search for: i3

Hi friends, Please can anyone help me with an easier solution of doing the below mentioned work. Suppose i have a dataset like this:--- i1 i2 i3 i4 i5 1 7 13 1 2 2 8 14 2 2 3 9 15 3 3 4 10 16 4 4 5 11 17 5 5 6 12 18 6 7 *i1,i2,i3,i4,i5 are my items.I am able to find all possible pairs i.e Say this dataframe is "item_pairs" **i1,i2 **i1,i3 **i1,i4 i1,i5 **i2,i1 **i2,i3 i2,i4 i2,i5 **i3,i1 **i3,i2 **i3,i4 **...

...select data from a bigger sample. I want to select n days before each index and n days after the index. Any clever way to do it. A for loop would do but I wanted to know if there is a moreR-friendly way to approach this Example # InitialIndices i2 = (90, 190, 290) # Indices I want to end up with i3 = c(85, 86, 87, 88, 89, 90, 91, 92, 93, 94, 95, 185, 186, 187, 188, 189, 190, 191, 192, 193, 194, 195 285, 286, 287, 288, 289, 290, 291, 292, 293, 294, 295) # A way to get Final Indices SampleWidth i3 = c(i2) for (j in seq(1, SampleWidth )) { i3 = c(i3, i2 + j ) i3 = c(i3, i2 - j ) } I tried to...

Hi Jin Gu Kang > 5 %0 = type { i3, i4, i3, i2 } each of the above struct fields occupies 1 byte. If you were hoping that an i3 field would occupy only 3 bits then I am sorry, it doesn't work that way. Ciao, Duncan.

...5 char d:2; 6 }; 7 8 struct test vm = {1, 2, 3, 1}; 9 10 int main(void) 11 { 12 int a; 13 vm.d = 1; 14 } Above source code is compiled using cross arm-llvm-gcc as following. 1 ; ModuleID = './temp2.c' 2 target datalayout = "e-p:32:32:32-i1:8:8-i8:8:8-i16:16:16-i32:32:32-i64:64:64-f32:32:32-f64:64:64-v64:64:64-v128:128:128-a0:0:64" 3 target triple = "armv5-none-linux-gnueabi" 4 5 %0 = type { i8, i8 } 6 %struct.test = type <{ i8, i8 }> 7 8 @vm = global %0 { i8 17, i8 11 } ; <%0*> [#uses=1] 9 10 define...

I am trying to multiply a 3D array of 4x4x4 by the 4 3D arrays of a 4D array with dimensions 4x4x4x4 (the last dimension being the one that I want to split by). (4x4x4 array) > hiaAry , , a1 i1 i2 i3 i4 h1 9.5936098 6.001040 0.08772 0.3138600 h2 1.2003500 1.454570 2.79248 0.0000000 h3 0.1346500 0.201220 0.39256 0.5464000 h4 0.0109000 0.012270 0.16417 0.2766900 , , a2 i1 i2 i3 i4 h1 195.08275 74.23508453 39.23165 14.89689 h2 1.80127 7.41649055 6.97972...

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...list(ls()) #index matrix t1<-c(1:15) t2<-c(1:5)%x%matrix(1,3,1) t3<-matrix(1,3,1)%x%c(1:5) t4<-c(1:3)%x%matrix(1,5,1) comb<-cbind(t1,t2,t3,t4) nu<-rchisq(15,4) gam<-matrix(rchisq(75,df=5),15,5) undint<-function(u){ prob<-function(i) { i1<-comb[i,1] i2<-comb[i,2] i3<-comb[i,3] i4<-comb[i,4] val1<-gam[((i3-1)*2+1):(i3*2),i4] vc1<-matrix(12,length(val1),1)-matrix(u,length(val1),1)-val1 distr1<-prod(pchisq(vc1,df=2)) p10<-distr1*1/(exp(-nu[i1])*(1+gam[i2,i4]))*(1-plogis((u+log(gam[i3,i4])-log(1+gam[i2,i4])),0,1)) p10 } val<-c(1:15) q<-sa...

...;- c() for (i1 in seq(n)) { for (i2 in seq(n)) { if (length(unique(c(i1,i2)))==1) {next} rl <- cbind(rl,matrix(c(i1,i2))) } } rl } bag3 <- function(n) { rl <- c() for (i1 in seq(n)) { for (i2 in seq(n)) { for (i3 in seq(n)) { if (length(unique(c(i1,i2,i3)))==1) {next} rl <- cbind(rl,matrix(c(i1,i2,i3))) } } } rl } But I think it should be somehow possible in R to use one general function for all sizes. Can someone help? I don't exactly know...

Hi, I'm having a problem with a step that should be pretty simple. I have a dataframe, d, with column names : gene s1 s2 s3. The column "gene" stores an Id; the rest of the columns store intensity data. I would like to extract the rows for gene Ids i1, i2, i3 ( I know a priori that those rows exist). So I do this: subset(d, gene %in% c(i1, i2, i3)). This does not give me the required data. Any ideas where I am going wrong? TIA, Anjan -- =================================== anjan purkayastha, phd. research associate fas center for systems biology, harva...

...visible by 5. > > No, that's not true, because we are dealing with arithmetic modulo > 2^64 here. Since 5 and 2^64 are relatively prime, you can prove that > given any i64 value R, there exists an i64 value X such that mul i64 > 5, X is equal to R. Here's an example using i3 arithmetic (possible > values: 0, ..., 7). > mul i3 5, 0 -> 0 > mul i3 5, 5 -> 1 (because 5 * 5 is 25; reduced modulo 8 that gives 1) > mul i3 5, 2 -> 2 > mul i3 5, 7 -> 3 > mul i3 5, 4 -> 4 > mul i3 5, 1 -> 5 > mul i3 5, 6 -> 6 > mul...

...th may be limited to 64 bits. I need such variables to represent instruction's operands, >> example is at the end this message. >> > > any standard compliant C compiler supports i5, believe it or not. > > Try this: > > #include <stdio.h> > > struct i3 { int i:3; }; > > int main(void) { > struct i3 A, B, C; > > A.i = 2; > B.i = 3; > C.i = A.i + B.i; > printf("%d + %d = %d\n", A.i, B.i, C.i); > return 0; > } > > I get: > 2 + 3 = -3 > which is correct for i3 arithmetic! > Ther...

Hello Everyone I have a dual-core Intel i3-370M processor and Windows 64 operating system. I have a 3GB RAM and my processor can support a maximum of 8GB RAM. I have virtual memory enabled on my computer. I am running a program in the 64 bit R which implements a MCMC on a large dataset and involves around 80000 iterations. The processor...

On 08/16/2013 02:42 AM, Star Tan wrote: > At 2013-08-16 12:44:02,"Tobias Grosser" <tobias at grosser.es> wrote: >> Hi, >> >> I tried to reproduce your findings, but could not do so. > > > Sorry, I did not put all code in my previous email because the code seems a little too long and complicated. > You can refer to the detailed C code and LLVM IR

Hi Duncan, Thank you a lot for your time to provide that great and informative explanation. Now the "undef" logic makes much more sense for me. >> /You are wrong to say that "div undef, %X" is folded to "undef" by InstructionSimplify, it is folded to zero./ My mistake. I meant to say "*f****div* undef, %X" is folded to "undef" (not

...width like 'i5 var', > the maximal bit-width may be limited to 64 bits. I need such variables to represent instruction's operands, > example is at the end this message. any standard compliant C compiler supports i5, believe it or not. Try this: #include <stdio.h> struct i3 { int i:3; }; int main(void) { struct i3 A, B, C; A.i = 2; B.i = 3; C.i = A.i + B.i; printf("%d + %d = %d\n", A.i, B.i, C.i); return 0; } I get: 2 + 3 = -3 which is correct for i3 arithmetic! There might be some problems with signed vs unsigned bitfields, depending on...

...; char b:4; char c:3; char d:2; }; struct testShort { short a:3; short b:4; short c:3; short d:2; }; struct testInt { int a:3; int b:4; int c:3; int d:2; }; In future, our IRs will be represented as following: %Char = type { c3, c4, c3, c2 } %Short = type { s3, s4, s3, s2 } %Int = type { i3, i4, i3, i2 } This was just concept and I didn't implement it yet. I usually try to modify original llvm system as less as possible, so I prefer adding new types to modifying original llvm types. I will report your opinion to my team members and we will discuss that. Thanks for your answer,...

...it. The IR module should be optional when serializing MI. The back-pointers from MI to IR are not required, and I can imagine many useful test cases that won’t need them. module: | define void @linkit(i8* %source) #0 { entry: %.b243 = load i1* @Pflag, align 1 %cond = select i1 %.b243, i32 (i8*, %struct.stat.6.13.20.64*)* @lstat, i32 (i8*, %struct.stat.6.13.20.64*)* @stat %call = call signext i32 %cond(i8* %source, %struct.stat.6.13.20.64* undef) #2 ret void } @Pflag = external unnamed_addr global i1 declare signext i32 @lstat(i8* nocapture, %struct.stat.6.13.20.64* no...

Dear all, Is there anyone who knows well i80 data type? Is there any corresponding data type for X86 processor? uint80_t or int80_t for gcc? PS. I found that the LLVM website is down! -- Hao Shen

...sponding data type > for X86 processor? uint80_t or int80_t for gcc? no, there is no native processor support for i80. GCC does not have a direct equivalent to i80. However if you declare a 80-bit wide C bitfield, then arithmetic on it is done in 80 bits. For example, here gcc should perform i3 arithmetic: #include <stdio.h> struct i3 { unsigned i:3; }; int main(void) { struct i3 A, B, C; A.i = 5; B.i = 5; C.i = A.i + B.i; printf("%d + %d = %d\n", A.i, B.i, C.i); return 0; }

Hi all, Anyone has installed i3 wm under CentOS7? Any samples? I have some doubts about to how to manage battery power, notifications, etc. Thanks -- Greetings, C. L. Martinez