search for: mpfr1

Displaying 3 results from an estimated 3 matches for "mpfr1".

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2016 Sep 10
3
c(<Matrix>, <Matrix>) / help(dotsMethods) etc
...equire(Rmpfr) > c(mpfr(3,100), 1/mpfr(7, 80)) ## works fine 2 'mpfr' numbers of precision 80 .. 100 bits [1] 3 0.14285714285714285714285708 > c(pi, 1/mpfr(7, 80)) ## "fails" even worse than in 'Matrix' case [[1]] [1] 3.141593 [[2]] 'mpfr1' 0.14285714285714285714285708 > Yes, it would be very nice if c(.) could be used to concatenate quite arbitrary R objects into one long atomic vector, but I don't see how to achieve this easily. The fact, that c() just builds a list of its arguments if it ("thinks" i...
2016 Sep 10
1
c(<Matrix>, <Matrix>) / help(dotsMethods) etc
...0 .. 100 bits >> [1] 3 0.14285714285714285714285708 >> >>> c(pi, 1/mpfr(7, 80)) ## "fails" even worse than in 'Matrix' case >> [[1]] >> [1] 3.141593 >> >> [[2]] >> 'mpfr1' 0.14285714285714285714285708 >> >>> >> >> >> Yes, it would be very nice if c(.) could be used to >> concatenate quite arbitrary R objects into one long atomic >> vector, but I don't see how to achieve this eas...
2016 Sep 10
0
c(<Matrix>, <Matrix>) / help(dotsMethods) etc
...80)) ## works fine > 2 'mpfr' numbers of precision 80 .. 100 bits > [1] 3 0.14285714285714285714285708 > >> c(pi, 1/mpfr(7, 80)) ## "fails" even worse than in 'Matrix' case > [[1]] > [1] 3.141593 > > [[2]] > 'mpfr1' 0.14285714285714285714285708 > >> > > > Yes, it would be very nice if c(.) could be used to > concatenate quite arbitrary R objects into one long atomic > vector, but I don't see how to achieve this easily. > > The fact, that c() just builds a list...