You first example is a list of 5 items, each item is a number
The second example is a list with one item: a vector with 5 elements.

You'll need c() to make a vector of the item to get the same result.
all.equal(list(c(0.8,0.9,1.0,1.1,1.2)), list(seq(0.8, 1.2, by = 1.1)))

ir. Thierry Onkelinx
Instituut voor natuur- en bosonderzoek / Research Institute for Nature and 
Forest
team Biometrie & Kwaliteitszorg / team Biometrics & Quality Assurance
Kliniekstraat 25
1070 Anderlecht
Belgium
+ 32 2 525 02 51
+ 32 54 43 61 85
thierry.onkel...@inbo.be
www.inbo.be

To call in the statistician after the experiment is done may be no more than 
asking him to perform a post-mortem examination: he may be able to say what the 
experiment died of.
~ Sir Ronald Aylmer Fisher

The plural of anecdote is not data.
~ Roger Brinner

The combination of some data and an aching desire for an answer does not ensure 
that a reasonable answer can be extracted from a given body of data.
~ John Tukey


-----Oorspronkelijk bericht-----
Van: r-help-boun...@r-project.org [mailto:r-help-boun...@r-project.org] Namens 
Anser Chen
Verzonden: dinsdag 30 oktober 2012 13:42
Aan: r-help@r-project.org
Onderwerp: [R] newbie: embeding seq in a list

Suppose I want to create a structure containing the following values:
0.8,0.9,1.0,1.1,1.2

If I use

env <- list(0.8,0.9,1.0,1.1,1.2)

then R returns

> env
[[1]]
[1] 0.8

[[2]]
[1] 0.9

[[3]]
[1] 1

[[4]]
[1] 1.1

[[5]]
[1] 1.2


But, if I try to 'save some key-strokes', and use


env <- list(seq(0.8,1.2,by=0.1))

then R returns

> env
[[1]]
[1] 0.8 0.9 1.0 1.1 1.2


I'd like the 'latter' to be equivalent to 'the former', but can't figure out 
how to achieve said aim.

Thanks in advance...

        [[alternative HTML version deleted]]

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