Fastest/cleanest way to pad data with zero-data

up vote
3
down vote

favorite

I want to pad a set of data in x, y pairs with x, 0, x, 0 on each side of each data point.

I know I can do this like so:

zeroPaddedData[xdata_, ydata_] :=

 With[padding = 
 Transpose[xdata, ConstantArray[0., Length@xdata]],
 Append[
 Riffle[
 Riffle[
 padding,
 Transpose[xdata, ydata]
 ],
 padding,
 3
 ],
 xdata[[-1]], 0.
 ]
 ]

which, for example, gives:

zpd = zeroPaddedData @@ Transpose@Table[x, Sin[x], x, 0, 2 Pi, .5]

0., 0., 0., 0., 0., 0., 0.5, 0., 0.5, 0.479426, 0.5, 
 0., 1., 0., 1., 0.841471, 1., 0., 1.5, 0., 1.5, 
 0.997495, 1.5, 0., 2., 0., 2., 0.909297, 2., 0., 2.5, 
 0., 2.5, 0.598472, 2.5, 0., 3., 0., 3., 0.14112, 3., 
 0., 3.5, 0., 3.5, -0.350783, 3.5, 0., 4., 
 0., 4., -0.756802, 4., 0., 4.5, 0., 4.5, -0.97753, 4.5, 
 0., 5., 0., 5., -0.958924, 5., 0., 5.5, 
 0., 5.5, -0.70554, 5.5, 0., 6., 0., 6., -0.279415, 6., 0.

Which will plot out like so:

zpd // ListLinePlot

enter image description here

But this is slow over larger data sets:

dats = Transpose@Table[x, Sin[x], x, 0, 2 Pi, .001];

zeroPaddedData @@ dats // RepeatedTiming // First

0.0039

How can I do this faster/cleaner?

edited 2 hours ago

asked 2 hours ago

b3m2a1

25k254147

Is zpd not ending with 6., 0. deliberate?
â€“Â J. M. is computer-lessâ™¦
2 hours ago

@J.M.iscomputer-less nope I just accidentally a result from before I added the Append
â€“Â b3m2a1
2 hours ago

add a commentÂ |Â

up vote
3
down vote

favorite

I want to pad a set of data in x, y pairs with x, 0, x, 0 on each side of each data point.

I know I can do this like so:

zeroPaddedData[xdata_, ydata_] :=

 With[padding = 
 Transpose[xdata, ConstantArray[0., Length@xdata]],
 Append[
 Riffle[
 Riffle[
 padding,
 Transpose[xdata, ydata]
 ],
 padding,
 3
 ],
 xdata[[-1]], 0.
 ]
 ]

which, for example, gives:

zpd = zeroPaddedData @@ Transpose@Table[x, Sin[x], x, 0, 2 Pi, .5]

0., 0., 0., 0., 0., 0., 0.5, 0., 0.5, 0.479426, 0.5, 
 0., 1., 0., 1., 0.841471, 1., 0., 1.5, 0., 1.5, 
 0.997495, 1.5, 0., 2., 0., 2., 0.909297, 2., 0., 2.5, 
 0., 2.5, 0.598472, 2.5, 0., 3., 0., 3., 0.14112, 3., 
 0., 3.5, 0., 3.5, -0.350783, 3.5, 0., 4., 
 0., 4., -0.756802, 4., 0., 4.5, 0., 4.5, -0.97753, 4.5, 
 0., 5., 0., 5., -0.958924, 5., 0., 5.5, 
 0., 5.5, -0.70554, 5.5, 0., 6., 0., 6., -0.279415, 6., 0.

Which will plot out like so:

zpd // ListLinePlot

enter image description here

But this is slow over larger data sets:

dats = Transpose@Table[x, Sin[x], x, 0, 2 Pi, .001];

zeroPaddedData @@ dats // RepeatedTiming // First

0.0039

How can I do this faster/cleaner?

edited 2 hours ago

asked 2 hours ago

b3m2a1

25k254147

Is zpd not ending with 6., 0. deliberate?
â€“Â J. M. is computer-lessâ™¦
2 hours ago

@J.M.iscomputer-less nope I just accidentally a result from before I added the Append
â€“Â b3m2a1
2 hours ago

add a commentÂ |Â

up vote
3
down vote

favorite

I want to pad a set of data in x, y pairs with x, 0, x, 0 on each side of each data point.

I know I can do this like so:

zeroPaddedData[xdata_, ydata_] :=

 With[padding = 
 Transpose[xdata, ConstantArray[0., Length@xdata]],
 Append[
 Riffle[
 Riffle[
 padding,
 Transpose[xdata, ydata]
 ],
 padding,
 3
 ],
 xdata[[-1]], 0.
 ]
 ]

which, for example, gives:

zpd = zeroPaddedData @@ Transpose@Table[x, Sin[x], x, 0, 2 Pi, .5]

0., 0., 0., 0., 0., 0., 0.5, 0., 0.5, 0.479426, 0.5, 
 0., 1., 0., 1., 0.841471, 1., 0., 1.5, 0., 1.5, 
 0.997495, 1.5, 0., 2., 0., 2., 0.909297, 2., 0., 2.5, 
 0., 2.5, 0.598472, 2.5, 0., 3., 0., 3., 0.14112, 3., 
 0., 3.5, 0., 3.5, -0.350783, 3.5, 0., 4., 
 0., 4., -0.756802, 4., 0., 4.5, 0., 4.5, -0.97753, 4.5, 
 0., 5., 0., 5., -0.958924, 5., 0., 5.5, 
 0., 5.5, -0.70554, 5.5, 0., 6., 0., 6., -0.279415, 6., 0.

Which will plot out like so:

zpd // ListLinePlot

enter image description here

But this is slow over larger data sets:

dats = Transpose@Table[x, Sin[x], x, 0, 2 Pi, .001];

zeroPaddedData @@ dats // RepeatedTiming // First

0.0039

How can I do this faster/cleaner?

edited 2 hours ago

asked 2 hours ago

b3m2a1

25k254147

I want to pad a set of data in x, y pairs with x, 0, x, 0 on each side of each data point.

I know I can do this like so:

zeroPaddedData[xdata_, ydata_] :=

 With[padding = 
 Transpose[xdata, ConstantArray[0., Length@xdata]],
 Append[
 Riffle[
 Riffle[
 padding,
 Transpose[xdata, ydata]
 ],
 padding,
 3
 ],
 xdata[[-1]], 0.
 ]
 ]

which, for example, gives:

zpd = zeroPaddedData @@ Transpose@Table[x, Sin[x], x, 0, 2 Pi, .5]

0., 0., 0., 0., 0., 0., 0.5, 0., 0.5, 0.479426, 0.5, 
 0., 1., 0., 1., 0.841471, 1., 0., 1.5, 0., 1.5, 
 0.997495, 1.5, 0., 2., 0., 2., 0.909297, 2., 0., 2.5, 
 0., 2.5, 0.598472, 2.5, 0., 3., 0., 3., 0.14112, 3., 
 0., 3.5, 0., 3.5, -0.350783, 3.5, 0., 4., 
 0., 4., -0.756802, 4., 0., 4.5, 0., 4.5, -0.97753, 4.5, 
 0., 5., 0., 5., -0.958924, 5., 0., 5.5, 
 0., 5.5, -0.70554, 5.5, 0., 6., 0., 6., -0.279415, 6., 0.

Which will plot out like so:

zpd // ListLinePlot

enter image description here

But this is slow over larger data sets:

dats = Transpose@Table[x, Sin[x], x, 0, 2 Pi, .001];

zeroPaddedData @@ dats // RepeatedTiming // First

0.0039

How can I do this faster/cleaner?

list-manipulation performance-tuning

edited 2 hours ago

asked 2 hours ago

b3m2a1

25k254147

edited 2 hours ago

asked 2 hours ago

b3m2a1

25k254147

edited 2 hours ago

asked 2 hours ago

b3m2a1

25k254147

asked 2 hours ago

b3m2a1

25k254147

asked 2 hours ago

b3m2a1

25k254147

Is zpd not ending with 6., 0. deliberate?
â€“Â J. M. is computer-lessâ™¦
2 hours ago

@J.M.iscomputer-less nope I just accidentally a result from before I added the Append
â€“Â b3m2a1
2 hours ago

add a commentÂ |Â

Is zpd not ending with 6., 0. deliberate?
â€“Â J. M. is computer-lessâ™¦
2 hours ago

@J.M.iscomputer-less nope I just accidentally a result from before I added the Append
â€“Â b3m2a1
2 hours ago

Is zpd not ending with 6., 0. deliberate?
â€“Â J. M. is computer-lessâ™¦
2 hours ago

@J.M.iscomputer-less nope I just accidentally a result from before I added the Append
â€“Â b3m2a1
2 hours ago

add a commentÂ |Â

2 Answers
2

active

oldest

votes

up vote
2
down vote

One reason for slowness is that the input data was not packed. Moreover Transpose is often faster than Riffle:

xdata, ydata = Transpose[Map[x [Function] x, Sin[x], Range[0., 2 Pi, .0001]]];

f[xdata_, ydata_] := Transpose[
 Flatten[Transpose[ConstantArray[xdata, 3]]],
 With[o = ConstantArray[0., Length[ydata]],
 Flatten[Transpose[o, ydata, o]]
 ]
 ]

a = zeroPaddedData[xdata, ydata]; // RepeatedTiming // First
b = f[xdata, ydata]; // RepeatedTiming // First
a == b

0.030

0.00165

True

answered 2 hours ago

Henrik Schumacher

42.3k261125

add a commentÂ |Â

up vote
1
down vote

Slower but simple

tab = Table[x, Sin[x], x, 0, 2 Pi, .5];
SequenceReplace[tab, a_,b_:>Sequence[a,0,a,b,a,0]] == zpd

True

Update: Using Upsample on the second argument gives a slight improvement over Henrik's f:

ClearAll[zPad]
zPad[xd_, yd_] := Transpose[Flatten[ConstantArray[xd, 3], 2, 1], Upsample[yd, 3, 2]];

xdata, ydata = Transpose[Map[x [Function] x, Sin[x], Range[0., 2 Pi, .0001]]]; 
a = zeroPaddedData[xdata, ydata]; // RepeatedTiming // First

0.024

b = f[xdata, ydata]; // RepeatedTiming // First

0.0017

c = zPad[xdata, ydata]; // RepeatedTiming // First

0.0016

a == b == c

True

Interesting to note that although Upsample is almost twice as fast as With[o = ConstantArray[0., Length[ydata]], Flatten[Transpose[o, ydata, o] ]] this advantage is not retained when combined with other steps:

r1 = Upsample[ydata, 3, 2] ; // RepeatedTiming // First

0.00061

r2 = With[o = ConstantArray[0., Length[ydata]], 
 Flatten[Transpose[o, ydata, o] ]]; // RepeatedTiming // First

0.0013

r1 == r2

True

edited 39 mins ago

answered 1 hour ago

kglr

166k8188389

add a commentÂ |Â

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2 Answers
2

active

oldest

votes

2 Answers
2

active

oldest

votes

up vote
2
down vote

One reason for slowness is that the input data was not packed. Moreover Transpose is often faster than Riffle:

xdata, ydata = Transpose[Map[x [Function] x, Sin[x], Range[0., 2 Pi, .0001]]];

f[xdata_, ydata_] := Transpose[
 Flatten[Transpose[ConstantArray[xdata, 3]]],
 With[o = ConstantArray[0., Length[ydata]],
 Flatten[Transpose[o, ydata, o]]
 ]
 ]

a = zeroPaddedData[xdata, ydata]; // RepeatedTiming // First
b = f[xdata, ydata]; // RepeatedTiming // First
a == b

0.030

0.00165

True

answered 2 hours ago

Henrik Schumacher

42.3k261125

add a commentÂ |Â

up vote
2
down vote

One reason for slowness is that the input data was not packed. Moreover Transpose is often faster than Riffle:

xdata, ydata = Transpose[Map[x [Function] x, Sin[x], Range[0., 2 Pi, .0001]]];

f[xdata_, ydata_] := Transpose[
 Flatten[Transpose[ConstantArray[xdata, 3]]],
 With[o = ConstantArray[0., Length[ydata]],
 Flatten[Transpose[o, ydata, o]]
 ]
 ]

a = zeroPaddedData[xdata, ydata]; // RepeatedTiming // First
b = f[xdata, ydata]; // RepeatedTiming // First
a == b

0.030

0.00165

True

answered 2 hours ago

Henrik Schumacher

42.3k261125

add a commentÂ |Â

up vote
2
down vote

One reason for slowness is that the input data was not packed. Moreover Transpose is often faster than Riffle:

xdata, ydata = Transpose[Map[x [Function] x, Sin[x], Range[0., 2 Pi, .0001]]];

f[xdata_, ydata_] := Transpose[
 Flatten[Transpose[ConstantArray[xdata, 3]]],
 With[o = ConstantArray[0., Length[ydata]],
 Flatten[Transpose[o, ydata, o]]
 ]
 ]

a = zeroPaddedData[xdata, ydata]; // RepeatedTiming // First
b = f[xdata, ydata]; // RepeatedTiming // First
a == b

0.030

0.00165

True

answered 2 hours ago

Henrik Schumacher

42.3k261125

One reason for slowness is that the input data was not packed. Moreover Transpose is often faster than Riffle:

xdata, ydata = Transpose[Map[x [Function] x, Sin[x], Range[0., 2 Pi, .0001]]];

f[xdata_, ydata_] := Transpose[
 Flatten[Transpose[ConstantArray[xdata, 3]]],
 With[o = ConstantArray[0., Length[ydata]],
 Flatten[Transpose[o, ydata, o]]
 ]
 ]

a = zeroPaddedData[xdata, ydata]; // RepeatedTiming // First
b = f[xdata, ydata]; // RepeatedTiming // First
a == b

0.030

0.00165

True

answered 2 hours ago

Henrik Schumacher

42.3k261125

answered 2 hours ago

Henrik Schumacher

42.3k261125

answered 2 hours ago

Henrik Schumacher

42.3k261125

answered 2 hours ago

Henrik Schumacher

42.3k261125

add a commentÂ |Â

up vote
1
down vote

Slower but simple

tab = Table[x, Sin[x], x, 0, 2 Pi, .5];
SequenceReplace[tab, a_,b_:>Sequence[a,0,a,b,a,0]] == zpd

True

Update: Using Upsample on the second argument gives a slight improvement over Henrik's f:

ClearAll[zPad]
zPad[xd_, yd_] := Transpose[Flatten[ConstantArray[xd, 3], 2, 1], Upsample[yd, 3, 2]];

xdata, ydata = Transpose[Map[x [Function] x, Sin[x], Range[0., 2 Pi, .0001]]]; 
a = zeroPaddedData[xdata, ydata]; // RepeatedTiming // First

0.024

b = f[xdata, ydata]; // RepeatedTiming // First

0.0017

c = zPad[xdata, ydata]; // RepeatedTiming // First

0.0016

a == b == c

True

r1 = Upsample[ydata, 3, 2] ; // RepeatedTiming // First

0.00061

r2 = With[o = ConstantArray[0., Length[ydata]], 
 Flatten[Transpose[o, ydata, o] ]]; // RepeatedTiming // First

0.0013

r1 == r2

True

edited 39 mins ago

answered 1 hour ago

kglr

166k8188389

add a commentÂ |Â

up vote
1
down vote

Slower but simple

tab = Table[x, Sin[x], x, 0, 2 Pi, .5];
SequenceReplace[tab, a_,b_:>Sequence[a,0,a,b,a,0]] == zpd

True

Update: Using Upsample on the second argument gives a slight improvement over Henrik's f:

ClearAll[zPad]
zPad[xd_, yd_] := Transpose[Flatten[ConstantArray[xd, 3], 2, 1], Upsample[yd, 3, 2]];

xdata, ydata = Transpose[Map[x [Function] x, Sin[x], Range[0., 2 Pi, .0001]]]; 
a = zeroPaddedData[xdata, ydata]; // RepeatedTiming // First

0.024

b = f[xdata, ydata]; // RepeatedTiming // First

0.0017

c = zPad[xdata, ydata]; // RepeatedTiming // First

0.0016

a == b == c

True

r1 = Upsample[ydata, 3, 2] ; // RepeatedTiming // First

0.00061

r2 = With[o = ConstantArray[0., Length[ydata]], 
 Flatten[Transpose[o, ydata, o] ]]; // RepeatedTiming // First

0.0013

r1 == r2

True

edited 39 mins ago

answered 1 hour ago

kglr

166k8188389

add a commentÂ |Â

up vote
1
down vote

Slower but simple

tab = Table[x, Sin[x], x, 0, 2 Pi, .5];
SequenceReplace[tab, a_,b_:>Sequence[a,0,a,b,a,0]] == zpd

True

Update: Using Upsample on the second argument gives a slight improvement over Henrik's f:

ClearAll[zPad]
zPad[xd_, yd_] := Transpose[Flatten[ConstantArray[xd, 3], 2, 1], Upsample[yd, 3, 2]];

xdata, ydata = Transpose[Map[x [Function] x, Sin[x], Range[0., 2 Pi, .0001]]]; 
a = zeroPaddedData[xdata, ydata]; // RepeatedTiming // First

0.024

b = f[xdata, ydata]; // RepeatedTiming // First

0.0017

c = zPad[xdata, ydata]; // RepeatedTiming // First

0.0016

a == b == c

True

r1 = Upsample[ydata, 3, 2] ; // RepeatedTiming // First

0.00061

r2 = With[o = ConstantArray[0., Length[ydata]], 
 Flatten[Transpose[o, ydata, o] ]]; // RepeatedTiming // First

0.0013

r1 == r2

True

edited 39 mins ago

answered 1 hour ago

kglr

166k8188389

Slower but simple

tab = Table[x, Sin[x], x, 0, 2 Pi, .5];
SequenceReplace[tab, a_,b_:>Sequence[a,0,a,b,a,0]] == zpd

True

Update: Using Upsample on the second argument gives a slight improvement over Henrik's f:

ClearAll[zPad]
zPad[xd_, yd_] := Transpose[Flatten[ConstantArray[xd, 3], 2, 1], Upsample[yd, 3, 2]];

xdata, ydata = Transpose[Map[x [Function] x, Sin[x], Range[0., 2 Pi, .0001]]]; 
a = zeroPaddedData[xdata, ydata]; // RepeatedTiming // First

0.024

b = f[xdata, ydata]; // RepeatedTiming // First

0.0017

c = zPad[xdata, ydata]; // RepeatedTiming // First

0.0016

a == b == c

True

r1 = Upsample[ydata, 3, 2] ; // RepeatedTiming // First

0.00061

r2 = With[o = ConstantArray[0., Length[ydata]], 
 Flatten[Transpose[o, ydata, o] ]]; // RepeatedTiming // First

0.0013

r1 == r2

True

edited 39 mins ago

answered 1 hour ago

kglr

166k8188389

edited 39 mins ago

answered 1 hour ago

kglr

166k8188389

answered 1 hour ago

kglr

166k8188389

answered 1 hour ago

kglr

166k8188389

add a commentÂ |Â

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