I am trying to plot the predictions of a lmer model with the following code:
p1 <- ggplot(Mac_Data_Tracking, aes(x = Rspan, y = SubjEff, colour = NsCond)) +
geom_point(size=3) +
geom_line(data=newdat, aes(y=predict(SubjEff.model,newdata=newdat)),lineend="round")
print(p1)
I get weird inflections at the end of each line, is there a way to remove them? I have changed the data in newdat, but the lines always have these inflections.
Lines with Inflections at ends:
Note that you have geom_line(data=newdat, aes(y=predict(SubjEff.model,newdata=newdat)). So you've fed newdat to geom_line as the data frame to use for plotting. But then for your y-value you provide a separate vector of predictions (based on newdat), when y should actually be just a column of newdat. I'm not sure why that's causing the inflections at the ends (probably there are, somehow, two different y-values being provided for each of the endpoint x-values), but that's probably the source of your problem.
Instead, you should create a column in newdat with the predictions (if you haven't already) and feed that column name to ggplot as the y in geom_line. To add a column of predictions, do the following:
newdat$pred = predict(SubjEff.model,newdata=newdat)
You should also give geom_line the x values that correspond to the y values in newdat. So your code would be:
geom_line(data=newdat, aes(y=pred, x=Rspan), lineend="round")
(Where Rspan will (automatically) be the Rspan column in newdat.)
It was a problem with having 2 x values, actually...it was having 2 subject values.
The linear mixed model is:
Mixed.model <- lmer(Outcome ~ NsCond + Rspan + (1|Subject), data=Data))
For newdat, I was intially using:
newdat <- expand.grid(Subject=c(min(Data$Subject),max(Data$Subject)),Rspan=c(min(Data$Rspan), max(Data$Rspan)),NsCond=unique(Data$NsCond))
Which gave me:
Subject Rspan NsCond
1 1 0.2916667 Pink
2 18 0.2916667 Pink
3 1 1.0000000 Pink
4 18 1.0000000 Pink
5 1 0.2916667 Babble
6 18 0.2916667 Babble
7 1 1.0000000 Babble
8 18 1.0000000 Babble
9 1 0.2916667 Loss
10 18 0.2916667 Loss
11 1 1.0000000 Loss
12 18 1.0000000 Loss
For each Rspan (x) there are 2 "Subjects" (1 and 18).
I changed newdat to:
newdat <- expand.grid(Subject=1,Rspan=c(min(Data$Rspan), max(Data$Rspan)),NsCond=unique(Data$NsCond))
Which results in:
Subject Rspan NsCond
1 1 0.2916667 Pink
2 1 1.0000000 Pink
3 1 0.2916667 Babble
4 1 1.0000000 Babble
5 1 0.2916667 Loss
6 1 1.0000000 Loss
Now it looks good
Related
I have a data (ge3_ratio) with 86 observations measuring the ratio of phonological errors produced by children. The data was collected from spontaneous speech transcripts from 3 children aged 21 to 65 months. The data look like this:
age
child
Transcripts
Process
Liq
n
ratio
21
Luana
1
posteriorização
N
1
1.0000000
23
Luana
1
apagamento de líquida
S
1
1.0000000
24
Leonardo
1
apagamento de líquida
S
1
1.0000000
24
Túlio
3
assimilação + apagamento de líquida
S
1
0.3333333
24
Túlio
3
metátese
S
1
0.3333333
24
Túlio
3
semivocalização de líquidas
S
1
0.3333333
24
Túlio
3
substituição de líquida
S
1
0.3333333
25
Túlio
2
anteriorização
N
1
0.5000000
25
Túlio
2
apagamento de líquida
S
1
0.5000000
26
Luana
1
apagamento de líquida
S
1
1.0000000
The ratio column was created dividing n by Transcripts. The column Liq describes whether the phonological error (Process) involves a liquid consonant (coded as S) or not (coded as N). I want to test the hypothesis that phonological errors involving liquid consonants last for longer than those not involving them. "Last for longer" does not mean only that the curve length will be greater for S than for N, but also that the ratio must be greater for S than for N as the child grows older. I have the following graphs showing this development path ("Liquida" just means "Liquid"):
ggplot(ge3_ratio, aes(age, ratio, color = Liq)) +
geom_smooth() + scale_color_manual(values = kb) +
labs(x = "age in months")
ggplot(ge3_ratio, aes(age, ratio, color = Liq)) +
geom_path() + scale_color_manual(values = kb) +
labs(x = "age in months")
geom_smooth:
geom_path:
I want help as how to proceed with the analysis (taking into account my hypothesis), which tests I can run and how to do that in R. I started reading about growth curve analysis, but I am unsure about how I can apply the concepts in my data.
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.
Is there a way to extract the values of the fitted line returned from stat_smooth?
The code I am using looks like this:
p <- ggplot(df1, aes(x=Days, y= Qty,group=Category,color=Category))
p <- p + stat_smooth(method=glm, fullrange=TRUE)+ geom_point())
This new r user would greatly appreciate any guidance.
Riffing off of #James example
p <- qplot(hp,wt,data=mtcars) + stat_smooth()
You can use the intermediate stages of the ggplot building process to pull out the plotted data. The results of ggplot_build is a list, one component of which is data which is a list of dataframes which contain the computed values to be plotted. In this case, the list is two dataframes since the original qplot creates one for points and the stat_smooth creates a smoothed one.
> ggplot_build(p)$data[[2]]
geom_smooth: method="auto" and size of largest group is <1000, so using loess. Use 'method = x' to change the smoothing method.
x y ymin ymax se PANEL group
1 52.00000 1.993594 1.149150 2.838038 0.4111133 1 1
2 55.58228 2.039986 1.303264 2.776709 0.3586695 1 1
3 59.16456 2.087067 1.443076 2.731058 0.3135236 1 1
4 62.74684 2.134889 1.567662 2.702115 0.2761514 1 1
5 66.32911 2.183533 1.677017 2.690049 0.2465948 1 1
6 69.91139 2.232867 1.771739 2.693995 0.2244980 1 1
7 73.49367 2.282897 1.853241 2.712552 0.2091756 1 1
8 77.07595 2.333626 1.923599 2.743652 0.1996193 1 1
9 80.65823 2.385059 1.985378 2.784740 0.1945828 1 1
10 84.24051 2.437200 2.041282 2.833117 0.1927505 1 1
11 87.82278 2.490053 2.093808 2.886297 0.1929096 1 1
12 91.40506 2.543622 2.145018 2.942225 0.1940582 1 1
13 94.98734 2.597911 2.196466 2.999355 0.1954412 1 1
14 98.56962 2.652852 2.249260 3.056444 0.1964867 1 1
15 102.15190 2.708104 2.303465 3.112744 0.1969967 1 1
16 105.73418 2.764156 2.357927 3.170385 0.1977705 1 1
17 109.31646 2.821771 2.414230 3.229311 0.1984091 1 1
18 112.89873 2.888224 2.478136 3.298312 0.1996493 1 1
19 116.48101 2.968745 2.531045 3.406444 0.2130917 1 1
20 120.06329 3.049545 2.552102 3.546987 0.2421773 1 1
21 123.64557 3.115893 2.573577 3.658208 0.2640235 1 1
22 127.22785 3.156368 2.601664 3.711072 0.2700548 1 1
23 130.81013 3.175495 2.625951 3.725039 0.2675429 1 1
24 134.39241 3.181411 2.645191 3.717631 0.2610560 1 1
25 137.97468 3.182252 2.658993 3.705511 0.2547460 1 1
26 141.55696 3.186155 2.670350 3.701961 0.2511175 1 1
27 145.13924 3.201258 2.687208 3.715308 0.2502626 1 1
28 148.72152 3.235698 2.721744 3.749652 0.2502159 1 1
29 152.30380 3.291766 2.782767 3.800765 0.2478037 1 1
30 155.88608 3.353259 2.857911 3.848607 0.2411575 1 1
31 159.46835 3.418409 2.938257 3.898561 0.2337596 1 1
32 163.05063 3.487074 3.017321 3.956828 0.2286972 1 1
33 166.63291 3.559111 3.092367 4.025855 0.2272319 1 1
34 170.21519 3.634377 3.165426 4.103328 0.2283065 1 1
35 173.79747 3.712729 3.242093 4.183364 0.2291263 1 1
36 177.37975 3.813399 3.347232 4.279565 0.2269509 1 1
37 180.96203 3.910849 3.447572 4.374127 0.2255441 1 1
38 184.54430 3.977051 3.517784 4.436318 0.2235917 1 1
39 188.12658 4.037302 3.583959 4.490645 0.2207076 1 1
40 191.70886 4.091635 3.645111 4.538160 0.2173882 1 1
41 195.29114 4.140082 3.700184 4.579981 0.2141624 1 1
42 198.87342 4.182676 3.748159 4.617192 0.2115424 1 1
43 202.45570 4.219447 3.788162 4.650732 0.2099688 1 1
44 206.03797 4.250429 3.819579 4.681280 0.2097573 1 1
45 209.62025 4.275654 3.842137 4.709171 0.2110556 1 1
46 213.20253 4.295154 3.855951 4.734357 0.2138238 1 1
47 216.78481 4.308961 3.861497 4.756425 0.2178456 1 1
48 220.36709 4.317108 3.859541 4.774675 0.2227644 1 1
49 223.94937 4.319626 3.851025 4.788227 0.2281358 1 1
50 227.53165 4.316548 3.836964 4.796132 0.2334829 1 1
51 231.11392 4.308435 3.818728 4.798143 0.2384117 1 1
52 234.69620 4.302276 3.802201 4.802351 0.2434590 1 1
53 238.27848 4.297902 3.787395 4.808409 0.2485379 1 1
54 241.86076 4.292303 3.772103 4.812503 0.2532567 1 1
55 245.44304 4.282505 3.754087 4.810923 0.2572576 1 1
56 249.02532 4.269040 3.733184 4.804896 0.2608786 1 1
57 252.60759 4.253361 3.710042 4.796680 0.2645121 1 1
58 256.18987 4.235474 3.684476 4.786473 0.2682509 1 1
59 259.77215 4.215385 3.656265 4.774504 0.2722044 1 1
60 263.35443 4.193098 3.625161 4.761036 0.2764974 1 1
61 266.93671 4.168621 3.590884 4.746357 0.2812681 1 1
62 270.51899 4.141957 3.553134 4.730781 0.2866658 1 1
63 274.10127 4.113114 3.511593 4.714635 0.2928472 1 1
64 277.68354 4.082096 3.465939 4.698253 0.2999729 1 1
65 281.26582 4.048910 3.415849 4.681971 0.3082025 1 1
66 284.84810 4.013560 3.361010 4.666109 0.3176905 1 1
67 288.43038 3.976052 3.301132 4.650972 0.3285813 1 1
68 292.01266 3.936392 3.235952 4.636833 0.3410058 1 1
69 295.59494 3.894586 3.165240 4.623932 0.3550782 1 1
70 299.17722 3.850639 3.088806 4.612473 0.3708948 1 1
71 302.75949 3.804557 3.006494 4.602619 0.3885326 1 1
72 306.34177 3.756345 2.918191 4.594499 0.4080510 1 1
73 309.92405 3.706009 2.823813 4.588205 0.4294926 1 1
74 313.50633 3.653554 2.723308 4.583801 0.4528856 1 1
75 317.08861 3.598987 2.616650 4.581325 0.4782460 1 1
76 320.67089 3.542313 2.503829 4.580796 0.5055805 1 1
77 324.25316 3.483536 2.384853 4.582220 0.5348886 1 1
78 327.83544 3.422664 2.259739 4.585589 0.5661643 1 1
79 331.41772 3.359701 2.128512 4.590891 0.5993985 1 1
80 335.00000 3.294654 1.991200 4.598107 0.6345798 1 1
Knowing a priori where the one you want is in the list isn't easy, but if nothing else you can look at the column names.
It is still better to do the smoothing outside the ggplot call, though.
EDIT:
It turns out replicating what ggplot2 does to make the loess is not as straightforward as I thought, but this will work. I copied it out of some internal functions in ggplot2.
model <- loess(wt ~ hp, data=mtcars)
xrange <- range(mtcars$hp)
xseq <- seq(from=xrange[1], to=xrange[2], length=80)
pred <- predict(model, newdata = data.frame(hp = xseq), se=TRUE)
y = pred$fit
ci <- pred$se.fit * qt(0.95 / 2 + .5, pred$df)
ymin = y - ci
ymax = y + ci
loess.DF <- data.frame(x = xseq, y, ymin, ymax, se = pred$se.fit)
ggplot(mtcars, aes(x=hp, y=wt)) +
geom_point() +
geom_smooth(aes_auto(loess.DF), data=loess.DF, stat="identity")
That gives a plot that looks identical to
ggplot(mtcars, aes(x=hp, y=wt)) +
geom_point() +
geom_smooth()
(which is the expanded form of the original p).
stat_smooth does produce output that you can use elsewhere, and with a slightly hacky way, you can put it into a variable in the global environment.
You enclose the output variable in .. on either side to use it. So if you add an aes in the stat_smooth call and use the global assign, <<-, to assign the output to a varible in the global environment you can get the the fitted values, or others - see below.
qplot(hp,wt,data=mtcars) + stat_smooth(aes(outfit=fit<<-..y..))
fit
[1] 1.993594 2.039986 2.087067 2.134889 2.183533 2.232867 2.282897 2.333626
[9] 2.385059 2.437200 2.490053 2.543622 2.597911 2.652852 2.708104 2.764156
[17] 2.821771 2.888224 2.968745 3.049545 3.115893 3.156368 3.175495 3.181411
[25] 3.182252 3.186155 3.201258 3.235698 3.291766 3.353259 3.418409 3.487074
[33] 3.559111 3.634377 3.712729 3.813399 3.910849 3.977051 4.037302 4.091635
[41] 4.140082 4.182676 4.219447 4.250429 4.275654 4.295154 4.308961 4.317108
[49] 4.319626 4.316548 4.308435 4.302276 4.297902 4.292303 4.282505 4.269040
[57] 4.253361 4.235474 4.215385 4.193098 4.168621 4.141957 4.113114 4.082096
[65] 4.048910 4.013560 3.976052 3.936392 3.894586 3.850639 3.804557 3.756345
[73] 3.706009 3.653554 3.598987 3.542313 3.483536 3.422664 3.359701 3.294654
The outputs you can obtain are:
y, predicted value
ymin, lower pointwise confidence interval around
the mean
ymax, upper pointwise confidence interval around the mean
se, standard error
Note that by default it predicts on 80 data points, which may not be aligned with your original data.
A more general approach could be to simply use the predict() function to predict any range of values that are interesting.
# define the model
model <- loess(wt ~ hp, data = mtcars)
# predict fitted values for each observation in the original dataset
modelFit <- data.frame(predict(model, se = TRUE))
# define data frame for ggplot
df <- data.frame(cbind(hp = mtcars$hp
, wt = mtcars$wt
, fit = modelFit$fit
, upperBound = modelFit$fit + 2 * modelFit$se.fit
, lowerBound = modelFit$fit - 2 * modelFit$se.fit
))
# build the plot using the fitted values from the predict() function
# geom_linerange() and the second geom_point() in the code are built using the values from the predict() function
# for comparison ggplot's geom_smooth() is also shown
g <- ggplot(df, aes(hp, wt))
g <- g + geom_point()
g <- g + geom_linerange(aes(ymin = lowerBound, ymax = upperBound))
g <- g + geom_point(aes(hp, fit, size = 1))
g <- g + geom_smooth(method = "loess")
g
# Predict any range of values and include the standard error in the output
predict(model, newdata = 100:300, se = TRUE)
If you want to bring in the power of the tidyverse, you can use the "broom" library to add the predicted values from the loess function to your original dataset. This is building on #phillyooo's solution.
library(tidyverse)
library(broom)
# original graph with smoother
ggplot(data=mtcars, aes(hp,wt)) +
stat_smooth(method = "loess", span = 0.75)
# Create model that will do the same thing as under the hood in ggplot2
model <- loess(wt ~ hp, data = mtcars, span = 0.75)
# Add predicted values from model to original dataset using broom library
mtcars2 <- augment(model, mtcars)
# Plot both lines
ggplot(data=mtcars2, aes(hp,wt)) +
geom_line(aes(hp, .fitted), color = "red") +
stat_smooth(method = "loess", span = 0.75)
Save the graph object and use ggplot_build() or layer_data() to obtain the elements/estimates for the layers. e.g.
pp<-ggplot(mtcars, aes(x=hp, y=wt)) + geom_point() + geom_smooth();
ggplot_build(pp)
I'm trying to plot a logistic regression in R, for a continuous independent variable and a dichotomous dependent variable. I have very limited experience with R, but my professor has asked me to add this graph to a paper I'm writing, and he said R would probably be the best way to create it. Anyway, I'm sure there are tons of mistakes here, but this is the sort of this previously suggested on StackOverflow:
ggplot(vvv, aes(x = vvv$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.
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)
})