Mixing
Test of integers? Or, round the number if the first two decimal numbers are sufficiently close to 0 or 1?
The following is a MWE, which explains my intention.
documentclass{article}
usepackage{tikz}
usetikzlibrary{math}
begin{document}
tikzmath{
integer = 4/2; decimal = 5/3;
integerB=1/3*3;
}
$integer$ is an integer, and it should be printed as 2.
And $decimal$ is a decimal number, I would like to round it to 1.7.
Another difficulty is that $integerB$ is an integer in fact,
and should be printed as 1.
end{document}
I wonder if it is possible to test a number to determine it is an integer. Alternatively, it would also be great if it is possible to determine whether the first two decimal numbers are sufficiently close to 0 or 1.
tables tikzmath decimal-number
edited Dec 31 '18 at 16:11
egreg
711k8618913174
asked Dec 31 '18 at 15:58
GlennGlenn
411
add a comment |
The following is a MWE, which explains my intention.
documentclass{article}
usepackage{tikz}
usetikzlibrary{math}
begin{document}
tikzmath{
integer = 4/2; decimal = 5/3;
integerB=1/3*3;
}
$integer$ is an integer, and it should be printed as 2.
And $decimal$ is a decimal number, I would like to round it to 1.7.
Another difficulty is that $integerB$ is an integer in fact,
and should be printed as 1.
end{document}
I wonder if it is possible to test a number to determine it is an integer. Alternatively, it would also be great if it is possible to determine whether the first two decimal numbers are sufficiently close to 0 or 1.
tables tikzmath decimal-number
edited Dec 31 '18 at 16:11
egreg
711k8618913174
asked Dec 31 '18 at 15:58
GlennGlenn
411
add a comment |
The following is a MWE, which explains my intention.
documentclass{article}
usepackage{tikz}
usetikzlibrary{math}
begin{document}
tikzmath{
integer = 4/2; decimal = 5/3;
integerB=1/3*3;
}
$integer$ is an integer, and it should be printed as 2.
And $decimal$ is a decimal number, I would like to round it to 1.7.
Another difficulty is that $integerB$ is an integer in fact,
and should be printed as 1.
end{document}
I wonder if it is possible to test a number to determine it is an integer. Alternatively, it would also be great if it is possible to determine whether the first two decimal numbers are sufficiently close to 0 or 1.
tables tikzmath decimal-number
edited Dec 31 '18 at 16:11
egreg
711k8618913174
asked Dec 31 '18 at 15:58
GlennGlenn
411
The following is a MWE, which explains my intention.
documentclass{article}
usepackage{tikz}
usetikzlibrary{math}
begin{document}
tikzmath{
integer = 4/2; decimal = 5/3;
integerB=1/3*3;
}
$integer$ is an integer, and it should be printed as 2.
And $decimal$ is a decimal number, I would like to round it to 1.7.
Another difficulty is that $integerB$ is an integer in fact,
and should be printed as 1.
end{document}
I wonder if it is possible to test a number to determine it is an integer. Alternatively, it would also be great if it is possible to determine whether the first two decimal numbers are sufficiently close to 0 or 1.
tables tikzmath decimal-number
tables tikzmath decimal-number
edited Dec 31 '18 at 16:11
egreg
711k8618913174
asked Dec 31 '18 at 15:58
GlennGlenn
411
edited Dec 31 '18 at 16:11
egreg
711k8618913174
asked Dec 31 '18 at 15:58
GlennGlenn
411
edited Dec 31 '18 at 16:11
egreg
711k8618913174
edited Dec 31 '18 at 16:11
egreg
711k8618913174
edited Dec 31 '18 at 16:11
egreg
711k8618913174
711k8618913174
asked Dec 31 '18 at 15:58
GlennGlenn
411
asked Dec 31 '18 at 15:58
GlennGlenn
411
asked Dec 31 '18 at 15:58
GlennGlenn
411
411
add a comment |
add a comment |
5 Answers
5
active
oldest
votes
One cannot say from its floating point representation whether the output of an arithmetic operation involving division or non rational operations is actually an integer.
You can consider the l3fp module of expl3, available through the package xfp.
documentclass{article}
usepackage{xfp}
begin{document}
$fpeval{5/3}$ is a decimal number, I would like to round it to
$fpeval{round(5/3,1)}$ or to $fpeval{round(5/3,2)}$
Another difficulty is that $fpeval{(1/3)*3}$ is an integer in fact,
and should be printed as $fpeval{round((1/3)*3,1)}$ or
$fpeval{round((1/3)*3,2)}$.
end{document}
answered Dec 31 '18 at 16:10
egregegreg
711k8618913174
add a comment |
If the accuracy of your numbers is important you might consider farming that out to a computer algebra system (CAS). The sagetex package relies on the CAS Sage; the documentation can be found on CTAN right here. Documentation on Sage is found here .Sage is not part of the LaTeX distribution (it's big) so it needs to be installed on your computer or, even easier, accessed through a free Cocalc account.
documentclass{article}
usepackage{sagetex}
usepackage{tikz}
usetikzlibrary{math}
begin{document}
begin{sagesilent}
a = 4/2
b = 5/3
c = 1/3*3
end{sagesilent}
$sage{a}$ is an integer, and it should be printed as $sage{a}$.
And $sage{b}$ is not an integer. As a decimal it is approximately
$sage{b.n(digits=6)}$. I would like to round it to $sage{b.n(digits=1)}$.
Another difficulty is that $sage{c}$ is an integer in fact,
and should be printed as $sage{c}$.
end{document}
The output, running in Cocalc, gives:
Notice that Sage interprets your numbers correctly: 4/2 is recognized as 2 and 1/3*3 is recognized as 1. It does need to know the format you want of non integers; but it recognizes that 5/3 is a fraction that can't be reduced and leaves it as a fraction. To force it into a decimal and to specify the number of digits we append .n(digits=6); the documentation is here.
answered Dec 31 '18 at 16:41
DJPDJP
7,09421630
add a comment |
Assuming that this is a question on how to do this with tikzmath, I'd do
documentclass{article}
usepackage{tikz}
usetikzlibrary{math}
begin{document}
tikzmath{
function myint(x) {
if abs(x-round(x)) < 0.1 then { print{x is an integernewline};
return int(round(x));
} else { print{x is not an integernewline};
return x;
}; };
integer = myint(4/2);
decimal = myint(5/3);
integerB= myint(1/3*3);
}
$integer$ is an integer, and it should be printed as 2.
And $pgfmathprintnumber[precision=1]{decimal}$ is a decimal number, I would like to round it to 1.7.
Another difficulty is that $integerB$ is an integer in fact,
and should be printed as 1.
end{document}
Note that I used pgfmathprintnumber to round to one digit after the dot. Of course, you can remove the prints.
answered Dec 31 '18 at 17:50
marmotmarmot
90.4k4104195
add a comment |
documentclass{article}
usepackage{xintexpr}% recommended package
newcommandtest[1]{xintifboolexpr{ifint(#1, 1, 0)}
{xinttheexpr reduce(#1)relaxspace (originally texttt{detokenize{#1}})
is an integer.}
{xinttheiexpr[2] #1relaxspace (originally texttt{detokenize{#1}}),
we rounded to two decimal places) is not an integer.}par
}
begin{document}
test{4/2}
test{5/3}
test{(1/3)*3}
test{(1/7 - 1/8 - 1/57)*3192}
test{(1/2 - 1/3 - 1/7 - 1/43 - 1/1807 - 1/3263443)*10650056950806}
end{document}
answered Jan 1 at 13:20
jfbujfbu
46.3k66148
strangely xintexpr hasifint(expression, YES, NO)function but notisint(expression)which would evaluate to 1 or 0 directly.
– jfbu
Jan 1 at 13:21
about "Alternatively, it would also be great if it is possible to determine whether the first two decimal numbers are sufficiently close to 0 or 1." this can be done by xintexpr of course as it computes exactly with arbitrarily big fractions. But I don't know exactly what is asked here.
– jfbu
Jan 1 at 13:23
for comparisonfpeval{(1/2 - 1/3 - 1/7 - 1/43 - 1/1807 - 1/3263443)*10650056950806}evaluates to0.9999211586333241, exactly likexintthefloatexpr...relax. But the latter afterxintDigits:=48;will evaluate to1.00000000000000000000000000000000000522205263811.
– jfbu
Jan 1 at 13:46
FWIW, using the default precision settings, Lua evaluates(1/2 - 1/3 - 1/7 - 1/43 - 1/1807 - 1/3263443)*10650056950806as1.000285363444. Of course,round{comp{(1/2 - 1/3 - 1/7 - 1/43 - 1/1807 - 1/3263443)*10650056950806}}{1}(whereroundandcompare defined in my answer) produces1. Whew!
– Mico
Jan 1 at 14:08
add a comment |
Here's a LuaLaTeX-based solution. integer (4/2) and integerB ((1/3)*3) evaluate to integers automatically according to Lua rules. The LaTeX macro round, which takes two arguments, lets users round numbers to a specified set of digits after the decimal marker.
% !TEX TS-program = lualatex
documentclass{article}
% Set up 2 auxilliary Lua functions to round numbers
directlua{
function math.round_int ( x )
return x>=0 and math.floor(x+0.5) or math.ceil(x-0.5)
end
function math.round ( x , n )
return ( math.round_int ( x*10^n ) / 10^n )
end
}
newcommandcomp[1]{directlua{tex.sprint(#1)}}
newcommandround[2]{directlua{tex.sprint(math.round(#1,#2))}}
newcommand{integer}{comp{4/2}}
newcommand{decimal}{comp{5/3}}
newcommand{integerB}{comp{(1/3)*3}}
begin{document}
integer is an integer.
integerB is also an integer.
decimal is a decimal number. I would like to round it to round{decimal}{1}.
end{document}
answered Jan 1 at 12:59
MicoMico
274k30371758
1
does(1/7-1/8-1/57)*3192evaluate to an integer in Lua? (just curious...)
– jfbu
Jan 1 at 13:37
1
in case answer is yes (perhaps from some distributivity done automatically), I have(1/7 - 1/8 - 1/57)*(3000+192)up my sleeve :)
– jfbu
Jan 1 at 13:43
@jfbu - Both expressions evaluate to0.99999999999998(13 nines followed by an 8), using the standard number of significant digits used intex.sprint. If I reduced that number by 1 or 2 digits, one is back to1(pun intended).
– Mico
Jan 1 at 13:46
2
I thinkxintthefloatexprworks withxintDigits:=2;but I am not sure with only 1 digit :) you are really quite a challenge!
– jfbu
Jan 1 at 13:53
add a comment |
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5 Answers
5
active
oldest
votes
5 Answers
5
active
oldest
votes
active
oldest
votes
active
oldest
votes
One cannot say from its floating point representation whether the output of an arithmetic operation involving division or non rational operations is actually an integer.
You can consider the l3fp module of expl3, available through the package xfp.
documentclass{article}
usepackage{xfp}
begin{document}
$fpeval{5/3}$ is a decimal number, I would like to round it to
$fpeval{round(5/3,1)}$ or to $fpeval{round(5/3,2)}$
Another difficulty is that $fpeval{(1/3)*3}$ is an integer in fact,
and should be printed as $fpeval{round((1/3)*3,1)}$ or
$fpeval{round((1/3)*3,2)}$.
end{document}
answered Dec 31 '18 at 16:10
egregegreg
711k8618913174
add a comment |
One cannot say from its floating point representation whether the output of an arithmetic operation involving division or non rational operations is actually an integer.
You can consider the l3fp module of expl3, available through the package xfp.
documentclass{article}
usepackage{xfp}
begin{document}
$fpeval{5/3}$ is a decimal number, I would like to round it to
$fpeval{round(5/3,1)}$ or to $fpeval{round(5/3,2)}$
Another difficulty is that $fpeval{(1/3)*3}$ is an integer in fact,
and should be printed as $fpeval{round((1/3)*3,1)}$ or
$fpeval{round((1/3)*3,2)}$.
end{document}
answered Dec 31 '18 at 16:10
egregegreg
711k8618913174
add a comment |
One cannot say from its floating point representation whether the output of an arithmetic operation involving division or non rational operations is actually an integer.
You can consider the l3fp module of expl3, available through the package xfp.
documentclass{article}
usepackage{xfp}
begin{document}
$fpeval{5/3}$ is a decimal number, I would like to round it to
$fpeval{round(5/3,1)}$ or to $fpeval{round(5/3,2)}$
Another difficulty is that $fpeval{(1/3)*3}$ is an integer in fact,
and should be printed as $fpeval{round((1/3)*3,1)}$ or
$fpeval{round((1/3)*3,2)}$.
end{document}
answered Dec 31 '18 at 16:10
egregegreg
711k8618913174
One cannot say from its floating point representation whether the output of an arithmetic operation involving division or non rational operations is actually an integer.
You can consider the l3fp module of expl3, available through the package xfp.
documentclass{article}
usepackage{xfp}
begin{document}
$fpeval{5/3}$ is a decimal number, I would like to round it to
$fpeval{round(5/3,1)}$ or to $fpeval{round(5/3,2)}$
Another difficulty is that $fpeval{(1/3)*3}$ is an integer in fact,
and should be printed as $fpeval{round((1/3)*3,1)}$ or
$fpeval{round((1/3)*3,2)}$.
end{document}
answered Dec 31 '18 at 16:10
egregegreg
711k8618913174
answered Dec 31 '18 at 16:10
egregegreg
711k8618913174
answered Dec 31 '18 at 16:10
egregegreg
711k8618913174
answered Dec 31 '18 at 16:10
egregegreg
711k8618913174
711k8618913174
add a comment |
add a comment |
If the accuracy of your numbers is important you might consider farming that out to a computer algebra system (CAS). The sagetex package relies on the CAS Sage; the documentation can be found on CTAN right here. Documentation on Sage is found here .Sage is not part of the LaTeX distribution (it's big) so it needs to be installed on your computer or, even easier, accessed through a free Cocalc account.
documentclass{article}
usepackage{sagetex}
usepackage{tikz}
usetikzlibrary{math}
begin{document}
begin{sagesilent}
a = 4/2
b = 5/3
c = 1/3*3
end{sagesilent}
$sage{a}$ is an integer, and it should be printed as $sage{a}$.
And $sage{b}$ is not an integer. As a decimal it is approximately
$sage{b.n(digits=6)}$. I would like to round it to $sage{b.n(digits=1)}$.
Another difficulty is that $sage{c}$ is an integer in fact,
and should be printed as $sage{c}$.
end{document}
The output, running in Cocalc, gives:
Notice that Sage interprets your numbers correctly: 4/2 is recognized as 2 and 1/3*3 is recognized as 1. It does need to know the format you want of non integers; but it recognizes that 5/3 is a fraction that can't be reduced and leaves it as a fraction. To force it into a decimal and to specify the number of digits we append .n(digits=6); the documentation is here.
answered Dec 31 '18 at 16:41
DJPDJP
7,09421630
add a comment |
If the accuracy of your numbers is important you might consider farming that out to a computer algebra system (CAS). The sagetex package relies on the CAS Sage; the documentation can be found on CTAN right here. Documentation on Sage is found here .Sage is not part of the LaTeX distribution (it's big) so it needs to be installed on your computer or, even easier, accessed through a free Cocalc account.
documentclass{article}
usepackage{sagetex}
usepackage{tikz}
usetikzlibrary{math}
begin{document}
begin{sagesilent}
a = 4/2
b = 5/3
c = 1/3*3
end{sagesilent}
$sage{a}$ is an integer, and it should be printed as $sage{a}$.
And $sage{b}$ is not an integer. As a decimal it is approximately
$sage{b.n(digits=6)}$. I would like to round it to $sage{b.n(digits=1)}$.
Another difficulty is that $sage{c}$ is an integer in fact,
and should be printed as $sage{c}$.
end{document}
The output, running in Cocalc, gives:
Notice that Sage interprets your numbers correctly: 4/2 is recognized as 2 and 1/3*3 is recognized as 1. It does need to know the format you want of non integers; but it recognizes that 5/3 is a fraction that can't be reduced and leaves it as a fraction. To force it into a decimal and to specify the number of digits we append .n(digits=6); the documentation is here.
answered Dec 31 '18 at 16:41
DJPDJP
7,09421630
add a comment |
If the accuracy of your numbers is important you might consider farming that out to a computer algebra system (CAS). The sagetex package relies on the CAS Sage; the documentation can be found on CTAN right here. Documentation on Sage is found here .Sage is not part of the LaTeX distribution (it's big) so it needs to be installed on your computer or, even easier, accessed through a free Cocalc account.
documentclass{article}
usepackage{sagetex}
usepackage{tikz}
usetikzlibrary{math}
begin{document}
begin{sagesilent}
a = 4/2
b = 5/3
c = 1/3*3
end{sagesilent}
$sage{a}$ is an integer, and it should be printed as $sage{a}$.
And $sage{b}$ is not an integer. As a decimal it is approximately
$sage{b.n(digits=6)}$. I would like to round it to $sage{b.n(digits=1)}$.
Another difficulty is that $sage{c}$ is an integer in fact,
and should be printed as $sage{c}$.
end{document}
The output, running in Cocalc, gives:
Notice that Sage interprets your numbers correctly: 4/2 is recognized as 2 and 1/3*3 is recognized as 1. It does need to know the format you want of non integers; but it recognizes that 5/3 is a fraction that can't be reduced and leaves it as a fraction. To force it into a decimal and to specify the number of digits we append .n(digits=6); the documentation is here.
answered Dec 31 '18 at 16:41
DJPDJP
7,09421630
If the accuracy of your numbers is important you might consider farming that out to a computer algebra system (CAS). The sagetex package relies on the CAS Sage; the documentation can be found on CTAN right here. Documentation on Sage is found here .Sage is not part of the LaTeX distribution (it's big) so it needs to be installed on your computer or, even easier, accessed through a free Cocalc account.
documentclass{article}
usepackage{sagetex}
usepackage{tikz}
usetikzlibrary{math}
begin{document}
begin{sagesilent}
a = 4/2
b = 5/3
c = 1/3*3
end{sagesilent}
$sage{a}$ is an integer, and it should be printed as $sage{a}$.
And $sage{b}$ is not an integer. As a decimal it is approximately
$sage{b.n(digits=6)}$. I would like to round it to $sage{b.n(digits=1)}$.
Another difficulty is that $sage{c}$ is an integer in fact,
and should be printed as $sage{c}$.
end{document}
The output, running in Cocalc, gives:
Notice that Sage interprets your numbers correctly: 4/2 is recognized as 2 and 1/3*3 is recognized as 1. It does need to know the format you want of non integers; but it recognizes that 5/3 is a fraction that can't be reduced and leaves it as a fraction. To force it into a decimal and to specify the number of digits we append .n(digits=6); the documentation is here.
answered Dec 31 '18 at 16:41
DJPDJP
7,09421630
answered Dec 31 '18 at 16:41
DJPDJP
7,09421630
answered Dec 31 '18 at 16:41
DJPDJP
7,09421630
answered Dec 31 '18 at 16:41
DJPDJP
7,09421630
7,09421630
add a comment |
add a comment |
Assuming that this is a question on how to do this with tikzmath, I'd do
documentclass{article}
usepackage{tikz}
usetikzlibrary{math}
begin{document}
tikzmath{
function myint(x) {
if abs(x-round(x)) < 0.1 then { print{x is an integernewline};
return int(round(x));
} else { print{x is not an integernewline};
return x;
}; };
integer = myint(4/2);
decimal = myint(5/3);
integerB= myint(1/3*3);
}
$integer$ is an integer, and it should be printed as 2.
And $pgfmathprintnumber[precision=1]{decimal}$ is a decimal number, I would like to round it to 1.7.
Another difficulty is that $integerB$ is an integer in fact,
and should be printed as 1.
end{document}
Note that I used pgfmathprintnumber to round to one digit after the dot. Of course, you can remove the prints.
answered Dec 31 '18 at 17:50
marmotmarmot
90.4k4104195
add a comment |
Assuming that this is a question on how to do this with tikzmath, I'd do
documentclass{article}
usepackage{tikz}
usetikzlibrary{math}
begin{document}
tikzmath{
function myint(x) {
if abs(x-round(x)) < 0.1 then { print{x is an integernewline};
return int(round(x));
} else { print{x is not an integernewline};
return x;
}; };
integer = myint(4/2);
decimal = myint(5/3);
integerB= myint(1/3*3);
}
$integer$ is an integer, and it should be printed as 2.
And $pgfmathprintnumber[precision=1]{decimal}$ is a decimal number, I would like to round it to 1.7.
Another difficulty is that $integerB$ is an integer in fact,
and should be printed as 1.
end{document}
Note that I used pgfmathprintnumber to round to one digit after the dot. Of course, you can remove the prints.
answered Dec 31 '18 at 17:50
marmotmarmot
90.4k4104195
add a comment |
Assuming that this is a question on how to do this with tikzmath, I'd do
documentclass{article}
usepackage{tikz}
usetikzlibrary{math}
begin{document}
tikzmath{
function myint(x) {
if abs(x-round(x)) < 0.1 then { print{x is an integernewline};
return int(round(x));
} else { print{x is not an integernewline};
return x;
}; };
integer = myint(4/2);
decimal = myint(5/3);
integerB= myint(1/3*3);
}
$integer$ is an integer, and it should be printed as 2.
And $pgfmathprintnumber[precision=1]{decimal}$ is a decimal number, I would like to round it to 1.7.
Another difficulty is that $integerB$ is an integer in fact,
and should be printed as 1.
end{document}
Note that I used pgfmathprintnumber to round to one digit after the dot. Of course, you can remove the prints.
answered Dec 31 '18 at 17:50
marmotmarmot
90.4k4104195
Assuming that this is a question on how to do this with tikzmath, I'd do
documentclass{article}
usepackage{tikz}
usetikzlibrary{math}
begin{document}
tikzmath{
function myint(x) {
if abs(x-round(x)) < 0.1 then { print{x is an integernewline};
return int(round(x));
} else { print{x is not an integernewline};
return x;
}; };
integer = myint(4/2);
decimal = myint(5/3);
integerB= myint(1/3*3);
}
$integer$ is an integer, and it should be printed as 2.
And $pgfmathprintnumber[precision=1]{decimal}$ is a decimal number, I would like to round it to 1.7.
Another difficulty is that $integerB$ is an integer in fact,
and should be printed as 1.
end{document}
Note that I used pgfmathprintnumber to round to one digit after the dot. Of course, you can remove the prints.
answered Dec 31 '18 at 17:50
marmotmarmot
90.4k4104195
edited Dec 31 '18 at 18:20
answered Dec 31 '18 at 17:50
marmotmarmot
90.4k4104195
answered Dec 31 '18 at 17:50
marmotmarmot
90.4k4104195
answered Dec 31 '18 at 17:50
marmotmarmot
90.4k4104195
90.4k4104195
add a comment |
add a comment |
documentclass{article}
usepackage{xintexpr}% recommended package
newcommandtest[1]{xintifboolexpr{ifint(#1, 1, 0)}
{xinttheexpr reduce(#1)relaxspace (originally texttt{detokenize{#1}})
is an integer.}
{xinttheiexpr[2] #1relaxspace (originally texttt{detokenize{#1}}),
we rounded to two decimal places) is not an integer.}par
}
begin{document}
test{4/2}
test{5/3}
test{(1/3)*3}
test{(1/7 - 1/8 - 1/57)*3192}
test{(1/2 - 1/3 - 1/7 - 1/43 - 1/1807 - 1/3263443)*10650056950806}
end{document}
answered Jan 1 at 13:20
jfbujfbu
46.3k66148
strangely xintexpr hasifint(expression, YES, NO)function but notisint(expression)which would evaluate to 1 or 0 directly.
– jfbu
Jan 1 at 13:21
about "Alternatively, it would also be great if it is possible to determine whether the first two decimal numbers are sufficiently close to 0 or 1." this can be done by xintexpr of course as it computes exactly with arbitrarily big fractions. But I don't know exactly what is asked here.
– jfbu
Jan 1 at 13:23
for comparisonfpeval{(1/2 - 1/3 - 1/7 - 1/43 - 1/1807 - 1/3263443)*10650056950806}evaluates to0.9999211586333241, exactly likexintthefloatexpr...relax. But the latter afterxintDigits:=48;will evaluate to1.00000000000000000000000000000000000522205263811.
– jfbu
Jan 1 at 13:46
FWIW, using the default precision settings, Lua evaluates(1/2 - 1/3 - 1/7 - 1/43 - 1/1807 - 1/3263443)*10650056950806as1.000285363444. Of course,round{comp{(1/2 - 1/3 - 1/7 - 1/43 - 1/1807 - 1/3263443)*10650056950806}}{1}(whereroundandcompare defined in my answer) produces1. Whew!
– Mico
Jan 1 at 14:08
add a comment |
documentclass{article}
usepackage{xintexpr}% recommended package
newcommandtest[1]{xintifboolexpr{ifint(#1, 1, 0)}
{xinttheexpr reduce(#1)relaxspace (originally texttt{detokenize{#1}})
is an integer.}
{xinttheiexpr[2] #1relaxspace (originally texttt{detokenize{#1}}),
we rounded to two decimal places) is not an integer.}par
}
begin{document}
test{4/2}
test{5/3}
test{(1/3)*3}
test{(1/7 - 1/8 - 1/57)*3192}
test{(1/2 - 1/3 - 1/7 - 1/43 - 1/1807 - 1/3263443)*10650056950806}
end{document}
answered Jan 1 at 13:20
jfbujfbu
46.3k66148
strangely xintexpr hasifint(expression, YES, NO)function but notisint(expression)which would evaluate to 1 or 0 directly.
– jfbu
Jan 1 at 13:21
about "Alternatively, it would also be great if it is possible to determine whether the first two decimal numbers are sufficiently close to 0 or 1." this can be done by xintexpr of course as it computes exactly with arbitrarily big fractions. But I don't know exactly what is asked here.
– jfbu
Jan 1 at 13:23
for comparisonfpeval{(1/2 - 1/3 - 1/7 - 1/43 - 1/1807 - 1/3263443)*10650056950806}evaluates to0.9999211586333241, exactly likexintthefloatexpr...relax. But the latter afterxintDigits:=48;will evaluate to1.00000000000000000000000000000000000522205263811.
– jfbu
Jan 1 at 13:46
FWIW, using the default precision settings, Lua evaluates(1/2 - 1/3 - 1/7 - 1/43 - 1/1807 - 1/3263443)*10650056950806as1.000285363444. Of course,round{comp{(1/2 - 1/3 - 1/7 - 1/43 - 1/1807 - 1/3263443)*10650056950806}}{1}(whereroundandcompare defined in my answer) produces1. Whew!
– Mico
Jan 1 at 14:08
add a comment |
documentclass{article}
usepackage{xintexpr}% recommended package
newcommandtest[1]{xintifboolexpr{ifint(#1, 1, 0)}
{xinttheexpr reduce(#1)relaxspace (originally texttt{detokenize{#1}})
is an integer.}
{xinttheiexpr[2] #1relaxspace (originally texttt{detokenize{#1}}),
we rounded to two decimal places) is not an integer.}par
}
begin{document}
test{4/2}
test{5/3}
test{(1/3)*3}
test{(1/7 - 1/8 - 1/57)*3192}
test{(1/2 - 1/3 - 1/7 - 1/43 - 1/1807 - 1/3263443)*10650056950806}
end{document}
answered Jan 1 at 13:20
jfbujfbu
46.3k66148
documentclass{article}
usepackage{xintexpr}% recommended package
newcommandtest[1]{xintifboolexpr{ifint(#1, 1, 0)}
{xinttheexpr reduce(#1)relaxspace (originally texttt{detokenize{#1}})
is an integer.}
{xinttheiexpr[2] #1relaxspace (originally texttt{detokenize{#1}}),
we rounded to two decimal places) is not an integer.}par
}
begin{document}
test{4/2}
test{5/3}
test{(1/3)*3}
test{(1/7 - 1/8 - 1/57)*3192}
test{(1/2 - 1/3 - 1/7 - 1/43 - 1/1807 - 1/3263443)*10650056950806}
end{document}
answered Jan 1 at 13:20
jfbujfbu
46.3k66148
edited Jan 1 at 13:40
answered Jan 1 at 13:20
jfbujfbu
46.3k66148
answered Jan 1 at 13:20
jfbujfbu
46.3k66148
answered Jan 1 at 13:20
jfbujfbu
46.3k66148
46.3k66148
strangely xintexpr hasifint(expression, YES, NO)function but notisint(expression)which would evaluate to 1 or 0 directly.
– jfbu
Jan 1 at 13:21
about "Alternatively, it would also be great if it is possible to determine whether the first two decimal numbers are sufficiently close to 0 or 1." this can be done by xintexpr of course as it computes exactly with arbitrarily big fractions. But I don't know exactly what is asked here.
– jfbu
Jan 1 at 13:23
for comparisonfpeval{(1/2 - 1/3 - 1/7 - 1/43 - 1/1807 - 1/3263443)*10650056950806}evaluates to0.9999211586333241, exactly likexintthefloatexpr...relax. But the latter afterxintDigits:=48;will evaluate to1.00000000000000000000000000000000000522205263811.
– jfbu
Jan 1 at 13:46
FWIW, using the default precision settings, Lua evaluates(1/2 - 1/3 - 1/7 - 1/43 - 1/1807 - 1/3263443)*10650056950806as1.000285363444. Of course,round{comp{(1/2 - 1/3 - 1/7 - 1/43 - 1/1807 - 1/3263443)*10650056950806}}{1}(whereroundandcompare defined in my answer) produces1. Whew!
– Mico
Jan 1 at 14:08
add a comment |
strangely xintexpr hasifint(expression, YES, NO)function but notisint(expression)which would evaluate to 1 or 0 directly.
– jfbu
Jan 1 at 13:21
about "Alternatively, it would also be great if it is possible to determine whether the first two decimal numbers are sufficiently close to 0 or 1." this can be done by xintexpr of course as it computes exactly with arbitrarily big fractions. But I don't know exactly what is asked here.
– jfbu
Jan 1 at 13:23
for comparisonfpeval{(1/2 - 1/3 - 1/7 - 1/43 - 1/1807 - 1/3263443)*10650056950806}evaluates to0.9999211586333241, exactly likexintthefloatexpr...relax. But the latter afterxintDigits:=48;will evaluate to1.00000000000000000000000000000000000522205263811.
– jfbu
Jan 1 at 13:46
FWIW, using the default precision settings, Lua evaluates(1/2 - 1/3 - 1/7 - 1/43 - 1/1807 - 1/3263443)*10650056950806as1.000285363444. Of course,round{comp{(1/2 - 1/3 - 1/7 - 1/43 - 1/1807 - 1/3263443)*10650056950806}}{1}(whereroundandcompare defined in my answer) produces1. Whew!
– Mico
Jan 1 at 14:08
strangely xintexpr has
ifint(expression, YES, NO) function but not isint(expression) which would evaluate to 1 or 0 directly.– jfbu
Jan 1 at 13:21
strangely xintexpr has
ifint(expression, YES, NO) function but not isint(expression) which would evaluate to 1 or 0 directly.– jfbu
Jan 1 at 13:21
about "Alternatively, it would also be great if it is possible to determine whether the first two decimal numbers are sufficiently close to 0 or 1." this can be done by xintexpr of course as it computes exactly with arbitrarily big fractions. But I don't know exactly what is asked here.
– jfbu
Jan 1 at 13:23
about "Alternatively, it would also be great if it is possible to determine whether the first two decimal numbers are sufficiently close to 0 or 1." this can be done by xintexpr of course as it computes exactly with arbitrarily big fractions. But I don't know exactly what is asked here.
– jfbu
Jan 1 at 13:23
for comparison
fpeval{(1/2 - 1/3 - 1/7 - 1/43 - 1/1807 - 1/3263443)*10650056950806} evaluates to 0.9999211586333241, exactly like xintthefloatexpr...relax. But the latter after xintDigits:=48; will evaluate to 1.00000000000000000000000000000000000522205263811.– jfbu
Jan 1 at 13:46
for comparison
fpeval{(1/2 - 1/3 - 1/7 - 1/43 - 1/1807 - 1/3263443)*10650056950806} evaluates to 0.9999211586333241, exactly like xintthefloatexpr...relax. But the latter after xintDigits:=48; will evaluate to 1.00000000000000000000000000000000000522205263811.– jfbu
Jan 1 at 13:46
FWIW, using the default precision settings, Lua evaluates
(1/2 - 1/3 - 1/7 - 1/43 - 1/1807 - 1/3263443)*10650056950806 as 1.000285363444. Of course, round{comp{(1/2 - 1/3 - 1/7 - 1/43 - 1/1807 - 1/3263443)*10650056950806}}{1} (where round and comp are defined in my answer) produces 1. Whew!– Mico
Jan 1 at 14:08
FWIW, using the default precision settings, Lua evaluates
(1/2 - 1/3 - 1/7 - 1/43 - 1/1807 - 1/3263443)*10650056950806 as 1.000285363444. Of course, round{comp{(1/2 - 1/3 - 1/7 - 1/43 - 1/1807 - 1/3263443)*10650056950806}}{1} (where round and comp are defined in my answer) produces 1. Whew!– Mico
Jan 1 at 14:08
add a comment |
Here's a LuaLaTeX-based solution. integer (4/2) and integerB ((1/3)*3) evaluate to integers automatically according to Lua rules. The LaTeX macro round, which takes two arguments, lets users round numbers to a specified set of digits after the decimal marker.
% !TEX TS-program = lualatex
documentclass{article}
% Set up 2 auxilliary Lua functions to round numbers
directlua{
function math.round_int ( x )
return x>=0 and math.floor(x+0.5) or math.ceil(x-0.5)
end
function math.round ( x , n )
return ( math.round_int ( x*10^n ) / 10^n )
end
}
newcommandcomp[1]{directlua{tex.sprint(#1)}}
newcommandround[2]{directlua{tex.sprint(math.round(#1,#2))}}
newcommand{integer}{comp{4/2}}
newcommand{decimal}{comp{5/3}}
newcommand{integerB}{comp{(1/3)*3}}
begin{document}
integer is an integer.
integerB is also an integer.
decimal is a decimal number. I would like to round it to round{decimal}{1}.
end{document}
answered Jan 1 at 12:59
MicoMico
274k30371758
1
does(1/7-1/8-1/57)*3192evaluate to an integer in Lua? (just curious...)
– jfbu
Jan 1 at 13:37
1
in case answer is yes (perhaps from some distributivity done automatically), I have(1/7 - 1/8 - 1/57)*(3000+192)up my sleeve :)
– jfbu
Jan 1 at 13:43
@jfbu - Both expressions evaluate to0.99999999999998(13 nines followed by an 8), using the standard number of significant digits used intex.sprint. If I reduced that number by 1 or 2 digits, one is back to1(pun intended).
– Mico
Jan 1 at 13:46
2
I thinkxintthefloatexprworks withxintDigits:=2;but I am not sure with only 1 digit :) you are really quite a challenge!
– jfbu
Jan 1 at 13:53
add a comment |
Here's a LuaLaTeX-based solution. integer (4/2) and integerB ((1/3)*3) evaluate to integers automatically according to Lua rules. The LaTeX macro round, which takes two arguments, lets users round numbers to a specified set of digits after the decimal marker.
% !TEX TS-program = lualatex
documentclass{article}
% Set up 2 auxilliary Lua functions to round numbers
directlua{
function math.round_int ( x )
return x>=0 and math.floor(x+0.5) or math.ceil(x-0.5)
end
function math.round ( x , n )
return ( math.round_int ( x*10^n ) / 10^n )
end
}
newcommandcomp[1]{directlua{tex.sprint(#1)}}
newcommandround[2]{directlua{tex.sprint(math.round(#1,#2))}}
newcommand{integer}{comp{4/2}}
newcommand{decimal}{comp{5/3}}
newcommand{integerB}{comp{(1/3)*3}}
begin{document}
integer is an integer.
integerB is also an integer.
decimal is a decimal number. I would like to round it to round{decimal}{1}.
end{document}
answered Jan 1 at 12:59
MicoMico
274k30371758
1
does(1/7-1/8-1/57)*3192evaluate to an integer in Lua? (just curious...)
– jfbu
Jan 1 at 13:37
1
in case answer is yes (perhaps from some distributivity done automatically), I have(1/7 - 1/8 - 1/57)*(3000+192)up my sleeve :)
– jfbu
Jan 1 at 13:43
@jfbu - Both expressions evaluate to0.99999999999998(13 nines followed by an 8), using the standard number of significant digits used intex.sprint. If I reduced that number by 1 or 2 digits, one is back to1(pun intended).
– Mico
Jan 1 at 13:46
2
I thinkxintthefloatexprworks withxintDigits:=2;but I am not sure with only 1 digit :) you are really quite a challenge!
– jfbu
Jan 1 at 13:53
add a comment |
Here's a LuaLaTeX-based solution. integer (4/2) and integerB ((1/3)*3) evaluate to integers automatically according to Lua rules. The LaTeX macro round, which takes two arguments, lets users round numbers to a specified set of digits after the decimal marker.
% !TEX TS-program = lualatex
documentclass{article}
% Set up 2 auxilliary Lua functions to round numbers
directlua{
function math.round_int ( x )
return x>=0 and math.floor(x+0.5) or math.ceil(x-0.5)
end
function math.round ( x , n )
return ( math.round_int ( x*10^n ) / 10^n )
end
}
newcommandcomp[1]{directlua{tex.sprint(#1)}}
newcommandround[2]{directlua{tex.sprint(math.round(#1,#2))}}
newcommand{integer}{comp{4/2}}
newcommand{decimal}{comp{5/3}}
newcommand{integerB}{comp{(1/3)*3}}
begin{document}
integer is an integer.
integerB is also an integer.
decimal is a decimal number. I would like to round it to round{decimal}{1}.
end{document}
answered Jan 1 at 12:59
MicoMico
274k30371758
Here's a LuaLaTeX-based solution. integer (4/2) and integerB ((1/3)*3) evaluate to integers automatically according to Lua rules. The LaTeX macro round, which takes two arguments, lets users round numbers to a specified set of digits after the decimal marker.
% !TEX TS-program = lualatex
documentclass{article}
% Set up 2 auxilliary Lua functions to round numbers
directlua{
function math.round_int ( x )
return x>=0 and math.floor(x+0.5) or math.ceil(x-0.5)
end
function math.round ( x , n )
return ( math.round_int ( x*10^n ) / 10^n )
end
}
newcommandcomp[1]{directlua{tex.sprint(#1)}}
newcommandround[2]{directlua{tex.sprint(math.round(#1,#2))}}
newcommand{integer}{comp{4/2}}
newcommand{decimal}{comp{5/3}}
newcommand{integerB}{comp{(1/3)*3}}
begin{document}
integer is an integer.
integerB is also an integer.
decimal is a decimal number. I would like to round it to round{decimal}{1}.
end{document}
answered Jan 1 at 12:59
MicoMico
274k30371758
answered Jan 1 at 12:59
MicoMico
274k30371758
answered Jan 1 at 12:59
MicoMico
274k30371758
answered Jan 1 at 12:59
MicoMico
274k30371758
274k30371758
1
does(1/7-1/8-1/57)*3192evaluate to an integer in Lua? (just curious...)
– jfbu
Jan 1 at 13:37
1
in case answer is yes (perhaps from some distributivity done automatically), I have(1/7 - 1/8 - 1/57)*(3000+192)up my sleeve :)
– jfbu
Jan 1 at 13:43
@jfbu - Both expressions evaluate to0.99999999999998(13 nines followed by an 8), using the standard number of significant digits used intex.sprint. If I reduced that number by 1 or 2 digits, one is back to1(pun intended).
– Mico
Jan 1 at 13:46
2
I thinkxintthefloatexprworks withxintDigits:=2;but I am not sure with only 1 digit :) you are really quite a challenge!
– jfbu
Jan 1 at 13:53
add a comment |
1
does(1/7-1/8-1/57)*3192evaluate to an integer in Lua? (just curious...)
– jfbu
Jan 1 at 13:37
1
in case answer is yes (perhaps from some distributivity done automatically), I have(1/7 - 1/8 - 1/57)*(3000+192)up my sleeve :)
– jfbu
Jan 1 at 13:43
@jfbu - Both expressions evaluate to0.99999999999998(13 nines followed by an 8), using the standard number of significant digits used intex.sprint. If I reduced that number by 1 or 2 digits, one is back to1(pun intended).
– Mico
Jan 1 at 13:46
2
I thinkxintthefloatexprworks withxintDigits:=2;but I am not sure with only 1 digit :) you are really quite a challenge!
– jfbu
Jan 1 at 13:53
1
1
does
(1/7-1/8-1/57)*3192 evaluate to an integer in Lua? (just curious...)– jfbu
Jan 1 at 13:37
does
(1/7-1/8-1/57)*3192 evaluate to an integer in Lua? (just curious...)– jfbu
Jan 1 at 13:37
1
1
in case answer is yes (perhaps from some distributivity done automatically), I have
(1/7 - 1/8 - 1/57)*(3000+192) up my sleeve :)– jfbu
Jan 1 at 13:43
in case answer is yes (perhaps from some distributivity done automatically), I have
(1/7 - 1/8 - 1/57)*(3000+192) up my sleeve :)– jfbu
Jan 1 at 13:43
@jfbu - Both expressions evaluate to
0.99999999999998 (13 nines followed by an 8), using the standard number of significant digits used in tex.sprint. If I reduced that number by 1 or 2 digits, one is back to 1 (pun intended).– Mico
Jan 1 at 13:46
@jfbu - Both expressions evaluate to
0.99999999999998 (13 nines followed by an 8), using the standard number of significant digits used in tex.sprint. If I reduced that number by 1 or 2 digits, one is back to 1 (pun intended).– Mico
Jan 1 at 13:46
2
2
I think
xintthefloatexpr works with xintDigits:=2; but I am not sure with only 1 digit :) you are really quite a challenge!– jfbu
Jan 1 at 13:53
I think
xintthefloatexpr works with xintDigits:=2; but I am not sure with only 1 digit :) you are really quite a challenge!– jfbu
Jan 1 at 13:53
add a comment |
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