Loop through list programatically

Question

I have a list in R that I want to loop through all the elements.

This is the structure of the object:

> str(AAPL.OPT[c])
List of 1
 $ jun.12.2020:List of 2
  ..$ calls:'data.frame':   52 obs. of  7 variables:
  .. ..$ Strike: num [1:52] 180 185 200 210 240 ...
  .. ..$ Last  : num [1:52] 123 118 131 120 85 ...
  .. ..$ Chg   : num [1:52] 0 0 7.61 9.48 0 ...
  .. ..$ Bid   : num [1:52] 149 144 129 119 89 ...
  .. ..$ Ask   : num [1:52] 153.3 148.5 133.5 123.7 93.5 ...
  .. ..$ Vol   : int [1:52] NA 15 16 2 1 1 3 36 1 2 ...
  .. ..$ OI    : int [1:52] 0 15 25 4 50 3 4 36 6 10 ...
  ..$ puts :'data.frame':   56 obs. of  7 variables:
  .. ..$ Strike: num [1:56] 150 165 170 180 185 190 195 200 205 210 ...
  .. ..$ Last  : num [1:56] 0.05 0.02 0.14 0.05 0.03 0.02 0.01 0.02 0.01 0.01 ...
  .. ..$ Chg   : num [1:56] 0 0 0 0 0 0 0 0 0 0 ...
  .. ..$ Bid   : num [1:56] NA 0 0 0 0 0 0 0 0 0 ...
  .. ..$ Ask   : num [1:56] 2.13 0.11 0.11 1.8 1.87 0.01 1.88 0.5 1.88 2.13 ...
  .. ..$ Vol   : int [1:56] NA 1 1 2 1 16 1 17 1 21 ...
  .. ..$ OI    : int [1:56] 1 10 7 9 76 201 113 314 92 264 ...

I cannot access the next level of the object programatically (by indexing the value)

I want to do something like this:

AAPL.OPT[c][1]

instead of this

AAPL.OPT[c]$jun.12.2020

Sample data of AAPL.OPT[c]

$`jun.12.2020`$`calls`
                    Strike   Last        Chg    Bid    Ask   Vol   OI
AAPL200612C00180000  180.0 123.29 0.00000000 149.00 153.35    NA    0
AAPL200612C00185000  185.0 117.60 0.00000000 144.00 148.50    15   15
AAPL200612C00200000  200.0 131.15 7.60999300 129.00 133.50    16   25
AAPL200612C00210000  210.0 119.95 9.47999600 119.30 123.65     2    4
....

I think AAPL.OPT[[c]][[1]] should give you AAPL.OPT[c]$jun.12.2020 — Ronak Shah
– Ronak Shah, Commented Jun 7, 2020 at 4:33

Ronak Shah · Accepted Answer · 2020-06-07 05:03:08Z

2

AAPL.OPT[c] gives a list of length 1 which has two other lists in them. If we use [[c]] it gives a list of length 2 andtTo access each dataframe you can subset them further using [[ so AAPL.OPT[[c]][[1]] and AAPL.OPT[[c]][[2]].

answered Jun 7, 2020 at 5:03

Ronak Shah

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akrun · Accepted Answer · 2020-06-07 18:21:17Z

1

We can use

 AAPL.OPT[[c]]$jun.12.2020

answered Jun 7, 2020 at 18:21

akrun

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1 Comment

the Tin Man Over a year ago

It helps more if you supply an explanation why this is the preferred solution and explain how it works. We want to educate, not just provide code.

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