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How to operate Moving Average on List of Lists?

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0

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

share|improve this question

asked Mar 28 at 4:33

rictuarrictuar

1

add a comment | 

0

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

share|improve this question

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

share|improve this question

asked Mar 28 at 4:33

rictuarrictuar

1

share|improve this question

asked Mar 28 at 4:33

rictuarrictuar

1

share|improve this question

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 Answer
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oldest

votes

1

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.

share|improve this answer

edited Mar 28 at 6:09

answered Mar 28 at 5:50

BillBill

2,56111 gold badge77 silver badges88 bronze badges

add a comment | 

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1

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.

share|improve this answer

edited Mar 28 at 6:09

answered Mar 28 at 5:50

BillBill

2,56111 gold badge77 silver badges88 bronze badges

add a comment | 

1

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.

share|improve this answer

edited Mar 28 at 6:09

answered Mar 28 at 5:50

BillBill

2,56111 gold badge77 silver badges88 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.

share|improve this answer

edited Mar 28 at 6:09

answered Mar 28 at 5:50

BillBill

2,56111 gold badge77 silver badges88 bronze badges

data=a, b, c, d, e,f, g, h, i, j;
Map[MovingAverage[#, 2]&, data]

(a+b)/2, (b+c)/2, (c+d)/2, (d+e)/2, (f+g)/2, (g+h)/2, (h+i)/2, (i+j)/2

share|improve this answer

edited Mar 28 at 6:09

answered Mar 28 at 5:50

BillBill

2,56111 gold badge77 silver badges88 bronze badges

answered Mar 28 at 5:50

BillBill

2,56111 gold badge77 silver badges88 bronze badges

answered Mar 28 at 5:50

BillBill

2,56111 gold badge77 silver badges88 bronze badges