r - How to generate random numbers in a data.frame with range -

i have data.frame want generate random numbers each list sequence.

i used sample function create random numbers created random numbers list [[1]], set [[2]] same numbers produced again. so, here how can create different random numbers set [[2]].

here simple code;

data.list <- lapply(1:2, function(x) { nrep <- 1 time <- rep(seq(90,54000,by=90),times=nrep)  mx <- rep(sort(sample(seq(0.012,-0.014,length.out = 600),replace=true)), times=nrep) <- rep(sort(sample(seq(0.02,-0.02,length.out = 600),replace=true)), times=nrep) mz <- rep(sort(sample(seq(-1,1,length.out=600),replace=true)), times=nrep) data.frame(time,mx,my,mz,set_nbr=x) }) 

this provide 5 first lines of each of datasets

[[1]]       time       mx                      mz       set_nbr 1      90 -1.391319e-02 -2.000000e-02 -1.000000000       1 2     180 -1.386978e-02 -1.986644e-02 -1.000000000       1 3     270 -1.386978e-02 -1.973289e-02 -0.996661102       1 4     360 -1.382638e-02 -1.973289e-02 -0.993322204       1 5     450 -1.382638e-02 -1.973289e-02 -0.979966611       1   ..     ..  ....            ....         ....           ...  [[2]]        time       mx                      mz       set_nbr 1      90 -1.395659e-02 -0.0200000000 -1.000000000       2 2     180 -1.391319e-02 -0.0199332220 -0.993322204       2 3     270 -1.386978e-02 -0.0199332220 -0.993322204       2 4     360 -1.386978e-02 -0.0199332220 -0.993322204       2 5     450 -1.382638e-02 -0.0199332220 -0.986644407       2 ..     ..  ....            ....         ....           ... 

edit 1:

regarding @bgoldst answer can produce different numbers

set.seed(1); data.list <- lapply(1:2, function(x) { nrep <- 1; time <- rep(seq(90,54000,by=90),times=nrep); mx <- rep(sort(runif(600,-0.014,0.012)),times=nrep); <- rep(sort(runif(600,-0.02,0.02)),times=nrep); mz <- rep(sort(runif(600,-1,1)),times=nrep); data.frame(time,mx,my,mz,set_nbr=x); }); 

on other hand when change nrep <- 3; same numbers created each nrep. thing want avoid beginning.

edit 2:

@bgoldst showed replicate job!

i think may have confusion how sample() works.

first, let's examine sample()'s behavior respect simple vector:

1:5; ## [1] 1 2 3 4 5 

when pass multi-element vector sample() randomizes order. means you'll different result every time, or rather, state more precisely, longer vector is, less same result twice:

set.seed(1); sample(1:5); sample(1:5); sample(1:5); ## [1] 2 5 4 3 1 ## [1] 5 4 2 3 1 ## [1] 2 1 3 4 5 

this means if sort after sampling, you'll same result every time. , if original vector sorted, result equal original vector. true regardless how sample() randomized order, because order restored sort():

set.seed(1); sort(sample(1:5)); sort(sample(1:5)); sort(sample(1:5)); ## [1] 1 2 3 4 5 ## [1] 1 2 3 4 5 ## [1] 1 2 3 4 5 

now if add replace=t (or rep=t if take advantage of partial matching concision, do), you're not randomizing order, you're selecting size elements replacement, size vector length if didn't provide size explicitly. means can repeated elements in result:

set.seed(1); sample(1:5,rep=t); sample(1:5,rep=t); sample(1:5,rep=t); ## [1] 2 2 3 5 2 ## [1] 5 5 4 4 1 ## [1] 2 1 4 2 4 

and so, if sort result, (likely) won't original vector, because elements have been repeated, , elements have been omitted:

set.seed(1); sort(sample(1:5,rep=t)); sort(sample(1:5,rep=t)); sort(sample(1:5,rep=t)); ## [1] 2 2 2 3 5 ## [1] 1 4 4 5 5 ## [1] 1 2 2 4 4 

that's happening code. output vectors are different between 2 list components, because you're sampling replacement before sorting, means different repetitions , omissions of elements occur each list component. since you're sampling same sequence , you're sorting result, you're bound similar-looking results each list component, though they're not identical.

i think might looking random deviates uniform distribution. can these runif():

set.seed(1); runif(5,-0.014,0.012); ## [1] -0.0070967748 -0.0043247786  0.0008941874  0.0096134025 -0.0087562698 set.seed(1); runif(5,-0.02,0.02); ## [1] -0.009379653 -0.005115044  0.002914135  0.016328312 -0.011932723 set.seed(1); runif(5,-1,1); ## [1] -0.4689827 -0.2557522  0.1457067  0.8164156 -0.5966361 

thus, code become:

set.seed(1); data.list <- lapply(1:2, function(x) {     nrep <- 1;     time <- rep(seq(90,54000,by=90),times=nrep);     mx <- rep(sort(runif(600,-0.014,0.012)),times=nrep);     <- rep(sort(runif(600,-0.02,0.02)),times=nrep);     mz <- rep(sort(runif(600,-1,1)),times=nrep);     data.frame(time,mx,my,mz,set_nbr=x); }); 

which gives:

lapply(data.list,head); ## [[1]] ##   time          mx                  mz set_nbr ## 1   90 -0.01395224 -0.01994741 -0.9967155       1 ## 2  180 -0.01394975 -0.01991923 -0.9933909       1 ## 3  270 -0.01378866 -0.01980934 -0.9905714       1 ## 4  360 -0.01371306 -0.01977090 -0.9854065       1 ## 5  450 -0.01371011 -0.01961713 -0.9850108       1 ## 6  540 -0.01365998 -0.01960718 -0.9846628       1 ## ## [[2]] ##   time          mx                  mz set_nbr ## 1   90 -0.01398426 -0.01997718 -0.9970438       2 ## 2  180 -0.01398293 -0.01989651 -0.9931286       2 ## 3  270 -0.01397330 -0.01988715 -0.9923425       2 ## 4  360 -0.01396455 -0.01957807 -0.9913645       2 ## 5  450 -0.01384501 -0.01939597 -0.9892001       2 ## 6  540 -0.01382531 -0.01931913 -0.9889356       2 

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edit: looked question wanted random numbers different between list components, say, between components generated 1:2 passed first argument lapply(). repetition of each random vector nrep times within each list component didn't appear relevant, partly because set nrep 1, there wasn't actual repetition.

but that's ok, can achieve requirement using replicate() instead of rep(), because replicate() actual runs expression argument once every repetition. have flatten result, because replicate() default returns matrix, , want straight vector:

set.seed(1); data.list <- lapply(1:2, function(x) {     nrep <- 2;     time <- rep(seq(90,54000,by=90),times=nrep);     mx <- c(replicate(nrep,sort(runif(600,-0.014,0.012))));     <- c(replicate(nrep,sort(runif(600,-0.02,0.02))));     mz <- c(replicate(nrep,sort(runif(600,-1,1))));     data.frame(time,mx,my,mz,set_nbr=x); }); lapply(data.list,function(x) x[c(1:6,601:606),]); ## [[1]] ##     time          mx                  mz set_nbr ## 1     90 -0.01395224 -0.01993431 -0.9988590       1 ## 2    180 -0.01394975 -0.01986782 -0.9948254       1 ## 3    270 -0.01378866 -0.01981143 -0.9943576       1 ## 4    360 -0.01371306 -0.01970813 -0.9789037       1 ## 5    450 -0.01371011 -0.01970022 -0.9697986       1 ## 6    540 -0.01365998 -0.01969326 -0.9659567       1 ## 601   90 -0.01396582 -0.01997579 -0.9970438       1 ## 602  180 -0.01394750 -0.01997375 -0.9931286       1 ## 603  270 -0.01387607 -0.01995893 -0.9923425       1 ## 604  360 -0.01385108 -0.01994546 -0.9913645       1 ## 605  450 -0.01375113 -0.01976155 -0.9892001       1 ## 606  540 -0.01374467 -0.01973125 -0.9889356       1 ## ## [[2]] ##     time          mx                  mz set_nbr ## 1     90 -0.01396979 -0.01999198 -0.9960861       2 ## 2    180 -0.01390373 -0.01995219 -0.9945237       2 ## 3    270 -0.01390252 -0.01991559 -0.9925640       2 ## 4    360 -0.01388905 -0.01978123 -0.9890171       2 ## 5    450 -0.01386718 -0.01967644 -0.9835435       2 ## 6    540 -0.01384351 -0.01958008 -0.9822988       2 ## 601   90 -0.01396739 -0.01989328 -0.9971255       2 ## 602  180 -0.01396433 -0.01985785 -0.9954987       2 ## 603  270 -0.01390700 -0.01984074 -0.9903196       2 ## 604  360 -0.01376890 -0.01982715 -0.9902251       2 ## 605  450 -0.01366110 -0.01979802 -0.9829480       2 ## 606  540 -0.01364868 -0.01977278 -0.9812671       2