I'm having difficulties about doing a CC analysis in R.
The assignment which I'm doing is from "Applied Multivariate Analysis" by Sharma, exercise 13.7, if you're familiar with it.
Basically, I'm asked to conduct a CCA on a set of variables. There are seven X variables, but only five Y variables, thus R complains that the dimensions are not compatible. See the image below for a visual representation of the data called CETNEW.
Edited (Changed from image to dput):
structure(list(...
1 = c("X1", "X2", "X3", "X4", "X5", "X6", "X7", "Y1", "Y2", "Y3", "Y4", "Y5"),
2 = c(2.72, 1.2, 0.82, 0.92, 1.19, 1, 1.45, 0.68, 0.98, 0.57, 1.07, 0.91), ...
3 = c(1.2, 3.78, 0.7, 1.04, 1.06, 1.32, 1.31, 0.56, 1, 0.79, 1.13, 1.38), ...
4 = c(0.82, 0.7, 1.7, 0.59, 0.83, 1.08, 1.01, 0.65, 0.78, 0.66, 0.93, 0.77), ...
5 = c(0.92, 1.04, 0.59, 3.09, 1.06, 0.93, 1.47, 0.62, 1.26, 0.51, 0.94, 0.85), ...
6 = c(1.19, 1.06, 0.83, 1.06, 2.94, 1.36, 1.66, 0.68, 1.16, 0.77, 1.37, 1.11), ...
7 = c(1, 1.32, 1.08, 0.93, 1.36, 2.94, 1.56, 0.9, 1.23, 0.78, 1.65, 1.31), ...
8 = c(1.45, 1.31, 1.01, 1.47, 1.66, 1.56, 3.11, 1.03, 1.7, 0.81, 1.63, 1.44), ...
9 = c(0.68, 0.56, 0.65, 0.62, 0.68, 0.9, 1.03, 1.71, 0.99, 0.65, 0.86, 0.72), ...
10 = c(0.98, 1, 0.78, 1.26, 1.16, 1.23, 1.7, 0.99, 3.07, 0.61, 1.43, 1.28), ...
11 = c(0.57, 0.79, 0.66, 0.51, 0.77, 0.78, 0.81, 0.65, 0.61, 2.83, 1.04, 0.84), ...
12 = c(1.07, 1.13, 0.93, 0.94, 1.37, 1.65, 1.63, 0.86, 1.43, 1.04, 2.83, 1.6), ...
13 = c(0.91, 1.38, 0.77, 0.85, 1.11, 1.31, 1.44, 0.72, 1.28, 0.84, 1.6, 4.01)),
row.names = c(NA, -12L), class = c("tbl_df", "tbl", "data.frame"))
What I've Done so Far
CETNEW <- CETNEW[,-1] #To remove the non-numeric values
Create two variables (criterion and predictor variables) as:
CETNEWx <- CETNEW[1:7,]
CETNEWy <- CETNEW[8:12,]
Then I've been using various packages such as CCA, CCP and candisk. From CCA:
ccCETNEW <- cc(CETNEWx,CETNEWy)
Yields the following error message:
Error in cov(X, Y, use = "pairwise") : incompatible dimensions
The matcor function also from CCA, yields the following error message:
Error in data.frame(..., check.names = FALSE) : arguments imply differing number of rows: 7, 5
Thus, it would seem that it all boils down to the different dimension problem. I've talked to my professor about it, but since he is using SAS, which apparently are compatible with this problem and could solve it, he could not help me.
Please, if you're familiar with canonical correlation and have had a similar problem before, any help regarding this topic is highly appreciated.
If you look at your data, notice the first column is divided into X and Y labels. That suggests to me that your data are transposed. If so, each column is an observation and the X and Y labels indicate various measurements taken on each observation. Canonical correlations are performed on two groups of measurements/variables from a single set of observations. First, here is the transposed data:
CETNEW.T <- structure(list(X1 = c(2.72, 1.2, 0.82, 0.92, 1.19, 1, 1.45, 0.68,
0.98, 0.57, 1.07, 0.91), X2 = c(1.2, 3.78, 0.7, 1.04, 1.06, 1.32,
1.31, 0.56, 1, 0.79, 1.13, 1.38), X3 = c(0.82, 0.7, 1.7, 0.59,
0.83, 1.08, 1.01, 0.65, 0.78, 0.66, 0.93, 0.77), X4 = c(0.92,
1.04, 0.59, 3.09, 1.06, 0.93, 1.47, 0.62, 1.26, 0.51, 0.94, 0.85
), X5 = c(1.19, 1.06, 0.83, 1.06, 2.94, 1.36, 1.66, 0.68, 1.16,
0.77, 1.37, 1.11), X6 = c(1, 1.32, 1.08, 0.93, 1.36, 2.94, 1.56,
0.9, 1.23, 0.78, 1.65, 1.31), X7 = c(1.45, 1.31, 1.01, 1.47,
1.66, 1.56, 3.11, 1.03, 1.7, 0.81, 1.63, 1.44), Y1 = c(0.68,
0.56, 0.65, 0.62, 0.68, 0.9, 1.03, 1.71, 0.99, 0.65, 0.86, 0.72
), Y2 = c(0.98, 1, 0.78, 1.26, 1.16, 1.23, 1.7, 0.99, 3.07, 0.61,
1.43, 1.28), Y3 = c(0.57, 0.79, 0.66, 0.51, 0.77, 0.78, 0.81,
0.65, 0.61, 2.83, 1.04, 0.84), Y4 = c(1.07, 1.13, 0.93, 0.94,
1.37, 1.65, 1.63, 0.86, 1.43, 1.04, 2.83, 1.6), Y5 = c(0.91,
1.38, 0.77, 0.85, 1.11, 1.31, 1.44, 0.72, 1.28, 0.84, 1.6, 4.01
)), class = "data.frame", row.names = c(NA, -12L))
Now the analysis runs fine:
library("CCA")
str(CETNEW.T)
# 'data.frame': 12 obs. of 12 variables:
# $ X1: num 2.72 1.2 0.82 0.92 1.19 1 1.45 0.68 0.98 0.57 ...
# $ X2: num 1.2 3.78 0.7 1.04 1.06 1.32 1.31 0.56 1 0.79 ...
# $ X3: num 0.82 0.7 1.7 0.59 0.83 1.08 1.01 0.65 0.78 0.66 ...
# $ X4: num 0.92 1.04 0.59 3.09 1.06 0.93 1.47 0.62 1.26 0.51 ...
# $ X5: num 1.19 1.06 0.83 1.06 2.94 1.36 1.66 0.68 1.16 0.77 ...
# $ X6: num 1 1.32 1.08 0.93 1.36 2.94 1.56 0.9 1.23 0.78 ...
# $ X7: num 1.45 1.31 1.01 1.47 1.66 1.56 3.11 1.03 1.7 0.81 ...
# $ Y1: num 0.68 0.56 0.65 0.62 0.68 0.9 1.03 1.71 0.99 0.65 ...
# $ Y2: num 0.98 1 0.78 1.26 1.16 1.23 1.7 0.99 3.07 0.61 ...
# $ Y3: num 0.57 0.79 0.66 0.51 0.77 0.78 0.81 0.65 0.61 2.83 ...
# $ Y4: num 1.07 1.13 0.93 0.94 1.37 1.65 1.63 0.86 1.43 1.04 ...
# $ Y5: num 0.91 1.38 0.77 0.85 1.11 1.31 1.44 0.72 1.28 0.84 ...
X <- CETNEW.T[, 1:7]
Y <- CETNEW.T[, 8:12]
ccCETNEW <- cc(X, Y)
ccCETNEW is list with 5 parts containing the results.
Related
I used a for loop to create a correlation matrix, because I needed to use polychor to generate polychoric correaltions and I was only able to get polychor to correlate two variables at a time. Anyway, I created my own correlation table with the following code:
for(i in 1:ncol(gd2)) {
for (j in 1:ncol(gd2)) {
corVal
The table looks like this:
head(dtnew)
Better Afraid Alive Bored Drop Empty Energy Happy Help Home Hope Memory Satis Spirit Worth TOT
1: 1.00 0.32 0.29 0.39 0.36 0.46 0.25 0.43 0.39 0.13 0.46 0.39 0.50 0.45 0.48 0.67
2: 0.32 1.00 0.25 0.20 0.24 0.30 0.23 0.30 0.43 0.15 0.44 0.28 0.31 0.29 0.34 0.62
3: 0.29 0.25 1.00 0.26 0.28 0.46 0.38 0.60 0.35 0.19 0.41 0.10 0.49 0.53 0.43 0.65
4: 0.39 0.20 0.26 1.00 0.36 0.56 0.31 0.36 0.39 0.16 0.32 0.23 0.39 0.35 0.44 0.67
5: 0.36 0.24 0.28 0.36 1.00 0.44 0.41 0.37 0.43 0.31 0.35 0.22 0.42 0.37 0.40 0.72
6: 0.46 0.30 0.46 0.56 0.44 1.00 0.32 0.55 0.51 0.18 0.45 0.17 0.62 0.52 0.64 0.75
>
But longer.
Here is the dput()
structure(list(Better = c(1, 0.32, 0.29, 0.39, 0.36, 0.46, 0.25,
0.43, 0.39, 0.13, 0.46, 0.39, 0.5, 0.45, 0.48, 0.67), Afraid = c(0.32,
1, 0.25, 0.2, 0.24, 0.3, 0.23, 0.3, 0.43, 0.15, 0.44, 0.28, 0.31,
0.29, 0.34, 0.62), Alive = c(0.29, 0.25, 1, 0.26, 0.28, 0.46,
0.38, 0.6, 0.35, 0.19, 0.41, 0.1, 0.49, 0.53, 0.43, 0.65), Bored = c(0.39,
0.2, 0.26, 1, 0.36, 0.56, 0.31, 0.36, 0.39, 0.16, 0.32, 0.23,
0.39, 0.35, 0.44, 0.67), Drop = c(0.36, 0.24, 0.28, 0.36, 1,
0.44, 0.41, 0.37, 0.43, 0.31, 0.35, 0.22, 0.42, 0.37, 0.4, 0.72
), Empty = c(0.46, 0.3, 0.46, 0.56, 0.44, 1, 0.32, 0.55, 0.51,
0.18, 0.45, 0.17, 0.62, 0.52, 0.64, 0.75), Energy = c(0.25, 0.23,
0.38, 0.31, 0.41, 0.32, 1, 0.48, 0.37, 0.36, 0.31, 0.14, 0.4,
0.43, 0.38, 0.74), Happy = c(0.43, 0.3, 0.6, 0.36, 0.37, 0.55,
0.48, 1, 0.45, 0.21, 0.49, 0.22, 0.69, 0.84, 0.49, 0.8), Help = c(0.39,
0.43, 0.35, 0.39, 0.43, 0.51, 0.37, 0.45, 1, 0.2, 0.51, 0.32,
0.5, 0.44, 0.6, 0.73), Home = c(0.13, 0.15, 0.19, 0.16, 0.31,
0.18, 0.36, 0.21, 0.2, 1, 0.23, 0.13, 0.13, 0.15, 0.26, 0.63),
Hope = c(0.46, 0.44, 0.41, 0.32, 0.35, 0.45, 0.31, 0.49,
0.51, 0.23, 1, 0.38, 0.48, 0.47, 0.59, 0.73), Memory = c(0.39,
0.28, 0.1, 0.23, 0.22, 0.17, 0.14, 0.22, 0.32, 0.13, 0.38,
1, 0.25, 0.24, 0.31, 0.66), Satis = c(0.5, 0.31, 0.49, 0.39,
0.42, 0.62, 0.4, 0.69, 0.5, 0.13, 0.48, 0.25, 1, 0.66, 0.6,
0.78), Spirit = c(0.45, 0.29, 0.53, 0.35, 0.37, 0.52, 0.43,
0.84, 0.44, 0.15, 0.47, 0.24, 0.66, 1, 0.51, 0.77), Worth = c(0.48,
0.34, 0.43, 0.44, 0.4, 0.64, 0.38, 0.49, 0.6, 0.26, 0.59,
0.31, 0.6, 0.51, 1, 0.77), TOT = c(0.67, 0.62, 0.65, 0.67,
0.72, 0.75, 0.74, 0.8, 0.73, 0.63, 0.73, 0.66, 0.78, 0.77,
0.77, 0.89)), row.names = c(NA, -16L), class = c("data.table",
"data.frame"), .internal.selfref = <pointer: 0x000001d7adc21ef0>)
</pre/>
I would like to generate a visual using corrplot. However, when I try, I get an error:
Error in is.finite(tmp) : default method not implemented for type 'list'
My data is indeed of type list. I have tried usuing 'unlist'. Not sure what else to try.
There is a problem with your dput() output, possibly because you have a data.table. I can read it by deleting ", .internal.selfref = <pointer: 0x000001d7adc21ef0>" from the last line so that it ends class = c("data.table", "data.frame")). Printing that out shows a problem with the last line/column (Tot). The bottom row in that column should be 1.00, but it is 0.89. We can trim that and use as.matrix (my mistake in the earlier comment) to convert the data frame:
gd3 <- gd2[-16, -16]
corrplot(as.matrix(gd3))
library(corrplot)
M <- cor(df)
head(round(M,2))
corrplot(M, method="number")
EDITED:
I have a dataframe that stores information about when particular assessment happened ('when'). This assessment happened at different times (t1 - t3) which vary by participant.
The dataframe also contains all the assessments ever completed by every participant (including the one referenced in the 'when' column). I only want the assessment information represented in the 'when' column. So if the number is 1, I want to keep all the data related to that assessment and remove all the data that was not collected at that assessment. Please note that I have many more variables in my actual data set than are represented in this shortened data set so any solution should not rely on repeating variable names.
Here's the best I can do. The problem with this solution is that it would have to be repeated for every variable name.
df2 <- mutate(.data = df,
a1G_when = if_else(when == 1, a1G_t1, NA_real_))
# here is what we start with
df <- structure(list(id = 1:10, when = c(1, 3, 2, 1, 2, 1, 3, 2, 3,
1), a1G_t1 = c(0.78, 0.21, 0.04, 0.87, 0.08, 0.25, 0.9, 0.77,
0.51, 0.5), Stqo_t1 = c(0.68, 0.77, 0.09, 0.66, 0.94, 0.05, 0.97,
0.92, 1, 0.04), Twcdz_t1 = c(0.95, 0.41, 0.29, 0.54, 0.06, 0.45,
0.6, 0.24, 0.17, 0.55), Kgh_t1 = c(0.25, 0.86, 0.37, 0.34, 0.97,
0.75, 0.73, 0.68, 0.37, 0.66), `2xWX_t1` = c(0.47, 0.52, 0.23,
0.5, 0.88, 0.71, 0.21, 0.98, 0.76, 0.21), `2IYnS_t1` = c(0.32,
0.75, 0.03, 0.46, 0.89, 0.71, 0.51, 0.83, 0.34, 0.32), a1G_t2 = c(0.97,
0.01, 0.58, 0.33, 0.58, 0.37, 0.76, 0.33, 0.39, 0.56), Stqo_t2 = c(0.78,
0.42, 0.5, 0.69, 0.09, 0.72, 0.84, 0.94, 0.46, 0.83), Twcdz_t2 = c(0.62,
0.34, 0.72, 0.62, 0.8, 0.26, 0.3, 0.88, 0.42, 0.53), Kgh_t2 = c(0.99,
0.66, 0.02, 0.17, 0.51, 0.03, 0.03, 0.74, 0.1, 0.26), `2xWX_t2` = c(0.68,
0.97, 0.56, 0.27, 0.66, 0.71, 0.96, 0.24, 0.37, 0.76), `2IYnS_t2` = c(0.24,
0.88, 0.58, 0.31, 0.8, 0.92, 0.91, 0.9, 0.55, 0.52), a1G_t3 = c(0.73,
0.6, 0.66, 0.06, 0.33, 0.34, 0.09, 0.44, 0.73, 0.56), Stqo_t3 = c(0.28,
0.88, 0.56, 0.75, 0.85, 0.33, 0.88, 0.4, 0.63, 0.61), Twcdz_t3 = c(0.79,
0.95, 0.41, 0.07, 0.99, 0.06, 0.74, 0.17, 0.89, 0.4), Kgh_t3 = c(0.06,
0.52, 0.35, 0.91, 0.43, 0.74, 0.72, 0.96, 0.39, 0.4), `2xWX_t3` = c(0.25,
0.09, 0.64, 0.32, 0.15, 0.14, 0.18, 0.33, 0.97, 0.6), `2IYnS_t3` = c(0.92,
0.49, 0.09, 0.95, 0.3, 0.83, 0.82, 0.56, 0.29, 0.36)), row.names = c(NA,
-10L), class = "data.frame")
# here is an example of what I want with the first column. I would also want all other repeating columns to look like this (Stq0_when, Twcdz, etc.)
id when a1G_when
1 1 1 0.78
2 2 3 0.88
3 3 2 0.58
4 4 1 0.87
5 5 2 0.58
6 6 1 0.25
7 7 3 0.09
8 8 2 0.33
9 9 3 0.73
10 10 1 0.50
Using data.table, you could do something like:
library(data.table)
cols <- unique(paste0(gsub("_.*", "", setdiff(names(df), c("id", "when"))), "_when"))
setDT(df)[
, (cols) := lapply(cols, function(x) paste0(gsub("_.*", "", x), "_t", when))][
, (cols) := lapply(cols, function(x) as.character(.SD[[get(x)]])), by = cols][
, (cols) := lapply(.SD, as.numeric), .SDcols = cols
]
Output (only first 10 rows and only relevant when columns):
a1G_when Stqo_when Twcdz_when Kgh_when 2xWX_when 2IYnS_when
1: 0.78 0.68 0.95 0.25 0.47 0.32
2: 0.60 0.88 0.95 0.52 0.09 0.49
3: 0.58 0.50 0.72 0.02 0.56 0.58
4: 0.87 0.66 0.54 0.34 0.50 0.46
5: 0.58 0.09 0.80 0.51 0.66 0.80
6: 0.25 0.05 0.45 0.75 0.71 0.71
7: 0.09 0.88 0.74 0.72 0.18 0.82
8: 0.33 0.94 0.88 0.74 0.24 0.90
9: 0.73 0.63 0.89 0.39 0.97 0.29
10: 0.50 0.04 0.55 0.66 0.21 0.32
Here is an opportunity to use the new tidyr::pivot_longer. We can use this to reshape the data so that var and t are in their own columns, filter to just the rows with the data we want (i.e. where t equals when) and then pivot the data back out to wide.
library(tidyverse)
df1 <- structure(list(ID = c(101, 102, 103, 104, 105), when = c(1, 2, 3, 1, 2), var1_t1 = c(5, 6, 4, 5, 6), var2_t1 = c(2, 3, 4, 2, 3), var1_t2 = c(7, 8, 9, 7, 8), var2_t2 = c(5, 4, 5, 4, 5), var1_t3 = c(3, 4, 3, 4, 3), var2_t3 = c(6, 7, 6, 7, 6)), row.names = c(NA, 5L), class = "data.frame")
df1 %>%
pivot_longer(
cols = starts_with("var"),
names_to = c("var", "t"),
names_sep = "_t",
values_to = "val",
col_ptypes = list(var = character(), t = numeric())
) %>%
filter(when == t) %>%
select(-t) %>%
pivot_wider(names_from = "var", values_from = "val")
#> # A tibble: 5 x 4
#> ID when var1 var2
#> <dbl> <dbl> <dbl> <dbl>
#> 1 101 1 5 2
#> 2 102 2 8 4
#> 3 103 3 3 6
#> 4 104 1 5 2
#> 5 105 2 8 5
Created on 2019-07-16 by the reprex package (v0.3.0)
I need to read the following matrix from a file. It's a symmetric correlation matrix, so half of it is omitted.
1.00
0.49 1.00
0.53 0.57 1.00
0.49 0.46 0.48 1.00
0.51 0.53 0.57 0.57 1.00
0.33 0.30 0.31 0.24 0.38 1.00
0.32 0.21 0.23 0.22 0.32 0.43 1.00
0.20 0.16 0.14 0.12 0.17 0.27 0.33 1.00
0.19 0.08 0.07 0.19 0.23 0.24 0.26 0.25 1.00
0.30 0.27 0.24 0.21 0.32 0.34 0.54 0.46 0.28 1.00
0.37 0.35 0.37 0.29 0.36 0.37 0.32 0.29 0.30 0.35 1.00
0.21 0.20 0.18 0.16 0.27 0.40 0.58 0.45 0.27 0.59 0.31 1.00
Currently, I'm using
data1 <- na.omit(as.vector(t(read.table('triangle-data.txt', fill = TRUE))))
pt <- 12
R <- matrix(0, nrow = pt , ncol = pt)
for(i in 1:pt){
R[i, 1:i] <- data1[(i*(i-1)/2 + 1): (i*(i+1)/2)]
}
R <- R + t(R) - diag(rep(1, pt))
R
The result is
> dput(R)
structure(c(1, 0.49, 0.53, 0.49, 0.51, 0.33, 0.32, 0.2, 0.19,
0.3, 0.37, 0.21, 0.49, 1, 0.57, 0.46, 0.53, 0.3, 0.21, 0.16,
0.08, 0.27, 0.35, 0.2, 0.53, 0.57, 1, 0.48, 0.57, 0.31, 0.23,
0.14, 0.07, 0.24, 0.37, 0.18, 0.49, 0.46, 0.48, 1, 0.57, 0.24,
0.22, 0.12, 0.19, 0.21, 0.29, 0.16, 0.51, 0.53, 0.57, 0.57, 1,
0.38, 0.32, 0.17, 0.23, 0.32, 0.36, 0.27, 0.33, 0.3, 0.31, 0.24,
0.38, 1, 0.43, 0.27, 0.24, 0.34, 0.37, 0.4, 0.32, 0.21, 0.23,
0.22, 0.32, 0.43, 1, 0.33, 0.26, 0.54, 0.32, 0.58, 0.2, 0.16,
0.14, 0.12, 0.17, 0.27, 0.33, 1, 0.25, 0.46, 0.29, 0.45, 0.19,
0.08, 0.07, 0.19, 0.23, 0.24, 0.26, 0.25, 1, 0.28, 0.3, 0.27,
0.3, 0.27, 0.24, 0.21, 0.32, 0.34, 0.54, 0.46, 0.28, 1, 0.35,
0.59, 0.37, 0.35, 0.37, 0.29, 0.36, 0.37, 0.32, 0.29, 0.3, 0.35,
1, 0.31, 0.21, 0.2, 0.18, 0.16, 0.27, 0.4, 0.58, 0.45, 0.27,
0.59, 0.31, 1), .Dim = c(12L, 12L))
This is too unwieldy, and I need to hard-code its size. Is there a more convenient way?
I used a combination of readLines and strsplit to read the file
a <- sapply(sapply(lapply(readLines("triangle.txt"),
function(x) strsplit(x, " ")), "[", 1),
function(x) na.omit(as.numeric(x)))
and rbind to cast it into a square matrix
A <- do.call("rbind", a)
Despite the warning, the lower part of the matrix is correctly read from the file, but the upper part is all messed up, which I fixed with a little dirty trick
A[upper.tri(A)] <- 0
A <- A + t(A) - diag(nrow(A))
EDIT
Another simpler solution based on the vector of the coefficients:
data1 <- na.omit(as.vector(t(read.table('triangle.txt', fill = TRUE))))
n <- Re(polyroot(c(-length(data1), 1/2, 1/2)))[1]
A <- matrix(0, n, n)
A[upper.tri(A, diag = T)] <- data1
A <- A + t(A) - diag(n)
Currently we compute and sort data of stocks (X1 to X10). Historical data is stored in Excel and R for the time period 1950-1980, 1980-1999 and for 1950-1999.
The dataset:
date X1 X2 X3 X4 X5 X6 X7 X8 X9 X10
1 1950-01-01 5.92 6.35 4.61 4.08 5.47 3.90 2.35 1.49 2.27 0.82
2 1950-02-01 2.43 2.16 2.10 1.58 -0.05 1.14 1.51 1.52 2.02 1.12
3 1950-03-01 -0.81 0.21 -1.67 -0.02 -0.79 0.18 -0.22 1.03 0.12 1.75
4 1950-04-01 5.68 6.45 5.41 5.94 6.10 5.87 3.82 3.34 3.44 3.97
5 1950-05-01 3.84 1.60 1.64 3.33 2.54 2.12 4.46 2.83 3.82 4.75
6 1950-06-01 -9.88 -10.56 -8.02 -7.86 -7.27 -7.44 -7.13 -7.76 -6.32 -5.04
7 1950-07-01 9.09 8.76 7.31 5.88 3.84 4.61 3.09 3.07 1.41 0.42
598 1999-10-01 -0.95 -1.88 -1.25 -0.52 1.65 0.72 5.41 4.38 5.58 6.59
599 1999-11-01 11.57 9.15 8.17 7.14 6.15 4.95 5.78 4.21 1.55 2.15
600 1999-12-01 12.32 14.97 9.29 11.77 11.09 5.89 11.88 11.26 6.23 5.64
The main question is, we would like to compute/plot efficient frontiers for these 4 time periods to see how the efficient frontier has evolved in 1 graph. Are there ways to do this in R?
The efficient frontier is the set of optimal portfolios that offers the highest expected return for a defined level of risk or the lowest risk for a given level of expected return.
In modern portfolio theory, the efficient frontier (or portfolio frontier) is an investment portfolio which occupies the 'efficient' parts of the risk-return spectrum. Formally, it is the set of portfolios which satisfy the condition that no other portfolio exists with a higher expected return but with the same standard deviation of return.
So, how would one go about computing this in R?
dput sample data (first 50 rows)
> dput(head(data,50))
structure(list(X__1 = structure(c(-631152000, -628473600, -626054400,
-623376000, -620784000, -618105600, -615513600, -612835200, -610156800,
-607564800, -604886400, -602294400, -599616000, -596937600, -594518400,
-591840000, -589248000, -586569600, -583977600, -581299200, -578620800,
-576028800, -573350400, -570758400, -568080000, -565401600, -562896000,
-560217600, -557625600, -554947200, -552355200, -549676800, -546998400,
-544406400, -541728000, -539136000, -536457600, -533779200, -531360000,
-528681600, -526089600, -523411200, -520819200, -518140800, -515462400,
-512870400, -510192000, -507600000, -504921600, -502243200), class = c("POSIXct",
"POSIXt"), tzone = "UTC"), X__2 = c(5.92, 2.43, -0.81, 5.68,
3.84, -9.88, 9.09, 4.93, 3.99, -0.5, 3.09, 15.77, 8.22, 0.36,
-7.36, 3.84, -2.81, -7.12, 3.57, 6.59, 1.04, -1.41, -1.42, -0.53,
1.86, -3.25, 0.68, -4.4, 0.57, 2.5, -0.36, -0.74, -1.11, -0.58,
3.22, 0.33, 5.01, 2.75, -1.25, -2.13, 1.3, -4.42, 0.25, -5.56,
-4.09, 2.71, 2.01, -3.15, 8.48, -0.16), X__3 = c(6.35, 2.16,
0.21, 6.45, 1.6, -10.56, 8.76, 4.63, 3.52, -1.2, 3.36, 10.98,
8.41, 0.81, -4.01, 3.56, -4.27, -6.11, 4.7, 5.3, 2.73, -3.07,
-0.13, 0.6, 1.1, -2.77, 2.37, -4.5, 1.87, 3.18, 1.51, 0.43, -1.91,
-1.52, 4.91, 1.43, 3.4, 3.03, -2.25, -2, 0.34, -4.75, 2.24, -6.53,
-1.87, 1.97, 1.78, -2.96, 7.38, 0.43), X__4 = c(4.61, 2.1, -1.67,
5.41, 1.64, -8.02, 7.31, 4.56, 5.18, -0.46, 3.52, 10.78, 8.46,
0.28, -4.88, 4.26, -3.25, -6.76, 6.78, 4.99, 3.86, -2.57, 0.59,
0.16, 1.75, -2.04, 2.49, -5.29, 1.76, 2.88, 0.76, 0.67, -1.67,
-1.45, 5.69, 2.95, 3.66, 1.15, -1.58, -2.34, 0.51, -3.82, 0.72,
-6.25, -2.33, 3.1, 2.19, -2.63, 7.3, 1.82), X__5 = c(4.08, 1.58,
-0.02, 5.94, 3.33, -7.86, 5.88, 4.68, 5.99, 0.75, 2.68, 9.29,
8, 1.08, -3.13, 4.21, -3.35, -5.01, 5.77, 4.85, 2.73, -3.44,
0.27, 1.56, 1.62, -2.35, 2.93, -4.62, 2.36, 2.56, 0.86, 0.16,
-1.8, -2.04, 5.12, 2.72, 3.21, 1.21, -2.17, -1.84, 0.32, -3.63,
1.47, -5.16, -0.65, 3.33, 1.34, -1.36, 6.24, 1.19), X__6 = c(5.47,
-0.05, -0.79, 6.1, 2.54, -7.27, 3.84, 6.29, 4.46, -0.24, 2.42,
6.12, 8.63, 0.88, -3.31, 4.56, -2.14, -5.62, 5.73, 5.36, 2.44,
-1.88, 0.83, 0.65, 1.47, -1.81, 2.31, -4.48, 2.56, 2.69, 0.9,
0.34, -0.62, -1.58, 6.59, 0.86, 3.58, 1.92, -1.85, -2.79, 0.7,
-3.4, 1.26, -5.26, -1.18, 4.26, 1.35, -0.97, 6.66, 1.77), X__7 = c(3.9,
1.14, 0.18, 5.87, 2.12, -7.44, 4.61, 4.57, 6.14, -0.84, 4.22,
8.37, 7.44, 0.69, -4.26, 4.13, -2.24, -6.75, 5.81, 4.35, 1.98,
-2.87, 0.93, 0.61, 1.27, -2.18, 2.97, -4.09, 2.27, 2.96, 1.16,
-0.38, -2.37, -0.71, 5.53, 2.45, 1.3, 0.31, -0.47, -2.03, 0.14,
-3.26, 1.79, -5.5, -1.47, 4.18, 1.96, -1.35, 7.06, 1.69), X__8 = c(2.35,
1.51, -0.22, 3.82, 4.46, -7.13, 3.09, 5.01, 5.84, -1.05, 3.81,
7.54, 6.46, 0.71, -3.56, 4.42, -1.87, -4.52, 7.3, 3.66, 2.11,
-2.92, 2.25, 2.17, 1.32, -1.71, 3.17, -4.63, 2.59, 3.89, 0.49,
0.21, -1.71, -1.18, 4.95, 3.21, 1.41, 0.89, -1.02, -2.89, 0.59,
-2.67, 1.47, -4.62, -0.69, 4.07, 2.83, -1.44, 6.11, 1.58), X__9 = c(1.49,
1.52, 1.03, 3.34, 2.83, -7.76, 3.07, 3.72, 6.21, -1.66, 3.46,
6.14, 7.17, 2.13, -3.19, 4.59, -2.65, -3.5, 7.43, 3.5, 2.41,
-2.73, 1.35, 1.97, 1.72, -1.8, 4.06, -5.35, 2.57, 3.14, 1.89,
-0.86, -1.73, -0.95, 6.07, 1.73, 1.09, 0.37, -1.34, -2.48, 0.31,
-3.2, 1.34, -4.99, -0.18, 4.35, 3.03, 0.09, 5.65, 2.39), X__10 = c(2.27,
2.02, 0.12, 3.44, 3.82, -6.32, 1.41, 4.54, 5.55, -0.97, 3.8,
5.69, 5.65, 1.78, -2.6, 4.21, -1.29, -2.63, 7.15, 3.52, 1.85,
-2.32, 0.96, 2.74, 1.9, -2.6, 3.83, -4.31, 3.15, 2.76, 0.93,
-0.39, -1.86, -1.57, 7.05, 2.36, -0.33, -0.23, -0.54, -2.6, 0.61,
-2.37, 2.12, -3.76, 0.47, 3.98, 3.03, 0.2, 5.63, 1.26), X__11 = c(0.82,
1.12, 1.75, 3.97, 4.75, -5.04, 0.42, 4.96, 4.32, 0.25, 2.26,
4.71, 5.05, 1.63, -1.53, 5.12, -2.59, -1.92, 6.89, 4.48, -0.09,
-2.49, 0.26, 4.03, 1.37, -2.82, 4.95, -5.1, 3.4, 4.29, 0.89,
-1.06, -2.18, -0.31, 5.76, 3.32, -1.04, -0.63, -1.78, -2.97,
0.55, -1.3, 2.75, -4.47, 0.48, 4.83, 2.85, 0.27, 4.4, 1.93)), .Names = c("date",
"X1", "X2", "X3", "X4", "X5", "X6", "X7", "X8",
"X9", "X10"), row.names = c(NA, 50L), class = c("tbl_df",
"tbl", "data.frame"))
After a few correpondence via the comments with #Jonathan, I widened the example data from 3 columns to 12 columns with some sampling. And the code at the "With short-selling" section at the blog scales well for 10K observations:
# using code at:
# https://www.r-bloggers.com/a-gentle-introduction-to-finance-using-r-efficient-frontier-and-capm-part-1/
# https://datashenanigan.wordpress.com/2016/05/24/a-gentle-introduction-to-finance-using-r-efficient-frontier-and-capm-part-1/
library(data.table)
calcEFParams <- function(rets)
{
retbar <- colMeans(rets, na.rm = T)
covs <- var(rets, na.rm = T) # calculates the covariance of the returns
invS <- solve(covs)
i <- matrix(1, nrow = length(retbar))
alpha <- t(i) %*% invS %*% i
beta <- t(i) %*% invS %*% retbar
gamma <- t(retbar) %*% invS %*% retbar
delta <- alpha * gamma - beta * beta
retlist <- list(alpha = as.numeric(alpha),
beta = as.numeric(beta),
gamma = as.numeric(gamma),
delta = as.numeric(delta))
return(retlist)
}
# load data
link <- "https://raw.githubusercontent.com/DavZim/Efficient_Frontier/master/data/mult_assets.csv"
df <- data.table(read.csv(link))
df2 <- df[,lapply(.SD, sample),]
df3 <- cbind(df, df2)
df4 <- df3[,lapply(.SD, sample),]
df5 <- cbind(df3, df4)
Now loading the microbenchmark package, the performance is as such:
> library(microbenchmark)
> microbenchmark(calcEFParams(df5), times = 10)
Unit: milliseconds
expr min lq mean median uq max neval
calcEFParams(df5) 2.692514 2.764053 2.795127 2.777547 2.805447 3.024349 10
It seems that David Zimmermann's code is scalable and efficient enough!
I am having trouble extracting rownames from an object.
When I type in rownames(object), I obtain "null", but if I type in object, I obtain the matrix of information. If it helps, when I type in class(object), it tells me that it is a list. What I am looking for is a method to obtain the row names on the side. Thanks!
>
alpha84 alpha91 alpha98 alpha105 alpha112 alpha119
YBR088C 1.08 0.27 0.04 -0.51 -0.80 -0.89
YDL003W 0.62 -0.01 -0.36 -0.04 -0.55 -0.55
YDR097C 0.64 0.18 -0.05 0.03 -0.76 -0.66
YDR507C 0.53 0.13 0.07 0.14 -0.56 -0.41
YER070W 0.73 0.20 0.00 0.11 -0.53 -0.72
YER095W 0.28 -0.05 -0.11 -0.13 -0.87 -0.90
YER111C 0.37 -0.19 -0.11 -0.54 -0.34 -0.47
YGR189C 0.81 0.12 0.15 -0.39 -0.60 -1.20
YKL045W 0.46 -0.27 -0.10 -0.23 -0.42 -1.21
YLR183C 0.96 0.14 0.28 -0.17 -0.14 -0.68
YML027W 0.50 -0.01 0.11 -0.33 -0.44 -0.94
YMR179W 0.42 0.04 -0.40 -0.47 -0.12 -0.61
YNL300W 0.79 0.33 0.54 -0.09 -0.31 -1.01
YOR074C 0.73 0.09 -0.27 -0.22 -0.62 -0.80
YPL163C 1.61 0.84 0.82 -0.09 -0.48 -0.97
YPL256C 1.10 0.56 0.18 -0.32 -0.38 -1.04
structure(list(4 = structure(list(alpha0 = c(-1.15, -1.22,
-0.72, -1.76, -1.46, -0.57, -1.21, -0.32, -0.8, -1.7, -1.72,
-1.3, -1.24, -1.14, -2.42, -1.41), alpha7 = c(-0.86, -0.74, -0.85,
-0.34, -0.76, 0.42, -0.26, -0.65, 0.01, -1.46, -0.66, 0.07, -0.78,
-0.31, -2.15, -0.69), alpha14 = c(1.21, 1.34, 0.54, 0.18, 1.08,
1.03, 1.36, 0.87, 0.86, 0.93, 1.73, 0.98, 0.31, 0.57, 0.66, 1.39
), alpha21 = c(1.62, 1.5, 1.04, 1.07, 1.5, 1.35, 1.37, 1.1, 0.84,
1.12, 1.29, 1.12, 1.46, 1.08, 1.98, 1.98), alpha28 = c(1.12,
0.63, 0.84, 0.37, 0.74, 0.64, 0.54, 1.17, 0.51, 0.91, 0.51, 0.13,
1.11, 1.17, 1.55, 0.74), alpha35 = c(0.16, 0.29, 0.24, 0.32,
0.47, 0.42, 0.18, 0.44, 0.14, 0.11, 0.28, 0.19, 0.62, 0.57, 0.78,
0.21), alpha42 = c(-0.44, -0.55, -0.64, -0.5, -0.7, -0.4, -0.85,
0.37, -0.4, 0, 0.23, -0.58, 0.07, 0.31, 0.14, -0.36), alpha49 = c(-0.93,
-0.65, -0.83, -0.25, -0.68, -0.9, -0.82, -0.93, -0.64, -0.73,
-0.55, -0.63, -0.23, -0.74, -0.94, -1.32), alpha56 = c(-1.23,
-0.76, -0.36, -0.48, -1.03, -0.73, -0.75, -1.45, -0.8, -0.9,
-0.97, -0.9, -0.58, -0.68, -1.03, -1.5), alpha63 = c(-0.62, -0.88,
-0.7, -0.25, -0.55, -0.47, 0.07, -0.57, 0.41, -0.46, -0.48, 0.09,
-1.01, -0.1, -1.5, -1.07), alpha70 = c(0.62, 0.69, 0.99, 0.79,
0.35, 0.2, 0.89, 0.15, 0.88, 0.85, 0.57, 0.54, -0.24, -0.38,
-0.03, 0.35), alpha77 = c(1.3, 1.25, 1.08, 0.97, 1.24, 0.78,
0.78, 0.92, 0.75, 0.93, 0.88, 1.44, 0.23, 0.75, 1.25, 1.57),
alpha84 = c(1.08, 0.62, 0.64, 0.53, 0.73, 0.28, 0.37, 0.81,
0.46, 0.96, 0.5, 0.42, 0.79, 0.73, 1.61, 1.1), alpha91 = c(0.27,
-0.01, 0.18, 0.13, 0.2, -0.05, -0.19, 0.12, -0.27, 0.14,
-0.01, 0.04, 0.33, 0.09, 0.84, 0.56), alpha98 = c(0.04, -0.36,
-0.05, 0.07, 0, -0.11, -0.11, 0.15, -0.1, 0.28, 0.11, -0.4,
0.54, -0.27, 0.82, 0.18), alpha105 = c(-0.51, -0.04, 0.03,
0.14, 0.11, -0.13, -0.54, -0.39, -0.23, -0.17, -0.33, -0.47,
-0.09, -0.22, -0.09, -0.32), alpha112 = c(-0.8, -0.55, -0.76,
-0.56, -0.53, -0.87, -0.34, -0.6, -0.42, -0.14, -0.44, -0.12,
-0.31, -0.62, -0.48, -0.38), alpha119 = c(-0.89, -0.55, -0.66,
-0.41, -0.72, -0.9, -0.47, -1.2, -1.21, -0.68, -0.94, -0.61,
-1.01, -0.8, -0.97, -1.04)), .Names = c("alpha0", "alpha7",
"alpha14", "alpha21", "alpha28", "alpha35", "alpha42", "alpha49",
"alpha56", "alpha63", "alpha70", "alpha77", "alpha84", "alpha91",
"alpha98", "alpha105", "alpha112", "alpha119"), row.names = c("YBR088C",
"YDL003W", "YDR097C", "YDR507C", "YER070W", "YER095W", "YER111C",
"YGR189C", "YKL045W", "YLR183C", "YML027W", "YMR179W", "YNL300W",
"YOR074C", "YPL163C", "YPL256C"), class = "data.frame")), .Names = "4")
You have a list of one element. This single element is a data.frame.
If you are after the rownames from this object, then index the list appropriately
rownames(object[[1]])
## [1] "YBR088C" "YDL003W" "YDR097C" "YDR507C" "YER070W" "YER095W" "YER111C" "YGR189C" "YKL045W" "YLR183C" "YML027W"
## [12] "YMR179W" "YNL300W" "YOR074C" "YPL163C" "YPL256C"
For a more general list of data.frames
# get rownames from all data.frames in a list
lapply(object, rownames)
If you want a data.frame, not a list containing a data.frame then you could simply assign the results from the first element to a separate element
object.df <- object[[1]]
If it is a list of data.frames, it probably is more idiomatically R to keep it in the list, and use lapply to work on each element.