I am trying to create a dummy variable that is equal to 1 if it is in certain states.
My code is not working and I don't see the dummy variable being generated. any help would be appreciated.
as.integer(df2$physicaladdress.stateorprovincecode %in%
c("NJ", "NC", "PA", "RI", "WA", "DE", "GA", "HI",
"ID", "MD", "MT", "NM", "SC", "TX", "UT", "LA", "OH"))
In the console for I receive this but no variable is generated in the dataframe with the dummy variable.
[1] 1 1 0 1 1 0 1 1 0 1 1 1 0 1 1 1 1 1 0 0 0 0 0 0
[25] 0 0 0 0 1 1 1 1 0 1 0 1 1 0 0 1 1 1 1 1 0 1 0 0
[49] 1 1 1 1 1 1 0 0 1 0 1 1 1 1 1 0 1 0 1 0 0 1 0 0
[73] 1 0 1 0 0 1 0 1 1 1 1 0 0 1 1 1 0 0 1 0 0 1 0 0
[97] 1 0 1 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0
[121] 0 0 0 0 0 1 0 1 0 0 0 0 1 1 0 1 1 1 1 1 0 0 0 0
[145] 0 0 0 1 1 1 1 1 0 1 0 0 0 1 0 0 0 1 0 1 1 1 0 1
[169] 0 1 0 1 0 1 0 1 1 0 0 1 0 1 0 0 0 0 0 0 0 0 0 0
[193] 0 1 0 1 0 0 1 0 1 1 1 0 1 0 0 0 0 1 0 0 0 1 1 1
[217] 1 0 1 0 0 0 0 1 0 0 1 1 0 1 0 0 1 0 1 0 0 0 0 0
[241] 0 0 1 0 0 0 0 0 0 1 0 0 0 1 1 1 1 0 0 0 0 0 0 1
[265] 1 0 0 0 0 0 1 0 0 1 1 1 1 0 0 0 1 0 1 0 1 0 1 0
[289] 1 0 1 1 1 0 0 1 0 0 0 1 1 1 1 0 1 0 0 0 0 0 0 1
[313] 0 0 0 0 1 0 0 1 0 0 0 0 0 1 0 1 0 0 0 1 1 0 0 0
[337] 0 0 1 1 0 1 0 0 0 0 1 0 0 1 0 0 0 1 0 0 0 0 1 1
[361] 0 0 0 1 0 0 0 0 0 1 1 0 0 0 0 0 1 1 0 0 1 1 0 0
[385] 1 0 0 0 1 1 1 0 1 1 0 0 0 1 1 0 1 0 1 1 0 1 1 0
[409] 0 0 1 1 0 1 0 0 0 0 0 1 1 0 0 0 0 0 0 0 1 1 0 0
[433] 0 1 0 0 0 0 0 0 0 0 1 0 1 0 1 0 0 0 1 1 0 0 1 1
[457] 0 0 1 1 1 0 0 0 0 1 1 1 0 0 0 1 1 0 1 0 1 0 1 1
[481] 0 0 1 0 0 1 0 0 1 0 0 0 0 0 0 1 1 0 1 0 0 0 1 1
[505] 0 1 0 0 0 1 0 0 0 0 1 1 0 1 0 1 0 1 0 1 0 1 1 0
[529] 0 1 1 0 1 1 0 1 0 0 0 0 1 0 0 0 0 1 0 0 0 1 0 0
[553] 0 1 1 1 1 1 1 1 1 1 1 0 0 0 0 1 1 0 0 0 0 0 1 0
[577] 0 0 0 0 1 0 0 1 1 0 0 0 0 0 1 0 0 0 0 0 0 0 1 1
[601] 0 1 0 1 1 0 1 0 1 1 1 1 0 1 0 0 0 0 1 1 0 0 1 0
[625] 0 0 0 0 0 0 0 0 0 0 0 1 0 0 0 0 1 0 0 0 0 1 1 0
[649] 1 0 1 0 0 0 0 0 0 0 0 0 0 0 0 1 0 0 0 0 0 0 0 1
[673] 0 0 1 0 0 0 0 1 0 0 0 1 0 0 1 0 0 0 0 1 1 1 0 0
[697] 0 0 0 0 0 0 0 0 0 1 0 1 0 0 0 1 0 0 0 0 0 1 0 0
[721] 1 1 0 0 0 0 1 0 0 0 0 0 1 1 1 1 1 1 0 0 1 1 1 1
[745] 0 1 1 0 0 1 0 1 1 0 0 0 1 1 0 1 0 0 0 0 0 1 0 1
[769] 0 1 1 1 1 0 1 0 0 0 0 1 0 1 0 1 1 1 1 0 1 0 0 0
[793] 0 0 1 0 0 0 0 1 0 0 1 0 0 1 0 1 0 0 0 0 1 0 1 0
[817] 0 0 0 0 1 0 0 0 1 0 1 0 0 0 0 0 0 0 1 1 1 0 0 0
[841] 1 1 0 0 0 0 1 0 0 1 1 1 1 0 1 0 0 0 0 1 0 1 0 0
[865] 0 1 1 1 0 1 1 0 1 0 0 0 1 0 0 1 0 1 0 0 0 1 1 0
[889] 1 0 1 0 1 1 1 1 0 1 0 1 0 1 0 0 0 0 1 0 1 0 0 0
[913] 0 0 0 0 0 1 1 1 0 1 0 1 1 1 0 0 1 1 1 0 0 0 0 0
[937] 1 0 1 1 0 0 0 0 0 0 0 1 0 1 0 1 1 1 1 0 0 1 0 0
[961] 0 0 0 0 0 0 0 0 0 1 1 0 0 0 0 0 0 0 0 1 0 0 1 0
[985] 1 0 0 0 0 0 0 0 1 1 1 0 0 1 1 0
[ reached getOption("max.print") -- omitted 547604 entries ]
Related
I have a termdocumentmatrix tdm1 and I put it in through this formula:
comparison.cloud(tdm1, random.order=FALSE,
colors = c("#00B2FF", "red", "#FF0099", "#6600CC", "green", "orange", "blue", "brown"),
title.size=1, max.words=50, scale=c(4, 0.5),rot.per=0.4)
However, I got an error which is "Error in strwidth(words[i], cex = size[i], ...) : invalid 'cex' value"
Not too sure what cex value am I missing.
The tdm1 is as follows:
Docs
Terms anger anticipation disgust fear joy sadness surprise trust
bag 1 0 0 0 0 1 0 0
choices 1 1 0 0 1 2 1 1
limited 1 0 0 0 0 1 0 0
plastic 1 0 0 0 0 1 0 0
provided 1 0 0 0 0 1 0 0
abit 0 1 0 0 1 1 1 1
ai 0 2 0 0 2 1 2 2
always 0 1 0 1 0 0 0 1
amazed 0 1 0 0 1 1 1 1
amount 0 1 0 0 1 0 1 1
app 0 2 0 1 2 1 2 2
area 0 1 0 0 1 0 1 1
areas 0 1 0 0 0 0 0 0
around 0 1 0 0 1 0 1 1
atmosphere 0 1 0 0 1 0 1 1
attended 0 1 0 0 1 1 1 1
back 0 1 0 0 1 1 1 0
basah 0 1 0 1 1 0 1 1
bought 0 1 0 0 0 0 0 0
brands 0 1 0 0 1 0 1 1
bras 0 1 0 1 1 0 1 1
breeze 0 1 0 0 1 1 1 1
buy 0 1 0 1 1 1 1 1
can 0 2 0 0 1 1 1 0
cant 0 1 0 0 0 0 0 0
cashiers 0 1 0 0 1 0 1 1
cbd 0 1 0 0 0 0 0 0
charged 0 1 0 0 1 1 1 1
choose 0 2 0 0 1 0 0 0
chopstick 0 1 0 0 0 0 0 0
classes 0 1 0 0 1 0 0 0
come 0 2 0 0 2 2 2 1
concept 0 4 0 0 3 0 3 3
confused 0 1 0 0 1 1 1 1
contains 0 1 0 0 1 1 1 1
convenient 0 8 0 0 5 1 4 4
cool 0 4 0 0 4 1 4 4
correct 0 1 0 0 1 1 1 1
cream 0 1 0 0 1 0 1 1
cup 0 1 0 0 0 0 0 0
curious 0 1 0 0 0 0 0 0
current 0 1 0 0 1 0 1 1
customer 0 1 0 0 0 0 0 0
cutlery 0 1 0 0 0 0 0 0
doesnt 0 1 0 0 1 0 1 1
dont 0 1 0 0 1 0 1 1
don’t 0 1 0 1 1 1 1 1
download 0 1 0 0 1 1 1 1
drinks 0 1 0 0 1 0 0 0
easy 0 3 0 0 2 0 2 2
eat 0 1 0 0 1 1 1 1
electronic 0 1 0 0 1 1 1 1
eleven 0 1 0 0 1 0 1 1
entering 0 1 0 0 1 1 1 1
ereciept 0 1 0 0 1 1 1 1
especially 0 1 0 0 1 0 1 1
even 0 2 0 2 1 1 1 1
exit 0 1 0 0 1 1 1 1
experience 0 3 0 1 2 1 2 2
explained 0 1 0 0 1 1 1 1
feel 0 1 0 0 1 0 1 1
first 0 1 0 0 1 1 1 1
found 0 1 0 0 1 0 1 1
free 0 1 0 0 1 0 1 1
friends 0 1 0 0 1 1 1 1
fussfree 0 1 0 0 1 0 1 1
gantry 0 1 0 0 1 0 1 1
get 0 2 0 1 2 1 1 1
go 0 3 0 1 3 1 3 3
good 0 2 0 0 2 0 2 2
goods 0 1 0 1 1 1 1 1
goto 0 1 0 0 0 0 0 0
great 0 4 0 0 3 0 3 3
greatly 0 1 0 0 1 0 1 1
hasslefree 0 1 0 0 0 0 0 1
history 0 1 0 0 1 1 1 1
hope 0 1 0 0 1 0 1 1
hour 0 1 0 0 1 0 1 1
hours 0 1 0 0 1 0 1 1
ice 0 1 0 0 1 0 1 1
im 0 1 0 0 1 1 1 0
inside 0 1 0 0 1 1 1 1
items 0 3 0 0 3 2 3 2
jiffy 0 1 0 0 1 1 1 1
just 0 4 0 0 3 2 3 2
large 0 1 0 0 1 0 1 1
leave 0 1 0 0 1 1 1 0
less 0 1 0 0 0 0 0 1
link 0 2 0 0 2 1 2 2
linked 0 1 0 0 1 0 1 1
lots 0 1 0 0 0 0 0 1
love 0 3 0 0 3 1 3 2
lovely 0 1 0 0 1 1 1 1
makes 0 1 0 0 1 0 1 1
making 0 1 0 0 0 0 0 1
many 0 1 0 0 0 0 0 0
method 0 1 0 0 1 0 1 1
methods 0 1 0 0 1 1 1 1
minute 0 2 0 0 1 1 1 2
mrt 0 1 0 1 1 0 1 1
muchneeded 0 1 0 0 1 0 1 1
near 0 1 0 0 1 0 1 1
nearby 0 1 0 0 1 0 1 1
new 0 1 0 0 1 0 1 1
newly 0 1 0 0 0 0 0 0
nice 0 1 0 0 0 0 0 0
noodles 0 1 0 0 0 0 0 0
number 0 1 0 0 1 1 1 1
offers 0 1 0 0 1 0 1 1
often 0 2 0 0 2 2 2 1
opened 0 1 0 0 0 0 0 0
operate 0 1 0 0 1 0 1 1
order 0 1 0 0 1 0 1 1
outlets 0 1 0 0 1 1 1 0
patiently 0 1 0 0 1 1 1 1
pay 0 1 0 0 1 0 1 1
payment 0 3 0 0 3 1 3 3
people 0 1 0 0 0 0 0 0
perfect 0 1 0 0 1 1 1 1
pick 0 4 0 0 3 0 3 3
picking 0 1 0 0 1 1 1 1
prepare 0 1 0 0 0 0 0 0
prices 0 1 0 0 1 0 1 1
product 0 2 0 0 2 2 2 2
products 0 7 0 0 5 1 5 4
promotion 0 1 0 1 1 0 1 1
promotions 0 1 0 0 0 0 0 1
quench 0 1 0 0 1 1 1 1
queue 0 2 0 0 2 1 2 2
quite 0 1 0 0 1 1 1 1
range 0 2 0 0 1 1 1 0
ready 0 1 0 0 1 1 1 1
reasonable 0 1 0 0 1 0 1 1
recieved 0 1 0 0 1 1 1 1
recommend 0 1 0 0 0 0 0 1
reduced 0 1 0 0 1 0 1 1
rush 0 1 0 0 1 1 1 0
salut 0 1 0 0 0 0 0 1
sandwiches 0 1 0 0 1 1 1 1
scan 0 1 0 0 1 0 1 1
see 0 3 0 0 2 0 2 2
seems 0 1 0 0 0 0 0 0
setup 0 1 0 0 0 0 0 1
shop 0 3 0 0 2 1 2 2
shopping 0 4 0 1 4 2 4 4
show 0 1 0 0 1 1 1 1
small 0 1 0 0 0 0 0 0
smu 0 2 0 0 2 0 2 2
snacks 0 3 0 0 3 1 1 1
spent 0 1 0 0 1 0 1 1
staff 0 3 0 1 3 3 3 3
stared 0 1 0 1 1 1 1 1
stop 0 1 0 0 1 1 1 1
store 0 14 0 0 10 5 9 9
stores 0 1 0 0 1 0 1 1
students 0 1 0 0 1 0 0 0
stuff 0 1 0 0 0 0 0 0
super 0 2 0 0 2 0 2 2
sure 0 1 0 0 1 1 1 1
sweets 0 1 0 0 1 0 0 0
take 0 1 0 0 1 1 1 0
technology 0 4 0 0 4 2 4 4
thankfully 0 1 0 0 1 1 1 1
theres 0 1 0 0 0 0 0 0
thirst 0 1 0 0 1 1 1 1
thought 0 1 0 0 0 0 0 0
time 0 2 0 0 2 1 2 2
took 0 1 0 0 0 0 0 1
truly 0 1 0 0 0 0 0 1
unmanned 0 1 0 0 1 1 1 1
use 0 1 0 0 1 0 1 1
used 0 1 0 0 1 0 1 1
useful 0 1 0 0 1 0 0 0
users 0 1 0 0 1 0 1 1
variety 0 4 0 0 4 0 3 3
wait 0 2 0 0 1 0 1 1
waited 0 1 0 0 1 1 1 1
walk 0 2 0 0 1 0 1 1
want 0 1 0 0 0 0 0 0
wanted 0 1 0 0 1 0 1 1
wasnt 0 1 0 0 1 1 1 1
whatever 0 1 0 0 1 0 1 1
wide 0 4 0 0 3 1 2 1
won’t 0 1 0 1 1 1 1 1
worry 0 1 0 1 1 1 1 1
avoid 0 0 0 1 0 0 0 0
away 0 0 0 1 0 0 0 0
better 0 0 0 1 0 0 0 0
choice 0 0 0 1 0 0 0 0
customers 0 0 0 1 0 0 0 0
deceptive 0 0 0 1 0 0 0 0
expired 0 0 0 1 0 0 0 0
listed 0 0 0 1 0 0 0 0
make 0 0 0 1 0 0 0 0
marketing 0 0 0 1 0 0 0 0
minutes 0 0 0 1 0 0 0 0
much 0 0 0 1 0 0 0 0
purchases 0 0 0 1 0 0 0 0
qr 0 0 0 1 0 0 0 0
resulting 0 0 0 1 0 0 0 0
scanner 0 0 0 1 0 0 0 0
screen 0 0 0 1 0 0 0 0
showing 0 0 0 1 0 0 0 0
still 0 0 0 1 0 0 0 0
takes 0 0 0 1 0 0 0 0
tries 0 0 0 1 0 0 0 0
trusted 0 0 0 1 0 0 0 0
trusting 0 0 0 1 0 0 0 0
works 0 0 0 1 0 0 0 0
Hence, not too sure what is the issue about since there is no NA!
Hope you can help. Thank you!
I use library(ergm) and library(igraph) and generate a ERGM network. But I want the adjacency matrix of that network. I am unable to find any function which can produce that.
library(ergm)
library(igraph)
g.use <- network(16,density=0.1,directed=FALSE)
#
# Starting from this network let's draw 3 realizations
# of a edges and 2-star network
#
g.sim <- simulate(~edges+kstar(2), nsim=3, coef=c(-1.8,0.03),
basis=g.use, control=control.simulate(
MCMC.burnin=1000,
MCMC.interval=100))
#g.sim[[3]]
summary(g.sim)
Is it possible to find the adjacency matrix from g.sim? and how?
EGRM package uses the network package and not the igraph package. You should maintain everythig in network and not load igraph as the two have some conflicting functions with same names.
In your case, you simulate 3 graphs thus you should have 3 adjacency matrices. The code is as below:
library(ergm)
g.use <- network(16,density=0.1,directed=FALSE)
g.sim <- simulate(~edges+kstar(2), nsim=3, coef=c(-1.8,0.03),
basis=g.use, control=control.simulate(
MCMC.burnin=1000,
MCMC.interval=100))
The code you want:
lapply(g.sim, as.matrix)
[[1]]
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16
1 0 0 0 0 1 0 0 0 0 0 0 0 0 0 0 0
2 0 0 0 0 0 1 0 1 0 0 0 0 0 1 0 0
3 0 0 0 1 1 0 1 0 0 0 0 0 1 0 0 1
4 0 0 1 0 0 0 0 0 1 0 0 0 0 0 0 0
5 1 0 1 0 0 0 0 0 0 0 0 0 1 1 0 0
6 0 1 0 0 0 0 0 0 0 1 0 0 0 0 0 0
7 0 0 1 0 0 0 0 0 0 0 1 0 0 0 0 1
8 0 1 0 0 0 0 0 0 0 1 1 1 1 0 1 0
9 0 0 0 1 0 0 0 0 0 0 0 0 0 0 0 1
10 0 0 0 0 0 1 0 1 0 0 0 0 1 0 0 0
11 0 0 0 0 0 0 1 1 0 0 0 0 1 0 0 0
12 0 0 0 0 0 0 0 1 0 0 0 0 0 0 0 0
13 0 0 1 0 1 0 0 1 0 1 1 0 0 0 0 1
14 0 1 0 0 1 0 0 0 0 0 0 0 0 0 0 0
15 0 0 0 0 0 0 0 1 0 0 0 0 0 0 0 0
16 0 0 1 0 0 0 1 0 1 0 0 0 1 0 0 0
[[2]]
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16
1 0 0 0 0 0 0 0 0 0 0 0 1 1 0 0 0
2 0 0 0 0 0 0 0 0 0 0 1 0 0 0 0 0
3 0 0 0 1 0 0 0 0 0 0 1 0 0 0 1 0
4 0 0 1 0 0 0 0 0 0 0 0 0 0 0 0 0
5 0 0 0 0 0 0 1 0 0 0 0 0 0 0 0 1
6 0 0 0 0 0 0 0 1 1 0 1 1 1 0 0 1
7 0 0 0 0 1 0 0 0 0 0 1 0 0 0 0 0
8 0 0 0 0 0 1 0 0 0 0 0 0 0 0 1 0
9 0 0 0 0 0 1 0 0 0 0 0 1 0 0 0 0
10 0 0 0 0 0 0 0 0 0 0 0 0 1 1 0 0
11 0 1 1 0 0 1 1 0 0 0 0 0 0 0 0 0
12 1 0 0 0 0 1 0 0 1 0 0 0 0 0 0 1
13 1 0 0 0 0 1 0 0 0 1 0 0 0 0 0 0
14 0 0 0 0 0 0 0 0 0 1 0 0 0 0 0 0
15 0 0 1 0 0 0 0 1 0 0 0 0 0 0 0 0
16 0 0 0 0 1 1 0 0 0 0 0 1 0 0 0 0
[[3]]
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16
1 0 0 0 0 0 0 0 0 0 0 0 0 1 0 0 1
2 0 0 0 0 0 1 0 0 0 0 0 1 1 0 0 0
3 0 0 0 0 1 0 0 0 0 1 1 0 0 0 1 0
4 0 0 0 0 0 0 0 0 0 0 0 0 1 0 0 0
5 0 0 1 0 0 0 0 0 0 0 0 0 0 0 0 0
6 0 1 0 0 0 0 1 0 1 0 0 0 1 0 1 0
7 0 0 0 0 0 1 0 0 0 0 0 0 0 1 1 0
8 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0
9 0 0 0 0 0 1 0 0 0 0 0 0 0 0 0 0
10 0 0 1 0 0 0 0 0 0 0 0 0 0 0 1 1
11 0 0 1 0 0 0 0 0 0 0 0 0 0 0 0 1
12 0 1 0 0 0 0 0 0 0 0 0 0 1 1 1 0
13 1 1 0 1 0 1 0 0 0 0 0 1 0 0 0 0
14 0 0 0 0 0 0 1 0 0 0 0 1 0 0 0 0
15 0 0 1 0 0 1 1 0 0 1 0 1 0 0 0 1
16 1 0 0 0 0 0 0 0 0 1 1 0 0 0 1 0
Suppose i have this matrix
0 1 2 3 4 5 6 98 183 385 419 420 422 423 469 470 35698 35709 35729 37415
0 0 1 1 1 0 0 1 0 0 1 0 0 0 0 1 0 0 0 0 1
1 1 0 1 0 0 1 1 0 0 1 0 0 0 0 1 0 0 0 0 0
2 1 1 0 1 1 0 0 0 0 0 1 0 1 0 1 0 0 0 0 0
3 1 0 1 0 1 1 0 1 1 0 1 1 1 1 0 0 1 0 0 1
4 0 0 1 1 0 1 1 1 0 0 1 1 1 0 0 1 0 1 1 0
5 0 1 0 1 1 0 1 1 0 0 0 1 0 0 0 1 0 0 1 0
6 1 1 0 0 1 1 0 1 0 1 0 0 0 0 0 0 0 0 0 0
98 0 0 0 1 1 1 1 0 0 0 0 1 0 0 0 1 0 0 1 0
183 0 0 0 1 0 0 0 0 0 0 0 0 1 1 0 0 0 0 0 1
385 1 1 0 0 0 0 1 0 0 0 0 0 0 0 0 0 0 0 0 0
419 0 0 1 1 1 0 0 0 0 0 0 1 1 0 0 0 1 1 0 0
420 0 0 0 1 1 1 0 1 0 0 1 0 0 0 0 0 1 1 0 0
422 0 0 1 1 1 0 0 0 1 0 1 0 0 1 1 0 0 0 0 1
423 0 0 0 1 0 0 0 0 1 0 0 0 1 0 0 0 0 0 0 1
469 1 1 1 0 0 0 0 0 0 0 0 0 1 0 0 0 0 0 0 1
470 0 0 0 0 1 1 0 1 0 0 0 0 0 0 0 0 0 0 1 0
35698 0 0 0 1 0 0 0 0 0 0 1 1 0 0 0 0 0 0 0 0
35709 0 0 0 0 1 0 0 0 0 0 1 1 0 0 0 0 0 0 0 0
35729 0 0 0 0 1 1 0 1 0 0 0 0 0 0 0 1 0 0 0 0
37415 1 0 0 1 0 0 0 0 1 0 0 0 1 1 1 0 0 0 0 0
I am getting a value from another program let us say
x=3.
I want to choose the name of rows where x == 1 i.e. where the value of 3 is 1.
Output will be : 0,2,4,5,98,183,419,420,422,423,35698,37415.
And I don't want to pass "3" directly into the command. I want to pass the variable x so that if this number varies I could get the output accordingly.
Can anyone help me, please? thanks in advance
x=matrix(c(1,1,2,5,6,6,5,7,7,8,3,3,1,9,20,20,4,7,9,5),4,5,dimnames = list(c(letters[1:4]),c(LETTERS[1:5])))
you'r requirement is row names then
rownames(x)[x[,"D"]==20]
here '20' is you'r input value and D is you'r searching column.
I want to create a histogram from my data set of the frequency of students who have had broken bones. The values are either 0 or 1.
I.E:
[1] 0 0 0 0 0 1 0 1 0 0 0 0 0 0 0 1 0 0 0 1 0 0 0 1 0 0 0 0 1 0 0 0 0 0 0 0 0
[38] 1 1 0 0 0 0 1 0 1 0 0 0 1 0 0 0 0 0 0 0 0 0 0 0 1 0 0 0 0 0 0 0 0 0 0 0 0
[75] 0 0 1 0 0 0 0 1 1 0 0 0 0 0 0 0 0 0 0 0 0 0 1 1 0 0 0 1 0 1 0 0 0 0 0 0 1
[112] 1 1 0 0 1 1 1 0 1 1 1 0 0 0 0 0 0 0 0 0 0 0 0 0 1 0 0 0 0 1 1 0 0 1 0 0 0
[149] 0 1 0 0 0 0 0 0 1 0 0 0 0 0 0 1 0 0 0 1 0 0 0 0 0 1 0 1 1 0 0 0 0 1 0 0 0
[186] 0 1 0 1 0 0 0 0 0 0 0 0 0 0 0 0 1 0 1 0 0 0 0 0 0 0 1 1 0 1 0 1 0 0 0 0 0
[223] 0 0 0 0 0 1 0 1 0 0 0 0 0 0 1 0 0 0 0 1 1 0 1 0 1 0 1 1 0 0 0 0 0 0 0 0 1
[260] 1 0 0 0 0 0 0 1 1 1 1 0 0 0 1 0 0 1 0 0 0 1 0 0 0 1 0 0 0 0 0 1 0 1 0 0 0
[297] 0 0 0 0 0 1 0 0 0 0 0 0 0 0 0 0 0 0 1 1 0 0 1 1 0 0 0 0 0 1 1 1 0 0 0 0 1
[334] 0 0 0 0 1 0 0 0 0 0 0 1 1 0 0 0 1 0 0 0 0 0 0 1 1 0 0 1 0 0 0 0 1 0 0 0 0
[371] 0 0 0 0 0 0 0 0 0 0 0 0
However the scale on the axis axis of the graph has increments of 0.2. I just want either 0 or 1 as the data is categorical. Would anyone please kindly tell me how to rectify this?
What you need is a combination of assigning the appropriate values to the breaks argument and the xaxp argument in ?hist. Consider:
# this just gives me your data:
my.data <- "
0 0 0 0 0 1 0 1 0 0 0 0 0 0 0 1 0 0 0 1 0 0 0 1 0 0 0 0 1 0 0 0 0 0 0 0 0
1 1 0 0 0 0 1 0 1 0 0 0 1 0 0 0 0 0 0 0 0 0 0 0 1 0 0 0 0 0 0 0 0 0 0 0 0
0 0 1 0 0 0 0 1 1 0 0 0 0 0 0 0 0 0 0 0 0 0 1 1 0 0 0 1 0 1 0 0 0 0 0 0 1
1 1 0 0 1 1 1 0 1 1 1 0 0 0 0 0 0 0 0 0 0 0 0 0 1 0 0 0 0 1 1 0 0 1 0 0 0
0 1 0 0 0 0 0 0 1 0 0 0 0 0 0 1 0 0 0 1 0 0 0 0 0 1 0 1 1 0 0 0 0 1 0 0 0
0 1 0 1 0 0 0 0 0 0 0 0 0 0 0 0 1 0 1 0 0 0 0 0 0 0 1 1 0 1 0 1 0 0 0 0 0
0 0 0 0 0 1 0 1 0 0 0 0 0 0 1 0 0 0 0 1 1 0 1 0 1 0 1 1 0 0 0 0 0 0 0 0 1
1 0 0 0 0 0 0 1 1 1 1 0 0 0 1 0 0 1 0 0 0 1 0 0 0 1 0 0 0 0 0 1 0 1 0 0 0
0 0 0 0 0 1 0 0 0 0 0 0 0 0 0 0 0 0 1 1 0 0 1 1 0 0 0 0 0 1 1 1 0 0 0 0 1
0 0 0 0 1 0 0 0 0 0 0 1 1 0 0 0 1 0 0 0 0 0 0 1 1 0 0 1 0 0 0 0 1 0 0 0 0
0 0 0 0 0 0 0 0 0 0 0 0"
my.data <- unlist(strsplit(my.data, " "))
my.data <- gsub("\\n", "", my.data)
my.data <- as.numeric(my.data)
hist(my.data, breaks=c(-.5, .5, 1.5), xaxp=c(0,1,1))
breaks is used to define exactly 2 bins, and xaxp is used to change the number and placement of the tick marks on the x axis (for more on how xaxp works, see this excellent answer: R, change the spacing of tick marks on the axis of a plot?) Here is the resulting figure:
On a different note, it is not clear how informative a histogram is for data like this (or perhaps even ever, see: assessing-approximate-distribution-of-data-based-on-a-histogram on stats.SE). You might just was well try:
> table(my.data)
my.data
0 1
296 86
If I have previously defined
n <- 200
maf <- .2
Is there any way to echo
snp <- rbinom(n,2,maf)
such that it displays as
snp <- rbinom(200,2,.2)
in the resulting knitr document.
You can use substitute like this :
```{r ,echo=FALSE}
substitute(snp <- rbinom(n,2,maf),list(n=n,maf=maf))
rbinom(n,2,maf)
```
## snp <- rbinom(200, 2, 0.2)
## [1] 0 0 1 1 1 1 0 0 0 0 0 0 0 1 1 0 0 0 0 1 0 0 0 0 0 2 0 1 1 2 1 0 0 0 0
## [36] 1 0 0 1 0 0 0 0 0 1 0 0 1 0 0 1 1 1 0 0 0 1 0 0 0 0 0 0 0 0 1 1 0 0 1
## [71] 0 1 0 0 0 0 0 1 0 1 0 1 1 0 0 1 0 1 0 0 1 0 1 1 1 2 0 1 0 1 1 0 1 0 0
## [106] 1 1 0 1 1 1 0 0 1 0 0 0 0 0 0 1 1 0 1 1 0 0 0 0 0 0 0 0 0 0 0 0 0 1 0
## [141] 1 0 0 1 0 0 1 0 2 0 0 0 1 0 0 0 1 1 0 0 0 0 0 2 1 1 2 0 1 0 0 0 0 0 1
## [176] 0 0 2 0 1 1 0 0 0 0 0 1 0 1 2 0 1 1 2 0 0 1 1 1 0