Problems with rqpd - CRE method - quantreg

I've been trying to run a quantile regression for panel data in R, using the 'rqpd' package and it works normally for PFE method (fixed effects). However, when I run the CRE method, it fails. Below are the codes I am using and the messages the program returns. Could someone give me a help, please?
> cre.form <- lnemp ~ lngt + lneduc + lndesp + lngdp | ID | lngt + lneduc + lndesp + lngdp
> crem.fit <- rqpd(cre.form, panel("cre"), data=Panel12)
Error messages:
Error in switch(ierr, "insufficient storage (work space) when calling extract\n", :
EXPR must be a length 1 vector
In addition: Warning messages:
1: In .local(x, ...) : singularity problem
2: In .local(x, ...) : singularity problem
3: In .local(x, ...) : singularity problem
4: In .local(x, ...) : singularity problem
5: In (fit$ierr != 0) && (fit$it < 5) :
'length(x) = 3 > 1' in coercion to 'logical(1)'
6: In (fit$ierr != 0) && (fit$it < 5) :
'length(x) = 3 > 1' in coercion to 'logical(1)'

Related

type Error: 'numpy.float64' object cannot be interpreted as an integer(function_base.py library issue)

TypeError: 'numpy.float64' object cannot be interpreted as an integer,all of these problems are happening in my numpy/core library i even tried to downgrade and uninstall numpy but still the problem is there the function_base.py
has this issue of type error
num = operator.index(num)
if num < 0:
raise ValueError("Number of samples, %s, must be non-negative." % num)
div = (num - 1) if endpoint else num

lua debug hook causing race condition?

I wanted to report some debug information for a parser I am writing in lua. I am using the debug hook facility for tracking but it seems like there is some form of race condition happening.
Here is my test code:
enters = 0
enters2 = 0
calls = 0
tailcalls = 0
returns = 0
lines = 0
other = 0
exits = 0
local function analyze(arg)
enters = enters + 1
enters2 = enters2 + 1
if arg == "call" then
calls = calls + 1
elseif arg == "tail call" then
tailcalls = tailcalls + 1
elseif arg == "return" then
returns = returns + 1
elseif arg == "line" then
lines = lines + 1
else
other = other + 1
end
exits = exits + 1
end
debug.sethook(analyze, "crl")
-- main code
print("enters = ", enters)
print("enters2 = ", enters2)
print("calls = ", calls)
print("tailcalls = ", tailcalls)
print("returns = ", returns)
print("lines = ", lines)
print("other = ", other)
print("exits = ", exits)
print("sum = ", calls + tailcalls + returns + lines + other)
and here is the result:
enters = 429988
enters2 = 429991
calls = 97433
tailcalls = 7199
returns = 97436
lines = 227931
other = 0
exits = 430009
sum = 430012
Why does none of this add up? I am running lua 5.4.2 on Ubuntu 20.04, no custom c libraries, no further messing with the debug library.

I found the problem...
The calls to the print function when printing the result also trigger the hook, which only affects the results that have not been printed yet.

Gradle Task - unable to execute fibonacci series in groovy

Facing problem in a question:
Write a gradle program to generate 10 fibonaci series, with task name as fibo, and variable name as num. Use command line argument for num.
For example, if a task name is test and I want to pass 10 as the input, use gradle test -Pnum=10.
I have created a function:
def fibo(n){
a = 0
b = 1
if (n == 1)
println a
else if
(n == 2)
println a + " " + b
else if (n > 2) {
print a + " " + b
i = 2
while (i <= n)
{
c = a + b
print " " + c
a = b
b = c
i = i + 1
}
}
}
My question is, how to link it with a task as I encounter error like:
FAILURE: Build failed with an exception.
* What went wrong:
Task 'fibo' not found in root project 'root'.
* Try:
Run gradle tasks to get a list of available tasks. Run with --stacktrace option to get the stack trace. Run with --info or --debug option to get more log output.
BUILD FAILED
Total time: 2.61 secs
or how to pass parameters in a gradle task?
Note: Please do not suggest optimization in fibonacci code, thats not a concern for now.

You can define a task like this:
def hello(name) {
println "Hello, $name"
}
task sayHello() {
doLast {
hello sayHelloTo
}
}
And call it like this:
% gradle sayHello -PsayHelloTo=World
> Task :sayHello
Hello, World
BUILD SUCCESSFUL in 518ms
1 actionable task: 1 executed

def fibo(num) {
if (num < 2) {
return 1
} else {
return fibo(num-2) + fib(num-1)
}
}
task (fibo) << {
println fibo(5)
}

Code Golf: Email Address Validation without Regular Expressions

Locked. This question and its answers are locked because the question is off-topic but has historical significance. It is not currently accepting new answers or interactions.
(Edit: What is Code Golf: Code Golf are challenges to solve a specific problem with the shortest amount of code by character count in whichever language you prefer. More info here on Meta StackOverflow. )
Code Golfers, here's a challenge on string operations.
Email Address Validation, but without regular expressions (or similar parsing library) of course. It's not so much about the email addresses but how short you can write the different string operations and constraints given below.
The rules are the following (yes, I know, this is not RFC compliant, but these are going to be the 5 rules for this challenge):
At least 1 character out of this group before the #:
A-Z, a-z, 0-9, . (period), _ (underscore)
# has to exist, exactly one time
john#smith.com
^
Period (.) has to exist exactly one time after the #
john#smith.com
^
At least 1 only [A-Z, a-z] character between # and the following . (period)
john#s.com
^
At least 2 only [A-Z, a-z] characters after the final . period
john#smith.ab
^^
Please post the method/function only, which would take a string (proposed email address) and then return a Boolean result (true/false) depending on the email address being valid (true) or invalid (false).
Samples:
b#w.org (valid/true) #w.org (invalid/false)
b#c#d.org (invalid/false) test#org (invalid/false)
test#%.org (invalid/false) s%p#m.org (invalid/false)
j_r#x.c.il (invalid/false) j_r#x.mil (valid/true)
r..t#x.tw (valid/true) foo#a%.com (invalid/false)
Good luck!

C89 (166 characters)
#define B(c)isalnum(c)|c==46|c==95
#define C(x)if(!v|*i++-x)return!1;
#define D(x)for(v=0;x(*i);++i)++v;
v;e(char*i){D(B)C(64)D(isalpha)C(46)D(isalpha)return!*i&v>1;}
Not re-entrant, but can be run multiple times. Test bed:
#include<stdio.h>
#include<assert.h>
main(){
assert(e("b#w.org"));
assert(e("r..t#x.tw"));
assert(e("j_r#x.mil"));
assert(!e("b#c#d.org"));
assert(!e("test#%.org"));
assert(!e("j_r#x.c.il"));
assert(!e("#w.org"));
assert(!e("test#org"));
assert(!e("s%p#m.org"));
assert(!e("foo#a%.com"));
puts("success!");
}

J
:[[/%^(:[[+-/^,&i|:[$[' ']^j+0__:k<3:]]

C89, 175 characters.
#define G &&*((a+=t+1)-1)==
#define H (t=strspn(a,A
t;e(char*a){char A[66]="_.0123456789Aa";short*s=A+12;for(;++s<A+64;)*s=s[-1]+257;return H))G 64&&H+12))G 46&&H+12))>1 G 0;}
I am using the standard library function strspn(), so I feel this answer isn't as "clean" as strager's answer which does without any library functions. (I also stole his idea of declaring a global variable without a type!)
One of the tricks here is that by putting . and _ at the start of the string A, it's possible to include or exclude them easily in a strspn() test: when you want to allow them, use strspn(something, A); when you don't, use strspn(something, A+12). Another is assuming that sizeof (short) == 2 * sizeof (char), and building up the array of valid characters 2 at a time from the "seed" pair Aa. The rest was just looking for a way to force subexpressions to look similar enough that they could be pulled out into #defined macros.
To make this code more "portable" (heh :-P) you can change the array-building code from
char A[66]="_.0123456789Aa";short*s=A+12;for(;++s<A+64;)*s=s[-1]+257;
to
char*A="_.0123456789ABCDEFGHIJKLMNOPQRSTUVWXYZabcdefghijklmnopqrstuvwxyz";
for a cost of 5 additional characters.

Python (181 characters including newlines)
def v(E):
import string as t;a=t.ascii_letters;e=a+"1234567890_.";t=e,e,"#",e,".",a,a,a,a,a,"",a
for c in E:
if c in t[0]:t=t[2:]
elif not c in t[1]:return 0>1
return""==t[0]
Basically just a state machine using obfuscatingly short variable names.

C (166 characters)
#define F(t,u)for(r=s;t=(*s-64?*s-46?isalpha(*s)?3:isdigit(*s)|*s==95?4:0:2:1);++s);if(s-r-1 u)return 0;
V(char*s){char*r;F(2<,<0)F(1=)F(3=,<0)F(2=)F(3=,<1)return 1;}
The single newline is required, and I've counted it as one character.

Python, 149 chars (after putting the whole for loop into one semicolon-separated line, which I haven't done here for "readability" purposes):
def v(s,t=0,o=1):
for c in s:
k=c=="#"
p=c=="."
A=c.isalnum()|p|(c=="_")
L=c.isalpha()
o&=[A,k|A,L,L|p,L,L,L][t]
t+=[1,k,1,p,1,1,0][t]
return(t>5)&o
Test cases, borrowed from strager's answer:
assert v("b#w.org")
assert v("r..t#x.tw")
assert v("j_r#x.mil")
assert not v("b#c#d.org")
assert not v("test#%.org")
assert not v("j_r#x.c.il")
assert not v("#w.org")
assert not v("test#org")
assert not v("s%p#m.org")
assert not v("foo#a%.com")
print "Yeah!"
Explanation: When iterating over the string, two variables keep getting updated.
t keeps the current state:
t = 0: We're at the beginning.
t = 1: We where at the beginning and have found at least one legal character (letter, number, underscore, period)
t = 2: We have found the "#"
t = 3: We have found at least on legal character (i.e. letter) after the "#"
t = 4: We have found the period in the domain name
t = 5: We have found one legal character (letter) after the period
t = 6: We have found at least two legal characters after the period
o as in "okay" starts as 1, i.e. true, and is set to 0 as soon as a character is found that is illegal in the current state.
Legal characters are:
In state 0: letter, number, underscore, period (change state to 1 in any case)
In state 1: letter, number, underscore, period, at-sign (change state to 2 if "#" is found)
In state 2: letter (change state to 3)
In state 3: letter, period (change state to 4 if period found)
In states 4 thru 6: letter (increment state when in 4 or 5)
When we have gone all the way through the string, we return whether t==6 (t>5 is one char less) and o is 1.

Whatever version of C++ MSVC2008 supports.
Here's my humble submission. Now I know why they told me never to do the things I did in here:
#define N return 0
#define I(x) &&*x!='.'&&*x!='_'
bool p(char*a) {
if(!isalnum(a[0])I(a))N;
char*p=a,*b=0,*c=0;
for(int d=0,e=0;*p;p++){
if(*p=='#'){d++;b=p;}
else if(*p=='.'){if(d){e++;c=p;}}
else if(!isalnum(*p)I(p))N;
if (d>1||e>1)N;
}
if(b>c||b+1>=c||c+2>=p)N;
return 1;
}

Not the greatest solution no doubt, and pretty darn verbose, but it is valid.
Fixed (All test cases pass now)
static bool ValidateEmail(string email)
{
var numbers = "1234567890";
var uppercase = "ABCDEFGHIJKLMNOPQRSTUVWXYZ";
var lowercase = uppercase.ToLower();
var arUppercase = uppercase.ToCharArray();
var arLowercase = lowercase.ToCharArray();
var arNumbers = numbers.ToCharArray();
var atPieces = email.Split(new string[] { "#"}, StringSplitOptions.RemoveEmptyEntries);
if (atPieces.Length != 2)
return false;
foreach (var c in atPieces[0])
{
if (!(arNumbers.Contains(c) || arLowercase.Contains(c) || arUppercase.Contains(c) || c == '.' || c == '_'))
return false;
}
if(!atPieces[1].Contains("."))
return false;
var dotPieces = atPieces[1].Split('.');
if (dotPieces.Length != 2)
return false;
foreach (var c in dotPieces[0])
{
if (!(arLowercase.Contains(c) || arUppercase.Contains(c)))
return false;
}
var found = 0;
foreach (var c in dotPieces[1])
{
if ((arLowercase.Contains(c) || arUppercase.Contains(c)))
found++;
else
return false;
}
return found >= 2;
}

C89 character set agnostic (262 characters)
#include <stdio.h>
/* the 'const ' qualifiers should be removed when */
/* counting characters: I don't like warnings :) */
/* also the 'int ' should not be counted. */
/* it needs only 2 spaces (after the returns), should be only 2 lines */
/* that's a total of 262 characters (1 newline, 2 spaces) */
/* code golf starts here */
#include<string.h>
int v(const char*e){
const char*s="0123456789._abcdefghijklmnopqrstuvwxyzABCDEFGHIJKLMNOPQRSTUVWXYZ";
if(e=strpbrk(e,s))
if(e=strchr(e+1,'#'))
if(!strchr(e+1,'#'))
if(e=strpbrk(e+1,s+12))
if(e=strchr(e+1,'.'))
if(!strchr(e+1,'.'))
if(strlen(e+1)>1)
return 1;
return 0;
}
/* code golf ends here */
int main(void) {
const char *t;
t = "b#w.org"; printf("%s ==> %d\n", t, v(t));
t = "r..t#x.tw"; printf("%s ==> %d\n", t, v(t));
t = "j_r#x.mil"; printf("%s ==> %d\n", t, v(t));
t = "b#c#d.org"; printf("%s ==> %d\n", t, v(t));
t = "test#%.org"; printf("%s ==> %d\n", t, v(t));
t = "j_r#x.c.il"; printf("%s ==> %d\n", t, v(t));
t = "#w.org"; printf("%s ==> %d\n", t, v(t));
t = "test#org"; printf("%s ==> %d\n", t, v(t));
t = "s%p#m.org"; printf("%s ==> %d\n", t, v(t));
t = "foo#a%.com"; printf("%s ==> %d\n", t, v(t));
return 0;
}
Version 2
Still C89 character set agnostic, bugs hopefully corrected (303 chars; 284 without the #include)
#include<string.h>
#define Y strchr
#define X{while(Y
v(char*e){char*s="0123456789_.abcdefghijklmnopqrstuvwxyzABCDEFGHIJKLMNOPQRSTUVWXYZ";
if(*e!='#')X(s,*e))e++;if(*e++=='#'&&!Y(e,'#')&&Y(e+1,'.'))X(s+12,*e))e++;if(*e++=='.'
&&!Y(e,'.')&&strlen(e)>1){while(*e&&Y(s+12,*e++));if(!*e)return 1;}}}return 0;}
That #define X is absolutely disgusting!
Test as for my first (buggy) version.

VBA/VB6 - 484 chars
Explicit off
usage: VE("b#w.org")
Function V(S, C)
V = True
For I = 1 To Len(S)
If InStr(C, Mid(S, I, 1)) = 0 Then
V = False: Exit For
End If
Next
End Function
Function VE(E)
VE = False
C1 = "abcdefghijklmnopqrstuvwxyzABCDEFGHILKLMNOPQRSTUVWXYZ"
C2 = "0123456789._"
P = Split(E, "#")
If UBound(P) <> 1 Then GoTo X
If Len(P(0)) < 1 Or Not V(P(0), C1 & C2) Then GoTo X
E = P(1): P = Split(E, ".")
If UBound(P) <> 1 Then GoTo X
If Len(P(0)) < 1 Or Not V(P(0), C1) Or Len(P(1)) < 2 Or Not V(P(1), C1) Then GoTo X
VE = True
X:
End Function

Java: 257 chars (not including the 3 end of lines for readability ;-)).
boolean q(char[]s){int a=0,b=0,c=0,d=0,e=0,f=0,g,y=-99;for(int i:s)
d=(g="#._0123456789QWERTYUIOPASDFGHJKLZXCVBNMqwertyuiopasdfghjklzxcvbnm".indexOf(i))<0?
y:g<1&&++e>0&(b<1|++a>1)?y:g==1&e>0&(c<1||f++>0)?y:++b>0&g>12?f>0?d+1:f<1&e>0&&++c>0?
d:d:d;return d>1;}
Passes all the tests (my older version was incorrect).

Erlang 266 chars:
-module(cg_email).
-export([test/0]).
%%% golf code begin %%%
-define(E,when X>=$a,X=<$z;X>=$A,X=<$Z).
-define(I(Y,Z),Y([X|L])?E->Z(L);Y(_)->false).
-define(L(Y,Z),Y([X|L])?E;X>=$0,X=<$9;X=:=$.;X=:=$_->Z(L);Y(_)->false).
?L(e,m).
m([$#|L])->a(L);?L(m,m).
?I(a,i).
i([$.|L])->l(L);?I(i,i).
?I(l,c).
?I(c,g).
g([])->true;?I(g,g).
%%% golf code end %%%
test() ->
true = e("b#w.org"),
false = e("b#c#d.org"),
false = e("test#%.org"),
false = e("j_r#x.c.il"),
true = e("r..t#x.tw"),
false = e("test#org"),
false = e("s%p#m.org"),
true = e("j_r#x.mil"),
false = e("foo#a%.com"),
ok.

Ruby, 225 chars.
This is my first Ruby program, so it's probably not very Ruby-like :-)
def v z;r=!a=b=c=d=e=f=0;z.chars{|x|case x when'#';r||=b<1||!e;e=!1 when'.'
e ?b+=1:(a+=1;f=e);r||=a>1||(c<1&&!e)when'0'..'9';b+=1;r|=!e when'A'..'Z','a'..'z'
e ?b+=1:f ?c+=1:d+=1;else r=1 if x!='_'||!e|!b+=1;end};!r&&d>1 end

'Using no regex':
PHP 47 Chars.
<?=filter_var($argv[1],FILTER_VALIDATE_EMAIL);

Haskell (GHC 6.8.2), 165 161 144C Characters
Using pattern matching, elem, span and all:
a=['A'..'Z']++['a'..'z']
e=f.span(`elem`"._0123456789"++a)
f(_:_,'#':d)=g$span(`elem`a)d
f _=False
g(_:_,'.':t#(_:_:_))=all(`elem`a)t
g _=False
The above was tested with the following code:
main :: IO ()
main = print $ and [
e "b#w.org",
e "r..t#x.tw",
e "j_r#x.mil",
not $ e "b#c#d.org",
not $ e "test#%.org",
not $ e "j_r#x.c.il",
not $ e "#w.org",
not $ e "test#org",
not $ e "s%p#m.org",
not $ e "foo#a%.com"
]

Using gdb, display multiple vars in one line?

How can I ask to display multiple vars in one line? So I want to get output like:
30 if(s[i] != '\0')
5: s[i] = 101 'e'
4: exp = 14
3: val = 123.45
2: sign = 1
1: i = 6
I've been typing in disp s[i] ENTER disp exp ENTER (etc, etc) and I just know there's got to be a better way to do this in one line of typing.

To establish multiple active "variable displays" without re-typing each of display i, display s[i], etc. every time you restart GDB, use a GDB "canned command sequence".
For example, add this to your ~/.gdbinit:
define disp_vars
disp i
disp sign
disp val
disp exp
disp s[i]
end
Now you can add all the displays at once by typing disp_vars at the GDB prompt.

Employed Russian gave the correct solution but for those that want see it used in an example see below. If you're not sure if you want to commit to putting the .gdbinit in your home directory, you can also put it in the directory you're executing the program from to experiment.
$ gcc -g atof_ex4.2.c
$ gdb ./a.out
(gdb) b 30
Breakpoint 1 at 0x1907: file atof_ex4.2.c, line 30.
(gdb) h user-defined
List of commands:
disp_vars -- User-defined
(gdb) disp_vars #this will enable the user defined canned sequence (but I haven't done run yet! So I'll this actually doesn't work yet.)
No symbol "i" in current context.
(gdb) r
Starting program: a.out
Breakpoint 1, atof (s=0xbffff028 "123.45e-6") at atof_ex4.2.c:30
30 if(s[i] != '\0')
(gdb) s # No disp_vars output yet because I have to do it AFTER 'run' command
32 if(s[i] == 'e' || s[i] == 'E')
(gdb) disp_vars # Now it will work ;)
(gdb) s
35 sign = (s[i] == '-') ? -1 : 1;
5: s[i] = 45 '-'
4: exp = 14
3: val = 123.45
2: sign = 1
1: i = 7
Of course 'r' is for run, 's' is for step, 'b' is for break, etc. I've also omitted some output. Notice that I had to enter the 'disp_vars' command again after 'run'. Thanks Employed Russian.

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