Stephen Choularton schrieb:
>Hi
>
>I am trying to produce a little table like this:
>
>0 1
>0 601 408
>1 290 2655
>
>but I cannot get the syntax right.
>
>Can anyone help.
>
>Stephen
>
>
>
> vec1= c(0,0,1)
> vec2= c(NA,601,290)
> vec3=
On Thu, 2005-09-29 at 06:57 +1000, Stephen Choularton wrote:
> Hi
>
> I am trying to produce a little table like this:
>
> 0 1
> 0 601 408
> 1 290 2655
>
> but I cannot get the syntax right.
>
> Can anyone help.
>
> Stephen
Do you hav
Hi
I am trying to produce a little table like this:
0 1
0 601 408
1 290 2655
but I cannot get the syntax right.
Can anyone help.
Stephen
--
27/09/2005
[[alternative HTML version deleted]]
___
ebashi yahoo.com> writes:
:
: I'll appreciate if some one can help me with the
: following loop. This is the logic of the loop,
: if we have the following data;
: > x.df
: x.dif
: ..
: ..
: 102 0.00
: 103 0.42
: 104 0.08
: 105 0.00
: 106 0.00
: 107 0.00
: 108 -0.16
: 109 -0.3
On 28-Nov-04 ebashi wrote:
> I'll appreciate if some one can help me with the
> following loop. This is the logic of the loop,
> if we have the following data;
>> x.df
> x.dif
> ..
> ..
> 102 0.00
> 103 0.42
> 104 0.08
> 105 0.00
> 106 0.00
> 107 0.00
> 108 -0.16
> 109 -0.34
> 1
ebashi wrote:
I'll appreciate if some one can help me with the
following loop. This is the logic of the loop,
if we have the following data;
x.df
x.dif
..
..
102 0.00
103 0.42
104 0.08
105 0.00
106 0.00
107 0.00
108 -0.16
109 -0.34
110 0.00
111 -0.17
112 -0.33
113 0.00
114 0.
I'll appreciate if some one can help me with the
following loop. This is the logic of the loop,
if we have the following data;
> x.df
x.dif
..
..
102 0.00
103 0.42
104 0.08
105 0.00
106 0.00
107 0.00
108 -0.16
109 -0.34
110 0.00
111 -0.17
112 -0.33
113 0.00
114 0.00
115 0.00
