I have matrix and and i want to do increment in a do loop and and i want to store in list using append to. It works fine for less values, But when i want to do it for values around 1 million the program is slow.
x = {{0.1},{0.2}};xm = {{0.4},{0.5}};
Do[datax[i] = {}; dataxm[i] = {};, {i, 1, 2}]
Do[
x = x + t;
xm = xm + t;
Do[AppendTo[datax[i], Flatten[{t, x[[i]]}]];, {i, 1, 2}];
Do[AppendTo[dataxm[i], Flatten[{t, xm[[i]]}]];, {i, 1, 2}], {t, 0, 1, 0.1}]
datax[1]
{{0., 0.1}, {0.1, 0.2}, {0.2, 0.4}, {0.3, 0.7}, {0.4, 1.1}, {0.5, 1.6}, {0.6, 2.2}, {0.7, 2.9}, {0.8, 3.7}, {0.9, 4.6}, {1., 5.6}}
datax[2]
{{0., 0.2}, {0.1, 0.3}, {0.2, 0.5}, {0.3, 0.8}, {0.4, 1.2}, {0.5,
1.7}, {0.6, 2.3}, {0.7, 3.}, {0.8, 3.8}, {0.9, 4.7}, {1., 5.7}}
dataxm[1]
{{0., 0.4}, {0.1, 0.5}, {0.2, 0.7}, {0.3, 1.}, {0.4,
1.4}, {0.5, 1.9}, {0.6, 2.5}, {0.7, 3.2}, {0.8, 4.}, {0.9,
4.9}, {1., 5.9}}
dataxm[2]
{{0., 0.5}, {0.1, 0.6}, {0.2, 0.8}, {0.3, 1.1}, {0.4,
1.5}, {0.5, 2.}, {0.6, 2.6}, {0.7, 3.3}, {0.8, 4.1}, {0.9, 5.}, {1.,
6.}}
Similar to the above I want to use reap and sow functions to speed it up but i can store only the last value. Why?
x = {{0.1},{0.2}};xm = {{0.4},{0.5}};
Do[
x = x + t;xm = xm + t;
Do[datax[i] = Reap[Sow[{t, x[[i]]}]][[2, 1]], {i, 1, 2}];
Do[dataxm[i] = Reap[Sow[{t, xm[[i]]}]][[2, 1]], {i, 1, 2}];
, {t, 0, 1, 0.1}]
datax[1]
{{1., {5.6}}}
datax[2]
{{1., {5.7}}}
datax[1]
{{1., {5.9}}}
datax[2]
{{1., {6.0}}
Can anyone help in fixing this issue? Thanks in advance
Answer
You need olny one Reap
for all Sow
s -- that's primarily the advantage of using them.
Moreover, you may use tags to obtain datax[1]
and datax[2]
:
x = {{0.1}, {0.2}};
ClearAll[datax];
datax = Association@Reap[
Do[
x = x + t;
Do[
Sow[
Flatten[{t, x[[tag]]}],
tag
],
{tag, 1, 2}],
{t, 0, 1, 0.1}],
_, Rule][[2]]
<|1 -> {{0., 0.1}, {0.1, 0.2}, {0.2, 0.4}, {0.3, 0.7}, {0.4, 1.1}, {0.5, 1.6}, {0.6, 2.2}, {0.7, 2.9}, {0.8, 3.7}, {0.9, 4.6}, {1., 5.6}}, 2 -> {{0., 0.2}, {0.1, 0.3}, {0.2, 0.5}, {0.3, 0.8}, {0.4, 1.2}, {0.5, 1.7}, {0.6, 2.3}, {0.7, 3.}, {0.8, 3.8}, {0.9, 4.7}, {1., 5.7}}|>
Now you can access datax[1]
and datax[2]
as before.
Because the question changed: This should allow you to assemble more than one list in one go:
data = Merge[
Reap[
Do[
x = x + t;
xm = xm + t;
Do[
Sow[Flatten[{t, x[[i]]}], "datax" -> i];
Sow[Flatten[{t, xm[[i]]}], "dataxm" -> i]
,
{i, 1, 2}], {t, 0, 1, 0.1}],
_, #1[[1]] -> Association[#1[[2]] -> #2] &][[2]]
, Association
];
Now data["datax"]
and data["dataxm"]
should behave like datax
and dataxm
.
You can obtain essentially the same result much more efficiently with arrays produced by Table
:
x = {0.1, 0.2};
xm = {0.4, 0.5};
{datax, dataxm} = Transpose[
Table[{{{t, t}, x += t}, {{t, t}, xm += t}}, {t, 0, 1, 0.1}],
{3, 1, 4, 2}];
Now you have (notice the double brackets):
datax[[1]]
datax[[2]]
dataxm[[1]]
dataxm[[2]]
Even faster for this particular problem is to avoid cursive addition and to use vectorized operations.
The Reap
/Sow
method from above:
n = 1000000;
x = {0.1, 0.2};
xm = {0.4, 0.5};
First@AbsoluteTiming[
data = Merge[
Reap[
Do[
x = x + t;
xm = xm + t;
Do[
Sow[Flatten[{t, x[[i]]}], "datax" -> i];
Sow[Flatten[{t, xm[[i]]}], "dataxm" -> i]
,
{i, 1, 2}], {t, 0., 1., 1./n}],
_, #1[[1]] -> Association[#1[[2]] -> #2] &][[2]]
, Association
];
]
11.4845
The Table
method from above:
x = {0.1, 0.2};
xm = {0.4, 0.5};
First@AbsoluteTiming[
{datax, dataxm} =
Transpose[
Table[{{{t, t}, x += t}, {{t, t}, xm += t}}, {t, 0., 1.,
1./n}], {3, 1, 4, 2}];
]
2.74062
Vectorized version:
x = {0.1, 0.2};
xm = {0.4, 0.5};
First@AbsoluteTiming[
tlist = Subdivide[0., 1., n];
slist = Accumulate[tlist];
datax2 = {
Transpose[{tlist, x[[1]] + slist}],
Transpose[{tlist, x[[2]] + slist}]
};
dataxm2 = {
Transpose[{tlist, xm[[1]] + slist}],
Transpose[{tlist, xm[[2]] + slist}]
};
]
0.061355
So, the latter is 44/187 time faster than the other methods. So you should take these considerations into account when using Reap
and Sow
.
Errors:
Max[Abs[data["datax"][1] - datax[[1]]]]
Max[Abs[data["datax"][2] - datax[[2]]]]
Max[Abs[data["dataxm"][1] - dataxm[[1]]]]
Max[Abs[data["dataxm"][2] - dataxm[[2]]]]
Max[Abs[datax - datax2]]
Max[Abs[dataxm - dataxm2]]
0.
0.
0.
0.
5.82077*10^-11
5.82077*10^-11
Comments
Post a Comment