I am using the function ada in R, and I'm having a little difficulty. I have training data that looks like this
V13 V15 V17 V19
1 0.017241379 0.471264368 0.01449275 0.24637681
2 0.255813953 0.011627907 0.06849315 0.05479452
3 0.040000000 0.400000000 0.06000000 0.10000000
4 0.500000000 0.000000000 0.05128205 0.00000000
5 0.102040816 0.367346939 0.05769231 0.19230769
6 0.561403509 0.105263158 0.11111111 0.00000000
7 0.300813008 0.048780488 0.12222222 0.03333333
8 0.000000000 0.714285714 0.14285714 0.07142857
9 0.328947368 0.013157895 0.01492537 0.00000000
10 0.536585366 0.060975610 0.16071429 0.03571429
11 0.338461538 0.030769231 0.11764706 0.03921569
12 0.033898305 0.322033898 0.11764706 0.21568627
This is what I have stored in the variable
matrix.x
Then I have the response variables y
y
1 1
2 -1
3 1
4 -1
5 1
6 -1
7 -1
8 1
9 -1
10 -1
11 -1
12 1
I simply run the following:
ada.obj = ada(matrix.x, matrix.y)
And then
ada.pred = predict(ada.obj, matrix.x)
And for some reason, I get a matrix with all 1s or all -1s. What am I doing wrong? Ideally, I want the ada.pred to spit out the exact classifications of the training data.
Thanks.
Also how would I go about using the AdabOost1.M1 function in caret package of R?
If I do DF$where <- tree$where after fitting an rpart object using DF as my data, will each row be mapped to its corresponding leaf through the column where?
Thanks!
As an example of how to demonstrate that this is possibly true (modulo my understanding of your question being correct), we work with the first example in ?rpart:
require(rpart)
fit <- rpart(Kyphosis ~ Age + Number + Start, data = kyphosis)
kyphosis$where <- fit$where
> str(kyphosis)
'data.frame': 81 obs. of 5 variables:
$ Kyphosis: Factor w/ 2 levels "absent","present": 1 1 2 1 1 1 1 1 1 2 ...
$ Age : int 71 158 128 2 1 1 61 37 113 59 ...
$ Number : int 3 3 4 5 4 2 2 3 2 6 ...
$ Start : int 5 14 5 1 15 16 17 16 16 12 ...
$ where : int 9 7 9 9 3 3 3 3 3 8 ...
> plot(fit)
> text(fit, use.n = TRUE)
And now look at some tables based on the 'where' vector and some logical tests:
First node:
> with(kyphosis, table(where, Start >= 8.5))
where FALSE TRUE
3 0 29
5 0 12
7 0 14
8 0 7
9 19 0 # so this is the row that describes that split
> fit$frame[9,]
var n wt dev yval complexity ncompete nsurrogate yval2.V1
3 <leaf> 19 19 8 2 0.01 0 0 2.0000000
yval2.V2 yval2.V3 yval2.V4 yval2.V5 yval2.nodeprob
3 8.0000000 11.0000000 0.4210526 0.5789474 0.2345679
Second node:
> with(kyphosis, table(where, Start >= 8.5, Start>=14.5))
, , = FALSE
where FALSE TRUE
3 0 0
5 0 12
7 0 14
8 0 7
9 19 0
, , = TRUE
where FALSE TRUE
3 0 29
5 0 0
7 0 0
8 0 0
9 0 0
And this is the row of fit$frame that describes the second split:
> fit$frame[3,]
var n wt dev yval complexity ncompete nsurrogate yval2.V1
4 <leaf> 29 29 0 1 0.01 0 0 1.0000000
yval2.V2 yval2.V3 yval2.V4 yval2.V5 yval2.nodeprob
4 29.0000000 0.0000000 1.0000000 0.0000000 0.3580247
So I would characterize the value of fit$where as describing the "terminal nodes" which are being labeled as '<leaf>', which may or not be what you were calling the "nodes".
> fit$frame
var n wt dev yval complexity ncompete nsurrogate yval2.V1
1 Start 81 81 17 1 0.17647059 2 1 1.00000000
2 Start 62 62 6 1 0.01960784 2 2 1.00000000
4 <leaf> 29 29 0 1 0.01000000 0 0 1.00000000
5 Age 33 33 6 1 0.01960784 2 2 1.00000000
10 <leaf> 12 12 0 1 0.01000000 0 0 1.00000000
11 Age 21 21 6 1 0.01960784 2 0 1.00000000
22 <leaf> 14 14 2 1 0.01000000 0 0 1.00000000
23 <leaf> 7 7 3 2 0.01000000 0 0 2.00000000
3 <leaf> 19 19 8 2 0.01000000 0 0 2.00000000
yval2.V2 yval2.V3 yval2.V4 yval2.V5 yval2.nodeprob
1 64.00000000 17.00000000 0.79012346 0.20987654 1.00000000
2 56.00000000 6.00000000 0.90322581 0.09677419 0.76543210
4 29.00000000 0.00000000 1.00000000 0.00000000 0.35802469
5 27.00000000 6.00000000 0.81818182 0.18181818 0.40740741
10 12.00000000 0.00000000 1.00000000 0.00000000 0.14814815
11 15.00000000 6.00000000 0.71428571 0.28571429 0.25925926
22 12.00000000 2.00000000 0.85714286 0.14285714 0.17283951
23 3.00000000 4.00000000 0.42857143 0.57142857 0.08641975
3 8.00000000 11.00000000 0.42105263 0.57894737 0.23456790
I am creating a scatter plot using ggplot2. The default gives me an x axis that has every value form 0 to 30. I'd prefer to have it go by 5s or something like that. I have been trying to use scale_x_continuous(), but I get this:
Error: Discrete value supplied to continuous scale
Here is the code that I am trying to work with:
Daily.Average.plot <- ggplot(data = Daily.average, aes(factor(Day), Mass))+
geom_point(aes(color = factor(Temp))) +
scale_x_continuous(breaks = seq(0,30,5))
Daily.Average.plot
When I run this without the scale_x_continuous I get a graph that looks fine with no errors, just the incorrect x axis. All of the columns in the data set are numeric when I check str(), if that matters. Do I have an error in my code, or should I be using something different to change the scale?
Here is a sample of my data set:
N Day Mass Temp
1 1 0.00000000 5
2 2 0.00000000 5
3 3 0.07692308 5
4 4 0.07692308 5
5 5 0.07692308 5
6 6 0.15384615 5
7 7 0.15384615 5
8 8 0.23076923 5
9 9 0.38461538 5
10 10 0.46153846 5
11 1 0.00000000 10
12 2 0.00000000 10
13 3 0.00000000 10
14 4 0.09090909 10
15 5 0.09090909 10
16 6 0.54545455 10
17 7 0.54545455 10
18 8 0.63636364 10
19 9 0.90909091 10
20 10 1.36363636 10
21 1 0.00000000 15
22 2 0.07692308 15
23 3 0.61538462 15
24 4 0.76923077 15
25 5 0.76923077 15
26 6 1.23076923 15
27 7 1.69230769 15
28 8 2.07692308 15
29 9 2.46153846 15
30 10 3.07692308 15
does anyone know if it is possible to exclude zero values from a barplot in ggplot?
I have a dataset that contains proportions as follows:
X5employf prop X5employff
1 increase 0.02272727
2 increase 0.59090909 1
3 increase 0.02272727 1 and 8
4 increase 0.02272727 2
5 increase 0.34090909 3
6 increase 0.00000000 4
7 increase 0.00000000 5
8 increase 0.00000000 6
9 increase 0.00000000 6 and 7
10 increase 0.00000000 6 and 7
11 increase 0.00000000 7
12 increase 0.00000000 8
13 decrease 0.00000000
14 decrease 0.00000000 1
15 decrease 0.00000000 1 and 8
16 decrease 0.00000000 2
17 decrease 0.00000000 3
18 decrease 0.10000000 4
19 decrease 0.50000000 5
20 decrease 0.20000000 6
21 decrease 0.00000000 6 and 7
22 decrease 0.00000000 6 and 7
23 decrease 0.10000000 7
24 decrease 0.10000000 8
25 same 0.00000000
26 same 0.00000000 1
27 same 0.00000000 1 and 8
28 same 0.00000000 2
29 same 0.00000000 3
30 same 0.21052632 4
31 same 0.31578947 5
32 same 0.26315789 6
33 same 0.15789474 6 and 7
34 same 0.00000000 6 and 7
35 same 0.05263158 7
36 same 0.00000000 8
as you can see in the 'prop' column there are a lot of zero values. I am producing a facetted bar plot with 'X5employf' column as the facet. But because of the zero values I end up with a lot of empty space on my plot(see below). Is there a way of forcing ggplot to not plot the zero values? Its not the case of dropping unused factors as these are not NA values but 0s. Any ideas??!
For your plot, simply use which to specify that you only want to use the subset of the dataframe containing non-zero proportions. This way you don't have to modify your original dataframe. Then, specify "free_x" in your scales argument within facet_grid to get rid of your empty space in your faceted plot.
plot <- ggplot(df[which(df$prop>0),], aes(X5employff, prop)) +
geom_bar(aes(fill=X5employff, stat="identity")) +
facet_grid( ~ X5employf, scales="free_x") +
theme_bw()
plot
Note that I replaced the blank fields with "blank" for the sake of quick import into R from Excel.
I'm unsure whether or not there is a way to set ignored values in ggplot. However you could consider simply recoding 0's to NA:
df[df$prop == 0] <- NA
I'm using MALLET for topic analysis which is outputting results in text files ("topics.txt") of several thousand rows and a hundred or so rows where each row consists of tab-separated variables like this:
Num1 text1 topic1 proportion1 topic2 proportion2 topic3 proportion3, etc.
Num2 text2 topic1 proportion1 topic2 proportion2 topic3 proportion3, etc.
Num3 text3 topic1 proportion1 topic2 proportion2 topic3 proportion3, etc.
Here's a snippet of the actual data:
> dat[1:5,1:10]
V1 V2 V3 V4 V5 V6 V7 V8 V9 V10
1 0 10.txt 27 0.4560785 23 0.3040853 20 0.1315621 21 0.03632624
2 1 1001.txt 20 0.2660085 12 0.2099153 8 0.1699586 13 0.16922928
3 2 1002.txt 16 0.3341721 2 0.1747023 10 0.1360454 12 0.07507119
4 3 1003.txt 12 0.5366148 8 0.2255179 18 0.1388561 0 0.01867091
5 4 1005.txt 16 0.2363206 0 0.2214441 24 0.1914769 7 0.17760521
I'm trying to use R to convert this output into a data table where the topics are column headers and each topic contains the values of the variable 'proportion' directly to the right hand side of each variable 'topic', for each value of 'text'. Like this:
topic1 topic2 topic3
text1 proportion1 proportion2 proportion3
text2 proportion1 proportion2 proportion3
or with the data snippet above, like so:
0 2 7 8 10 12 13 16 18 20 21 23 24 27
10.txt 0 0 0 0 0 0 0 0 0 0.1315621 0.03632624 0.3040853 0 0.4560785
1001.txt 0 0 0 0.1699586 0 0.2099153 0.1692292 0 0 0.2660085 0 0 0 0
1002.txt 0 0.1747023 0 0 0.1360454 0.0750711 0 0.3341721 0 0 0 0 0 0
1003.txt 0.0186709 0 0 0.2255179 0 0.5366148 0 0 0.138856 0 0 0 0 0
1005.txt 0.2214441 0 0.1776052 0 0 0 0 0.2363206 0 0 0 0 0.1914769 0
This is the R code I've got to do the job, sent from a friend, but it doesn't work for me (and I don't know enough about it to fix it myself):
##########################################
dat<-read.table("topics.txt", header=F, sep="\t")
datnames<-subset(dat, select=2)
dat2<-subset(dat, select=3:length(dat))
y <- data.frame(topic=character(0),proportion=character(0),text=character(0))
for(i in seq(1, length(dat2), 2)){
z<-i+1
x<-dat2[,i:z]
x<-cbind(x, datnames)
colnames(x)<-c("topic","proportion", "text")
y<-rbind(y, x)
}
# Right at this step at the end of the block
# I get this message that may indicate the problem:
# Error in c(in c("topic", "proportion", "text") : unused argument(s) ("text")
y[is.na(y)] <- 0
xdat<-xtabs(proportion ~ text+topic, data=y)
write.table(xdat, file="topicMatrix.txt", sep="\t", eol = "\n", quote=TRUE, col.names=TRUE, row.names=TRUE)
##########################################
I'd be most grateful for any suggestions on how I can get this code working. My problem may be related to this one and possibly this one also, but I don't yet have the skills to make immediate use of the answers to those questions.
Here is one approach to your problem
dat <-read.table(as.is = TRUE, header = FALSE, textConnection(
"Num1 text1 topic1 proportion1 topic2 proportion2 topic3 proportion3
Num2 text2 topic1 proportion1 topic2 proportion2 topic3 proportion3
Num3 text3 topic1 proportion1 topic2 proportion2 topic3 proportion3"))
NTOPICS = 3
nam <- c('num', 'text',
paste(c('topic', 'proportion'), rep(1:NTOPICS, each = 2), sep = ""))
dat_l <- reshape(setNames(dat, nam), varying = 3:length(nam), direction = 'long',
sep = "")
reshape2::dcast(dat_l, num + text ~ topic, value_var = 'proportion')
num text topic1 topic2 topic3
1 Num1 text1 proportion1 proportion2 proportion3
2 Num2 text2 proportion1 proportion2 proportion3
3 Num3 text3 proportion1 proportion2 proportion3
EDIT. This will work irrespective of whether the proportions are text or numbers. You can also modify NTOPICS to suit the number of topics you have
You can get this into a long format but to go further required real data.
EDITED after data offered. Still not sure about the overall structure of what is coming from MALLET, but at least the R functions are demonstrated. This approach has the "feature" that proportions are summed if there are overlapping topics. Depending on the data layout that may be an advantage or not.
dat <-read.table(textConnection(" V1 V2 V3 V4 V5 V6 V7 V8 V9 V10
1 0 10.txt 27 0.4560785 23 0.3040853 20 0.1315621 21 0.03632624
2 1 1001.txt 20 0.2660085 12 0.2099153 8 0.1699586 13 0.16922928
3 2 1002.txt 16 0.3341721 2 0.1747023 10 0.1360454 12 0.07507119
4 3 1003.txt 12 0.5366148 8 0.2255179 18 0.1388561 0 0.01867091
5 4 1005.txt 16 0.2363206 0 0.2214441 24 0.1914769 7 0.17760521
"),
header=TRUE)
ldat <- reshape(dat, idvar=1:2, varying=list(topics=c("V3", "V5", "V7", "V9"),
props=c("V4", "V6", "V8", "V10")),
direction="long")
####------------------####
> ldat
V1 V2 time V3 V4
0.10.txt.1 0 10.txt 1 27 0.45607850
1.1001.txt.1 1 1001.txt 1 20 0.26600850
2.1002.txt.1 2 1002.txt 1 16 0.33417210
3.1003.txt.1 3 1003.txt 1 12 0.53661480
4.1005.txt.1 4 1005.txt 1 16 0.23632060
0.10.txt.2 0 10.txt 2 23 0.30408530
1.1001.txt.2 1 1001.txt 2 12 0.20991530
2.1002.txt.2 2 1002.txt 2 2 0.17470230
3.1003.txt.2 3 1003.txt 2 8 0.22551790
4.1005.txt.2 4 1005.txt 2 0 0.22144410
0.10.txt.3 0 10.txt 3 20 0.13156210
1.1001.txt.3 1 1001.txt 3 8 0.16995860
2.1002.txt.3 2 1002.txt 3 10 0.13604540
3.1003.txt.3 3 1003.txt 3 18 0.13885610
4.1005.txt.3 4 1005.txt 3 24 0.19147690
0.10.txt.4 0 10.txt 4 21 0.03632624
1.1001.txt.4 1 1001.txt 4 13 0.16922928
2.1002.txt.4 2 1002.txt 4 12 0.07507119
3.1003.txt.4 3 1003.txt 4 0 0.01867091
4.1005.txt.4 4 1005.txt 4 7 0.17760521
Now can show you how to use xtabs() since those "proportions" are "numeric". Something like this may eventually be what you want. I was surprised that the topics were also integers but perhaps there is a mapping from topic numbers to topic names?:
> xtabs(V4 ~ V3 + V2, data=ldat)
V2
V3 10.txt 1001.txt 1002.txt 1003.txt 1005.txt
0 0.00000000 0.00000000 0.00000000 0.01867091 0.22144410
2 0.00000000 0.00000000 0.17470230 0.00000000 0.00000000
7 0.00000000 0.00000000 0.00000000 0.00000000 0.17760521
8 0.00000000 0.16995860 0.00000000 0.22551790 0.00000000
10 0.00000000 0.00000000 0.13604540 0.00000000 0.00000000
12 0.00000000 0.20991530 0.07507119 0.53661480 0.00000000
13 0.00000000 0.16922928 0.00000000 0.00000000 0.00000000
16 0.00000000 0.00000000 0.33417210 0.00000000 0.23632060
18 0.00000000 0.00000000 0.00000000 0.13885610 0.00000000
20 0.13156210 0.26600850 0.00000000 0.00000000 0.00000000
21 0.03632624 0.00000000 0.00000000 0.00000000 0.00000000
23 0.30408530 0.00000000 0.00000000 0.00000000 0.00000000
24 0.00000000 0.00000000 0.00000000 0.00000000 0.19147690
27 0.45607850 0.00000000 0.00000000 0.00000000 0.00000000
Returning to this problem, I've found that the reshape function is far too demanding on memory, so I use a data.table method instead. A few more steps, but a huge amount faster and substantially less memory intensive.
dat <- read.table(text = "V1 V2 V3 V4 V5 V6 V7 V8 V9 V10
1 0 10.txt 27 0.4560785 23 0.3040853 20 0.1315621 21 0.03632624
2 1 1001.txt 20 0.2660085 12 0.2099153 8 0.1699586 13 0.16922928
3 2 1002.txt 16 0.3341721 2 0.1747023 10 0.1360454 12 0.07507119
4 3 1003.txt 12 0.5366148 8 0.2255179 18 0.1388561 0 0.01867091
5 4 1005.txt 16 0.2363206 0 0.2214441 24 0.1914769 7 0.17760521")
dat$V11 <- rep(NA, 5) # my real data has this extra unwanted col
dat <- data.table(dat)
# get document number
docnum <- dat$V1
# get text number
txt <- dat$V2
# remove doc num and text num so we just have topic and props
dat1 <- dat[ ,c("V1","V2", paste0("V", ncol(dat))) := NULL]
# get topic numbers
n <- ncol(dat1)
tops <- apply(dat1, 1, function(i) i[seq(1, n, 2)])
# get props
props <- apply(dat1, 1, function(i) i[seq(2, n, 2)])
# put topics and props together
tp <- lapply(1:ncol(tops), function(i) data.frame(tops[,i], props[,i]))
names(tp) <- txt
# make into long table
dt <- data.table::rbindlist(tp)
dt$doc <- unlist(lapply(txt, function(i) rep(i, ncol(dat1)/2)))
dt$docnum <- unlist(lapply(docnum, function(i) rep(i, ncol(dat1)/2)))
# reshape to wide
library(data.table)
setkey(dt, tops...i., doc)
out <- dt[CJ(unique(tops...i.), unique(doc))][, as.list(props...i.), by=tops...i.]
setnames(out, c("topic", as.character(txt)))
# transpose to have table of docs (rows) and columns (topics)
tout <- data.table(t(out))
setnames(tout, unname(as.character(tout[1,])))
tout <- tout[-1,]
row.names(tout) <- txt
# replace NA with zero
tout[is.na(tout)] <- 0
And here's the output, docs as rows, topics as columns, doc names are in the rownames, which are not printed, but available for later use.
tout
0 2 7 8 10 12 13 16 18
1: 0.00000000 0.0000000 0.0000000 0.0000000 0.0000000 0.00000000 0.0000000 0.0000000 0.0000000
2: 0.00000000 0.0000000 0.0000000 0.1699586 0.0000000 0.20991530 0.1692293 0.0000000 0.0000000
3: 0.00000000 0.1747023 0.0000000 0.0000000 0.1360454 0.07507119 0.0000000 0.3341721 0.0000000
4: 0.01867091 0.0000000 0.0000000 0.2255179 0.0000000 0.53661480 0.0000000 0.0000000 0.1388561
5: 0.22144410 0.0000000 0.1776052 0.0000000 0.0000000 0.00000000 0.0000000 0.2363206 0.0000000
20 21 23 24 27
1: 0.1315621 0.03632624 0.3040853 0.0000000 0.4560785
2: 0.2660085 0.00000000 0.0000000 0.0000000 0.0000000
3: 0.0000000 0.00000000 0.0000000 0.0000000 0.0000000
4: 0.0000000 0.00000000 0.0000000 0.0000000 0.0000000
5: 0.0000000 0.00000000 0.0000000 0.1914769 0.0000000