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Editorial Team
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Asked: 2026-06-08T02:24:38+00:00

There is a data.frame > x date open high low close volume adjusted 1

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There is a data.frame

> x

        date  open  high   low close   volume adjusted  
1 2010-01-04 24.52 24.58 23.68 23.71 24192200    23.71  
2 2010-01-05 23.75 23.90 22.75 23.30 55649900    23.30  
3 2010-01-06 23.25 23.25 22.72 22.90 41214300    22.90  
4 2010-01-07 22.90 23.05 22.40 22.65 35533600    22.65  
5 2010-01-08 22.50 22.75 22.35 22.60 28854300    22.60  
6 2010-01-11 23.50 23.68 22.28 22.60 44284600    22.60

> is.vector(x[,1])
[1] FALSE  
> is.vector(x[,2])
[1] TRUE  
> is.vector(x[,1])
[1] FALSE 
> is.vector(x[,3])
[1] TRUE  
> is.vector(x[,4])
[1] TRUE  
> is.vector(x[,5])  
[1] TRUE  
> is.vector(x[,6])
[1] TRUE  
> is.vector(x[,7])
[1] TRUE

I want to know why is.vector(x[,1]) is not TRUE?

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1 Answer

  1. Editorial Team

    From ?is.vector :

    is.vector returns TRUE if x is a vector of the specified mode having
    no attributes other than names. It returns FALSE otherwise.

    Your first column is likely stored as a date class. Thus if you get its attributes (using str()) you will see that it has some attributes other than just names.

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