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Asked: 2026-05-21T16:48:29+00:00

I am new to R. I have an output of a function which when

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I am new to R. I have an output of a function which when captured in a variable claims it is a list. 

> typeof(fc)
[1] "list"

Here are the first few lines of how fc looks like

  Point       Forecast       Lo 80     Hi 80       Lo 95     Hi 95
  10229      -2.237542 -5.81116452  1.336080 -7.70292589  3.227841
  10230       1.683324 -3.19974731  6.566396 -5.78468918  9.151337
  10231       3.893685 -1.32257692  9.109948 -4.08389942 11.871270
  .......

How to I make a vector with only say the Forecast column from this type. I googled around and figured that I should say fc$Forecast. But when I do that I get NULL printed. Much appreciate your time/help. Thanks.

As requested: The last few lines from dput output:

Names = c("method", "model", "level", 
"mean", "lower", "upper", "x", "xname", "fitted", "residuals"
), class = "forecast")

And the first few lines. 

structure(list(method = "ARIMA(4,0,2) with non-zero mean", model 
= structure(list(coef = structure(c(0.261848480125606, 0.55212561713038,       
0.00823985719608051, -0.051126398268002, -0.369141509818343, 
1.60863444159457, -0.0928857946719395, -0.0901626797717415, 
0.0526652165617547, -0.242259088064732, 0.164509902811429, 
-0.149744064351169, 0.108818836027556, -0.0143040675162776
), .Names = c("ar1", "ar2", "ar3", "ar4", "ma1", "ma2", "intercept", 
"S1", "C1", "S2", "C2", "S3", "C3", "S4", "C4")), sigma2 = 1.3269817639389, 
var.coef = structure(c(0.494

Inbetween it is just a wall of 3 tuples.

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1 Answer

  1. Editorial Team

    You are getting the output of print.forecast which is not what the forecast-object actually looks like internally (as can be seen from the fragments of str(fc). AND … The name of the first column is not “Forecast” but rather `Point Forecast” (which will need quoting because of the embedded spaces). So after looking at print.forecast with getAnywhere I predicted you get your desired vector with the obvious modification of this (following execution of the example in help(forecast.Arima):

    > as.data.frame(forecast(fit))$`Point Forecast`
 [1] -0.7725430  0.4591482  0.8732595  1.0124943  1.0593135  1.0750607  1.0803599  1.0821453
 [9]  1.0827486  1.0829539   ### "works"

    Therefore… try:

    as.data.frame(fc)$`Point Forecast`
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