Dear Mark,
many thanks for your suggestions !
best,
yichih
--
Yichih Hsieh
e-mail : yichih.hs...@gmail.com
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PLEASE do re
Dear baptiste,
many thanks for your suggestions !
best,
yichih
2009/9/5, baptiste auguie :
>
> Hi,
>
> you have two problems in your first scenario,
>
> 1- Wrong operator precedence. For example,
>
> > 1 == 2 | 3
> [1] TRUE
>
> where 1==2 is tested as FALSE, but 1 is not tested against 3 for e
Yichih,
Answer 2 is "correct," because your indexing specification for 1 is wrong.
You also seem to have left out a comma.
##
mu1990$wage[mu1990$edu==2|mu1990$edu==3|mu1990$edu==4, ] ## like this
mu1990$wage[mu1990$edu%in%2:4, ]
You really could have worked this out for yourself by looking at t
Hi,
you have two problems in your first scenario,
1- Wrong operator precedence. For example,
> 1 == 2 | 3
[1] TRUE
where 1==2 is tested as FALSE, but 1 is not tested against 3 for equality as
it would be using,
> 1 == 2 | 1 == 3
[1] FALSE
or using %in% 2:3
Instead, R evaluates "FALSE | 3", a
Dear all,
I got another problem:
if education have five levels
edu=1
edu=2
edu=3
edu=4
edu=5
If I want to appoint y=edu2~4 in 1990
which programs is correct?
I tried this two programs, they both work, but two results is different.
1.
fig2b<-reldist(y=mu1990$wage[mu1990$edu==2|3|4],..)
2.
Dear Petr,
your suggestion is useful
many thanks for your help !
best,
Yichih
2009/9/3 Petr PIKAL
> Hi
>
> use any of suitable selection ways that are in R.
>
> E.g.
>
> data[data$gender==1, ]
>
> selects only female values
>
> data$wage[(data$gender==1) & (data$race=1)] selects black femal
Dear all,
I have 1980~1990 eleven datas,
every year have three variables,
wage
gender(1=female, 2=male)
race(1=black, 2=white)
My original commands is:
fig2b<-reldist(y=mu1990$wage,yo=mu1980$wage,...)
I have three questions:
1. If I want to appoint y=women's wage in 1990
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