟随this tutorial,使您使用Python。

但是,一方面,我在计算每日收益时遇到问题。实际上,由于以下原因,我拥有了所有数据:

url <- 'https://www.stoxx.com/document/Indices/Current/HistoricalData/hbrbcpe.txt'
df <- read.table(url, sep = ';', skip = 4, stringsAsFactors = FALSE)
names(df) <- c('Date','SX5P','SX5E','SXXP','SXXE','SXXF','SXXA','DK5F','DKXF')
df$Date <- as.Date(sub('(.{2}).(.{2}).(.{4})', "\\3-\\2-\\1", df$Date))

然后我必须计算它们。我已经看到there is a function, Delt表示能够在两列之间进行操作。但是我必须在每个单元之间做出区别。我不知道该怎么做:

new = df[2:9]
# How to calculate the returns ?
Delt(df.a_given_day,df.a_given_day_plus_1,k=0:2)  #... Delt do it 0,1 y 2 periods between two columns.

之后,我可以使用cov_matrix_df <- cov(data.matrix(new, rownames.force = NA))计算协方差,然后继续搜索以计算风险。

另一方面,我不知道如何修改它以确定在每个月末 的风险来决定在每个月末 分配。

我的尝试:

this answer中,我尝试了以下答案:

dr_df = cbind(df[-1,1],apply(df[,-1],2,function(x) diff(x)/head(x,-1)))

哪个返回:

> head(dr_df)
                  SX5P         SX5E         SXXP         SXXE         SXXF         SXXA
[1,] 6209  0.000000000  0.000000000  0.000000000  0.000000000  0.000000000  0.000000000
[2,] 6210 -0.005303226 -0.010035301 -0.002295795 -0.007912355 -0.006424638  0.004488850
[3,] 6213  0.001297202  0.007344861  0.003027734  0.008179959  0.007800472 -0.003027245
[4,] 6214  0.005220951  0.004441575  0.005554214  0.005983773  0.006517975  0.004916136
[5,] 6215  0.006817713 -0.003513166  0.006003842 -0.002318782 -0.002124861  0.015683453
[6,] 6216 -0.004595435 -0.013101262 -0.003103366 -0.011014551 -0.009531535  0.005949851
              DK5F          DKXF
[1,]  0.0000000000  0.0000000000
[2,]  0.0036574404  0.0038133008
[3,]  0.0035823477  0.0001519526
[4,]  0.0039234391  0.0036463081
[5,] -0.0007969471  0.0021192855
[6,] -0.0098164026 -0.0087613293

哪个看起来不错,但是我不理解代码:/当我尝试创建协方差矩阵时,我遇到了一些问题:

> cov(dr_df[2:8])
Error in cov(dr_df[2:8]) : supply both 'x' and 'y' or a matrix-like 'x'
> cov(dr_df)
             SX5P SX5E SXXP SXXE SXXF SXXA DK5F DKXF
     9886513   NA   NA   NA   NA   NA   NA   NA   NA
SX5P      NA   NA   NA   NA   NA   NA   NA   NA   NA
SX5E      NA   NA   NA   NA   NA   NA   NA   NA   NA
SXXP      NA   NA   NA   NA   NA   NA   NA   NA   NA
SXXE      NA   NA   NA   NA   NA   NA   NA   NA   NA
SXXF      NA   NA   NA   NA   NA   NA   NA   NA   NA
SXXA      NA   NA   NA   NA   NA   NA   NA   NA   NA
DK5F      NA   NA   NA   NA   NA   NA   NA   NA   NA
DKXF      NA   NA   NA   NA   NA   NA   NA   NA   NA

索伦的尝试

似乎我对SX5P - SX5P1d的二进制运算符使用了非数字参数:

> library(lubridate)
Attaching package: ‘lubridate’

The following objects are masked from ‘package:data.table’:

    hour, mday, month, quarter, wday, week, yday, year

The following object is masked from ‘package:base’:

    date

> library(data.table)
> 
> 
> url <- 'https://www.stoxx.com/document/Indices/Current/HistoricalData/hbrbcpe.txt'
> df <- read.table(url, sep = ';', skip = 4, stringsAsFactors = FALSE)
> names(df) <- c('Date','SX5P','SX5E','SXXP','SXXE','SXXF','SXXA','DK5F','DKXF')
> df$Date <- dmy(df$Date)
> df$End_month_date <- ceiling_date(df$Date,unit="month") - days(1)
> 
> dt <- as.data.table(df)
> 
> #daily returns
> dt[, c("last_date",'SX5P1d','SX5E1d','SXXP1d','SXXE1d','SXXF1d','SXXA1d','DK5F1d','DKXF1d') := shift(.SD[,c("Date",'SX5P','SX5E','SXXP','SXXE','SXXF','SXXA','DK5F','DKXF')], n=1, fill=NA, type=c("lag")),]
Warning messages:
1: In `[.data.table`(dt, , `:=`(c("last_date", "SX5P1d", "SX5E1d",  :
  Supplied 9 items to be assigned to 7673 items of column 'last_date' (recycled leaving remainder of 5 items).
2: In `[.data.table`(dt, , `:=`(c("last_date", "SX5P1d", "SX5E1d",  :
  Supplied 9 items to be assigned to 7673 items of column 'SX5P1d' (recycled leaving remainder of 5 items).
3: In `[.data.table`(dt, , `:=`(c("last_date", "SX5P1d", "SX5E1d",  :
  Supplied 9 items to be assigned to 7673 items of column 'SX5E1d' (recycled leaving remainder of 5 items).
4: In `[.data.table`(dt, , `:=`(c("last_date", "SX5P1d", "SX5E1d",  :
  Supplied 9 items to be assigned to 7673 items of column 'SXXP1d' (recycled leaving remainder of 5 items).
5: In `[.data.table`(dt, , `:=`(c("last_date", "SX5P1d", "SX5E1d",  :
  Supplied 9 items to be assigned to 7673 items of column 'SXXE1d' (recycled leaving remainder of 5 items).
6: In `[.data.table`(dt, , `:=`(c("last_date", "SX5P1d", "SX5E1d",  :
  Supplied 9 items to be assigned to 7673 items of column 'SXXF1d' (recycled leaving remainder of 5 items).
7: In `[.data.table`(dt, , `:=`(c("last_date", "SX5P1d", "SX5E1d",  :
  Supplied 9 items to be assigned to 7673 items of column 'SXXA1d' (recycled leaving remainder of 5 items).
8: In `[.data.table`(dt, , `:=`(c("last_date", "SX5P1d", "SX5E1d",  :
  Supplied 9 items to be assigned to 7673 items of column 'DK5F1d' (recycled leaving remainder of 5 items).
9: In `[.data.table`(dt, , `:=`(c("last_date", "SX5P1d", "SX5E1d",  :
  Supplied 9 items to be assigned to 7673 items of column 'DKXF1d' (recycled leaving remainder of 5 items).
> dt[,`:=`(SX5P_r=SX5P-SX5P1d,
+          SX5E_r=SX5E-SX5E1d,
+          SXXP_r=SXXP-SXXP1d,
+          SXXE_r=SXXE-SXXE1d,
+          SXXF_r=SXXF-SXXF1d,
+          SXXA_r=SXXA-SXXA1d,
+          DK5F_r=DK5F-DK5F1d,
+          DKXF_r=DKXF-DKXF1d)]
Error in SX5P - SX5P1d : non-numeric argument to binary operator
> #monthly returns
> returns <- dt[,list(SX5P=sum(SX5P_r,na.rm=T),
+                     SX5E=sum(SX5E_r,na.rm=T),
+                     SXXP=sum(SXXP_r,na.rm=T),
+                     SXXE=sum(SXXE_r,na.rm=T),
+                     SXXF=sum(SXXF_r,na.rm=T),
+                     SXXA=sum(SXXA_r,na.rm=T),
+                     DK5F=sum(DK5F_r,na.rm=T),
+                     DKXF=sum(DKXF_r,na.rm=T)),by="End_month_date"]
Error in `[.data.table`(dt, , list(SX5P = sum(SX5P_r, na.rm = T), SX5E = sum(SX5E_r,  : 
  object 'SX5P_r' not found

dt操作之后的shift生成了警告消息:

> head(dt)
         Date   SX5P   SX5E  SXXP  SXXE  SXXF  SXXA   DK5F  DKXF End_month_date
1: 1986-12-31 775.00 900.82 82.76 98.58 98.06 69.06 645.26 65.56     1986-12-31
2: 1987-01-01 775.00 900.82 82.76 98.58 98.06 69.06 645.26 65.56     1987-01-31
3: 1987-01-02 770.89 891.78 82.57 97.80 97.43 69.37 647.62 65.81     1987-01-31
4: 1987-01-05 771.89 898.33 82.82 98.60 98.19 69.16 649.94 65.82     1987-01-31
5: 1987-01-06 775.92 902.32 83.28 99.19 98.83 69.50 652.49 66.06     1987-01-31
6: 1987-01-07 781.21 899.15 83.78 98.96 98.62 70.59 651.97 66.20     1987-01-31
   last_date SX5P1d SX5E1d SXXP1d SXXE1d SXXF1d SXXA1d DK5F1d DKXF1d
1:        NA     NA     NA     NA     NA     NA     NA     NA     NA
2:      Date   Date   Date   Date   Date   Date   Date   Date   Date
3:      SX5P   SX5P   SX5P   SX5P   SX5P   SX5P   SX5P   SX5P   SX5P
4:      SX5E   SX5E   SX5E   SX5E   SX5E   SX5E   SX5E   SX5E   SX5E
5:      SXXP   SXXP   SXXP   SXXP   SXXP   SXXP   SXXP   SXXP   SXXP
6:      SXXE   SXXE   SXXE   SXXE   SXXE   SXXE   SXXE   SXXE   SXXE

2 个答案:

答案 0 :(得分:1)

数据在日期2016-03-25和2016-03-28中具有一些不稳定的值。

library(dplyr)
df <- filter(df, SX5P>0)            # drop erratic data points
percent_change <- function(x) (x - lag(x)) / lag(x) # function that calculates percentage change 
daily_return <- df %>% 
  mutate_at(vars(-Date), percent_change) %>%     # for each column excluding Date, apply percent_change function
  filter(complete.cases(.)) %>%                  # filter out NAs
  select(-Date) %>%                              # drop Date variable 
  as.matrix()                                    # convert to matrix                   

head(daily_return, 5)
#             SX5P         SX5E         SXXP         SXXE         SXXF         SXXA          DK5F         DKXF
#[1,]  0.000000000  0.000000000  0.000000000  0.000000000  0.000000000  0.000000000  0.0000000000 0.0000000000
#[2,] -0.005303226 -0.010035301 -0.002295795 -0.007912355 -0.006424638  0.004488850  0.0036574404 0.0038133008
#[3,]  0.001297202  0.007344861  0.003027734  0.008179959  0.007800472 -0.003027245  0.0035823477 0.0001519526
#[4,]  0.005220951  0.004441575  0.005554214  0.005983773  0.006517975  0.004916136  0.0039234391 0.0036463081
#[5,]  0.006817713 -0.003513166  0.006003842 -0.002318782 -0.002124861  0.015683453 -0.0007969471 0.0021192855

cov(daily_return)  
#             SX5P         SX5E         SXXP         SXXE         SXXF         SXXA         DK5F         DKXF
#SX5P 0.0001458898 0.0001531675 0.0001339905 0.0001400356 0.0001335696 0.0001283412 0.0001355236 0.0001410957
#SX5E 0.0001531675 0.0001781671 0.0001431415 0.0001622366 0.0001519764 0.0001252829 0.0001497803 0.0001561299
#SXXP 0.0001339905 0.0001431415 0.0001267415 0.0001328073 0.0001265858 0.0001210988 0.0001314346 0.0001359420
#SXXE 0.0001400356 0.0001622366 0.0001328073 0.0001502001 0.0001410354 0.0001165071 0.0001412857 0.0001471070
#SXXF 0.0001335696 0.0001519764 0.0001265858 0.0001410354 0.0001343114 0.0001130397 0.0001380515 0.0001432671
#SXXA 0.0001283412 0.0001252829 0.0001210988 0.0001165071 0.0001130397 0.0001257977 0.0001221743 0.0001254364
#DK5F 0.0001355236 0.0001497803 0.0001314346 0.0001412857 0.0001380515 0.0001221743 0.0001914781 0.0001946354
#DKXF 0.0001410957 0.0001561299 0.0001359420 0.0001471070 0.0001432671 0.0001254364 0.0001946354 0.0002103559 

每月回报

library(lubridate)
percent_change2 <- function(x)last(x)/first(x) - 1
monthly_return <- df %>% 
  group_by(gr = floor_date(Date, unit = "month")) %>%
  summarize_at(vars(-Date, -gr), percent_change2) %>%
  ungroup() %>%
  select(-gr) %>% 
  as.matrix()  
head(monthly_return, 5)

            SX5P         SX5E        SXXP         SXXE         SXXF       SXXA        DK5F        DKXF
[1,]  0.00000000  0.000000000 0.000000000  0.000000000  0.000000000 0.00000000  0.00000000  0.00000000
[2,] -0.01089032 -0.046335561 0.005316578 -0.025867316 -0.025494595 0.04170287 -0.02977095 -0.01281269
[3,]  0.03167912 -0.009493186 0.032518367 -0.011141476 -0.011708861 0.07918740  0.05577361  0.04355828
[4,]  0.02633308  0.031731340 0.025284157  0.027359491  0.027197099 0.02322630  0.04121760  0.03157433
[5,]  0.02660200 -0.002816901 0.023347620 -0.003767437 -0.002362366 0.05061867  0.03758165  0.03917672

cov(monthly_return)
            SX5P        SX5E        SXXP        SXXE        SXXF        SXXA        DK5F        DKXF
SX5P 0.002068415 0.002243488 0.002011784 0.002160762 0.002076261 0.001867744 0.002282369 0.002381529
SX5E 0.002243488 0.002712719 0.002225923 0.002605715 0.002448324 0.001857319 0.002549326 0.002671546
SXXP 0.002011784 0.002225923 0.002025003 0.002182308 0.002095078 0.001873543 0.002321951 0.002407614
SXXE 0.002160762 0.002605715 0.002182308 0.002548197 0.002399266 0.001826243 0.002514475 0.002629281
SXXF 0.002076261 0.002448324 0.002095078 0.002399266 0.002291523 0.001797954 0.002458753 0.002558314
SXXA 0.001867744 0.001857319 0.001873543 0.001826243 0.001797954 0.001927949 0.002134767 0.002189677
DK5F 0.002282369 0.002549326 0.002321951 0.002514475 0.002458753 0.002134767 0.003414248 0.003523391
DKXF 0.002381529 0.002671546 0.002407614 0.002629281 0.002558314 0.002189677 0.003523391 0.003813587

data.table版本

library(data.table)
per_change <- function(x)x/shift(x) - 1
setDT(df)
df <- df[SX5P>0]
daily <- df[, lapply(.SD, per_change), .SDcols=-"Date"][-1, ]
daily
cov(daily)
monthly <- df[, lapply(.SD, percent_change2), by = .(gr=floor_date(Date, unit = "month")), .SDcols=-"Date"][-1, -"gr" ]
cov(monthly)

答案 1 :(得分:0)

以下解决方案将值移动一个日期(这可能比一个字面上的日期日期更好,以适应缺少的日期,周末,假期等)。它将查找“今天”和“昨天”的差作为收益,并求和整个月的总计,并按一个月的最后一天进行汇总。

library(lubridate)
library(data.table)


url <- 'https://www.stoxx.com/document/Indices/Current/HistoricalData/hbrbcpe.txt'
df <- read.table(url, sep = ';', skip = 4, stringsAsFactors = FALSE)
names(df) <- c('Date','SX5P','SX5E','SXXP','SXXE','SXXF','SXXA','DK5F','DKXF')
df$Date <- dmy(df$Date)
df$End_month_date <- ceiling_date(df$Date,unit="month") - days(1)

dt <- as.data.table(df)

#daily returns
dt[, c("last_date",'SX5P1d','SX5E1d','SXXP1d','SXXE1d','SXXF1d','SXXA1d','DK5F1d','DKXF1d') := shift(.SD[,c("Date",'SX5P','SX5E','SXXP','SXXE','SXXF','SXXA','DK5F','DKXF')], n=1, fill=NA, type=c("lag")),]
dt[,`:=`(SX5P_r=SX5P-SX5P1d,
         SX5E_r=SX5E-SX5E1d,
         SXXP_r=SXXP-SXXP1d,
         SXXE_r=SXXE-SXXE1d,
         SXXF_r=SXXF-SXXF1d,
         SXXA_r=SXXA-SXXA1d,
         DK5F_r=DK5F-DK5F1d,
         DKXF_r=DKXF-DKXF1d)]
#monthly returns
returns <- dt[,list(SX5P=sum(SX5P_r,na.rm=T),
                    SX5E=sum(SX5E_r,na.rm=T),
                    SXXP=sum(SXXP_r,na.rm=T),
                    SXXE=sum(SXXE_r,na.rm=T),
                    SXXF=sum(SXXF_r,na.rm=T),
                    SXXA=sum(SXXA_r,na.rm=T),
                    DK5F=sum(DK5F_r,na.rm=T),
                    DKXF=sum(DKXF_r,na.rm=T)),by="End_month_date"]

结果:

> returns
     End_month_date   SX5P    SX5E   SXXP   SXXE   SXXF   SXXA    DK5F   DKXF
  1:     1986-12-31   0.00    0.00   0.00   0.00   0.00   0.00    0.00   0.00
  2:     1987-01-31  -8.44  -41.74   0.44  -2.55  -2.50   2.88  -19.21  -0.84
  3:     1987-02-28  21.55  -18.11   2.53  -1.95  -1.87   6.15   41.03   3.32
  4:     1987-03-31  24.13   28.47   2.67   2.80   2.62   2.53   31.40   2.53
  5:     1987-04-30  25.96   12.02   2.33   0.96   0.82   3.44   22.66   2.64
 ---                                                                         
355:     2016-06-30 -94.12 -198.74 -17.57 -20.95 -20.18 -13.80 -384.00 -22.60
356:     2016-07-31  64.29  126.02  12.01  15.55  16.22   8.23  219.13  18.83
357:     2016-08-31 -14.71   32.37   1.64   3.98   2.69  -0.67  -26.23  -4.42
358:     2016-09-30 -19.74  -20.89  -0.61  -0.45  -0.79  -0.76  -48.93  -0.71
359:     2016-10-31  27.89   27.26   3.18   2.42   3.14   3.83   96.24   5.45