Dear R-users,
I want to model a proportional variable bounded by [0,1] (the % of land
fertilized). A high percentage of the data contains 0s (60%), a smaller
percentage contains 1s (10%), and all the rest falls in between.
I want to compare different models with each other to see their
performa
Dear colleagues,
I am trying to find a simple code to demean
1) only certain values of a dataset,
2) by group
3) and in a weighted fasion.
Currently, I can only demean all the numeric variables in the dataset:
Data[,sapply(Data, is.numeric)] <- apply(Data[sapply(Data,
is.numeric)], 2, function(
8075 -1.818075
>
> 4: 2005 28954 -1.818075 -1.818075
>
> 5: 2005 28955 -1.818075NA
>
> 6: 2007 28955 -1.818075 -1.818075
>
>
>
> Is it what you intended?
>
> Cheers
>
> Petr
>
>
>
>
>
> *From:* Janka VANSCHOENWINKEL [mailto:janka.vanschoe
year = c(2005, 2006, 2007, 2008), t1 = c(-1.81807494163513,
> -1.81807494163513, -1.81807494163513, -1.81807494163513
> )), .Names = c("year", "t1"), row.names = c(17L, 18L,
> 30L, 44L), class = "data.frame")), .Names = c("28954", &
Hi!
I have the following dataset with the variables ID (this is a unique ID per
farmer), year, and another variable t1.
I now would like to have a fourth variable which is the lag value of t1 for
each farm ID.
I found a code on the internet that does exactly what I need, but it does
not work for
Thank you all very much. A combination of the solutions suggested solved my
problem!
2015-08-16 22:31 GMT+02:00 David Winsemius :
>
> On Aug 16, 2015, at 8:57 AM, Janka VANSCHOENWINKEL wrote:
>
> > Hi David,
> >
> > Thanks for your comment. I'll explain w
92617989, 0, 0,
> > 65.5172407627106, 0, 61.904764175415, 34.4827562570572, 7.95454531908035,
> > 75, 0, 0, 0, 0, 0, 0, 5.26393800973892, 0, 0, 0, 0, 0, 0, 0,
> > 0, 0, 74.6153831481934, 84.6153914928436, 0, 5.09554147720337,
> > 0, 0, 0, 21.0884347558022, 18.4549376368523, 6.1
ot;, "BG", "CZ", "EE", "HU", "LT", "LV",
"PL", "RO", "SI", "SK", "b48", "b50", "irrigation", "awc_class",
"sys02", "se025"), row.names = c(
r
> 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
>
> 2015-08-11 14:57 GMT+02:00 Janka Vanschoenwinkel <
> janka.vanschoenwin...@uhasselt.be>:
>
&g
If you supply a single number to the breaks parameter of cut I think it is
> the number of intervals.
>
>
> On 11/08/2015 13:57, Janka Vanschoenwinkel wrote:
>
>> Hi Thierry!
>>
>> Thanks for your answer. I tried this, but I get this error:
>>
>> "E
d 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
>
> 2015-08-11 14:38 GMT+02:
Dear list members,
I have a loop where I want to do several calculations for different samples
and save the results for each sample. These samples are for each loop
different. I want to use the "i" in the loop to cut the samples.
So for instance:
- In loop 1 (i=1), I have a sample from 0-1 an
Dear list members,
I am building a model such as:
Y1 = Y2*X1 + X2
Y2 = Y1*X1 + X2
X2 is the exogenous variable
Z1 is the instrument of Y1
Z2 is the instrument of Y2
This is a simultaneous equation model. I know how to build a simultaneous
equation model without interaction terms:
library(syste
Dear R-colleagues,
I am looking for a way to test whether one regression has significant
different coefficients and overall results for 10 groups (grouping variable
is "irr").
*What I have*
The regression is:
Depend = temp + temp² + perc + perc² + conti è split up for multiple groups
of irr
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