I ran this code to find the norm of some fundamnetal units of a biqaudratic number field, but I faced following problem
for (q=5, 200, for(p=q+1, 200, if (isprime(p)==1 && isprime(q)==1 ,k1=bnfinit(y^2-2*p,1); k2=bnfinit(y^2-q,1); k3=bnfinit(y^2-2*p*q,1); ep1=k1[8][5][1]; ep2=k2[8][5][1]; ep3=k3[8][5][1]; normep1=nfeltnorm(k1,ep1); normep2=nfeltnorm(k2,ep2); normep3=nfeltnorm(k3,ep3); li=[[q,p], [normep1, normep2, normep3]]; lis4=concat(lis4,[li]))))
and it works for small p and q. However, when I ran that for p and q greater than 150, it gives the following error:
First, I didn't use the flag=1 for bnf, but after adding that, still I get the same error.
Please, do not use indexing like ...[8][5][1] to get the fundamental units (FU). It seems that bnfinit omits FU matrix for some p and q. Instead, use the member function fu to receive FU. Please, find the example below:
> [q, p] = [23, 109];
> k = bnfinit(y^2 - 2*p*q, 1);
> k[8][5]
[;]
> k[8][5][1] \\ you will get the error here trying to index the empty matrix.
...
incorrect type in _[_] OCcompo1 [not a vector] (t_MAT).
> k.fu
[Mod(-355285121749346859670064114879166870*y - 25157598731408198132266996072608016699, y^2 - 5014)]
> norm(k.fu[1])
1
Related
I am getting error mentioned in the title and didn't find yet a solution.
X = train[feats].values
y = train['Target'].values
cv = StratifiedKFold(n_splits=3, random_state=2021, shuffle=True)
model = LogisticRegression(solver='liblinear')
scores = []
for train_idx, test_idx in cv.split(X, y):
model.fit(X[train_idx], y[train_idx])
y_pred = model.predict(X[test_idx])
score = mean_absolute_error(y[test_idx], y_pred )
scores.append(score)
print(np.mean(scores), np.std(scores))
fig = plt.figure(figsize=(15,6));
ax1 = fig.add_subplot(121)
ax2 = fig.add_subplot(122)
skplt.metrics.plot_confusion_matrix(y, y_pred, ax = ax1) #error line
skplt.metrics.plot_roc(y, y_pred, ax = ax2)
ValueError: Found input variables with inconsistent numbers of samples: [32561, 10853]
I checked the code, read many threads on this error. Somebody suggested me as a solution to put the cross-validation in a loop, but I don't know how to manage this with code (and also which part of operation to put in a loop, and how to write a condition that should be ending this loop). Please, help me with a specific answer that will help me to easily fix problem with my current level of advancement.
This is my first mathmatica code,
I defined the functions:
\[Beta] := v/c
\[Gamma] := 1/Sqrt[1 - \[Beta]^2]
TotalE[\[Gamma][\[Beta]]] := \[Gamma]mc^2
KE := TotalE[\[Gamma][\[Beta]]] - mc^2
No i want to make a series expansion of KE at β → 0 up to order 2,
I tried:
Series[KE, {\[Beta], 1, 2}]
But i got the error massage:
General::ivar: v/c is not a valid variable.
I also wanted to define Ekin as function of β,
so i used Solve function to get the inverse function, β[Ekin]:
Solve[KE, \[Beta]]
The same errors arises again:
Solve::ivar: v/c is not a valid variable.
Try this
Clear[\[Gamma],\[Beta],mc,KE,s,v,c]
\[Gamma] = 1/Sqrt[1 - \[Beta]^2];
TotalE[\[Gamma]*\[Beta]] = \[Gamma]*mc^2;
KE = TotalE[\[Gamma]*\[Beta]] - mc^2;
s=Normal[Series[KE, {\[Beta], 1, 2}]]/.\[Beta]->v/c
Reduce[KE==0, \[Beta]]/.\[Beta]->v/c
which returns
O-mc^2 + mc^2/(Sqrt[2]*Sqrt[1 - v/c]) -
(mc^2*(-1 + v/c))/(4*Sqrt[2]*Sqrt[1 - v/c]) +
(3*mc^2*(-1 + v/c)^2)/(32*Sqrt[2]*Sqrt[1 - v/c])
and
(mc != 0 && v/c == 0)||(-1+v^2/c^2 !=0 && mc == 0)
What that is trying to do is do your calculations with the simple variable beta, before you turn that into v/c and after the calculations replace beta with v/c.
But there are still things about the way you have written that which worry me. You are kind of writing TotalE like it is a function, but that is not the way to define a Mathematica function and I am concerned this may be going to get you into trouble.
Please let me know if I have misunderstood some of what you are trying to do and explain what I've done wrong and I will try to find a way to fix that.
I have read a lot about the pain of replicate the easy robust option from STATA to R to use robust standard errors. I replicated following approaches: StackExchange and Economic Theory Blog. They work but the problem I face is, if I want to print my results using the stargazer function (this prints the .tex code for Latex files).
Here is the illustration to my problem:
reg1 <-lm(rev~id + source + listed + country , data=data2_rev)
stargazer(reg1)
This prints the R output as .tex code (non-robust SE) If i want to use robust SE, i can do it with the sandwich package as follow:
vcov <- vcovHC(reg1, "HC1")
if I now use stargazer(vcov) only the output of the vcovHC function is printed and not the regression output itself.
With the package lmtest() it is possible to print at least the estimator, but not the observations, R2, adj. R2, Residual, Residual St.Error and the F-Statistics.
lmtest::coeftest(reg1, vcov. = sandwich::vcovHC(reg1, type = 'HC1'))
This gives the following output:
t test of coefficients:
Estimate Std. Error t value Pr(>|t|)
(Intercept) -2.54923 6.85521 -0.3719 0.710611
id 0.39634 0.12376 3.2026 0.001722 **
source 1.48164 4.20183 0.3526 0.724960
country -4.00398 4.00256 -1.0004 0.319041
---
Signif. codes: 0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1
How can I add or get an output with the following parameters as well?
Residual standard error: 17.43 on 127 degrees of freedom
Multiple R-squared: 0.09676, Adjusted R-squared: 0.07543
F-statistic: 4.535 on 3 and 127 DF, p-value: 0.00469
Did anybody face the same problem and can help me out?
How can I use robust standard errors in the lm function and apply the stargazer function?
You already calculated robust standard errors, and there's an easy way to include it in the stargazeroutput:
library("sandwich")
library("plm")
library("stargazer")
data("Produc", package = "plm")
# Regression
model <- plm(log(gsp) ~ log(pcap) + log(pc) + log(emp) + unemp,
data = Produc,
index = c("state","year"),
method="pooling")
# Adjust standard errors
cov1 <- vcovHC(model, type = "HC1")
robust_se <- sqrt(diag(cov1))
# Stargazer output (with and without RSE)
stargazer(model, model, type = "text",
se = list(NULL, robust_se))
Solution found here: https://www.jakeruss.com/cheatsheets/stargazer/#robust-standard-errors-replicating-statas-robust-option
Update I'm not so much into F-Tests. People are discussing those issues, e.g. https://stats.stackexchange.com/questions/93787/f-test-formula-under-robust-standard-error
When you follow http://www3.grips.ac.jp/~yamanota/Lecture_Note_9_Heteroskedasticity
"A heteroskedasticity-robust t statistic can be obtained by dividing an OSL estimator by its robust standard error (for zero null hypotheses). The usual F-statistic, however, is invalid. Instead, we need to use the heteroskedasticity-robust Wald statistic."
and use a Wald statistic here?
This is a fairly simple solution using coeftest:
reg1 <-lm(rev~id + source + listed + country , data=data2_rev)
cl_robust <- coeftest(reg1, vcov = vcovCL, type = "HC1", cluster = ~
country)
se_robust <- cl_robust[, 2]
stargazer(reg1, reg1, cl_robust, se = list(NULL, se_robust, NULL))
Note that I only included cl_robust in the output as a verification that the results are identical.
I am trying to perform multiple sample comparison and Tukey HSD using the statsmodels module, but I keep getting this error message, "ValueError: v must be > 1 when p >= .9". I have tried looking this up on the internet for a possible solution, but no avail. Any chance anyone familiar with this module could help me out decipher what I am doing wrong to prompt this error. I use Python version 2.7x and spyder. Below is a sample of my data and the print statement. Thanks!
import numpy as np
from statsmodels.stats.multicomp import (pairwise_tukeyhsd,MultiComparison)
###--- Here are the data I am using:
data1 = np.array([ 1, 1, 1, 1, 976, 24, 1, 1, 15, 15780])
data2 = np.array(['lau15', 'gr17', 'fri26', 'bays29', 'dantzig4', 'KAT38','HARV50', 'HARV10', 'HARV20', 'HARV41'], dtype='|S8')
####--- Here's my print statement code:
print pairwise_tukeyhsd(data1, data2, alpha=0.05)
Seems you have to provide more data than a single observation per group, in order for the test to work.
Minimal example:
from statsmodels.stats.multicomp import pairwise_tukeyhsd,MultiComparison
data=[1,2,3]
groups=['a','b','c']
print("1st try:")
try:
print(pairwise_tukeyhsd(data,groups, alpha=0.05))
except ValueError as ve:
print("whoops!", ve)
data.append(2)
groups.append('a')
print("2nd try:")
try:
print( pairwise_tukeyhsd(data, groups, alpha=0.05))
except ValueError as ve:
print("whoops!", ve)
Output:
1st try:
/home/user/.local/lib/python3.7/site-packages/numpy/core/fromnumeric.py:3367: RuntimeWarning: Degrees of freedom <= 0 for slice
**kwargs)
/home/user/.local/lib/python3.7/site-packages/numpy/core/_methods.py:132: RuntimeWarning: invalid value encountered in double_scalars
ret = ret.dtype.type(ret / rcount)
whoops! v must be > 1 when p >= .9
2nd try:
Multiple Comparison of Means - Tukey HSD, FWER=0.05
====================================================
group1 group2 meandiff p-adj lower upper reject
----------------------------------------------------
a b 0.5 0.1 -16.045 17.045 False
a c 1.5 0.1 -15.045 18.045 False
b c 1.0 0.1 -18.1046 20.1046 False
----------------------------------------------------
I came cross a function of graphing cumulative return of a strategy and the peaks of the return in a great example of combining shiny and quantstrat, thanks to Simon Otziger. The source code is here. The code works fine most of time, but for some data it won't graph the peaks properly.
The code is simplified but the key logic is not changed. I ran the code with three set of data (cumPNL1, cumPNL2, cumPNL3) copied from three example strategies, in which the first data will cause the code to fail to graph peaks properly.
I ran the following codes with cumPNL1, cumPNL2, cumPNL3 separately. with both cumPNL2 and cumPNL3 the code can produce cumulative return line and peak points successfully. however, with cumPNL1 the code can only produce line, but peaks are not at the right positions.
I noticed that both peakIndex based on cumPNL2 and cumPNL3 have their first value being TRUE, so when I change the code by adding a line peakIndex[1] <- TRUE, cumPNL1 will work fine with the modified code.
Though now it works with modified code, I have no idea why it is behaving like this. Could anyone have a look? Thanks
cumPNL1 <- c(-193,-345,-406,-472,-562,-543,-450,-460,-544,-659,-581,-342,-384,276,-858,-257.99)
cumPNL2 <- c(35.64,4.95,-2.97,-6.93,11.88,-19.8,-26.73,-39.6,-49.5,-50.49,-51.48,-48.51,-50.49,-55.44,143.55,770.22,745.47,691.02,847.44,1141.47,1007.82,1392.93,1855.26,1863.18,2536.38,2778.93,2811.6,2859.12,2417.58)
cumPNL3 <- c(35.64,4.95,-2.97,-6.93,11.88,-19.8,-26.73,-39.6,-49.5,-50.49,-51.48,-48.51,-50.49,-55.44,143.55,770.22,745.47,691.02,847.44,1141.47,1007.82,1392.93,1855.26,1863.18,2536.38,2778.93,2811.6,2859.12,2417.58)
peakIndex <- c(cumPNL3[1] > 0, diff(cummax(cumPNL3)) > 0)
# peakIndex[1] <- TRUE
dev.new()
plot(cumPNL3, type='n', xlab="index of trades", ylab="returns in cash", main="cumulative returns and peaks")
grid()
lines(cumPNL3)
points(cbind(1 : length(cumPNL3), cumPNL3)[peakIndex, ],
pch=19, col='green', cex=0.6)
legend(
x='bottomright', inset=0.1,
legend=c('Net Profit','Peaks'),
lty=c(1, NA), pch=c(NA, 19),
col=c('black','green')
)
cumPNL1 has a single peak and R reduces the dimension from a numerical matrix to a numerical vector of length 2. The points function plots the two numerical vector values on the y-axis using the x-axis index 1 and 2:
peakIndex1 <- c(cumPNL1[1] > 0, diff(cummax(cumPNL1)) > 0)
peakIndex3 <- c(cumPNL3[1] > 0, diff(cummax(cumPNL3)) > 0)
str(cbind(1 : length(cumPNL1), cumPNL1)[peakIndex1,])
str(cbind(1 : length(cumPNL3), cumPNL3)[peakIndex3,])
Output:
> str(cbind(1 : length(cumPNL1), cumPNL1)[peakIndex1,])
num [1:12, 1:2] 1 15 16 19 20 22 23 24 25 26 ...
- attr(*, "dimnames")=List of 2
..$ : NULL
..$ : chr [1:2] "" "cumPNL1"
> str(cbind(1 : length(cumPNL3), cumPNL3)[peakIndex3,])
Named num [1:2] 14 276
- attr(*, "names")= chr [1:2] "" "cumPNL3"
Usually setting plot = FALSE preserves the object, e.g., str(cbind(1 : length(cumPNL3), cumPNL3)[peakIndex3, drop = FALSE]), which somehow does not work in this case. However, changing the points line to the following fixes the problem:
points(seq_along(cumPNL3)[peakIndex], cumPNL3[peakIndex], pch = 19,
col = 'green', cex = 0.6)
Thanks for reporting the issue. I will push the fix to GitHub tomorrow.