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Formatting table with a lua for loop

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.everyoneloves__top-leaderboard:empty,.everyoneloves__mid-leaderboard:empty,.everyoneloves__bot-mid-leaderboard:empty margin-bottom:0;

3

I was trying to get a table like this

 1
2 3 4
5 6 7
8 9 10

with this function

function myTable()
 t = 1, 2, 3, 4, 5, 6, 7, 8, 9, 10
 for k = 1, #t do
 if k == 1 then 
 tex.print(string.format([["& %d & \"]], t[1]))
 else
 tex.print(string.format([["%d & %d & %d \"]], 
 t[k], t[k + 1], t[k + 2]))
 k = k + 2
 end 
 end
end

but it says:

ua:25: bad argument #4 to 'format' (number expected, got nil)

How to do that? Looks like lua's got a weird way of implementing for loop, am I wrong?

In my document I've:

documentclassarticle
usepackageharfload,fontspec,amsmath
setmainfontKalpurush[RawFeature=mode=harf]
parindent 0pt
directluarequire("test.lua")

begindocument

begintable[h]
 centering
 begintabularccc
 directluamyTable()
 endtabular
endtable

enddocument

luatex lua harftex

share|improve this question

edited Jun 3 at 4:46

Emon Haque

asked Jun 3 at 4:03

Emon HaqueEmon Haque

1116

add a comment | 

3

I was trying to get a table like this

 1
2 3 4
5 6 7
8 9 10

with this function

function myTable()
 t = 1, 2, 3, 4, 5, 6, 7, 8, 9, 10
 for k = 1, #t do
 if k == 1 then 
 tex.print(string.format([["& %d & \"]], t[1]))
 else
 tex.print(string.format([["%d & %d & %d \"]], 
 t[k], t[k + 1], t[k + 2]))
 k = k + 2
 end 
 end
end

but it says:

ua:25: bad argument #4 to 'format' (number expected, got nil)

How to do that? Looks like lua's got a weird way of implementing for loop, am I wrong?

In my document I've:

documentclassarticle
usepackageharfload,fontspec,amsmath
setmainfontKalpurush[RawFeature=mode=harf]
parindent 0pt
directluarequire("test.lua")

begindocument

begintable[h]
 centering
 begintabularccc
 directluamyTable()
 endtabular
endtable

enddocument

luatex lua harftex

share|improve this question

edited Jun 3 at 4:46

Emon Haque

asked Jun 3 at 4:03

Emon HaqueEmon Haque

1116

add a comment | 

I was trying to get a table like this

 1
2 3 4
5 6 7
8 9 10

with this function

function myTable()
 t = 1, 2, 3, 4, 5, 6, 7, 8, 9, 10
 for k = 1, #t do
 if k == 1 then 
 tex.print(string.format([["& %d & \"]], t[1]))
 else
 tex.print(string.format([["%d & %d & %d \"]], 
 t[k], t[k + 1], t[k + 2]))
 k = k + 2
 end 
 end
end

but it says:

ua:25: bad argument #4 to 'format' (number expected, got nil)

How to do that? Looks like lua's got a weird way of implementing for loop, am I wrong?

In my document I've:

documentclassarticle
usepackageharfload,fontspec,amsmath
setmainfontKalpurush[RawFeature=mode=harf]
parindent 0pt
directluarequire("test.lua")

begindocument

begintable[h]
 centering
 begintabularccc
 directluamyTable()
 endtabular
endtable

enddocument

luatex lua harftex

share|improve this question

edited Jun 3 at 4:46

Emon Haque

asked Jun 3 at 4:03

Emon HaqueEmon Haque

1116

 1
2 3 4
5 6 7
8 9 10

function myTable()
 t = 1, 2, 3, 4, 5, 6, 7, 8, 9, 10
 for k = 1, #t do
 if k == 1 then 
 tex.print(string.format([["& %d & \"]], t[1]))
 else
 tex.print(string.format([["%d & %d & %d \"]], 
 t[k], t[k + 1], t[k + 2]))
 k = k + 2
 end 
 end
end

ua:25: bad argument #4 to 'format' (number expected, got nil)

documentclassarticle
usepackageharfload,fontspec,amsmath
setmainfontKalpurush[RawFeature=mode=harf]
parindent 0pt
directluarequire("test.lua")

begindocument

begintable[h]
 centering
 begintabularccc
 directluamyTable()
 endtabular
endtable

enddocument

share|improve this question

edited Jun 3 at 4:46

Emon Haque

asked Jun 3 at 4:03

Emon HaqueEmon Haque

1116

share|improve this question

edited Jun 3 at 4:46

Emon Haque

asked Jun 3 at 4:03

Emon HaqueEmon Haque

1116

asked Jun 3 at 4:03

Emon HaqueEmon Haque

1116

asked Jun 3 at 4:03

Emon HaqueEmon Haque

1116

1 Answer
1

active

oldest

votes

5

From the lua.org site:

... you should never change the value of the control variable [of a for loop]: The effect of such changes is unpredictable.

In short, it's not a good idea to increment the value of k (the control variable of your for loop) "by hand".

I therefore think that you should use a while loop instead of a for loop. Then, increment k by 1 if k==1 is true and by 3 if it is not. Finally, you should omit the " symbols from the first argument of string.format; alternatively, omit the [[ and ]] directives and replace \ with \\.)

documentclassarticle
usepackageluacode
beginluacode
function myTable()
 t = 1, 2, 3, 4, 5, 6, 7, 8, 9, 10
 k=1
 while k<#t do
 if k == 1 then 
 tex.print(string.format([[ & %d & \]], t[k]))
 k=k+1
 else
 tex.print(string.format([[%d & %d & %d \]], 
 t[k], t[k + 1], t[k + 2]))
 k = k + 3
 end 
 end
end
endluacode

begindocument
begintabularccc
 directluamyTable()
endtabular
enddocument

share|improve this answer

edited Jun 3 at 7:12

answered Jun 3 at 4:17

MicoMico

294k32406797

add a comment | 

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5

From the lua.org site:

... you should never change the value of the control variable [of a for loop]: The effect of such changes is unpredictable.

In short, it's not a good idea to increment the value of k (the control variable of your for loop) "by hand".

I therefore think that you should use a while loop instead of a for loop. Then, increment k by 1 if k==1 is true and by 3 if it is not. Finally, you should omit the " symbols from the first argument of string.format; alternatively, omit the [[ and ]] directives and replace \ with \\.)

documentclassarticle
usepackageluacode
beginluacode
function myTable()
 t = 1, 2, 3, 4, 5, 6, 7, 8, 9, 10
 k=1
 while k<#t do
 if k == 1 then 
 tex.print(string.format([[ & %d & \]], t[k]))
 k=k+1
 else
 tex.print(string.format([[%d & %d & %d \]], 
 t[k], t[k + 1], t[k + 2]))
 k = k + 3
 end 
 end
end
endluacode

begindocument
begintabularccc
 directluamyTable()
endtabular
enddocument

share|improve this answer

edited Jun 3 at 7:12

answered Jun 3 at 4:17

MicoMico

294k32406797

add a comment | 

5

From the lua.org site:

... you should never change the value of the control variable [of a for loop]: The effect of such changes is unpredictable.

In short, it's not a good idea to increment the value of k (the control variable of your for loop) "by hand".

I therefore think that you should use a while loop instead of a for loop. Then, increment k by 1 if k==1 is true and by 3 if it is not. Finally, you should omit the " symbols from the first argument of string.format; alternatively, omit the [[ and ]] directives and replace \ with \\.)

documentclassarticle
usepackageluacode
beginluacode
function myTable()
 t = 1, 2, 3, 4, 5, 6, 7, 8, 9, 10
 k=1
 while k<#t do
 if k == 1 then 
 tex.print(string.format([[ & %d & \]], t[k]))
 k=k+1
 else
 tex.print(string.format([[%d & %d & %d \]], 
 t[k], t[k + 1], t[k + 2]))
 k = k + 3
 end 
 end
end
endluacode

begindocument
begintabularccc
 directluamyTable()
endtabular
enddocument

share|improve this answer

edited Jun 3 at 7:12

answered Jun 3 at 4:17

MicoMico

294k32406797

add a comment | 

From the lua.org site:

... you should never change the value of the control variable [of a for loop]: The effect of such changes is unpredictable.

In short, it's not a good idea to increment the value of k (the control variable of your for loop) "by hand".

I therefore think that you should use a while loop instead of a for loop. Then, increment k by 1 if k==1 is true and by 3 if it is not. Finally, you should omit the " symbols from the first argument of string.format; alternatively, omit the [[ and ]] directives and replace \ with \\.)

documentclassarticle
usepackageluacode
beginluacode
function myTable()
 t = 1, 2, 3, 4, 5, 6, 7, 8, 9, 10
 k=1
 while k<#t do
 if k == 1 then 
 tex.print(string.format([[ & %d & \]], t[k]))
 k=k+1
 else
 tex.print(string.format([[%d & %d & %d \]], 
 t[k], t[k + 1], t[k + 2]))
 k = k + 3
 end 
 end
end
endluacode

begindocument
begintabularccc
 directluamyTable()
endtabular
enddocument

share|improve this answer

edited Jun 3 at 7:12

answered Jun 3 at 4:17

MicoMico

294k32406797

documentclassarticle
usepackageluacode
beginluacode
function myTable()
 t = 1, 2, 3, 4, 5, 6, 7, 8, 9, 10
 k=1
 while k<#t do
 if k == 1 then 
 tex.print(string.format([[ & %d & \]], t[k]))
 k=k+1
 else
 tex.print(string.format([[%d & %d & %d \]], 
 t[k], t[k + 1], t[k + 2]))
 k = k + 3
 end 
 end
end
endluacode

begindocument
begintabularccc
 directluamyTable()
endtabular
enddocument

share|improve this answer

edited Jun 3 at 7:12

answered Jun 3 at 4:17

MicoMico

294k32406797

answered Jun 3 at 4:17

MicoMico

294k32406797

answered Jun 3 at 4:17

MicoMico

294k32406797