How to programmatically specify multiple iterators?

up vote
3
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How to use Row[Table[B[i],0,i,i,0,2],","] directly in Do command? I mean that the following command

Row[Table[B[i],0,i,i,0,2],","]

gives

B[0], 0, 0,B[1], 0, 1,B[2],0,2

But, the following command returns the error

"Do::nliter: Non-list iterator Row[Table[B[i], 0, i, i, 0, 2], ,] at position 2 does not evaluate to a real numeric value."

Do[Print[B[0]+B[1]+B[2]],Row[Table[B[i],0,i,i,0,2],","]]

Of course, one can type instead by hand the following:

Do[Print[B[0]+B[1]+B[2]],B[0],0,0,B[1],0,1,B[2],0,2]

But, I feel that typing an output by hand again is not really an optimal method.

edited 15 mins ago

Carl Woll

56.7k272147

asked 51 mins ago

veo

1254

New contributor

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up vote
3
down vote

favorite

How to use Row[Table[B[i],0,i,i,0,2],","] directly in Do command? I mean that the following command

Row[Table[B[i],0,i,i,0,2],","]

gives

B[0], 0, 0,B[1], 0, 1,B[2],0,2

But, the following command returns the error

"Do::nliter: Non-list iterator Row[Table[B[i], 0, i, i, 0, 2], ,] at position 2 does not evaluate to a real numeric value."

Do[Print[B[0]+B[1]+B[2]],Row[Table[B[i],0,i,i,0,2],","]]

Of course, one can type instead by hand the following:

Do[Print[B[0]+B[1]+B[2]],B[0],0,0,B[1],0,1,B[2],0,2]

But, I feel that typing an output by hand again is not really an optimal method.

edited 15 mins ago

Carl Woll

56.7k272147

asked 51 mins ago

veo

1254

New contributor

add a commentÂ |Â

up vote
3
down vote

favorite

How to use Row[Table[B[i],0,i,i,0,2],","] directly in Do command? I mean that the following command

Row[Table[B[i],0,i,i,0,2],","]

gives

B[0], 0, 0,B[1], 0, 1,B[2],0,2

But, the following command returns the error

"Do::nliter: Non-list iterator Row[Table[B[i], 0, i, i, 0, 2], ,] at position 2 does not evaluate to a real numeric value."

Do[Print[B[0]+B[1]+B[2]],Row[Table[B[i],0,i,i,0,2],","]]

Of course, one can type instead by hand the following:

Do[Print[B[0]+B[1]+B[2]],B[0],0,0,B[1],0,1,B[2],0,2]

But, I feel that typing an output by hand again is not really an optimal method.

edited 15 mins ago

Carl Woll

56.7k272147

asked 51 mins ago

veo

1254

New contributor

How to use Row[Table[B[i],0,i,i,0,2],","] directly in Do command? I mean that the following command

Row[Table[B[i],0,i,i,0,2],","]

gives

B[0], 0, 0,B[1], 0, 1,B[2],0,2

But, the following command returns the error

"Do::nliter: Non-list iterator Row[Table[B[i], 0, i, i, 0, 2], ,] at position 2 does not evaluate to a real numeric value."

Do[Print[B[0]+B[1]+B[2]],Row[Table[B[i],0,i,i,0,2],","]]

Of course, one can type instead by hand the following:

Do[Print[B[0]+B[1]+B[2]],B[0],0,0,B[1],0,1,B[2],0,2]

But, I feel that typing an output by hand again is not really an optimal method.

list-manipulation row do

edited 15 mins ago

Carl Woll

56.7k272147

asked 51 mins ago

veo

1254

New contributor

edited 15 mins ago

Carl Woll

56.7k272147

asked 51 mins ago

veo

1254

New contributor

edited 15 mins ago

Carl Woll

56.7k272147

edited 15 mins ago

Carl Woll

56.7k272147

edited 15 mins ago

Carl Woll

56.7k272147

asked 51 mins ago

veo

1254

New contributor

asked 51 mins ago

veo

1254

asked 51 mins ago

veo

1254

New contributor

veo is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.

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3 Answers
3

active

oldest

votes

up vote
4
down vote

accepted

What you actually want is to create a Sequence from the Table to be used as your iterators.

You can do this with

Do[Print[B[0] + B[1] + B[2]], Sequence @@ Table[B[i], 0, i, i, 0, 2] // 
Evaluate]

(*0
 1
 2
 1
 2
 3*)

Or, so you don't have to force evaluation,

Do[Print[B[0] + B[1] + B[2]], ##] & @@ Table[B[i], 0, i, i, 0, 2]

edited 37 mins ago

answered 42 mins ago

That Gravity Guy

42625

add a commentÂ |Â

up vote
2
down vote

Do[Print[B[0] + B[1] + B[2]], 
 Evaluate[Sequence @@ First@ Row[Table[B[i], 0, i, i, 0, 2], ","]]]

row = Row[Table[B[i], 0, i, i, 0, 2], ","];
Do[Print[B[0] + B[1] + B[2]], Evaluate[Sequence @@ row[[1]]]]

edited 36 mins ago

answered 42 mins ago

kglr

160k8184384

add a commentÂ |Â

up vote
1
down vote

Perhaps you can avoid Do and instead use Tuples:

Tuples @ Range[0, 0, 1, 2]

0, 0, 0, 0, 0, 1, 0, 0, 2, 0, 1, 0, 0, 1, 1, 0, 1, 2

You can then use Total to sum each tuple:

Total[
 Tuples @ Range[0, 0, 1, 2],
 2
]

0, 1, 2, 1, 2, 3

answered 16 mins ago

Carl Woll

56.7k272147

add a commentÂ |Â

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3 Answers
3

active

oldest

votes

3 Answers
3

active

oldest

votes

up vote
4
down vote

accepted

What you actually want is to create a Sequence from the Table to be used as your iterators.

You can do this with

Do[Print[B[0] + B[1] + B[2]], Sequence @@ Table[B[i], 0, i, i, 0, 2] // 
Evaluate]

(*0
 1
 2
 1
 2
 3*)

Or, so you don't have to force evaluation,

Do[Print[B[0] + B[1] + B[2]], ##] & @@ Table[B[i], 0, i, i, 0, 2]

edited 37 mins ago

answered 42 mins ago

That Gravity Guy

42625

add a commentÂ |Â

up vote
4
down vote

accepted

What you actually want is to create a Sequence from the Table to be used as your iterators.

You can do this with

Do[Print[B[0] + B[1] + B[2]], Sequence @@ Table[B[i], 0, i, i, 0, 2] // 
Evaluate]

(*0
 1
 2
 1
 2
 3*)

Or, so you don't have to force evaluation,

Do[Print[B[0] + B[1] + B[2]], ##] & @@ Table[B[i], 0, i, i, 0, 2]

edited 37 mins ago

answered 42 mins ago

That Gravity Guy

42625

add a commentÂ |Â

up vote
4
down vote

accepted

What you actually want is to create a Sequence from the Table to be used as your iterators.

You can do this with

Do[Print[B[0] + B[1] + B[2]], Sequence @@ Table[B[i], 0, i, i, 0, 2] // 
Evaluate]

(*0
 1
 2
 1
 2
 3*)

Or, so you don't have to force evaluation,

Do[Print[B[0] + B[1] + B[2]], ##] & @@ Table[B[i], 0, i, i, 0, 2]

edited 37 mins ago

answered 42 mins ago

That Gravity Guy

42625

What you actually want is to create a Sequence from the Table to be used as your iterators.

You can do this with

Do[Print[B[0] + B[1] + B[2]], Sequence @@ Table[B[i], 0, i, i, 0, 2] // 
Evaluate]

(*0
 1
 2
 1
 2
 3*)

Or, so you don't have to force evaluation,

Do[Print[B[0] + B[1] + B[2]], ##] & @@ Table[B[i], 0, i, i, 0, 2]

edited 37 mins ago

answered 42 mins ago

That Gravity Guy

42625

edited 37 mins ago

answered 42 mins ago

That Gravity Guy

42625

answered 42 mins ago

That Gravity Guy

42625

answered 42 mins ago

That Gravity Guy

42625

add a commentÂ |Â

up vote
2
down vote

Do[Print[B[0] + B[1] + B[2]], 
 Evaluate[Sequence @@ First@ Row[Table[B[i], 0, i, i, 0, 2], ","]]]

row = Row[Table[B[i], 0, i, i, 0, 2], ","];
Do[Print[B[0] + B[1] + B[2]], Evaluate[Sequence @@ row[[1]]]]

edited 36 mins ago

answered 42 mins ago

kglr

160k8184384

add a commentÂ |Â

up vote
2
down vote

Do[Print[B[0] + B[1] + B[2]], 
 Evaluate[Sequence @@ First@ Row[Table[B[i], 0, i, i, 0, 2], ","]]]

row = Row[Table[B[i], 0, i, i, 0, 2], ","];
Do[Print[B[0] + B[1] + B[2]], Evaluate[Sequence @@ row[[1]]]]

edited 36 mins ago

answered 42 mins ago

kglr

160k8184384

add a commentÂ |Â

up vote
2
down vote

Do[Print[B[0] + B[1] + B[2]], 
 Evaluate[Sequence @@ First@ Row[Table[B[i], 0, i, i, 0, 2], ","]]]

row = Row[Table[B[i], 0, i, i, 0, 2], ","];
Do[Print[B[0] + B[1] + B[2]], Evaluate[Sequence @@ row[[1]]]]

edited 36 mins ago

answered 42 mins ago

kglr

160k8184384

Do[Print[B[0] + B[1] + B[2]], 
 Evaluate[Sequence @@ First@ Row[Table[B[i], 0, i, i, 0, 2], ","]]]

row = Row[Table[B[i], 0, i, i, 0, 2], ","];
Do[Print[B[0] + B[1] + B[2]], Evaluate[Sequence @@ row[[1]]]]

edited 36 mins ago

answered 42 mins ago

kglr

160k8184384

edited 36 mins ago

answered 42 mins ago

kglr

160k8184384

answered 42 mins ago

kglr

160k8184384

answered 42 mins ago

kglr

160k8184384

add a commentÂ |Â

up vote
1
down vote

Perhaps you can avoid Do and instead use Tuples:

Tuples @ Range[0, 0, 1, 2]

0, 0, 0, 0, 0, 1, 0, 0, 2, 0, 1, 0, 0, 1, 1, 0, 1, 2

You can then use Total to sum each tuple:

Total[
 Tuples @ Range[0, 0, 1, 2],
 2
]

0, 1, 2, 1, 2, 3

answered 16 mins ago

Carl Woll

56.7k272147

add a commentÂ |Â

up vote
1
down vote

Perhaps you can avoid Do and instead use Tuples:

Tuples @ Range[0, 0, 1, 2]

0, 0, 0, 0, 0, 1, 0, 0, 2, 0, 1, 0, 0, 1, 1, 0, 1, 2

You can then use Total to sum each tuple:

Total[
 Tuples @ Range[0, 0, 1, 2],
 2
]

0, 1, 2, 1, 2, 3

answered 16 mins ago

Carl Woll

56.7k272147

add a commentÂ |Â

up vote
1
down vote

Perhaps you can avoid Do and instead use Tuples:

Tuples @ Range[0, 0, 1, 2]

0, 0, 0, 0, 0, 1, 0, 0, 2, 0, 1, 0, 0, 1, 1, 0, 1, 2

You can then use Total to sum each tuple:

Total[
 Tuples @ Range[0, 0, 1, 2],
 2
]

0, 1, 2, 1, 2, 3

answered 16 mins ago

Carl Woll

56.7k272147

Perhaps you can avoid Do and instead use Tuples:

Tuples @ Range[0, 0, 1, 2]

0, 0, 0, 0, 0, 1, 0, 0, 2, 0, 1, 0, 0, 1, 1, 0, 1, 2

You can then use Total to sum each tuple:

Total[
 Tuples @ Range[0, 0, 1, 2],
 2
]

0, 1, 2, 1, 2, 3

answered 16 mins ago

Carl Woll

56.7k272147

answered 16 mins ago

Carl Woll

56.7k272147

answered 16 mins ago

Carl Woll

56.7k272147

answered 16 mins ago

Carl Woll

56.7k272147

add a commentÂ |Â

veo is a new contributor. Be nice, and check out our Code of Conduct.

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