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How to operate Moving Average on List of Lists?
What is in your Mathematica tool bag?Generate a list in Mathematica with a conditional tested for each elementManipulate, checkboxes from the listHow do I find Waldo with Mathematica?How to do Tally-like operation on list based on elements' total in MathematicaDo operations for the second component of each element in a list in MathematicaFast extraction of elements from nested listsHow to smooth a histogram3D in mathematica using moving average?Moving Average plotHow to set an initial moving average value in Java?
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I have a list of lists of numbers in mathematica, i.e.
a, b, c, d, e,f, g, h, i, j
and I want to operate MovingAverage on each sublist.
However,
MovingAverage[listOfLists,movingAverageElementChoice,movingAverageElementChoice] simply returns a one dimmensional list, which is the number of elements in the jth column divided by numberOfSublists
I have tried a lot, please help!
I have tried MovingAverage[listOfLists,movingAverageElementChoice]
kineticsData8219K6Averaged =
MovingAverage[kineticsData8219K6, 10, 10, 10, 10, 10, 10];
I want it to output each sublist operated on by MovingAverage
wolfram-mathematica moving-average
asked Mar 28 at 4:33
rictuarrictuar
1
add a comment |
I have a list of lists of numbers in mathematica, i.e.
a, b, c, d, e,f, g, h, i, j
and I want to operate MovingAverage on each sublist.
However,
MovingAverage[listOfLists,movingAverageElementChoice,movingAverageElementChoice] simply returns a one dimmensional list, which is the number of elements in the jth column divided by numberOfSublists
I have tried a lot, please help!
I have tried MovingAverage[listOfLists,movingAverageElementChoice]
kineticsData8219K6Averaged =
MovingAverage[kineticsData8219K6, 10, 10, 10, 10, 10, 10];
I want it to output each sublist operated on by MovingAverage
wolfram-mathematica moving-average
asked Mar 28 at 4:33
rictuarrictuar
1
add a comment |
I have a list of lists of numbers in mathematica, i.e.
a, b, c, d, e,f, g, h, i, j
and I want to operate MovingAverage on each sublist.
However,
MovingAverage[listOfLists,movingAverageElementChoice,movingAverageElementChoice] simply returns a one dimmensional list, which is the number of elements in the jth column divided by numberOfSublists
I have tried a lot, please help!
I have tried MovingAverage[listOfLists,movingAverageElementChoice]
kineticsData8219K6Averaged =
MovingAverage[kineticsData8219K6, 10, 10, 10, 10, 10, 10];
I want it to output each sublist operated on by MovingAverage
wolfram-mathematica moving-average
asked Mar 28 at 4:33
rictuarrictuar
1
I have a list of lists of numbers in mathematica, i.e.
a, b, c, d, e,f, g, h, i, j
and I want to operate MovingAverage on each sublist.
However,
MovingAverage[listOfLists,movingAverageElementChoice,movingAverageElementChoice] simply returns a one dimmensional list, which is the number of elements in the jth column divided by numberOfSublists
I have tried a lot, please help!
I have tried MovingAverage[listOfLists,movingAverageElementChoice]
kineticsData8219K6Averaged =
MovingAverage[kineticsData8219K6, 10, 10, 10, 10, 10, 10];
I want it to output each sublist operated on by MovingAverage
wolfram-mathematica moving-average
wolfram-mathematica moving-average
asked Mar 28 at 4:33
rictuarrictuar
1
asked Mar 28 at 4:33
rictuarrictuar
1
asked Mar 28 at 4:33
rictuarrictuar
1
asked Mar 28 at 4:33
rictuarrictuar
1
asked Mar 28 at 4:33
rictuarrictuar
1
1
add a comment |
add a comment |
1 Answer
1
active
oldest
votes
You want to do the same thing to every element of a list. You should think of Map in almost every situation like this. Each of your elements just happen to be lists, but that isn't an issue. So
data=a, b, c, d, e,f, g, h, i, j;
Map[MovingAverage[#, 2]&, data]
gives you the result
(a+b)/2, (b+c)/2, (c+d)/2, (d+e)/2, (f+g)/2, (g+h)/2, (h+i)/2, (i+j)/2
And you can provide additional arguments inside that MovingAverage to modify the behavior to match what you need. The # will be replaced one at a time by each list inside data and the result of each of those MovingAverage will be put into a list and returned to you.
answered Mar 28 at 5:50
BillBill
2,5611 gold badge7 silver badges8 bronze badges
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1 Answer
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1 Answer
1
active
oldest
votes
active
oldest
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active
oldest
votes
You want to do the same thing to every element of a list. You should think of Map in almost every situation like this. Each of your elements just happen to be lists, but that isn't an issue. So
data=a, b, c, d, e,f, g, h, i, j;
Map[MovingAverage[#, 2]&, data]
gives you the result
(a+b)/2, (b+c)/2, (c+d)/2, (d+e)/2, (f+g)/2, (g+h)/2, (h+i)/2, (i+j)/2
And you can provide additional arguments inside that MovingAverage to modify the behavior to match what you need. The # will be replaced one at a time by each list inside data and the result of each of those MovingAverage will be put into a list and returned to you.
answered Mar 28 at 5:50
BillBill
2,5611 gold badge7 silver badges8 bronze badges
add a comment |
You want to do the same thing to every element of a list. You should think of Map in almost every situation like this. Each of your elements just happen to be lists, but that isn't an issue. So
data=a, b, c, d, e,f, g, h, i, j;
Map[MovingAverage[#, 2]&, data]
gives you the result
(a+b)/2, (b+c)/2, (c+d)/2, (d+e)/2, (f+g)/2, (g+h)/2, (h+i)/2, (i+j)/2
And you can provide additional arguments inside that MovingAverage to modify the behavior to match what you need. The # will be replaced one at a time by each list inside data and the result of each of those MovingAverage will be put into a list and returned to you.
answered Mar 28 at 5:50
BillBill
2,5611 gold badge7 silver badges8 bronze badges
add a comment |
You want to do the same thing to every element of a list. You should think of Map in almost every situation like this. Each of your elements just happen to be lists, but that isn't an issue. So
data=a, b, c, d, e,f, g, h, i, j;
Map[MovingAverage[#, 2]&, data]
gives you the result
(a+b)/2, (b+c)/2, (c+d)/2, (d+e)/2, (f+g)/2, (g+h)/2, (h+i)/2, (i+j)/2
And you can provide additional arguments inside that MovingAverage to modify the behavior to match what you need. The # will be replaced one at a time by each list inside data and the result of each of those MovingAverage will be put into a list and returned to you.
answered Mar 28 at 5:50
BillBill
2,5611 gold badge7 silver badges8 bronze badges
You want to do the same thing to every element of a list. You should think of Map in almost every situation like this. Each of your elements just happen to be lists, but that isn't an issue. So
data=a, b, c, d, e,f, g, h, i, j;
Map[MovingAverage[#, 2]&, data]
gives you the result
(a+b)/2, (b+c)/2, (c+d)/2, (d+e)/2, (f+g)/2, (g+h)/2, (h+i)/2, (i+j)/2
And you can provide additional arguments inside that MovingAverage to modify the behavior to match what you need. The # will be replaced one at a time by each list inside data and the result of each of those MovingAverage will be put into a list and returned to you.
answered Mar 28 at 5:50
BillBill
2,5611 gold badge7 silver badges8 bronze badges
edited Mar 28 at 6:09
answered Mar 28 at 5:50
BillBill
2,5611 gold badge7 silver badges8 bronze badges
answered Mar 28 at 5:50
BillBill
2,5611 gold badge7 silver badges8 bronze badges
answered Mar 28 at 5:50
BillBill
2,5611 gold badge7 silver badges8 bronze badges
2,5611 gold badge7 silver badges8 bronze badges
add a comment |
add a comment |
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