Frequency parameter of `ts()` and `findfrequency()` in R

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Is the result obtained in findfrequency() function of forecast package and the frequency parameter of ts() the same?

r time-series forecasting

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Is the result obtained in findfrequency() function of forecast package and the frequency parameter of ts() the same?

r time-series forecasting

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edited 3 hours ago

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Is the result obtained in findfrequency() function of forecast package and the frequency parameter of ts() the same?

r time-series forecasting

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edited 3 hours ago

SecretAgentMan

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user224318

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Is the result obtained in findfrequency() function of forecast package and the frequency parameter of ts() the same?

r time-series forecasting

r time-series forecasting

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edited 3 hours ago

SecretAgentMan

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asked 3 hours ago

user224318

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edited 3 hours ago

SecretAgentMan

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edited 3 hours ago

SecretAgentMan

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edited 3 hours ago

SecretAgentMan

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edited 3 hours ago

SecretAgentMan

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533120

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user224318

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asked 3 hours ago

user224318

61

asked 3 hours ago

user224318

61

61

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user224318 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
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New contributor

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It probably should be, but there are no guarantees that it has to be.

findfrequency will return the period with the maximum spectral amplitude of the signal provided as an input. It is relative straight-forward calculation using the results from spec.ar where we estimate the spectral density associated with each candidate frequency of a time-series model from auto-regressive fit. Assuming that our raw signal, x, having assumed frequency 1, we use it as input to spec.ar:

specARresults <- spec.ar(x)

then then results from findfrequency are simply:

1 / specARresults$freq[which.max(specARresults$spec)]

i.e. the inverse of the frequency having the highest spectral density. And as the inverse of a frequency is a period we get our result. :) (Side-note: Why findfrequency returns a period eludes me.)

We can theoretically use this result as input to ts's argument about frequency but we do not have to. The frequency we use to ts is probably better to relate to our actual problem. For example, we might have daily expenditure data that happen to have a strong weekly pattern (cause people spend more on weekends). If we do not care for a weekly pattern though, it might be more relevant for us to analyse them assuming a period of 365 days.

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answered 2 hours ago

usεr11852

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1 Answer
1

active

oldest

votes

1 Answer
1

active

oldest

votes

active

oldest

votes

active

oldest

votes

up vote
2
down vote

It probably should be, but there are no guarantees that it has to be.

findfrequency will return the period with the maximum spectral amplitude of the signal provided as an input. It is relative straight-forward calculation using the results from spec.ar where we estimate the spectral density associated with each candidate frequency of a time-series model from auto-regressive fit. Assuming that our raw signal, x, having assumed frequency 1, we use it as input to spec.ar:

specARresults <- spec.ar(x)

then then results from findfrequency are simply:

1 / specARresults$freq[which.max(specARresults$spec)]

i.e. the inverse of the frequency having the highest spectral density. And as the inverse of a frequency is a period we get our result. :) (Side-note: Why findfrequency returns a period eludes me.)

We can theoretically use this result as input to ts's argument about frequency but we do not have to. The frequency we use to ts is probably better to relate to our actual problem. For example, we might have daily expenditure data that happen to have a strong weekly pattern (cause people spend more on weekends). If we do not care for a weekly pattern though, it might be more relevant for us to analyse them assuming a period of 365 days.

share|cite|improve this answer

answered 2 hours ago

usεr11852

17k13670

• 
  
  
  

  
  

  
  
  
  
  
  

  
  Welcome to the CV community!
  – usεr11852
  2 hours ago
  

  
  
  
  

add a comment | 

up vote
2
down vote

It probably should be, but there are no guarantees that it has to be.

findfrequency will return the period with the maximum spectral amplitude of the signal provided as an input. It is relative straight-forward calculation using the results from spec.ar where we estimate the spectral density associated with each candidate frequency of a time-series model from auto-regressive fit. Assuming that our raw signal, x, having assumed frequency 1, we use it as input to spec.ar:

specARresults <- spec.ar(x)

then then results from findfrequency are simply:

1 / specARresults$freq[which.max(specARresults$spec)]

i.e. the inverse of the frequency having the highest spectral density. And as the inverse of a frequency is a period we get our result. :) (Side-note: Why findfrequency returns a period eludes me.)

We can theoretically use this result as input to ts's argument about frequency but we do not have to. The frequency we use to ts is probably better to relate to our actual problem. For example, we might have daily expenditure data that happen to have a strong weekly pattern (cause people spend more on weekends). If we do not care for a weekly pattern though, it might be more relevant for us to analyse them assuming a period of 365 days.

share|cite|improve this answer

answered 2 hours ago

usεr11852

17k13670

• 
  
  
  

  
  

  
  
  
  
  
  

  
  Welcome to the CV community!
  – usεr11852
  2 hours ago
  

  
  
  
  

add a comment | 

up vote
2
down vote

up vote
2
down vote

It probably should be, but there are no guarantees that it has to be.

findfrequency will return the period with the maximum spectral amplitude of the signal provided as an input. It is relative straight-forward calculation using the results from spec.ar where we estimate the spectral density associated with each candidate frequency of a time-series model from auto-regressive fit. Assuming that our raw signal, x, having assumed frequency 1, we use it as input to spec.ar:

specARresults <- spec.ar(x)

then then results from findfrequency are simply:

1 / specARresults$freq[which.max(specARresults$spec)]

i.e. the inverse of the frequency having the highest spectral density. And as the inverse of a frequency is a period we get our result. :) (Side-note: Why findfrequency returns a period eludes me.)

We can theoretically use this result as input to ts's argument about frequency but we do not have to. The frequency we use to ts is probably better to relate to our actual problem. For example, we might have daily expenditure data that happen to have a strong weekly pattern (cause people spend more on weekends). If we do not care for a weekly pattern though, it might be more relevant for us to analyse them assuming a period of 365 days.

share|cite|improve this answer

answered 2 hours ago

usεr11852

17k13670

It probably should be, but there are no guarantees that it has to be.

findfrequency will return the period with the maximum spectral amplitude of the signal provided as an input. It is relative straight-forward calculation using the results from spec.ar where we estimate the spectral density associated with each candidate frequency of a time-series model from auto-regressive fit. Assuming that our raw signal, x, having assumed frequency 1, we use it as input to spec.ar:

specARresults <- spec.ar(x)

then then results from findfrequency are simply:

1 / specARresults$freq[which.max(specARresults$spec)]

i.e. the inverse of the frequency having the highest spectral density. And as the inverse of a frequency is a period we get our result. :) (Side-note: Why findfrequency returns a period eludes me.)

We can theoretically use this result as input to ts's argument about frequency but we do not have to. The frequency we use to ts is probably better to relate to our actual problem. For example, we might have daily expenditure data that happen to have a strong weekly pattern (cause people spend more on weekends). If we do not care for a weekly pattern though, it might be more relevant for us to analyse them assuming a period of 365 days.

share|cite|improve this answer

answered 2 hours ago

usεr11852

17k13670

share|cite|improve this answer

share|cite|improve this answer

answered 2 hours ago

usεr11852

17k13670

answered 2 hours ago

usεr11852

17k13670

answered 2 hours ago

usεr11852

17k13670

17k13670

• 
  
  
  

  
  

  
  
  
  
  
  

  
  Welcome to the CV community!
  – usεr11852
  2 hours ago
  

  
  
  
  

add a comment | 

• 
  
  
  

  
  

  
  
  
  
  
  

  
  Welcome to the CV community!
  – usεr11852
  2 hours ago
  

  
  
  
  

Welcome to the CV community!
– usεr11852
2 hours ago

Welcome to the CV community!
– usεr11852
2 hours ago

add a comment | 

user224318 is a new contributor. Be nice, and check out our Code of Conduct.

 

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