R 区分股票价格矩阵：删除并应用
R Differencing a matrix of stock prices: Delt and apply
我有一个数据框，其中每一列都是股票的价格历史记录。我想从中创建这些价格的百分比变化矩阵，因此我尝试使用 quantmod 中的 Delt 函数。当我尝试使用下面的代码 apply Delt 函数时，结果中有不正确的更改。当我将 Delt 分别应用于每一列时，会出现正确的更改。我做错了什么？
这里是 data
> library(quantmod)
> m<-read.csv(file="C:/Users/Desktop/prices.csv", header=TRUE, sep=",", row.names=1)
> head(apply(m, 2, Delt, k=1))
A B C D E
[1,] NA NA NA NA NA
[2,] 0.013883218 0.02402696 -0.0028845060 0.10537081 0.065129843
[3,] -0.261075312 -0.27656116 0.0064643504 -0.03748629 0.003693638
[4,] 0.100354272 0.28212656 -0.0006447236 0.01986472 0.022154153
[5,] -0.033372547 0.11482705 -0.0023028136 -0.06562413 0.093813308
[6,] -0.004419677 -0.18058097 0.0007338350 -0.03160490 -0.010180472
> head(Delt(m[,1]))
Delt.1.arithmetic
[1,] NA
[2,] -0.009326258
[3,] 0.015094952
[4,] 0.015403475
[5,] -0.017321925
[6,] 0.019550238
这里是m的前30行：
A B C D E
6/1/2006 186.57 15.16001 12.88992 128.04 63.23111
7/1/2006 184.83 13.93923 12.87614 127.91 63.17918
8/1/2006 187.62 14.40092 12.86571 128.26 63.33122
9/1/2006 190.51 13.87444 12.85661 128.34 63.60919
10/1/2006 187.21 12.84720 12.83384 126.74 62.97229
11/1/2006 190.87 13.52174 12.85892 127.57 64.05329
12/1/2006 198.13 13.81215 12.86256 133.37 64.81722
1/1/2007 195.92 14.24603 12.85711 132.01 65.18905
2/1/2007 196.68 13.95810 12.81181 130.22 65.17205
3/1/2007 195.84 13.74382 12.79951 131.86 65.47931
4/1/2007 196.78 13.76690 12.79787 133.54 65.93696
5/1/2007 199.87 14.22677 12.78429 136.04 65.77217
6/1/2007 198.24 14.10437 12.80410 134.49 65.31252
7/1/2007 200.87 14.19849 12.79181 135.41 65.34666
8/1/2007 203.25 14.08153 12.77221 136.67 65.90654
9/1/2007 201.71 13.99091 12.82413 136.30 65.55228
10/1/2007 204.35 14.62587 12.87034 142.31 67.68648
11/1/2007 208.01 15.15703 12.89241 144.25 68.97027
12/1/2007 205.63 14.69767 12.84175 146.33 69.04170
1/1/2008 198.64 14.58832 12.82002 145.92 69.43480
2/1/2008 196.51 13.58696 12.82676 148.02 70.73637
3/1/2008 198.91 12.77139 12.85116 151.79 71.66404
4/1/2008 197.59 12.52536 12.83977 156.14 72.56894
5/1/2008 197.48 13.14181 12.82874 154.74 73.44301
6/1/2008 198.23 13.14060 12.81345 155.54 73.41066
7/1/2008 199.49 12.67957 12.82051 157.93 73.46999
8/1/2008 197.50 13.85022 12.81246 155.64 72.86505
9/1/2008 180.14 12.89408 12.81132 146.17 70.21979
10/1/2008 176.97 12.13371 12.87266 140.09 69.55070
11/1/2008 160.76 10.22495 12.90323 127.26 67.40361
使用lapply
可能会更好
lst <- lapply(m, Delt, k=1)
head(data.frame(lst))
#Delt.1.arithmetic Delt.1.arithmetic.1 Delt.1.arithmetic.2 Delt.1.arithmetic.3 Delt.1.arithmetic.4
#1 NA NA NA NA NA
#2 -0.009326258 -0.08052690 -0.0010687188 -0.001015308 -0.0008213299
#3 0.015094952 0.03312212 -0.0008105454 0.002736299 0.0024065873
#4 0.015403475 -0.03655914 -0.0007071123 0.000623733 0.0043890342
#5 -0.017321925 -0.07403839 -0.0017710656 -0.012466885 -0.0100125948
#6 0.019550238 0.05250490 0.0019545514 0.006548840 0.0171662816
或者，如果我们使用 apply，则使用 as.vector
转换为 vector
head(apply(m, 2, function(x) Delt(as.vector(x), k = 1)))
# A B C D E
#[1,] NA NA NA NA NA
#[2,] -0.009326258 -0.08052690 -0.0010687188 -0.001015308 -0.0008213299
#[3,] 0.015094952 0.03312212 -0.0008105454 0.002736299 0.0024065873
#[4,] 0.015403475 -0.03655914 -0.0007071123 0.000623733 0.0043890342
#[5,] -0.017321925 -0.07403839 -0.0017710656 -0.012466885 -0.0100125948
#[6,] 0.019550238 0.05250490 0.0019545514 0.006548840 0.0171662816
取消命名整个数据集不会给出正确的结果
head(apply(unname(m), 2, quantmod::Delt))
# [,1] [,2] [,3] [,4] [,5]
#[1,] NA NA NA NA NA
#[2,] 0.013883218 0.02402696 -0.0028845060 0.10537081 0.065129843
#[3,] -0.261075312 -0.27656116 0.0064643504 -0.03748629 0.003693638
#[4,] 0.100354272 0.28212656 -0.0006447236 0.01986472 0.022154153
#[5,] -0.033372547 0.11482705 -0.0023028136 -0.06562413 0.093813308
#[6,] -0.004419677 -0.18058097 0.0007338350 -0.03160490 -0.010180472
但是，如果我们 unname 向量，它会给出正确的输出
head(apply(m, 2, function(x) Delt(unname(x), k=1)))
# A B C D E
#[1,] NA NA NA NA NA
#[2,] -0.009326258 -0.08052690 -0.0010687188 -0.001015308 -0.0008213299
#[3,] 0.015094952 0.03312212 -0.0008105454 0.002736299 0.0024065873
#[4,] 0.015403475 -0.03655914 -0.0007071123 0.000623733 0.0043890342
#[5,] -0.017321925 -0.07403839 -0.0017710656 -0.012466885 -0.0100125948
#[6,] 0.019550238 0.05250490 0.0019545514 0.006548840 0.0171662816
本质上，Delt 需要一个没有任何行名属性的 vector。 apply 中的 as.vector 和 unname 删除这些属性，而 lapply 默认删除它。
确认一下
head(apply(`row.names<-`(m, NULL), 2, Delt, k=1))
# A B C D E
#[1,] NA NA NA NA NA
#[2,] -0.009326258 -0.08052690 -0.0010687188 -0.001015308 -0.0008213299
#[3,] 0.015094952 0.03312212 -0.0008105454 0.002736299 0.0024065873
#[4,] 0.015403475 -0.03655914 -0.0007071123 0.000623733 0.0043890342
#[5,] -0.017321925 -0.07403839 -0.0017710656 -0.012466885 -0.0100125948
#[6,] 0.019550238 0.05250490 0.0019545514 0.006548840 0.0171662816
我有一个数据框，其中每一列都是股票的价格历史记录。我想从中创建这些价格的百分比变化矩阵，因此我尝试使用 quantmod 中的 Delt 函数。当我尝试使用下面的代码 apply Delt 函数时，结果中有不正确的更改。当我将 Delt 分别应用于每一列时，会出现正确的更改。我做错了什么？
这里是 data
> library(quantmod)
> m<-read.csv(file="C:/Users/Desktop/prices.csv", header=TRUE, sep=",", row.names=1)
> head(apply(m, 2, Delt, k=1))
A B C D E
[1,] NA NA NA NA NA
[2,] 0.013883218 0.02402696 -0.0028845060 0.10537081 0.065129843
[3,] -0.261075312 -0.27656116 0.0064643504 -0.03748629 0.003693638
[4,] 0.100354272 0.28212656 -0.0006447236 0.01986472 0.022154153
[5,] -0.033372547 0.11482705 -0.0023028136 -0.06562413 0.093813308
[6,] -0.004419677 -0.18058097 0.0007338350 -0.03160490 -0.010180472
> head(Delt(m[,1]))
Delt.1.arithmetic
[1,] NA
[2,] -0.009326258
[3,] 0.015094952
[4,] 0.015403475
[5,] -0.017321925
[6,] 0.019550238
这里是m的前30行：
A B C D E
6/1/2006 186.57 15.16001 12.88992 128.04 63.23111
7/1/2006 184.83 13.93923 12.87614 127.91 63.17918
8/1/2006 187.62 14.40092 12.86571 128.26 63.33122
9/1/2006 190.51 13.87444 12.85661 128.34 63.60919
10/1/2006 187.21 12.84720 12.83384 126.74 62.97229
11/1/2006 190.87 13.52174 12.85892 127.57 64.05329
12/1/2006 198.13 13.81215 12.86256 133.37 64.81722
1/1/2007 195.92 14.24603 12.85711 132.01 65.18905
2/1/2007 196.68 13.95810 12.81181 130.22 65.17205
3/1/2007 195.84 13.74382 12.79951 131.86 65.47931
4/1/2007 196.78 13.76690 12.79787 133.54 65.93696
5/1/2007 199.87 14.22677 12.78429 136.04 65.77217
6/1/2007 198.24 14.10437 12.80410 134.49 65.31252
7/1/2007 200.87 14.19849 12.79181 135.41 65.34666
8/1/2007 203.25 14.08153 12.77221 136.67 65.90654
9/1/2007 201.71 13.99091 12.82413 136.30 65.55228
10/1/2007 204.35 14.62587 12.87034 142.31 67.68648
11/1/2007 208.01 15.15703 12.89241 144.25 68.97027
12/1/2007 205.63 14.69767 12.84175 146.33 69.04170
1/1/2008 198.64 14.58832 12.82002 145.92 69.43480
2/1/2008 196.51 13.58696 12.82676 148.02 70.73637
3/1/2008 198.91 12.77139 12.85116 151.79 71.66404
4/1/2008 197.59 12.52536 12.83977 156.14 72.56894
5/1/2008 197.48 13.14181 12.82874 154.74 73.44301
6/1/2008 198.23 13.14060 12.81345 155.54 73.41066
7/1/2008 199.49 12.67957 12.82051 157.93 73.46999
8/1/2008 197.50 13.85022 12.81246 155.64 72.86505
9/1/2008 180.14 12.89408 12.81132 146.17 70.21979
10/1/2008 176.97 12.13371 12.87266 140.09 69.55070
11/1/2008 160.76 10.22495 12.90323 127.26 67.40361
使用lapply
lst <- lapply(m, Delt, k=1)
head(data.frame(lst))
#Delt.1.arithmetic Delt.1.arithmetic.1 Delt.1.arithmetic.2 Delt.1.arithmetic.3 Delt.1.arithmetic.4
#1 NA NA NA NA NA
#2 -0.009326258 -0.08052690 -0.0010687188 -0.001015308 -0.0008213299
#3 0.015094952 0.03312212 -0.0008105454 0.002736299 0.0024065873
#4 0.015403475 -0.03655914 -0.0007071123 0.000623733 0.0043890342
#5 -0.017321925 -0.07403839 -0.0017710656 -0.012466885 -0.0100125948
#6 0.019550238 0.05250490 0.0019545514 0.006548840 0.0171662816
或者，如果我们使用 apply，则使用 as.vector
vector
head(apply(m, 2, function(x) Delt(as.vector(x), k = 1)))
# A B C D E
#[1,] NA NA NA NA NA
#[2,] -0.009326258 -0.08052690 -0.0010687188 -0.001015308 -0.0008213299
#[3,] 0.015094952 0.03312212 -0.0008105454 0.002736299 0.0024065873
#[4,] 0.015403475 -0.03655914 -0.0007071123 0.000623733 0.0043890342
#[5,] -0.017321925 -0.07403839 -0.0017710656 -0.012466885 -0.0100125948
#[6,] 0.019550238 0.05250490 0.0019545514 0.006548840 0.0171662816
取消命名整个数据集不会给出正确的结果
head(apply(unname(m), 2, quantmod::Delt))
# [,1] [,2] [,3] [,4] [,5]
#[1,] NA NA NA NA NA
#[2,] 0.013883218 0.02402696 -0.0028845060 0.10537081 0.065129843
#[3,] -0.261075312 -0.27656116 0.0064643504 -0.03748629 0.003693638
#[4,] 0.100354272 0.28212656 -0.0006447236 0.01986472 0.022154153
#[5,] -0.033372547 0.11482705 -0.0023028136 -0.06562413 0.093813308
#[6,] -0.004419677 -0.18058097 0.0007338350 -0.03160490 -0.010180472
但是，如果我们 unname 向量，它会给出正确的输出
head(apply(m, 2, function(x) Delt(unname(x), k=1)))
# A B C D E
#[1,] NA NA NA NA NA
#[2,] -0.009326258 -0.08052690 -0.0010687188 -0.001015308 -0.0008213299
#[3,] 0.015094952 0.03312212 -0.0008105454 0.002736299 0.0024065873
#[4,] 0.015403475 -0.03655914 -0.0007071123 0.000623733 0.0043890342
#[5,] -0.017321925 -0.07403839 -0.0017710656 -0.012466885 -0.0100125948
#[6,] 0.019550238 0.05250490 0.0019545514 0.006548840 0.0171662816
本质上，Delt 需要一个没有任何行名属性的 vector。 apply 中的 as.vector 和 unname 删除这些属性，而 lapply 默认删除它。
确认一下
head(apply(`row.names<-`(m, NULL), 2, Delt, k=1))
# A B C D E
#[1,] NA NA NA NA NA
#[2,] -0.009326258 -0.08052690 -0.0010687188 -0.001015308 -0.0008213299
#[3,] 0.015094952 0.03312212 -0.0008105454 0.002736299 0.0024065873
#[4,] 0.015403475 -0.03655914 -0.0007071123 0.000623733 0.0043890342
#[5,] -0.017321925 -0.07403839 -0.0017710656 -0.012466885 -0.0100125948
#[6,] 0.019550238 0.05250490 0.0019545514 0.006548840 0.0171662816