Re: proportion of states life expectancy > 70

["Douglas Bates" <bates@stat.wisc.edu>]

On Nov 27, 2007 10:38 PM, bhavin toprani <b_toprani@hotmail.com> wrote:

> Dear all,

>  Could you please suggest an easy command eg, if, for ,etc to find the
> proportion of states having life expectancy greater than 70?

>  Thanks for your help in advance.

>  example of dataframe
>              Life.Exp
>  Alabama   69.05
>  Alaska      69.31
>  Arizona     70.55
>  Arkansas   70.66
>  California  71.71
>  Colorado   72.06

I haven't seen a response to this.  The general way to approach this
is to sum the logical indicators.  When the TRUE/FALSE values are used
in an arithmetic expression they are converted to 1/0 values.
Suppose that your data frame is called mydf.  Then the expression

sum(mydf$Life.Exp > 70)/nrow(mydf$Life.Exp)

should do it.  If you want to be careful about the possibility of
missing data you should use

sum(mydf$Life.Exp > 70, na.rm = TRUE)/sum(!is.na(mydf$Life.Exp))

instead.