Unable to evaluate simple expression in Julia - julia

I am trying to plot portfolio mean vs standard deviation (scatter plots) for several different values of the correlation between stock returns. However for each different value of the correlation coefficient, I can't seem to be able to get a different value of the standard deviation vector.
rA = .05; sA = .00; lA = "A" # mean and std.dev. and label of stock A returns
rB = .10; sB = .38; lB = "B" # mean and std.dev. and label of stock B returns
# several possible values of the correlation (There should be one curve for each value)
rrab1 = 1
rrab2 = .5
rrab3 = 0 # and other such values
# Create portfolio weight vectors (Needed to generate the curves)
x = collect(0 : .05 : 1) # weight on stock A (a vector)
y = 1 .-x # weight on stock B (a vector)
# Evaluate the portfolio mean and the portfolio std.dev.
rab1 = rA*x + rB*y # Portfolio return
sab1 = sqrt.(sA^2*x.^2 + sB^2*y.^2 - 2*rrab1*sA*sB .*x .* y) # Portfolio std.dev. for correl = 1
rab2 = rA*x + rB*y
sab2 = sqrt.(sA^2*x.^2 + sB^2*y.^2 - 2*rrab2*sA*sB .*x .* y ) # Portfolio std.dev. for correl = .5
rab3 = rA*x + rB*y
sab3 = sqrt.(sA^2*x.^2 + sB^2*y.^2 - 2*rrab3*sA*sB .*x .* y ) # Portfolio std.dev. for correl = 1
and so on for other possible values of correlation.
I want to plot the all the 3 curves in the same r-s axes using PGFPlotsX. But I am getting just a single plot, that for rrab1.
If I check my std.dev. vectors I find that I am unable to get different vectors.
sab1 == sab2 == sab3 # outputs true
using DataFrames
s = DataFrame()
s.x = sab1
s.y = sab3
s
outputs:
21 rows × 2 columns
x y
Float64 Float64
1 0.38 0.38
2 0.361 0.361
3 0.342 0.342
4 0.323 0.323
5 0.304 0.304
6 0.285 0.285
7 0.266 0.266
8 0.247 0.247
9 0.228 0.228
10 0.209 0.209
11 0.19 0.19
12 0.171 0.171
13 0.152 0.152
14 0.133 0.133
15 0.114 0.114
16 0.095 0.095
17 0.076 0.076
18 0.057 0.057
19 0.038 0.038
20 0.019 0.019
21 0.0 0.0
That is, all the s vectors are the same although the correlations are different.
I am new to Julia and must be missing something elementary.

I think Bogumil's comment should be an answer - this isn't really a Julia issue but follows directly from your assumptions. You start from:
sA = .00
and then calculate:
sab1 = sqrt.(sA^2*x.^2 + sB^2*y.^2 - 2*rrab1*sA*sB .*x .* y)
sab2 = sqrt.(sA^2*x.^2 + sB^2*y.^2 - 2*rrab2*sA*sB .*x .* y )
sab3 = sqrt.(sA^2*x.^2 + sB^2*y.^2 - 2*rrab3*sA*sB .*x .* y )
Simplifying this by using the fact that sA = 0.0 and therefore all terms multiplied by sA vanish:
sab1 = sqrt.(sB^2*y.^2)
sab2 = sqrt.(sB^2*y.^2)
sab3 = sqrt.(sB^2*y.^2)
and indeed
julia> sqrt.(sB^2 * y.^2)
21-element Vector{Float64}:
0.38
0.361
0.342
0.323
0.304
0.28500000000000003
0.26599999999999996
0.24700000000000003
0.22799999999999998
0.20900000000000002
0.19
0.17099999999999999
0.152
0.13299999999999998
0.11400000000000002
0.095
0.07599999999999998
0.05700000000000001
0.03799999999999999
0.019000000000000017
0.0

Related

How to get a scientific p-value using the cmprsk package?

Hi stackoverflow community,
I am a recent R starter and today I tried several hours to figure out how to get a scientific p-value (e.g. 3*e-1) from a competing risk analysis using the cmprsk package.
I used:
sumary_J1<-crr(ftime, fstatus, cov1, failcode=2)
summary(sumary_J1)
And got
Call:
crr(ftime = ftime, fstatus = fstatus, cov1 = cov1, failcode = 2)
coef exp(coef) se(coef) z p-value
group1 0.373 1.45 0.02684 13.90 0.00
age 0.122 1.13 0.00384 31.65 0.00
sex 0.604 1.83 0.04371 13.83 0.00
bmi 0.012 1.01 0.00611 1.96 0.05
exp(coef) exp(-coef) 2.5% 97.5%
group1 1.45 0.689 1.38 1.53
age 1.13 0.886 1.12 1.14
sex 1.83 0.546 1.68 1.99
bmi 1.01 0.988 1.00 1.02
Num. cases = 470690 (1900 cases omitted due to missing values)
Pseudo Log-likelihood = -28721
Pseudo likelihood ratio test = 2229 on 4 df,
I can see the p-value column,but I only get two decimal places. I would like to see as many decimal places as possible or print those p-values in the format e.g. 3.0*e-3.
I tried all of those, but nothing worked so far:
summary(sumary_J1, digits=max(options()$digits - 5,10))
print.crr(sumary_J1, digits = 20)
print.crr(sumary_J1, digits = 3, scipen = -2)
print.crr(sumary_J1, format = "e", digits = 3)
Maybe someone is able to help me! Thanks!
Best,
Carolin
The use of digits=2 limits the number of digits to the right of the decimal point when used as an argument to a .summary value. The digits parameter does affect how results are displayed for summary.crr.
summary(z, digits=3) # using first example in `?cmprsk::crr`
#----------------------
#Competing Risks Regression
Call:
crr(ftime = ftime, fstatus = fstatus, cov1 = cov)
coef exp(coef) se(coef) z p-value
x1 0.2668 1.306 0.421 0.633 0.526
x2 -0.0557 0.946 0.381 -0.146 0.884
x3 0.2805 1.324 0.381 0.736 0.462
exp(coef) exp(-coef) 2.5% 97.5%
x1 1.306 0.766 0.572 2.98
x2 0.946 1.057 0.448 2.00
x3 1.324 0.755 0.627 2.79
Num. cases = 200
Pseudo Log-likelihood = -320
Pseudo likelihood ratio test = 1.02 on 3 df,
You can use formatC to control format:
formatC( summary(z, digits=5)$coef , format="e")
#------------>
coef exp(coef) se(coef) z p-value
x1 "2.6676e-01" "1.3057e+00" "4.2115e-01" "6.3340e-01" "5.2647e-01"
x2 "-5.5684e-02" "9.4584e-01" "3.8124e-01" "-1.4606e-01" "8.8387e-01"
x3 "2.8049e-01" "1.3238e+00" "3.8098e-01" "7.3622e-01" "4.6159e-01"
You also might search on [r] very small p-value
Here's the first of over 100 hits on that topic, which despite not very much attention, still has very useful information and coding examples: Reading a very small p-value in R
By looking at the function that prints the output of crr() (cmprsk::print.crr) you can see what is done to create the p-values displayed in the summary. The code below is taken from that function.
x <- sumary_J1
v <- sqrt(diag(x$var))
signif(v, 4) # Gives you the one-sided p-values.
v <- 2 * (1 - pnorm(abs(x$coef)/v))
signif(v, 4) # Gibes you the two-sided p-values.

How do I include p-value and R-square for the estimates in semPaths?

I am using semPaths (semPlot package) to draw my structural equation models. After some trial and error, I have a pretty good script to show what I want. Except, I haven’t been able to figure out how to include the p-value/significance levels of the estimates/regression coefficients in the figure.
Can/how can I include significance levels either as e.g. p-value in the edge labels below the estimate or as a broken line for insignificance or …?
I am also interested in including the R-square, but not as critically as the significance level.
This is the script I am using so far:
semPaths(fitmod.bac.class2,
what = "std",
whatLabels = "std",
style="ram",
edge.label.cex = 1.3,
layout = 'tree',
intercepts=FALSE,
residuals=FALSE,
nodeLabels = c("Negati-\nvicutes","cand_class\n_MB_A2_108", "CO2", "Bacilli","Ignavi-\nbacteria","C/N", "pH","Water\ncontent"),
sizeMan=7 )
Example of one of the SemPath outputs
In this example the following are not significant:
Ignavibacteria -> First_C_CO2_ugC_gC_day, p = 0.096
pH -> Ignavibacteria, p = 0.151
cand_class_MB_A2_108 <-> Bacilli correlation, p = 0.054
I am a R-user and not really a coder, so I might just be missing a crucial point in the arguments.
I am testing a lot of different models at the moment, and would really like not to have to draw them all up by hand.
update:
Using semPlotModel: Am I right in understanding that semPlotModel doesn’t include the significance levels from the sem function (see my script and output below)? I am specifically looking to include the P(>|z|) for regressions and covariance.
Is it just me that is missing that, or is it not included? If it is not included, my solution is simply just to custom the edge labels.
{model.NA.UP.bac.class2 <- '
#LATANT VARIABLES
#REGRESSIONS
#soil organic carbon quality
c_Negativicutes ~ CN
#microorganisms
First_C_CO2_ugC_gC_day ~ c_Bacilli
First_C_CO2_ugC_gC_day ~ c_Ignavibacteria
First_C_CO2_ugC_gC_day ~ c_cand_class_MB_A2_108
First_C_CO2_ugC_gC_day ~ c_Negativicutes
#pH
c_Bacilli ~pH
c_Ignavibacteria ~pH
c_cand_class_MB_A2_108~pH
c_Negativicutes ~pH
#COVARIANCE
initial_water ~~ CN
c_cand_class_MB_A2_108 ~~ c_Bacilli
'
fitmod.bac.class2 <- sem(model.NA.UP.bac.class2, data=datapNA.UP.log, missing="ml", meanstructure=TRUE, fixed.x=FALSE, std.lv=FALSE, std.ov=FALSE)
summary(fitmod.bac.class2, standardized=TRUE, fit.measures=TRUE, rsq=TRUE)
out <- capture.output(summary(fitmod.bac.class2, standardized=TRUE, fit.measures=TRUE, rsq=TRUE))
}
Output:
lavaan 0.6-5 ended normally after 188 iterations
Estimator ML
Optimization method NLMINB
Number of free parameters 28
Number of observations 30
Number of missing patterns 1
Model Test User Model:
Test statistic 17.816
Degrees of freedom 16
P-value (Chi-square) 0.335
Model Test Baseline Model:
Test statistic 101.570
Degrees of freedom 28
P-value 0.000
User Model versus Baseline Model:
Comparative Fit Index (CFI) 0.975
Tucker-Lewis Index (TLI) 0.957
Loglikelihood and Information Criteria:
Loglikelihood user model (H0) 472.465
Loglikelihood unrestricted model (H1) 481.373
Akaike (AIC) -888.930
Bayesian (BIC) -849.697
Sample-size adjusted Bayesian (BIC) -936.875
Root Mean Square Error of Approximation:
RMSEA 0.062
90 Percent confidence interval - lower 0.000
90 Percent confidence interval - upper 0.185
P-value RMSEA <= 0.05 0.414
Standardized Root Mean Square Residual:
SRMR 0.107
Parameter Estimates:
Information Observed
Observed information based on Hessian
Standard errors Standard
Regressions:
Estimate Std.Err z-value P(>|z|) Std.lv Std.all
c_Negativicutes ~
CN 0.419 0.143 2.939 0.003 0.419 0.416
c_cand_class_MB_A2_108 ~
CN -0.433 0.160 -2.707 0.007 -0.433 -0.394
First_C_CO2_ugC_gC_day ~
c_Bacilli 0.525 0.128 4.092 0.000 0.525 0.496
c_Ignavibacter 0.207 0.124 1.667 0.096 0.207 0.195
c_c__MB_A2_108 0.310 0.125 2.475 0.013 0.310 0.301
c_Negativicuts 0.304 0.137 2.220 0.026 0.304 0.271
c_Bacilli ~
pH 0.624 0.135 4.604 0.000 0.624 0.643
c_Ignavibacteria ~
pH 0.245 0.171 1.436 0.151 0.245 0.254
c_cand_class_MB_A2_108 ~
pH 0.393 0.151 2.597 0.009 0.393 0.394
c_Negativicutes ~
pH 0.435 0.129 3.361 0.001 0.435 0.476
Covariances:
Estimate Std.Err z-value P(>|z|) Std.lv Std.all
CN ~~
initial_water 0.001 0.000 2.679 0.007 0.001 0.561
.c_cand_class_MB_A2_108 ~~
.c_Bacilli -0.000 0.000 -1.923 0.054 -0.000 -0.388
Intercepts:
Estimate Std.Err z-value P(>|z|) Std.lv Std.all
.c_Negativicuts 0.145 0.198 0.734 0.463 0.145 3.826
.c_c__MB_A2_108 1.038 0.226 4.594 0.000 1.038 25.076
.Frs_C_CO2_C_C_ -0.346 0.233 -1.485 0.137 -0.346 -8.115
.c_Bacilli 0.376 0.135 2.778 0.005 0.376 9.340
.c_Ignavibacter 0.754 0.170 4.424 0.000 0.754 18.796
CN 0.998 0.007 145.158 0.000 0.998 26.502
pH 0.998 0.008 131.642 0.000 0.998 24.034
initial_water 0.998 0.008 125.994 0.000 0.998 23.003
Variances:
Estimate Std.Err z-value P(>|z|) Std.lv Std.all
.c_Negativicuts 0.001 0.000 3.873 0.000 0.001 0.600
.c_c__MB_A2_108 0.001 0.000 3.833 0.000 0.001 0.689
.Frs_C_CO2_C_C_ 0.001 0.000 3.873 0.000 0.001 0.408
.c_Bacilli 0.001 0.000 3.873 0.000 0.001 0.586
.c_Ignavibacter 0.002 0.000 3.873 0.000 0.002 0.936
CN 0.001 0.000 3.873 0.000 0.001 1.000
initial_water 0.002 0.000 3.873 0.000 0.002 1.000
pH 0.002 0.000 3.873 0.000 0.002 1.000
R-Square:
Estimate
c_Negativicuts 0.400
c_c__MB_A2_108 0.311
Frs_C_CO2_C_C_ 0.592
c_Bacilli 0.414
c_Ignavibacter 0.064
Warning message:
In lav_model_hessian(lavmodel = lavmodel, lavsamplestats = lavsamplestats, :
lavaan WARNING: Hessian is not fully symmetric. Max diff = 5.15131396241486e-05
This example is taken from ?semPaths since we don't have your object.
library('semPlot')
modFile <- tempfile(fileext = '.OUT')
download.file('http://sachaepskamp.com/files/mi1.OUT', modFile)
Use semPlotModel to get the object without plotting. There you can inspect what is to be plotted. I just dug around without reading the docs until I found what it seems to be using.
After you run semPlotModel, the object has an element x#Pars which contains the edges, nodes, and the std which is being used for the edge labels in your case. semPaths also has an argument that allows you to make custom edge labels, so you can take the data you need from x#Pars and add your p-values:
x <- semPlotModel(modFile)
x#Pars
# label lhs edge rhs est std group fixed par
# 1 lambda[11]^{(y)} perfIQ -> pc 1.000 0.6219648 Group 1 TRUE 0
# 2 lambda[21]^{(y)} perfIQ -> pa 0.923 0.5664888 Group 1 FALSE 1
# 3 lambda[31]^{(y)} perfIQ -> oa 1.098 0.6550159 Group 1 FALSE 2
# 4 lambda[41]^{(y)} perfIQ -> ma 0.784 0.4609990 Group 1 FALSE 3
# 5 theta[11]^{(epsilon)} pc <-> pc 5.088 0.6131598 Group 1 FALSE 5
# 10 theta[22]^{(epsilon)} pa <-> pa 5.787 0.6790905 Group 1 FALSE 6
# 15 theta[33]^{(epsilon)} oa <-> oa 5.150 0.5709541 Group 1 FALSE 7
# 20 theta[44]^{(epsilon)} ma <-> ma 7.311 0.7874800 Group 1 FALSE 8
# 21 psi[11] perfIQ <-> perfIQ 3.210 1.0000000 Group 1 FALSE 4
# 22 tau[1]^{(y)} int pc 10.500 NA Group 1 FALSE 9
# 23 tau[2]^{(y)} int pa 10.374 NA Group 1 FALSE 10
# 24 tau[3]^{(y)} int oa 10.663 NA Group 1 FALSE 11
# 25 tau[4]^{(y)} int ma 10.371 NA Group 1 FALSE 12
# 11 lambda[11]^{(y)} perfIQ -> pc 1.000 0.6515609 Group 2 TRUE 0
# 27 lambda[21]^{(y)} perfIQ -> pa 0.923 0.5876948 Group 2 FALSE 1
# 31 lambda[31]^{(y)} perfIQ -> oa 1.098 0.6981974 Group 2 FALSE 2
# 41 lambda[41]^{(y)} perfIQ -> ma 0.784 0.4621919 Group 2 FALSE 3
# 51 theta[11]^{(epsilon)} pc <-> pc 5.006 0.5754684 Group 2 FALSE 14
# 101 theta[22]^{(epsilon)} pa <-> pa 5.963 0.6546148 Group 2 FALSE 15
# 151 theta[33]^{(epsilon)} oa <-> oa 4.681 0.5125204 Group 2 FALSE 16
# 201 theta[44]^{(epsilon)} ma <-> ma 8.356 0.7863786 Group 2 FALSE 17
# 211 psi[11] perfIQ <-> perfIQ 3.693 1.0000000 Group 2 FALSE 13
# 221 tau[1]^{(y)} int pc 10.500 NA Group 2 FALSE 9
# 231 tau[2]^{(y)} int pa 10.374 NA Group 2 FALSE 10
# 241 tau[3]^{(y)} int oa 10.663 NA Group 2 FALSE 11
# 251 tau[4]^{(y)} int ma 10.371 NA Group 2 FALSE 12
# 26 alpha[1] int perfIQ -2.469 NA Group 2 FALSE 18
As you can see there are more edge labels than ones that are plotted, and I have no idea how it chooses which to use, so I am just taking the first four from each group (since there are four edges shown and the stds match those. Maybe there is an option to plot all of them or select which ones you need--I haven't read the docs.
## take first four stds from each group, generate some p-values
l <- sapply(split(x#Pars$std, x#Pars$group), function(x) head(x, 4))
set.seed(1)
l <- sprintf('%.3f, p=%s', l, format.pval(runif(length(l)), digits = 2))
l
# [1] "0.622, p=0.27" "0.566, p=0.37" "0.655, p=0.57" "0.461, p=0.91" "0.652, p=0.20" "0.588, p=0.90" "0.698, p=0.94" "0.462, p=0.66"
Then you can plot the object with your new labels, edgeLabels = l
layout(1:2)
semPaths(
x,
edgeLabels = l,
ask = FALSE, title = FALSE,
what = 'std',
whatLabels = 'std',
style = 'ram',
edge.label.cex = 1.3,
layout = 'tree',
intercepts = FALSE,
residuals = FALSE,
sizeMan = 7
)
With the help from #rawr, I have worked it out. If anybody else needs to include estimates and p-value from Lavaan in their semPaths, here is how it can be done.
#extracting the parameters from the sem model and selecting the interactions relevant for the semPaths (here, I need 12 estimates and p-values)
table2<-parameterEstimates(fitmod.bac.class2,standardized=TRUE) %>% head(12)
#turning the chosen parameters into text
b<-gettextf('%.3f \n p=%.3f', table2$std.all, digits=table2$pvalue)
I can honestly say that I do not understand how the last bit of script works. This is copied from rawr's answer before a lot of trial and error until it worked. There might (quite possibly) be a nicer way to write it, but it works :)
#putting that list into edgeLabels in sempaths
semPaths(fitmod.bac.class2,
what = "std",
edgeLabels = b,
style="ram",
edge.label.cex = 1,
layout = 'tree',
intercepts=FALSE,
residuals=FALSE,
nodeLabels = c("Negati-\nvicutes","cand_class\n_MB_A2_108", "CO2", "Bacilli","Ignavi-\nbacteria","C/N", "pH","Water\ncontent"),
sizeMan=7
)
Just a small, but relevant detail for an improvement for the above answer.
The above code requires an inspection of the parameter table to count how many lines to maintain to specify as in %>%head(4).
We can exclude from the extracted parameter table those lines which lhs and rhs are not equal.
#extracting the parameters from the sem model and selecting the interactions relevant for the semPaths
table2<-parameterEstimates(fitmod.bac.class2,standardized=TRUE)%>%as.dataframe()
table2<-table2[!table2$lhs==table2$rhs,]
If the formula comprised also extra lines as those with ':=' those also will comprise the parameter table, and should be removed.
The remaining keeps the same...
#turning the chosen parameters into text
b<-gettextf('%.3f \n p=%.3f', table2$std.all, digits=table2$pvalue)
#putting that list into edgeLabels in sempaths
semPaths(fitmod.bac.class2,
what = "std",
edgeLabels = b,
style="ram",
edge.label.cex = 1,
layout = 'tree',
intercepts=FALSE,
residuals=FALSE,
nodeLabels = c("Negati-\nvicutes","cand_class\n_MB_A2_108", "CO2", "Bacilli","Ignavi-\nbacteria","C/N", "pH","Water\ncontent"),
sizeMan=7
)

How to calculate the Bonferroni Lower and Upper limits in R?

With the following data, I am trying to calculate the Chi Square and Bonferroni lower and upper Confidence intervals. The column "Data_No" identifies the dataset (as calculations needs to be done separately for each dataset).
Data_No Area Observed
1 3353 31
1 2297 2
1 1590 15
1 1087 16
1 817 2
1 847 10
1 1014 28
1 872 29
1 1026 29
1 1215 21
2 3353 31
2 2297 2
2 1590 15
3 1087 16
3 817 2
The code I used is
library(dplyr)
setwd("F:/GIS/July 2019/")
total_data <- read.csv("test.csv")
result_data <- NULL
for(i in unique(total_data$Data_No)){
data <- total_data[which(total_data$Data_No == i),] data <- data %>%
mutate(RelativeArea = Area/sum(Area), Expected = RelativeArea*sum(Observed), OminusE = Observed-Expected, O2 = OminusE^2, O2divE = O2/Expected, APU = Observed/sum(Observed), Alpha = 0.05/2*count(Data_No),
Zvalue = qnorm(Alpha,lower.tail=FALSE), lower = APU-Zvalue*sqrt(APU*(1-APU)/sum(Observed)), upper = APU+Zvalue*sqrt(APU*(1-APU)/sum(Observed)))
result_data <- rbind(result_data,data) }
write.csv(result_data,file='final_result.csv')
And the error message I get is:
Error in UseMethod("summarise_") : no applicable method for
'summarise_' applied to an object of class "c('integer', 'numeric')"
The column that I am calling "Alpha" is the alpha value of 0.05/2k, where K is the number of categories - in my example, I have 10 categories ("Data_No" column) for the first dataset, so "Alpha" needs to be 0.05/20 = 0.0025, and it's corresponding Z value is 2.807. The second dataset has 3 categories (so 0.05/6) and the third has 2 categories (0.05/4) in my example table (Data_No" column). Using the values from the newly calculated "Alpha" column, I then need to calculate the ZValue column (Zvalue = qnorm(Alpha,lower.tail=FALSE)) which I then use to calculate the lower and upper confidence intervals.
From the above data, here are the results that I should get, but note that I have had to manually calculate Alpha column and Zvalue, rather than insert those calculations within the R code:
Data_No Area Observed RelativeArea Alpha Z value lower upper
1 3353 31 0.237 0.003 2.807 0.092 0.247
1 2297 2 0.163 0.003 2.807 -0.011 0.033
1 1590 15 0.113 0.003 2.807 0.025 0.139
1 1087 16 0.077 0.003 2.807 0.029 0.146
1 817 2 0.058 0.003 2.807 -0.011 0.033
1 847 10 0.060 0.003 2.807 0.007 0.102
1 1014 28 0.072 0.003 2.807 0.078 0.228
1 872 29 0.062 0.003 2.807 0.083 0.234
1 1026 29 0.073 0.003 2.807 0.083 0.234
1 1215 21 0.086 0.003 2.807 0.049 0.181
2 3353 31 0.463 0.008 2.394 0.481 0.811
2 2297 2 0.317 0.008 2.394 -0.027 0.111
2 1590 15 0.220 0.008 2.394 0.152 0.473
3 1087 16 0.571 0.013 2.241 0.723 1.055
3 817 2 0.429 0.013 2.241 -0.055 0.277
Please note that I only included some of the columns generated from the code.
# You need to check the closing bracket for lower c.f. sqrt value. Following code should work.
data <- read.csv("test.csv")
data <- data %>% mutate(RelativeArea =
Area/sum(Area), Expected = RelativeArea*sum(Observed), OminusE =
Observed-Expected, O2 = OminusE^2, O2divE = O2/Expected, APU =
Observed/sum(Observed), lower =
APU-2.394*sqrt(APU*(1-APU)/sum(Observed)), upper =
APU+2.394*sqrt(APU*(1-APU)/sum(Observed)))
#Answer to follow-up question.
#Sample Data
Data_No Area Observed
1 3353 31
1 2297 2
2 1590 15
2 1087 16
#Code to run
total_data <- read.csv("test.csv")
result_data <- NULL
for(i in unique(total_data$Data_No)){
data <- total_data[which(total_data$Data_No == i),]
data <- data %>% mutate(RelativeArea =
Area/sum(Area), Expected = RelativeArea*sum(Observed), OminusE =
Observed-Expected, O2 = OminusE^2, O2divE = O2/Expected, APU =
Observed/sum(Observed), lower =
APU-2.394*sqrt(APU*(1-APU)/sum(Observed)), upper =
APU+2.394*sqrt(APU*(1-APU)/sum(Observed)))
result_data <- rbind(result_data,data)
}
write.csv(result_data,file='final_result.csv')
#Issue in calculating Alpha. I have updated the code.
library(dplyr)
setwd("F:/GIS/July 2019/")
total_data <- read.csv("test.csv")
#Creating the NO_OF_CATEGORIES column based on your question.
total_data$NO_OF_CATEGORIES <- 0
total_data[which(total_data$Data_No==1),]$NO_OF_CATEGORIES <- 10
total_data[which(total_data$Data_No==2),]$NO_OF_CATEGORIES <- 3
total_data[which(total_data$Data_No==3),]$NO_OF_CATEGORIES <- 2
#Actual code
result_data <- NULL
for(i in unique(total_data$Data_No)){
data <- total_data[which(total_data$Data_No == i),]
data <- data %>%
mutate(RelativeArea = Area/sum(Area), Expected = RelativeArea*sum(Observed), OminusE = Observed-Expected, O2 = OminusE^2, O2divE = O2/Expected, APU = Observed/sum(Observed), Alpha = 0.05/(2*(unique(data$NO_OF_CATEGORIES))),
Zvalue = qnorm(Alpha,lower.tail=FALSE), lower = APU-Zvalue*sqrt(APU*(1-APU)/sum(Observed)), upper = APU+Zvalue*sqrt(APU*(1-APU)/sum(Observed)))
result_data <- rbind(result_data,data) }
write.csv(result_data,file='final_result.csv')

How to get the overlap of two missing variables in R (similar to correlation matrix)

I want to use a simpler way to get the overlap of two missing variables and construct a heatmap similar to correlation matrix. The data I have is as below:
set.seed(123)
data = data.frame(id = 1:1000, age_missing = sample(c(0,1),1000, replace = TRUE), salary_missing = sample(c(0,1),1000, replace = TRUE),
address_missing = sample(c(0,1),1000, replace = TRUE),
gender_missing =sample(c(0,1),1000, replace = TRUE) )
The ideal output is
|var1 | var2| Missing Percent|
------------------------------
age age 0.5
age gender 0.05
age address 0.08
gender gender 0.15
gender age 0.05
Maybe something along the lines of
dd <- as.matrix(data[,2:5])
crossprod(dd) / nrow(dd)
which yields
age_missing salary_missing address_missing
age_missing 0.493 0.231 0.251
salary_missing 0.231 0.497 0.248
address_missing 0.251 0.248 0.494
gender_missing 0.244 0.271 0.247
gender_missing
age_missing 0.244
salary_missing 0.271
address_missing 0.247
gender_missing 0.506

Replicating the Example of Error Correction Model (ECM) in R?

I have used EViews and run Error Correction Model (ECM) and obtained some result. Now I would like to replicate that example in R and obtain the same result like I did in EViews.
The data I have used are as follows:
gdp<-c(6592.694,7311.75,7756.11,8374.175,9169.984,9994.071,10887.682,11579.432,12440.625,13582.799,15261.26,17728.673,21899.262,29300.921,34933.51,39768.017,42647.701,51144.915,61554.743,73407.498,81467.464,70500.215,70682.449,71496.768,67403.443,68781.085,98203.625,123083.47,131969.428,131738.237,164753.092,172008.565,193073.835,188423.703,201444.061,238561.784,234676.457,207826.099,213329.585,212301.777,192070.75,191678.678,207537.337,253945.777,291430.382,304983.602,324954.402,375041.784,414173.646,381775.165,376575.382)
life<-c(68.58560976,69.57731707,69.3095122,69.44365854,69.92195122,69.72219512,70.04585366,69.91780488,70.05756098,69.83317073,69.89073171,70.06926829,70.41365854,70.97926829,70.96243902,71.08414634,71.55121951,71.89536585,71.96707317,72.28731707,72.42365854,72.75804878,72.89707317,72.96853659,73.52756098,73.74512195,74.22292683,74.66926829,75.14414634,75.24804878,75.53,75.56780488,75.85536585,76.10634146,76.45707317,76.71560976,76.98365854,77.38756098,77.57317073,77.77560976,78.02682927,78.52682927,78.67804878,78.63170732,79.1804878,79.33170732,79.83170732,79.98292683,80.23414634,80.08292683,80.38292683)
The result I obtained in EViews and would like to replicate in R is shown in the following table:
I used package apt and its function ecmAsyFit(gdp, life, lag = 1, split = TRUE,model = "linear", thresh = 0)
The code I have run is as follows:
df <- ts(cbind(gdp, life), start = 1950, freq = 1)
fit <- ecmAsyFit(df[, 1], df[, 2], lag = 1, split = TRUE, model = "linear", thresh = 0)
summary(fit)
The result in R I obtained is shown under:
DepVar IndVar estimate error t.value p.value signif
1 diff.df[, 2].t_0 | (Intercept) 0.324 0.063 5.135 0.000 ***
2 | X.diff.df[, 2].t_1.pos -0.458 0.155 -2.954 0.005 ***
3 | X.diff.df[, 2].t_1.neg 0.443 0.546 0.811 0.422
4 | X.diff.df[, 1].t_1.pos 0.000 0.000 1.410 0.166
5 | X.diff.df[, 1].t_1.neg 0.000 0.000 -1.475 0.148 .
6 | X.ECT.t_1.pos 0.000 0.000 -1.819 0.076 *
7 | X.ECT.t_1.neg 0.000 0.000 -0.420 0.677
8 diff.df[, 1].t_0 - (Intercept) 3793.752 4912.683 0.772 0.444
9 - X.diff.df[, 2].t_1.pos -4510.643 12060.505 -0.374 0.710
10 - X.diff.df[, 2].t_1.neg -21884.942 42483.319 -0.515 0.609
11 - X.diff.df[, 1].t_1.pos 0.576 0.190 3.031 0.004 ***
12 - X.diff.df[, 1].t_1.neg 0.055 0.369 0.148 0.883
13 - X.ECT.t_1.pos -0.318 0.145 -2.193 0.034 **
14 - X.ECT.t_1.neg -0.175 0.130 -1.354 0.183
Problem:
The result I obtained in EViews (shown table) is not the same like it is R (shown right above)
Questions:
Is there a way to use ecmAsyFit function to obtain result like it is shown in the table?
Is there another function (or set of functions) which can run Error Correction Model and provide the result shown in the table?
Could someone share the code for application of Error Correction Model, so I could apply it?
I think that it should be the following:
df <- ts(cbind(gdp, life), start = 1950, freq = 1)
# Symmetrical VECM
fit1<- ecmSymFit(df[, 1], df[, 2], lag = 1)
#Setting split=FALSE makes the co-integration symmetrical.
fit2<- ecmAsyFit(df[, 1], df[, 2], lag = 1, split =FALSE, model = "linear", thresh = 0)
summary(fit)

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