I'm trying to plot a logistic regression in R, for a continuous independent variable and a dichotomous dependent variable. I have very limited experience with R, but my professor has asked me to add this graph to a paper I'm writing, and he said R would probably be the best way to create it. Anyway, I'm sure there are tons of mistakes here, but this is the sort of this previously suggested on StackOverflow:
ggplot(vvv, aes(x = vvv$V1, y=vvv$V2)) + geom_point() + stat_smooth(method="glm", family="binomial", se=FALSE)
curve(predict(ggg, data.frame(V1=x), type="response"), add=TRUE)
where vvv is the name of my csv file (31 obs. of 2 variables), V1 is the continuous variable, and V2 is the dichotomous one. Also, ggg (List of 30?) is the following:
ggg<- glm(formula = vvv$V2 ~ vvv$V1, family = "binomial", data = vvv)
The ggplot function produces a graph of my data points, but no logistic regression curve. The curve function results in the following error:
"Error in curve(predict(ggg, data.frame(V1 = x), type = "resp"), add = TRUE) : 'expr' did not evaluate to an object of length 'n'
In addition: Warning message:'newdata' had 101 rows but variables found have 31 rows"
I'm not sure what the problem is, and I'm having trouble finding resources for this specific issue. Can anybody help? It would be greatly appreciated :)
Edit: Thanks to anyone who responded! My data, vvv, is the following, where the percent was the initial probability for presence/absence of a species in a specific area, and the 1 and 0 indicate whether or not a species ended up being observed.:
V1 V2
1 95.00% 1
2 95.00% 0
3 95.00% 1
4 92.00% 1
5 92.00% 1
6 92.00% 1
7 92.00% 1
8 92.00% 1
9 92.00% 1
10 92.00% 1
11 85.00% 1
12 85.00% 1
13 85.00% 1
14 85.00% 1
15 85.00% 1
16 80.00% 1
17 80.00% 0
18 77.00% 1
19 77.00% 1
20 75.00% 0
21 70.00% 1
22 70.00% 0
23 70.00% 0
24 70.00% 1
25 70.00% 0
26 69.00% 1
27 65.00% 0
28 60.00% 1
29 50.00% 1
30 35.00% 0
31 25.00% 0
As #MrFlick commented, V1 is probably a factor. So, first you have to change it to numeric class. This just substitutes "%" for nothing and divides by 100, so you will have proportions as numeric class:
vvv$V1<-as.numeric(sub("%","",vvv$V1))/100
Doing this, you can use your own code and you will have a plot for a logistic regression:
ggplot(vvv, aes(x = vvv$V1, y=vvv$V2)) + geom_point() + stat_smooth(method="glm", family="binomial", se=F)
This should print not only the points, but also the logistic regression curve. I don't understand what is the point of using curves. From what I could understand from your question, this is enough for what you need.
Related
My general problem: I tend to struggle using ggplot, because it's very data-frame-centric but the objects I work with seem to fit matrices better than data frames. Here is an example (adapted a little).
I have a quantity x that can assume values 0:5, and a "context" that can have values 0 or 1. For each context I have 7 different frequency distributions over the quantity x. (More generally I could have more than two "contexts", more values of x, and more frequency distributions.)
I can represent these 7×2 frequency distributions as a list freqs of two matrices, say:
> freqs
$`context0`
x0 x1 x2 x3 x4 x5
sample1 20 10 10 21 37 2
sample2 34 40 6 10 1 8
sample3 52 4 1 2 17 25
sample4 16 32 25 11 5 10
sample5 28 2 10 4 21 35
sample6 22 13 35 12 13 5
sample7 9 5 43 29 4 10
$`context1`
x0 x1 x2 x3 x4 x5
sample1 15 21 14 15 14 21
sample2 27 8 6 5 29 25
sample3 13 7 5 26 48 0
sample4 33 3 18 11 13 22
sample5 12 23 40 11 2 11
sample6 5 51 2 28 5 9
sample7 3 1 21 10 63 2
or a 3D array.
Or I could use a data.table tablefreqs like this one:
> tablefreqs
context x0 x1 x2 x3 x4 x5
1: 0 20 10 10 21 37 2
2: 0 34 40 6 10 1 8
3: 0 52 4 1 2 17 25
4: 0 16 32 25 11 5 10
5: 0 28 2 10 4 21 35
6: 0 22 13 35 12 13 5
7: 0 9 5 43 29 4 10
8: 1 15 21 14 15 14 21
9: 1 27 8 6 5 29 25
10: 1 13 7 5 26 48 0
11: 1 33 3 18 11 13 22
12: 1 12 23 40 11 2 11
13: 1 5 51 2 28 5 9
14: 1 3 1 21 10 63 2
Now I'd like to draw the following line plot (there's a reason why I need line plots and not, say, histograms or bar plots):
The 7 frequency distributions for context 0, with x as x-axis and the frequency as y-axis, all in the same line plot (with some alpha).
The 7 frequency distributions for context 1, again with x as x-axis and the frequency as y-axis, all in the same line plot (with alpha), but displayed upside-down below the plot for context 0.
Ggplot would surely do this very nicely, but it seems to require some acrobatics with data tables:
– If I use the data table tablefreqs it's not clear to me how to plot all its rows having context==0 in the same plot: ggplot seems to only think column-wise, not row-wise. I could use the six values of x as table rows, but then the "context" values would also end up in a row, and I'm not sure I can subset a data table by values in a row, rather than in a column.
– If I use the matrix freqs, I could create a mini-data-table having x as one column and one frequency distribution as another column, input that into ggplot+geom_line, then go over all 7 frequency distributions in a for-loop maybe. Not clear to me how to tell ggplot to keep the previous plots in this case. Then another for-loop over the two "contexts".
I'd be grateful for suggestions on how to approach this problem in particular, and more generally on what objects to choose for storing this kind of data: matrices? data tables, maybe with a different structure than shown here? some other formats?
I would suggest to familiarize yourself with the concept of what is known as Tidy Data, which are principles for data handling and storage that are adopted by ggplot2 and a number of other packages.
You are free to use a matrix or list of matrices to store your data; however, you can certainly store the data as you describe it (and as I understand it) in a data frame or single table following the following convention of columns:
context | sample | x | freq
I'll show you how I would convert the tablefreqs dataset you shared with us into that format, then how I would go about creating a plot as you are describing it in your question. I'm assuming in this case you only have the two values for context, although you allude to there being more. I'm going to try to interpret correctly what you stated in your question.
Create the Tidy Data frame
Your data frame as shown contains columns x1 through x5 that have values for x spread across more than one column, when you really need these to be converted in the format shown above. This is called "gathering" your data, and you can do that with tidyr::gather().
First, I also need to replicate the naming of your samples according to the matrix dataset, so I'll do that and gather your data:
library(dplyr)
library(tidyr)
library(ggplot2)
# create the sample names
tablefreqs$sample <- rep(paste0('sample',1:7), 2)
# gather the columns together
df <- tablefreqs %>%
gather(key='x', value='freq', -c(context, sample))
Note that in the gather() function, we have to specify to leave alone the two columns df$context and df$sample, as they are not part of the gathering effort. But now we are left with df$x containing character vectors. We can't plot that, because we want the to be in the form of a number (at least... I'm assuming you do). For that, we'll convert using:
df$x <- as.numeric(gsub("[^[:digit:].]", "", df$x))
That extracts the number from each value in df$x and represents it as a number, not a character. We have the opposite issue with df$context, which is actually a discrete factor, and we should represent it as such in order to make plotting a bit easier:
df$context <- factor(df$context)
Create the Plot
Now we're ready to create the plot. From your description, I may not have this perfectly right, but it seems that you want a plot containing both context = 1 and context = 0, and when context = 1 the data should be "upside down". By that, I'm assuming you are talking about plotting df$freq when df$context == 0 and -df$freq when df$context == 1. We could do that using some fancy logic in the ggplot() call, but I find it's easier just to create a new column in your dataset to represent what we will be plotting on the y axis. We'll call this column df$freq_adj and use that for plotting:
df$freq_adj <- ifelse(df$context==1, -df$freq, df$freq)
Then we create the plot. I'll explain a bit below the result:
ggplot(df, aes(x=x, y=freq_adj)) +
geom_line(
aes(color=context, linetype=sample)
) +
geom_hline(yintercept=0, color='gray50') +
scale_x_continuous(expand=expansion(mult=0)) +
theme_bw()
Without some clearer description or picture of what you were looking to do, I took some liberties here. I used color to discriminate between the two values for context, and I'm using linetype to discriminate the different samples. I also added a line at 0, since it seemed appropriate to do so here, and the scale_x_continuous() command is removing the extra white space that is put in place at the extreme ends of the data.
An alternative that is maybe closer to your description would be to physically have a separation between the two plots, and represent context = 1 as a physically separate plot compared to context = 0, with one over top of the other.
Here's the code and plot:
ggplot(df, aes(x=x, y=freq_adj)) +
geom_line(aes(group=sample), alpha=0.3) +
facet_grid(context ~ ., scales='free_y') +
scale_x_continuous(expand=expansion(mult=0)) +
theme_bw()
There the use of aes(group=sample) is quite important, since I want all the lines for each sample to be the same (alpha setting and color), yet ggplot2 needs to know that the connections between the points should be based on "sample". This is done using the group= aesthetic. The scales='free_y' argument on facet_grid() allows the y axis scale to shrink and fit the data according to each facet.
I am trying to compute robust/cluster standard errors after using mlogit() to fit a Multinomial Logit (MNL) in a Discrete Choice problem. Unfortunately, I suspect I am having problems with it because I am using data in long format (this is a must in my case), and getting the error #Error in ef/X : non-conformable arrays after sandwich::vcovHC( , "HC0").
The Data
For illustration, please gently consider the following data. It represents data from 5 individuals (id_ind ) that choose among 3 alternatives (altern). Each of the five individuals chose three times; hence we have 15 choice situations (id_choice). Each alternative is represented by two generic attributes (x1 and x2), and the choices are registered in y (1 if selected, 0 otherwise).
df <- read.table(header = TRUE, text = "
id_ind id_choice altern x1 x2 y
1 1 1 1 1.586788801 0.11887832 1
2 1 1 2 -0.937965347 1.15742493 0
3 1 1 3 -0.511504401 -1.90667519 0
4 1 2 1 1.079365680 -0.37267925 0
5 1 2 2 -0.009203032 1.65150370 1
6 1 2 3 0.870474033 -0.82558651 0
7 1 3 1 -0.638604013 -0.09459502 0
8 1 3 2 -0.071679538 1.56879334 0
9 1 3 3 0.398263302 1.45735788 1
10 2 4 1 0.291413453 -0.09107974 0
11 2 4 2 1.632831160 0.92925495 0
12 2 4 3 -1.193272276 0.77092623 1
13 2 5 1 1.967624379 -0.16373709 1
14 2 5 2 -0.479859282 -0.67042130 0
15 2 5 3 1.109780885 0.60348187 0
16 2 6 1 -0.025834772 -0.44004183 0
17 2 6 2 -1.255129594 1.10928280 0
18 2 6 3 1.309493274 1.84247199 1
19 3 7 1 1.593558740 -0.08952151 0
20 3 7 2 1.778701074 1.44483791 1
21 3 7 3 0.643191170 -0.24761157 0
22 3 8 1 1.738820924 -0.96793288 0
23 3 8 2 -1.151429915 -0.08581901 0
24 3 8 3 0.606695064 1.06524268 1
25 3 9 1 0.673866953 -0.26136206 0
26 3 9 2 1.176959443 0.85005871 1
27 3 9 3 -1.568225496 -0.40002252 0
28 4 10 1 0.516456176 -1.02081089 1
29 4 10 2 -1.752854918 -1.71728381 0
30 4 10 3 -1.176101700 -1.60213536 0
31 4 11 1 -1.497779616 -1.66301234 0
32 4 11 2 -0.931117325 1.50128532 1
33 4 11 3 -0.455543630 -0.64370825 0
34 4 12 1 0.894843784 -0.69859139 0
35 4 12 2 -0.354902281 1.02834859 0
36 4 12 3 1.283785176 -1.18923098 1
37 5 13 1 -1.293772990 -0.73491317 0
38 5 13 2 0.748091387 0.07453705 1
39 5 13 3 -0.463585127 0.64802031 0
40 5 14 1 -1.946438667 1.35776140 0
41 5 14 2 -0.470448172 -0.61326604 1
42 5 14 3 1.478763383 -0.66490028 0
43 5 15 1 0.588240775 0.84448489 1
44 5 15 2 1.131731049 -1.51323232 0
45 5 15 3 0.212145247 -1.01804594 0
")
The problem
Consequently, we can fit an MNL using mlogit() and extract their robust variance-covariance as follows:
library(mlogit)
library(sandwich)
mo <- mlogit(formula = y ~ x1 + x2|0 ,
method ="nr",
data = df,
idx = c("id_choice", "altern"))
sandwich::vcovHC(mo, "HC0")
#Error in ef/X : non-conformable arrays
As we can see there is an error produced by sandwich::vcovHC, which says that ef/X is non-conformable. Where X <- model.matrix(x) and ef <- estfun(x, ...). After looking through the source code on the mirror on GitHub I spot the problem which comes from the fact that, given that the data is in long format, ef has dimensions 15 x 2 and X has 45 x 2.
My workaround
Given that the show must continue, I am computing the robust and cluster standard errors manually using some functions that I borrow from sandwich and I adjusted to accommodate the Stata's output.
> Robust Standard Errors
These lines are inspired on the sandwich::meat() function.
psi<- estfun(mo)
k <- NCOL(psi)
n <- NROW(psi)
rval <- (n/(n-1))* crossprod(as.matrix(psi))
vcov(mo) %*% rval %*% vcov(mo)
# x1 x2
# x1 0.23050261 0.09840356
# x2 0.09840356 0.12765662
Stata Equivalent
qui clogit y x1 x2 ,group(id_choice) r
mat li e(V)
symmetric e(V)[2,2]
y: y:
x1 x2
y:x1 .23050262
y:x2 .09840356 .12765662
> Clustered Standard Errors
Here, given that each individual answers 3 questions is highly likely that there is some degree of correlation among individuals; hence cluster corrections should be preferred in such situations. Below I compute the cluster correction in this case and I show the equivalence with the Stata output of clogit , cluster().
id_ind_collapsed <- df$id_ind[!duplicated(mo$model$idx$id_choice,)]
psi_2 <- rowsum(psi, group = id_ind_collapsed )
k_cluster <- NCOL(psi_2)
n_cluster <- NROW(psi_2)
rval_cluster <- (n_cluster/(n_cluster-1))* crossprod(as.matrix(psi_2))
vcov(mo) %*% rval_cluster %*% vcov(mo)
# x1 x2
# x1 0.1766707 0.1007703
# x2 0.1007703 0.1180004
Stata equivalent
qui clogit y x1 x2 ,group(id_choice) cluster(id_ind)
symmetric e(V)[2,2]
y: y:
x1 x2
y:x1 .17667075
y:x2 .1007703 .11800038
The Question:
I would like to accommodate my computations within the sandwich ecosystem, meaning not computing the matrices manually but actually using the sandwich functions. Is it possible to make it work with models in long format like the one described here? For example, providing the meat and bread objects directly to perform the computations? Thanks in advance.
PS: I noted that there is a dedicated bread function in sandwich for mlogit, but I could not spot something like meat for mlogit, but anyways I am probably missing something here...
Why vcovHC does not work for mlogit
The class of HC covariance estimators can just be applied in models with a single linear predictor where the score function aka estimating function is the product of so-called "working residuals" and a regressor matrix. This is explained in some detail in the Zeileis (2006) paper (see Equation 7), provided as vignette("sandwich-OOP", package = "sandwich") in the package. The ?vcovHC also pointed to this but did not explain it very well. I have improved this in the documentation at http://sandwich.R-Forge.R-project.org/reference/vcovHC.html now:
The function meatHC is the real work horse for estimating the meat of HC sandwich estimators - the default vcovHC method is a wrapper calling sandwich and bread. See Zeileis (2006) for more implementation details. The theoretical background, exemplified for the linear regression model, is described below and in Zeileis (2004). Analogous formulas are employed for other types of models, provided that they depend on a single linear predictor and the estimating functions can be represented as a product of “working residual” and regressor vector (Zeileis 2006, Equation 7).
This means that vcovHC() is not applicable to multinomial logit models as they generally use separate linear predictors for the separate response categories. Similarly, two-part or hurdle models etc. are not supported.
Basic "robust" sandwich covariance
Generally, for computing the basic Eicker-Huber-White sandwich covariance matrix estimator, the best strategy is to use the sandwich() function and not the vcovHC() function. The former works for any model with estfun() and bread() methods.
For linear models sandwich(..., adjust = FALSE) (default) and sandwich(..., adjust = TRUE) correspond to HC0 and HC1, respectively. In a model with n observations and k regression coefficients the former standardizes with 1/n and the latter with 1/(n-k).
Stata, however, divides by 1/(n-1) in logit models, see:
Different Robust Standard Errors of Logit Regression in Stata and R. To the best of my knowledge there is no clear theoretical reason for using specifically one or the other adjustment. And already in moderately large samples, this makes no difference anyway.
Remark: The adjustment with 1/(n-1) is not directly available in sandwich() as an option. However, coincidentally, it is the default in vcovCL() without specifying a cluster variable (i.e., treating each observation as a separate cluster). So this is a convenient "trick" if you want to get exactly the same results as Stata.
Clustered covariance
This can be computed "as usual" via vcovCL(..., cluster = ...). For mlogit models you just have to consider that the cluster variable just needs to be provided once (as opposed to stacked several times in long format).
Replicating Stata results
With the data and model from your post:
vcovCL(mo)
## x1 x2
## x1 0.23050261 0.09840356
## x2 0.09840356 0.12765662
vcovCL(mo, cluster = df$id_choice[1:15])
## x1 x2
## x1 0.1766707 0.1007703
## x2 0.1007703 0.1180004
I'm having some trouble using coxph(). I've two categorical variables:"tecnologia" and "pais", and I want to evaluate the possible interaction effect of "pais" on "tecnologia"."tecnologia" is a variable factor with 2 levels: gps and convencional. And "pais" as 2 levels: PT and ES. I have no idea why this warning keeps appearing.
Here's the code and the output:
cox_AC<-coxph(Surv(dados_temp$dias_seg,dados_temp$status)~tecnologia*pais,data=dados_temp)
Warning message:
In coxph(Surv(dados_temp$dias_seg, dados_temp$status) ~ tecnologia * :
X matrix deemed to be singular; variable 3
> cox_AC
Call:
coxph(formula = Surv(dados_temp$dias_seg, dados_temp$status) ~
tecnologia * pais, data = dados_temp)
coef exp(coef) se(coef) z p
tecnologiagps -0.152 0.859 0.400 -0.38 7e-01
paisPT 1.469 4.345 0.406 3.62 3e-04
tecnologiagps:paisPT NA NA 0.000 NA NA
Likelihood ratio test=23.8 on 2 df, p=6.82e-06 n= 127, number of events= 64
I'm opening another question about this subject, although I made a similar one some months ago, because I'm facing the same problem again, with other data. And this time I'm sure it's not a data related problem.
Can somebody help me?
Thank you
UPDATE:
The problem does not seem to be a perfect classification
> xtabs(~status+tecnologia,data=dados)
tecnologia
status conv doppler gps
0 39 6 24
1 30 3 34
> xtabs(~status+pais,data=dados)
pais
status ES PT
0 71 8
1 49 28
> xtabs(~tecnologia+pais,data=dados)
pais
tecnologia ES PT
conv 69 0
doppler 1 8
gps 30 28
Here's a simple example which seems to reproduce your problem:
> library(survival)
> (df1 <- data.frame(t1=seq(1:6),
s1=rep(c(0, 1), 3),
te1=c(rep(0, 3), rep(1, 3)),
pa1=c(0,0,1,0,0,0)
))
t1 s1 te1 pa1
1 1 0 0 0
2 2 1 0 0
3 3 0 0 1
4 4 1 1 0
5 5 0 1 0
6 6 1 1 0
> (coxph(Surv(t1, s1) ~ te1*pa1, data=df1))
Call:
coxph(formula = Surv(t1, s1) ~ te1 * pa1, data = df1)
coef exp(coef) se(coef) z p
te1 -23 9.84e-11 58208 -0.000396 1
pa1 -23 9.84e-11 100819 -0.000229 1
te1:pa1 NA NA 0 NA NA
Now lets look for 'perfect classification' like so:
> (xtabs( ~ s1+te1, data=df1))
te1
s1 0 1
0 2 1
1 1 2
> (xtabs( ~ s1+pa1, data=df1))
pa1
s1 0 1
0 2 1
1 3 0
Note that a value of 1 for pa1 exactly predicts having a status s1 equal to 0. That is to say, based on your data, if you know that pa1==1 then you can be sure than s1==0. Thus fitting Cox's model is not appropriate in this setting and will result in numerical errors.
This can be seen with
> coxph(Surv(t1, s1) ~ pa1, data=df1)
giving
Warning message:
In fitter(X, Y, strats, offset, init, control, weights = weights, :
Loglik converged before variable 1 ; beta may be infinite.
It's important to look at these cross tables before fitting models. Also it's worth starting with simpler models before considering those involving interactions.
If we add the interaction term to df1 manually like this:
> (df1 <- within(df1,
+ te1pa1 <- te1*pa1))
t1 s1 te1 pa1 te1pa1
1 1 0 0 0 0
2 2 1 0 0 0
3 3 0 0 1 0
4 4 1 1 0 0
5 5 0 1 0 0
6 6 1 1 0 0
Then check it with
> (xtabs( ~ s1+te1pa1, data=df1))
te1pa1
s1 0
0 3
1 3
We can see that it's a useless classifier, i.e. it does not help predict status s1.
When combining all 3 terms, the fitter does manage to produce a numerical value for te1 and pe1 even though pe1 is a perfect predictor as above. However a look at the values for the coefficients and their errors shows them to be implausible.
Edit #JMarcelino: If you look at the warning message from the first coxph model in the example, you'll see the warning message:
2: In coxph(Surv(t1, s1) ~ te1 * pa1, data = df1) :
X matrix deemed to be singular; variable 3
Which is likely the same error you're getting and is due to this problem of classification. Also, your third cross table xtabs(~ tecnologia+pais, data=dados) is not as important as the table of status by interaction term. You could add the interaction term manually first as in the example above then check the cross table. Or you could say:
> with(df1,
table(s1, pa1te1=pa1*te1))
pa1te1
s1 0
0 3
1 3
That said, I notice one of the cells in your third table has a zero (conv, PT) meaning you have no observations with this combination of predictors. This is going to cause problems when trying to fit.
In general, the outcome should be have some values for all levels of the predictors and the predictors should not classify the outcome as exactly all or nothing or 50/50.
Edit 2 #user75782131 Yes, generally speaking xtabs or a similar cross-table should be performed in models where the outcome and predictors are discrete i.e. have a limited no. of levels. If 'perfect classification' is present then a predictive model / regression may not be appropriate. This is true for example for logistic regression (outcome is binary) as well as Cox's model.
Suppose my dataset is a 100 x 3 matrix filled with categorical variables. I would like to do binary classification on the response variable. Let's make up a dataset with following code:
set.seed(2013)
y <- as.factor(round(runif(n=100,min=0,max=1),0))
var1 <- rep(c("red","blue","yellow","green"),each=25)
var2 <- rep(c("shortest","short","tall","tallest"),25)
df <- data.frame(y,var1,var2)
The data looks like this:
> head(df)
y var1 var2
1 0 red shortest
2 1 red short
3 1 red tall
4 1 red tallest
5 0 red shortest
6 1 red short
I tried to do random forest and adaboost on this data with two different approaches. The first approach is to use the data as it is:
> library(randomForest)
> randomForest(y~var1+var2,data=df,ntrees=500)
Call:
randomForest(formula = y ~ var1 + var2, data = df, ntrees = 500)
Type of random forest: classification
Number of trees: 500
No. of variables tried at each split: 1
OOB estimate of error rate: 44%
Confusion matrix:
0 1 class.error
0 29 22 0.4313725
1 22 27 0.4489796
----------------------------------------------------
> library(ada)
> ada(y~var1+var2,data=df)
Call:
ada(y ~ var1 + var2, data = df)
Loss: exponential Method: discrete Iteration: 50
Final Confusion Matrix for Data:
Final Prediction
True value 0 1
0 34 17
1 16 33
Train Error: 0.33
Out-Of-Bag Error: 0.33 iteration= 11
Additional Estimates of number of iterations:
train.err1 train.kap1
10 16
The second approach is to transform the dataset into wide format and treat each category as a variable. The reason I am doing this is because my actual dataset has 500+ factors in var1 and var2, and as a result, tree partitioning will always divide the 500 categories into 2 splits. A lot of information is lost by doing that. To transform the data:
id <- 1:100
library(reshape2)
tmp1 <- dcast(melt(cbind(id,df),id.vars=c("id","y")),id+y~var1,fun.aggregate=length)
tmp2 <- dcast(melt(cbind(id,df),id.vars=c("id","y")),id+y~var2,fun.aggregate=length)
df2 <- merge(tmp1,tmp2,by=c("id","y"))
The new data looks like this:
> head(df2)
id y blue green red yellow short shortest tall tallest
1 1 0 0 0 2 0 0 2 0 0
2 10 1 0 0 2 0 2 0 0 0
3 100 0 0 2 0 0 0 0 0 2
4 11 0 0 0 2 0 0 0 2 0
5 12 0 0 0 2 0 0 0 0 2
6 13 1 0 0 2 0 0 2 0 0
I apply random forest and adaboost to this new dataset:
> library(randomForest)
> randomForest(y~blue+green+red+yellow+short+shortest+tall+tallest,data=df2,ntrees=500)
Call:
randomForest(formula = y ~ blue + green + red + yellow + short + shortest + tall + tallest, data = df2, ntrees = 500)
Type of random forest: classification
Number of trees: 500
No. of variables tried at each split: 2
OOB estimate of error rate: 39%
Confusion matrix:
0 1 class.error
0 32 19 0.3725490
1 20 29 0.4081633
----------------------------------------------------
> library(ada)
> ada(y~blue+green+red+yellow+short+shortest+tall+tallest,data=df2)
Call:
ada(y ~ blue + green + red + yellow + short + shortest + tall +
tallest, data = df2)
Loss: exponential Method: discrete Iteration: 50
Final Confusion Matrix for Data:
Final Prediction
True value 0 1
0 36 15
1 20 29
Train Error: 0.35
Out-Of-Bag Error: 0.33 iteration= 26
Additional Estimates of number of iterations:
train.err1 train.kap1
5 10
The results from two approaches are different. The difference is more obvious as we introduce more levels in each variable, i.e., var1 and var2. My question is, since we are using exactly the same data, why is the result different? How should we interpret the results from both approaches? Which is more reliable?
While these two models look identical, they are fundamentally different from one another- On the second model, you implicitly include the possibility that a given observation may have multiple colors and multiple heights. The correct choice between the two model formulations will depend on the characteristics of your real-world observations. If these characters are exclusive (i.e., each observation is of a single color and height), the first formulation of the model will be the right one to use. However, if an observation may be both blue and green, or any other color combination, you may use the second formulation. From a hunch looking at your original data, it seems like the first one is most appropriate (i.e., how would an observation have multiple heights??).
Also, why did you code your logical variable columns in df2 as 0s and 2s instead of 0/1? I wander if that will have any impact on the fit depending on how the data is being coded as factor or numerical.
I'm a PhD student of genetics and I am trying do association analysis of some genetic data using linear regression. In the table below I'm regressing each 'trait' against each 'SNP' There is also a interaction term include as 'var'
I've only used R for 2 weeks and I don't have any programming background so please explain any help provided as I want to understand.
This is a sample of my data:
Sample ID var trait 1 trait 2 trait 3 SNP1 SNP2 SNP3
77856517 2 188 3 2 1 0 0
375689755 8 17 -1 -1 1 -1 -1
392513415 8 28 14 4 1 1 1
393612038 8 85 14 6 1 1 0
401623551 8 152 11 -1 1 0 0
348466144 7 -74 11 6 1 0 0
77852806 4 81 16 6 1 1 0
440614343 8 -93 8 0 0 1 0
77853193 5 3 6 5 1 1 1
and this is the code I've been using for a single regression:
result1 <-lm(trait1~SNP1+var+SNP1*var, na.action=na.exclude)
I want to run a loop where every trait is tested against each SNP.
I've been trying to modify codes I've found online but I always run into some error that I don't understand how to solve.
Thank you for any and all help.
Personally I don't find the problem so easy. Specially for an R novice.
Here a solution based on creating dynamically the regression formula.
The idea is to use paste function to create different formula terms, y~ x + var + x * var then coercing the result string tp a formula using as.formula. Here y and x are the formula dynamic terms: y in c(trait1,trai2,..) and x in c(SNP1,SNP2,...). Of course here I use lapply to loop.
lapply(1:3,function(i){
y <- paste0('trait',i)
x <- paste0('SNP',i)
factor1 <- x
factor2 <- 'var'
factor3 <- paste(x,'var',sep='*')
listfactor <- c(factor1,factor2,factor3)
form <- as.formula(paste(y, "~",paste(listfactor,collapse="+")))
lm(formula = form, data = dat)
})
I hope someone come with easier solution, ore more R-ish one:)
EDIT
Thanks to #DWin comment , we can simplify the formula to just y~x*var since it means y is modeled by x,var and x*var
So the code above will be simplified to :
lapply(1:3,function(i){
y <- paste0('trait',i)
x <- paste0('SNP',i)
LHS <- paste(x,'var',sep='*')
form <- as.formula(paste(y, "~",LHS)
lm(formula = form, data = dat)
})