Dear all,
please could you advise about a method to scale 2 plots of LOESS curves.
More specifically, we do have 2 sets of 5C data, and the loess plots
reflect the relationship between INTENSITY and DISTANCE (please see the R
code below).
I am looking for a method/formula to scale these 2 LOESS p
That you choose to interpret an income level of NA as zero could mean that the
value was encoded incorrectly or could mean that that the income of that person
was never reviewed and theirs might have represented a significant additional
amount of income.
If you believe it was encoded improperly
I think given the previous confusion and lack of clarity in this
description as well, you really need to give us
a. sample data using dput()
b. what you expect the result to be
c. what code you've tried, including how you've attempted using the
sum() function that several of us have recommended.
S
Hello,
I need to first apologize for the error in my first question
dataframe$A = 20 dataframe$B = NA
dataframe$A + dataframe$B actually returns NA
You quite understand my point of view. This is a household level data where you
need to compute the total income of each household member before
I think I find the way to extract smoothed probabilities from msmFit().
In the MSM.lm-class, we could use mod.msm@Fit@smoProb to extract the smoothed
probabilities, also, we could use mod.msm@Fit@filtProb to extract the filter
probabilities.
Best Regards
Paul Wang
pywan...@yahoo.com.tw
# it's not clear what your question is, but here's a stab in the dark at a
solution!
ind<- !is.na(dataframe$A) & !is.na(dataframe$B)
dataframe$A[ind] + dataframe$B[ind]
- Dan
P.S. I'm sure there are ways to do this using one of R's functions for
automatically removing NA's (na.rm = T), but unle
Sorry I misunderstood questions .
I think original poster wants result to be 0 , right ?
Then
dataframe$A[ is.na(dataframe$B) ] <- 0
dataframe$B[ is.na(dataframe$B) ] <- 0
is this case you would lose dataframe$A data or
dataframe$A + ( dataframe$B[ is.na(dataframe$B) ] <- -1*dataframe$A )
So you have decided that NA==0 and Inf == 0... if that is really what you want
then it looks like that is what you got. If you don't like the fact that you
are mucking with your data, then make a copy of the data first and muck with
that.
Blegh.
-
On 07/09/15 10:22, Olu Ola via R-help wrote:
Hello, I am currently working with a dataframe which has some missing
values represented by "NA". whenever, I add two columns in which at
least one of the pair of an observation is "NA", the sum returns
zero. That is for the same observation, if
dataf
Please use dput() to provide data, rather than expecting people to
open random attachments. Besides, there are multiple options for
getting data into R, and we need to know exactly what you did. dput()
is faster and easier.
What have you tried? Did you look at aggregate() as I suggested?
Sarah
O
I'm not quite sure how you get zero from that situation. Do you expect
the answer to be 20?
How about:
> dataframe <- data.frame(A=20, B=NA)
> dataframe$A + dataframe$B
[1] NA
> ?sum
> sum(dataframe$A, dataframe$B, na.rm=TRUE)
[1] 20
Sarah
On Sun, Sep 6, 2015 at 6:48 PM, ce wrote:
>
>
> I
That is not how R works. 20+NA is NA, which is not the same as zero. This is
not optional behaviour.
I notice that you put quotes around the NA if those really are there then
you should be getting an error.
You need to assemble a reproducible example, such as is described at [1]. By
doing
I use something like :
dataframe[ is.na(dataframe) ] <- 0
dataframe[ is.nan(dataframe) ] <- 0
dataframe[ is.infinite(dataframe) ] <- 0
-Original Message-
From: "Olu Ola via R-help" [r-help@r-project.org]
Date: 09/06/2015 06:24 PM
To: r-help@r-project.org
Subject: [R] Handling "NA" in
Sorry for repeating the message owing to previous uncorrect html
version delivered (Thank you bert Gunter for the alert).
Marc Schwartz enabled me to order a "two factors" interaction boxplot
with median associated to one factor alone: thanks.
I tried further to generate facets plot (3x2 boxplots
Hello,
I am currently working with a dataframe which has some missing values
represented by "NA". whenever, I add two columns in which at least one of the
pair of an observation is "NA", the sum returns zero. That is for the same
observation, if
dataframe$A = 20
dataframe$B = NA
dataframe$A +
Please stop posting in HTML and use dput() to provide data (as the
posting guide requests).
Cheers,
Bert Gunter
Bert Gunter
"Data is not information. Information is not knowledge. And knowledge
is certainly not wisdom."
-- Clifford Stoll
On Sun, Sep 6, 2015 at 8:37 AM, Sergio Fonda wrote:
Dear group,
I have the following data frame df
Measure_id i j id value1 1 5 1 1 2.02 1 5 2 1 2.03
1 5 1 2 1.54 1 5 2 2 1.55 1 5 1 3 0.06
1 5 2 3 0.07 1 5 1 4 0.08 1 5 2 4 1.09
1
I insist: as a trial, apply function "resample" (package signal) I think
it's worth doing a test ( it performs resampling through bandlimited
interpolation)
SF
Il 06/set/2015 18:52, "Jeff Newmiller" ha
scritto:
> There are lots of them. You might be having trouble searching because you
> don't kn
There are lots of them. You might be having trouble searching because you don't
know how to spell "interpolate".
?approx
?spline
Also look at the Time Series task view on CRAN.
---
Jeff NewmillerThe
Have you tried signal package?
SF
Il 06/set/2015 17:14, "AltShift" ha scritto:
> I need a function for regularising the time base of electronically acquired
> signals (i.e. vectors of samples with a nominally constant time base).
>
> For example, the accelerometer in my smartphone can deliver dat
Thanks to Marc Schwartz I was able to order a "two factors"
interaction boxplot with median associated to one factor alone.
I tried further to generate facets plot (3x2 boxplots in ggplot2) for the
dataframe reported at bottom and I'm not able to reach a correct plot.
The dataframe is a simulation
This uses a regular expression but is shorter:
> gsub("(.).", "\\1", "ABCDEFG")
[1] "ACEG"
It replaces each successive pair of characters with the first of that
pair. If there is an odd number of characters then the last character is
not matched and therefore kept -- thus it works properly for b
Dear all,
I need your urgent help J
I’m naïve, and I’m pretty sure my doubt is very simple to solve, but I’m not
getting it.
I used the following code to produce my research graphs, nonetheless, is this
problem, I do not have 6 graphs (1 – 6),
# 3 4 5
#2
#
I need a function for regularising the time base of electronically acquired
signals (i.e. vectors of samples with a nominally constant time base).
For example, the accelerometer in my smartphone can deliver data at about 50
Hz, but the sampling rate varies by about 5% throughout a recording. I ha
hi,
I am new to R using package RWeka.Currently i am using J48 classifier for
classification and prediction. My doubt is after building tree using J48,
when i use the same for prediction on my test set it gives the predicted
class names. How could i calculate accuracy, precision, recall and also
o
Suppose I had the following string, which has length of integer multiple
of some value n. So, say n=2, and the example string has a length of
(2x4) = 8 characters.
str <- "ABCDEFGH"
What I'm trying to figure out is a simple, base-R coded way (which I
heuristically call StrSubset in the follo
Thanks to Marc Schwartz I was able to order a "two factors"
interaction boxplot with median associated to one factor alone.
I tried further to generate facets plot (3x2 boxplots in ggplot2) for the
dataframe reported at bottom and I'm not able to reach a correct plot.
The dataframe is a simulation
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