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.
Related
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
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 ]
I am new to raster plots, and I am not sure which one is the fastest and more appropriate way to create a raster plot with my data.
I have a dataframe with 64 rows (location) and 202 columns (time). The values of the dataframe can be 0, 1 or 2. I would like to create a raster plot (with time as x axis, and location as y axes) in which I can visualise the values with 0 as white rectangles, the values with 1 as black rectangles and the values with 2 as grey rectangles.
X 0 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37 38 39 40 41 42 43 44 45 46 47 48 49 50 51 52 53 54 55 56 57 58 59 60 61 62 63 64 65 66 67 68 69 70 71 72 73 74 75 76 77 78 79 80 81 82 83 84 85 86 87 88 89 90 91 92 93 94 95 96 97 98 99 100 101 102 103 104 105 106 107 108 109 110 111 112 113 114 115 116 117 118 119 120 121 122 123 124 125 126 127 128 129 130 131 132 133 134 135 136 137 138 139 140 141 142 143 144 145 146 147 148 149 150 151 152 153 154 155 156 157 158 159 160 161 162 163 164 165 166 167 168 169 170 171 172 173 174 175 176 177 178 179 180 181 182 183 184 185 186 187 188 189 190 191 192 193 194 195 196 197 198 199 200
fp1 0 0 0 2 2 2 2 2 2 2 2 2 2 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0
fp2 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0
f3 0 0 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 0 0 0 0 0 0 0 0 0 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0
f4 0 0 0 0 0 0 0 1 1 1 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0
c3 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0
c4 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0
p3 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0
p4 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1
o1 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0
02 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 2 2 2 2 2 2 2 2 2 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0
library(raster)
r <- raster(as.matrix(mtcars))
r[0:95] <- 0
r[96:180] <- 2
r[181:352] <- 1
breakpoints <- c(0, 1, 2)
colors <- c("white","black","grey")
plot(r, breaks=breakpoints, col=colors, axes = FALSE, legend = FALSE)
Using R I have a table, lets say 'locations'
head(locations, n=10)
apillar fender fwheel fdoor compart rdoor rwheel boot
1 0 0 0 0 0 0 0 1
2 0 0 0 1 0 0 0 0
3 0 0 0 0 1 0 0 0
4 0 1 0 0 0 0 0 0
5 1 0 1 0 0 0 0 0
6 1 0 0 1 0 0 0 0
7 0 0 0 0 0 0 0 0
8 0 0 0 0 1 0 0 0
9 0 0 0 1 0 0 0 0
10 0 0 0 0 0 1 0 0
now i want to create a new variable "cat" which groups the impacts into category locations.
I have been using if, elseif and else command, but I cannot get it to work.
The command is:
cat <- lapply(locations, function(x) if (apillar|fender|fwheel == 1)print("front") else if (fdoor|compart|rdoor == 1)print("middle") else if(rwheel|boot ==1)print("rear") else print("NA")
such that cat should read rear, middle, middle, middle, front etc
When vectors of TRUE or FALSE statements are involved, I usually prefer not to work with if to avoid loops. I find conditional referencing to be more elegant in this case. See below.
locations <- read.table(header=TRUE, text=
"apillar fender fwheel fdoor compart rdoor rwheel boot
1 0 0 0 0 0 0 0 1
2 0 0 0 1 0 0 0 0
3 0 0 0 0 1 0 0 0
4 0 1 0 0 0 0 0 0
5 1 0 1 0 0 0 0 0
6 1 0 0 1 0 0 0 0
7 0 0 0 0 0 0 0 0
8 0 0 0 0 1 0 0 0
9 0 0 0 1 0 0 0 0
10 0 0 0 0 0 1 0 0")
locations$cat <- NA
within(locations,{
cat[apillar|fender|fwheel] <- "front"
cat[fdoor|compart|rdoor] <- "middle"
cat[rwheel|boot] <- "rear"
})
Result:
apillar fender fwheel fdoor compart rdoor rwheel boot cat
1 0 0 0 0 0 0 0 1 rear
2 0 0 0 1 0 0 0 0 middle
3 0 0 0 0 1 0 0 0 middle
4 0 1 0 0 0 0 0 0 front
5 1 0 1 0 0 0 0 0 front
6 1 0 0 1 0 0 0 0 middle
7 0 0 0 0 0 0 0 0 <NA>
8 0 0 0 0 1 0 0 0 middle
9 0 0 0 1 0 0 0 0 middle
10 0 0 0 0 0 1 0 0 middle
Cheers!
Corrected your own code:
locations$cat= with(locations, ifelse(apillar|fender|fwheel, "front", ifelse(fdoor|compart|rdoor,"middle",ifelse(rwheel|boot, "rear", "NA"))) )
> locations
apillar fender fwheel fdoor compart rdoor rwheel boot cat
1 0 0 0 0 0 0 0 1 rear
2 0 0 0 1 0 0 0 0 middle
3 0 0 0 0 1 0 0 0 middle
4 0 1 0 0 0 0 0 0 front
5 1 0 1 0 0 0 0 0 front
6 1 0 0 1 0 0 0 0 front
7 0 0 0 0 0 0 0 0 NA
8 0 0 0 0 1 0 0 0 middle
9 0 0 0 1 0 0 0 0 middle
10 0 0 0 0 0 1 0 0 middle
>
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