How to populate a table with results from a while function
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up vote
5
down vote
favorite
Does anybody know how to save the intermediate results inside a while function in a table?
Here is my code. I would like to save the x,y,counter values in a table in every step that while takes. Thank you!
counter = 0;
While[counter < 5,
counter = counter + 1;
sol = NSolve[2 x + y - counter == 0,
3 y + 5 x - 2 counter == 0, x, y];
]
replacement table infinite-loop
add a comment |Â
up vote
5
down vote
favorite
Does anybody know how to save the intermediate results inside a while function in a table?
Here is my code. I would like to save the x,y,counter values in a table in every step that while takes. Thank you!
counter = 0;
While[counter < 5,
counter = counter + 1;
sol = NSolve[2 x + y - counter == 0,
3 y + 5 x - 2 counter == 0, x, y];
]
replacement table infinite-loop
add a comment |Â
up vote
5
down vote
favorite
up vote
5
down vote
favorite
Does anybody know how to save the intermediate results inside a while function in a table?
Here is my code. I would like to save the x,y,counter values in a table in every step that while takes. Thank you!
counter = 0;
While[counter < 5,
counter = counter + 1;
sol = NSolve[2 x + y - counter == 0,
3 y + 5 x - 2 counter == 0, x, y];
]
replacement table infinite-loop
Does anybody know how to save the intermediate results inside a while function in a table?
Here is my code. I would like to save the x,y,counter values in a table in every step that while takes. Thank you!
counter = 0;
While[counter < 5,
counter = counter + 1;
sol = NSolve[2 x + y - counter == 0,
3 y + 5 x - 2 counter == 0, x, y];
]
replacement table infinite-loop
replacement table infinite-loop
asked Sep 8 at 9:20
harazogo
483
483
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add a comment |Â
4 Answers
4
active
oldest
votes
up vote
2
down vote
accepted
counter = 0;
sol = ;
While[counter < 5, counter = counter + 1;
AppendTo[sol, Flatten[counter, x, y /.
NSolve[2 x + y - counter == 0, 3 y + 5 x - 2 counter == 0, x, y]]]]
sol
1, 1., -1., 2, 2., -2., 3, 3., -3., 4, 4., -4., 5, 5., -5.
or
counter = 0;
sol2 = ConstantArray[0, 5];
While[counter < 5, counter = counter + 1;
sol2[[counter]] = Flatten@counter, x, y /.
NSolve[2 x + y - counter == 0, 3 y + 5 x - 2 counter == 0, x, y]]
sol2
1, 1., -1., 2, 2., -2., 3, 3., -3., 4, 4., -4., 5, 5., -5.
add a comment |Â
up vote
7
down vote
Use Sow
and Reap
:
counter = 0;
Reap @ While[counter < 5, counter = counter + 1;
sol = NSolve[2 x + y - counter == 0,
3 y + 5 x - 2 counter == 0, x, y];
Sow[x, y, counter /. sol]]
add a comment |Â
up vote
4
down vote
I think other answer take your question too literally. You do not really need While
. IMHO Mathematica way of solving this is to use Table
.
Assuming, that your equation always gives a single solution:
Table[
x, y, counter /.
First@NSolve[2 x + y - counter == 0,
3 y + 5 x - 2 counter == 0, x, y], counter, 5] // TableForm
add a comment |Â
up vote
3
down vote
To reach the same end, as you mentioned Table
(a digression: if you know Python, you can compare Wolfram's Table
with its list comprehensions, which is attached with more conciseness and readability.), the most direct method is just to use it:
Table[
Append[
NSolve[2 x + y - counter == 0, 3 y + 5 x - 2 counter == 0, x, y],
counter
],
counter, 4
]
x -> 1., y -> -1., 1, x -> 2., y -> -2., 2,
x -> 3., y -> -3., 3, x -> 4., y -> -4., 4
2
UsingAppend
you abuse the list of solutions fromNSolve
. I would make aNSolve,counter
instead ofAppend[NSolve,counter]
.
â Johu
Sep 8 at 11:44
add a comment |Â
4 Answers
4
active
oldest
votes
4 Answers
4
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
2
down vote
accepted
counter = 0;
sol = ;
While[counter < 5, counter = counter + 1;
AppendTo[sol, Flatten[counter, x, y /.
NSolve[2 x + y - counter == 0, 3 y + 5 x - 2 counter == 0, x, y]]]]
sol
1, 1., -1., 2, 2., -2., 3, 3., -3., 4, 4., -4., 5, 5., -5.
or
counter = 0;
sol2 = ConstantArray[0, 5];
While[counter < 5, counter = counter + 1;
sol2[[counter]] = Flatten@counter, x, y /.
NSolve[2 x + y - counter == 0, 3 y + 5 x - 2 counter == 0, x, y]]
sol2
1, 1., -1., 2, 2., -2., 3, 3., -3., 4, 4., -4., 5, 5., -5.
add a comment |Â
up vote
2
down vote
accepted
counter = 0;
sol = ;
While[counter < 5, counter = counter + 1;
AppendTo[sol, Flatten[counter, x, y /.
NSolve[2 x + y - counter == 0, 3 y + 5 x - 2 counter == 0, x, y]]]]
sol
1, 1., -1., 2, 2., -2., 3, 3., -3., 4, 4., -4., 5, 5., -5.
or
counter = 0;
sol2 = ConstantArray[0, 5];
While[counter < 5, counter = counter + 1;
sol2[[counter]] = Flatten@counter, x, y /.
NSolve[2 x + y - counter == 0, 3 y + 5 x - 2 counter == 0, x, y]]
sol2
1, 1., -1., 2, 2., -2., 3, 3., -3., 4, 4., -4., 5, 5., -5.
add a comment |Â
up vote
2
down vote
accepted
up vote
2
down vote
accepted
counter = 0;
sol = ;
While[counter < 5, counter = counter + 1;
AppendTo[sol, Flatten[counter, x, y /.
NSolve[2 x + y - counter == 0, 3 y + 5 x - 2 counter == 0, x, y]]]]
sol
1, 1., -1., 2, 2., -2., 3, 3., -3., 4, 4., -4., 5, 5., -5.
or
counter = 0;
sol2 = ConstantArray[0, 5];
While[counter < 5, counter = counter + 1;
sol2[[counter]] = Flatten@counter, x, y /.
NSolve[2 x + y - counter == 0, 3 y + 5 x - 2 counter == 0, x, y]]
sol2
1, 1., -1., 2, 2., -2., 3, 3., -3., 4, 4., -4., 5, 5., -5.
counter = 0;
sol = ;
While[counter < 5, counter = counter + 1;
AppendTo[sol, Flatten[counter, x, y /.
NSolve[2 x + y - counter == 0, 3 y + 5 x - 2 counter == 0, x, y]]]]
sol
1, 1., -1., 2, 2., -2., 3, 3., -3., 4, 4., -4., 5, 5., -5.
or
counter = 0;
sol2 = ConstantArray[0, 5];
While[counter < 5, counter = counter + 1;
sol2[[counter]] = Flatten@counter, x, y /.
NSolve[2 x + y - counter == 0, 3 y + 5 x - 2 counter == 0, x, y]]
sol2
1, 1., -1., 2, 2., -2., 3, 3., -3., 4, 4., -4., 5, 5., -5.
edited Sep 8 at 10:07
answered Sep 8 at 9:24
kglr
161k8185384
161k8185384
add a comment |Â
add a comment |Â
up vote
7
down vote
Use Sow
and Reap
:
counter = 0;
Reap @ While[counter < 5, counter = counter + 1;
sol = NSolve[2 x + y - counter == 0,
3 y + 5 x - 2 counter == 0, x, y];
Sow[x, y, counter /. sol]]
add a comment |Â
up vote
7
down vote
Use Sow
and Reap
:
counter = 0;
Reap @ While[counter < 5, counter = counter + 1;
sol = NSolve[2 x + y - counter == 0,
3 y + 5 x - 2 counter == 0, x, y];
Sow[x, y, counter /. sol]]
add a comment |Â
up vote
7
down vote
up vote
7
down vote
Use Sow
and Reap
:
counter = 0;
Reap @ While[counter < 5, counter = counter + 1;
sol = NSolve[2 x + y - counter == 0,
3 y + 5 x - 2 counter == 0, x, y];
Sow[x, y, counter /. sol]]
Use Sow
and Reap
:
counter = 0;
Reap @ While[counter < 5, counter = counter + 1;
sol = NSolve[2 x + y - counter == 0,
3 y + 5 x - 2 counter == 0, x, y];
Sow[x, y, counter /. sol]]
answered Sep 8 at 9:27
LouisB
4,1541616
4,1541616
add a comment |Â
add a comment |Â
up vote
4
down vote
I think other answer take your question too literally. You do not really need While
. IMHO Mathematica way of solving this is to use Table
.
Assuming, that your equation always gives a single solution:
Table[
x, y, counter /.
First@NSolve[2 x + y - counter == 0,
3 y + 5 x - 2 counter == 0, x, y], counter, 5] // TableForm
add a comment |Â
up vote
4
down vote
I think other answer take your question too literally. You do not really need While
. IMHO Mathematica way of solving this is to use Table
.
Assuming, that your equation always gives a single solution:
Table[
x, y, counter /.
First@NSolve[2 x + y - counter == 0,
3 y + 5 x - 2 counter == 0, x, y], counter, 5] // TableForm
add a comment |Â
up vote
4
down vote
up vote
4
down vote
I think other answer take your question too literally. You do not really need While
. IMHO Mathematica way of solving this is to use Table
.
Assuming, that your equation always gives a single solution:
Table[
x, y, counter /.
First@NSolve[2 x + y - counter == 0,
3 y + 5 x - 2 counter == 0, x, y], counter, 5] // TableForm
I think other answer take your question too literally. You do not really need While
. IMHO Mathematica way of solving this is to use Table
.
Assuming, that your equation always gives a single solution:
Table[
x, y, counter /.
First@NSolve[2 x + y - counter == 0,
3 y + 5 x - 2 counter == 0, x, y], counter, 5] // TableForm
edited Sep 8 at 13:42
answered Sep 8 at 11:51
Johu
3,3381033
3,3381033
add a comment |Â
add a comment |Â
up vote
3
down vote
To reach the same end, as you mentioned Table
(a digression: if you know Python, you can compare Wolfram's Table
with its list comprehensions, which is attached with more conciseness and readability.), the most direct method is just to use it:
Table[
Append[
NSolve[2 x + y - counter == 0, 3 y + 5 x - 2 counter == 0, x, y],
counter
],
counter, 4
]
x -> 1., y -> -1., 1, x -> 2., y -> -2., 2,
x -> 3., y -> -3., 3, x -> 4., y -> -4., 4
2
UsingAppend
you abuse the list of solutions fromNSolve
. I would make aNSolve,counter
instead ofAppend[NSolve,counter]
.
â Johu
Sep 8 at 11:44
add a comment |Â
up vote
3
down vote
To reach the same end, as you mentioned Table
(a digression: if you know Python, you can compare Wolfram's Table
with its list comprehensions, which is attached with more conciseness and readability.), the most direct method is just to use it:
Table[
Append[
NSolve[2 x + y - counter == 0, 3 y + 5 x - 2 counter == 0, x, y],
counter
],
counter, 4
]
x -> 1., y -> -1., 1, x -> 2., y -> -2., 2,
x -> 3., y -> -3., 3, x -> 4., y -> -4., 4
2
UsingAppend
you abuse the list of solutions fromNSolve
. I would make aNSolve,counter
instead ofAppend[NSolve,counter]
.
â Johu
Sep 8 at 11:44
add a comment |Â
up vote
3
down vote
up vote
3
down vote
To reach the same end, as you mentioned Table
(a digression: if you know Python, you can compare Wolfram's Table
with its list comprehensions, which is attached with more conciseness and readability.), the most direct method is just to use it:
Table[
Append[
NSolve[2 x + y - counter == 0, 3 y + 5 x - 2 counter == 0, x, y],
counter
],
counter, 4
]
x -> 1., y -> -1., 1, x -> 2., y -> -2., 2,
x -> 3., y -> -3., 3, x -> 4., y -> -4., 4
To reach the same end, as you mentioned Table
(a digression: if you know Python, you can compare Wolfram's Table
with its list comprehensions, which is attached with more conciseness and readability.), the most direct method is just to use it:
Table[
Append[
NSolve[2 x + y - counter == 0, 3 y + 5 x - 2 counter == 0, x, y],
counter
],
counter, 4
]
x -> 1., y -> -1., 1, x -> 2., y -> -2., 2,
x -> 3., y -> -3., 3, x -> 4., y -> -4., 4
edited Sep 8 at 10:09
answered Sep 8 at 9:40
ÃÂûÃÂþñýôÃÂÿàÃÂõóó
2,016721
2,016721
2
UsingAppend
you abuse the list of solutions fromNSolve
. I would make aNSolve,counter
instead ofAppend[NSolve,counter]
.
â Johu
Sep 8 at 11:44
add a comment |Â
2
UsingAppend
you abuse the list of solutions fromNSolve
. I would make aNSolve,counter
instead ofAppend[NSolve,counter]
.
â Johu
Sep 8 at 11:44
2
2
Using
Append
you abuse the list of solutions from NSolve
. I would make a NSolve,counter
instead of Append[NSolve,counter]
.â Johu
Sep 8 at 11:44
Using
Append
you abuse the list of solutions from NSolve
. I would make a NSolve,counter
instead of Append[NSolve,counter]
.â Johu
Sep 8 at 11:44
add a comment |Â
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