e may be other datasets where you have more columns than rows and
>never
>want to sample each column more than once, in which case you would set
>replace=FALSE.
>
>Best Regards.
>
>
>
>On Wed, Mar 1, 2017 at 5:38 AM, peter dalgaard
>wrote:
>
>>
>> array[cbi
Dear all,
I have to pick one value per row from an array, but from row to row from
a different column. The column positions of the values for each row are
stored in a vector.
array: 999 rows, 48 columns
vector: 999 values (each between 1 and 48) indicating for each row which
value to pick from t
[2,]11 4
> [3,]323
> [4,]241
>
> Or, brute-force-and-ignorance:
>
>> sapply(l, function(e) e[, 2])
> [,1] [,2] [,3]
> [1,]432
> [2,]114
> [3,]323
> [4,]241
>
>
&g
myarray) <- c(nrow(mylist$a), ncol(mylist$a), length(mylist))
> dimnames(myarray) <- list(hour = rownames(mylist$a),
> week = colnames(mylist$a),
> other = names(mylist))
> # now you can do:
> mean(myarray[, "week1", &quo
Hello,
I have a list of 7 data frames, each data frame having 24 rows (hour of
the day) and 5 columns (weeks) with a total of 5 x 24 values
I would like to combine all 7 columns of week 1 (and 2 ...) in a
separate data frame for hourly calculations, e.g.
> apply(new.data.frame,1,mean)
In some wa
Hi all,
after browsing the archives for hours I'm still not sure about the proper
analysis for my dataset.
I subjected each of about 50 critters (about 10 each in 5 distinct
populations) to 4 consecutive treatments (exposure to increasing
concentrations), with one measurement per treatment and
Hi all,
the following problem is still beyond my R-knowledge:
I have one data vector containing the signal from 4 channels that are measured
subsequently and in repeating cycles (with one factor vector for cycle and
one for channel identification).
To extract the mean of each channel during ea
Thanks for your comments!
> Yes. You are fitting by least-squares on two different scales:
> differences in y and differences in log(y) are not comparable.
>
> Both are correct solutions to different problems. Since we have no idea
> what 'x' and 'y' are, we cannot even guess which is more appro
Dear all,
I did a non-linear least square model fit
y ~ a * x^b
(a) > nls(y ~ a * x^b, start=list(a=1,b=1))
to obtain the coefficients a & b.
I did the same with the linearized formula, including a linear model
log(y) ~ log(a) + b * log(x)
(b) > nls(log10(y) ~ log10(a) + b*log10(x), start=l
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