I am new to plotly and not very good with R. I am trying to do stack plots and ended up with a very cumbersome code, that I am sure could be simplify using RColorbrewer and perhaps ggplot2 to group my stacked bar plots, but I am unsure on how to do it.
Below is the data I used, which is in a data.frame called data2
Nation glider radar AUV ROV USV corer towed_eq Seismic_eq Drill_rig Manned_sub Other clean
1 Belgium 0 0 1 1 1 3 0 0 0 0 0 6
2 Bulgaria 0 0 0 0 0 0 1 0 0 1 0 2
3 Croatia 0 2 1 2 0 0 0 0 0 0 0 5
4 Cyprus 3 0 0 0 0 0 0 0 0 0 0 3
5 Estonia 0 0 0 1 0 0 0 0 0 0 0 1
6 Finland 1 0 0 0 0 0 0 0 0 0 0 1
7 France 11 2 3 1 0 1 1 3 0 1 0 23
8 Germany 18 3 3 4 0 0 1 4 2 1 0 36
9 Greece 1 0 0 3 0 0 0 0 0 0 0 4
10 Ireland 0 0 0 2 0 0 0 0 0 0 0 2
11 Italy 10 8 3 2 4 0 0 1 0 0 0 28
12 Malta 0 2 0 0 0 0 0 0 0 0 0 2
13 Netherlands 0 2 0 0 0 0 0 0 0 0 0 2
14 Norway 17 3 1 3 0 1 3 1 0 0 1 30
15 Poland 0 0 0 1 0 0 0 0 0 0 0 1
16 Portugal 0 3 6 6 4 2 1 0 0 2 1 25
17 Romania 0 0 0 1 0 0 0 0 0 0 0 1
18 Slovenia 0 1 0 0 0 0 0 0 0 0 0 1
19 Spain 12 17 2 1 0 0 0 2 0 0 0 34
20 Sweden 0 2 1 3 0 0 0 0 0 0 0 6
21 Turkey 0 0 0 0 0 0 0 0 0 2 0 2
22 United Kingdom 0 0 13 4 1 11 4 2 1 0 4 40
23 Unknown 5 0 0 0 0 0 0 0 0 0 0 5
And this is the code I used
fig <- plot_ly(data2, x = ~Nation, y = ~glider, type = 'bar', name = 'Glider')
fig <- fig %>% add_trace(y = ~radar, name = 'Radar', marker=list(color='rgb(26, 118, 255)'))
fig <- fig %>% add_trace(y = ~AUV, name = 'AUV',marker=list(color='rgb(255, 128, 0)'))
fig <- fig %>% add_trace(y = ~ROV, name = 'ROV',marker=list(color='rgb(204, 0, 0)'))
fig <- fig %>% add_trace(y = ~USV, name = 'USV',marker=list(color='rgb(51, 255, 153)'))
fig <- fig %>% add_trace(y = ~corer, name = 'Corer',marker=list(color='rgb(204, 0, 204)'))
fig <- fig %>% add_trace(y = ~towed_eq, name = 'Towed equipment',marker=list(color='rgb(255, 255, 51)'))
fig <- fig %>% add_trace(y = ~Seismic_eq, name = 'Seismic equipment',marker=list(color='rgb(255, 204, 229)'))
fig <- fig %>% add_trace(y = ~Drill_rig, name = 'Drill rig',marker=list(color='rgb(102, 255, 255)'))
fig <- fig %>% add_trace(y = ~Manned_sub, name = 'Manned submersible',marker=list(color='rgb(128, 255, 0)'))
fig <- fig %>% add_trace(y = ~Other, name = 'Other equipment',marker=list(color='rgb(153, 153, 0)'))
fig <- fig %>% layout(xaxis = list(title = "",tickfont = list(size = 14)), yaxis = list(title = 'Number of assets',tickfont = list(size = 14)), barmode = 'stack')
fig
Is there an easier way to code this by using Rcolorbrewer instead of coding each color? and is it possible to group my stacked barplots Group1 (glider, auv, rov, usv), Group 2 (corer,towed_ew, seismic_eq, drill_rig) and Group 3 (radar, manned_sub, Other)?stack_plot
You can try this approach by melting the data:
library(dplyr)
library(plotly)
library(tidyr)
library(RColorBrewer)
#Data
data <- structure(list(Nation = c("Belgium", "Bulgaria", "Croatia", "Cyprus",
"Estonia", "Finland", "France", "Germany", "Greece", "Ireland",
"Italy", "Malta", "Netherlands", "Norway", "Poland", "Portugal",
"Romania", "Slovenia", "Spain", "Sweden", "Turkey", "United Kingdom",
"Unknown"), glider = c(0, 0, 0, 3, 0, 1, 11, 18, 1, 0, 10, 0,
0, 17, 0, 0, 0, 0, 12, 0, 0, 0, 5), radar = c(0, 0, 2, 0, 0,
0, 2, 3, 0, 0, 8, 2, 2, 3, 0, 3, 0, 1, 17, 2, 0, 0, 0), AUV = c(1,
0, 1, 0, 0, 0, 3, 3, 0, 0, 3, 0, 0, 1, 0, 6, 0, 0, 2, 1, 0, 13,
0), ROV = c(1, 0, 2, 0, 1, 0, 1, 4, 3, 2, 2, 0, 0, 3, 1, 6, 1,
0, 1, 3, 0, 4, 0), USV = c(1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 4, 0,
0, 0, 0, 4, 0, 0, 0, 0, 0, 1, 0), corer = c(3, 0, 0, 0, 0, 0,
1, 0, 0, 0, 0, 0, 0, 1, 0, 2, 0, 0, 0, 0, 0, 11, 0), towed_eq = c(0,
1, 0, 0, 0, 0, 1, 1, 0, 0, 0, 0, 0, 3, 0, 1, 0, 0, 0, 0, 0, 4,
0), Seismic_eq = c(0, 0, 0, 0, 0, 0, 3, 4, 0, 0, 1, 0, 0, 1,
0, 0, 0, 0, 2, 0, 0, 2, 0), Drill_rig = c(0, 0, 0, 0, 0, 0, 0,
2, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 1, 0), Manned_sub = c(0,
1, 0, 0, 0, 0, 1, 1, 0, 0, 0, 0, 0, 0, 0, 2, 0, 0, 0, 0, 2, 0,
0), Other = c(0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 1, 0, 1,
0, 0, 0, 0, 0, 4, 0), clean = c(6, 2, 5, 3, 1, 1, 23, 36, 4,
2, 28, 2, 2, 30, 1, 25, 1, 1, 34, 6, 2, 40, 5)), row.names = c(NA,
-23L), class = "data.frame")
Now the code:
#First reshape
df2 <- pivot_longer(data,cols = -Nation)
#Plot
p <- plot_ly(df2, x = df2$Nation,
y = df2$value,
type = 'bar',
name = df2$name,
text = df2$value,
color = df2$name,
colors = brewer.pal(length(unique(df2$name)),
"Paired"))%>%
layout(barmode = 'stack',hoverlabel = list(bgcolor= 'white') ,bargap = 0.5) %>%
layout(xaxis = list(categoryorder = 'array',
categoryarray = df2$Nation), showlegend = T)
The output:
Related
I have survey data that has a binary 1, 0 (indicating peak or off-peak) variable with the related peak or off-peak numbers in two separate columns.
structure(list(q9_jul_2019 = c(1, 0, 1, 0, 1, 0), q9_aug_2019 = c(1,
0, 1, 0, 1, 0), q9_sep_2019 = c(1, 0, 1, 0, 1, 0), q9_oct_2019 = c(0,
0, 1, 0, 1, 0), q9_nov_2019 = c(0, 0, 1, 0, 1, 0), q9_dec_2019 = c(0,
0, 1, 0, 0, 0), q9_jan_2020 = c(0, 0, 1, 0, 0, 0), q9_feb_2020 = c(0,
1, 0, 1, 0, 0), q9_mar_2020 = c(1, 1, 0, 1, 0, 0), q9_apr_2020 = c(1,
1, 1, 1, 0, 1), q9_may_2020 = c(0, 1, 0, 0, 0, 0), q9_jun_2020 = c(0,
0, 0, 0, 0, 0), q15 = c(1, 10, 30, 0, 2, 0), q22 = c(0, 10, 6,
0, 0, 0)), row.names = c(NA, 6L), class = "data.frame")
I have created new monthly columns that have the associated visitation numbers in that column but I'm sure there must be a neater way to do it using across(). I haven't been able to make it work though, so at the moment I'm stuck at the following:
survey <- survey %>%
mutate(visitation_jul_19 = if_else(q9_jul_2019 == 1, q15, q22),
visitation_aug_19 = if_else(q9_aug_2019 == 1, q15, q22),
visitation_sep_19 = if_else(q9_sep_2019 == 1, q15, q22),
visitation_oct_19 = if_else(q9_oct_2019 == 1, q15, q22),
visitation_nov_19 = if_else(q9_nov_2019 == 1, q15, q22),
visitation_dec_19 = if_else(q9_dec_2019 == 1, q15, q22),
visitation_jan_20 = if_else(q9_jan_2020 == 1, q15, q22),
visitation_feb_20 = if_else(q9_feb_2020 == 1, q15, q22),
visitation_mar_20 = if_else(q9_mar_2020 == 1, q15, q22),
visitation_apr_20 = if_else(q9_apr_2020 == 1, q15, q22),
visitation_may_20 = if_else(q9_may_2020 == 1, q15, q22),
visitation_jun_20 = if_else(q9_jun_2020 == 1, q15, q22))
You may try
library(dplyr)
survey %>%
mutate(across(q9_jul_2019:q9_jun_2020, ~ ifelse(.x == 1, q15, q22)))
q9_jul_2019 q9_aug_2019 q9_sep_2019 q9_oct_2019 q9_nov_2019 q9_dec_2019 q9_jan_2020 q9_feb_2020 q9_mar_2020 q9_apr_2020
1 1 1 1 0 0 0 0 0 1 1
2 10 10 10 10 10 10 10 10 10 10
3 30 30 30 30 30 30 30 6 6 30
4 0 0 0 0 0 0 0 0 0 0
5 2 2 2 2 2 0 0 0 0 0
6 0 0 0 0 0 0 0 0 0 0
q9_may_2020 q9_jun_2020 q15 q22
1 0 0 1 0
2 10 10 10 10
3 6 6 30 6
4 0 0 0 0
5 0 0 2 0
6 0 0 0 0
require(gtsummary)
test <- structure(list(`1` = c(0, 0, 0, 0, 0, 0, 0, 0, 1, 0), `2` = c(1,0, 0, 0, 0, 1, 0, 1, 0, 0), `3` = c(0, 0, 0, 0, 0, 0, 0, 0, 0,0), `4` = c(1, 1, 0, 0, 1, 0, 0, 0, 0, 0), `5` = c(1, 0, 1, 1,0, 1, 1, 0, 0, 0), `6` = c(0, 0, 0, 1, 0, 0, 1, 0, 0, 0), `7` = c(0,0, 0, 0, 0, 0, 0, 0, 0, 0), `8` = c(0, 0, 0, 0, 0, 0, 0, 0, 0,0), `9` = c(0, 0, 0, 0, 0, 0, 0, 0, 0, 0), `10` = c(0, 0, 0,0, 0, 0, 0, 0, 0, 1)), row.names = c(NA, -10L), class = c("tbl_df","tbl", "data.frame"))
In this example data, I have 10 categorical variables.
`1` `2` `3` `4` `5` `6` `7` `8` `9` `10`
<dbl> <dbl> <dbl> <dbl> <dbl> <dbl> <dbl> <dbl> <dbl> <dbl>
1 0 1 0 1 1 0 0 0 0 0
2 0 0 0 1 0 0 0 0 0 0
3 0 0 0 0 1 0 0 0 0 0
4 0 0 0 0 1 1 0 0 0 0
5 0 0 0 1 0 0 0 0 0 0
6 0 1 0 0 1 0 0 0 0 0
7 0 0 0 0 1 1 0 0 0 0
8 0 1 0 0 0 0 0 0 0 0
9 1 0 0 0 0 0 0 0 0 0
10 0 0 0 0 0 0 0 0 0 1
Since they can overlap each other, I have put them in different columns,
using 0 and 1, indicatting "yes" or "no" to having (or not having) the categorical variable.
When test %>% tbl_summary(), it creates:
I would like to sort this by frequency, but
test %>% tbl_summary(sort = list(everything() ~ "frequency"))
does not work.
Is there anyway to do this?
Thank you in advance.
The tbl_summary(sort=) argument sorts levels within a variable, not the order the variables appear in the table. Variables are appear in the table in the same order they appear in the data frame.
We can update the order in the data frame using the code below.
library(gtsummary)
#> #Uighur
packageVersion("gtsummary")
#> [1] '1.5.0'
test <- structure(list(`1` = c(0, 0, 0, 0, 0, 0, 0, 0, 1, 0), `2` = c(1,0, 0, 0, 0, 1, 0, 1, 0, 0), `3` = c(0, 0, 0, 0, 0, 0, 0, 0, 0,0), `4` = c(1, 1, 0, 0, 1, 0, 0, 0, 0, 0), `5` = c(1, 0, 1, 1,0, 1, 1, 0, 0, 0), `6` = c(0, 0, 0, 1, 0, 0, 1, 0, 0, 0), `7` = c(0,0, 0, 0, 0, 0, 0, 0, 0, 0), `8` = c(0, 0, 0, 0, 0, 0, 0, 0, 0,0), `9` = c(0, 0, 0, 0, 0, 0, 0, 0, 0, 0), `10` = c(0, 0, 0,0, 0, 0, 0, 0, 0, 1)), row.names = c(NA, -10L), class = c("tbl_df","tbl", "data.frame"))
# order variables by prevelence
prev <- purrr::map_dbl(test, mean) %>% sort(decreasing = TRUE)
test %>%
select(all_of(names(prev))) %>%
tbl_summary() %>%
as_kable() # convert to kable for SO
Characteristic
N = 10
5
5 (50%)
2
3 (30%)
4
3 (30%)
6
2 (20%)
1
1 (10%)
10
1 (10%)
3
0 (0%)
7
0 (0%)
8
0 (0%)
9
0 (0%)
Created on 2021-12-10 by the reprex package (v2.0.1)
I have a paneldata dataframe structure, something like this:
df <- data.frame("id" = c(1, 1, 1, 1, 2, 2, 2, 2, 3, 3, 3, 3),
"Status_2014" = c(1, 1, 1, 1, 0, 0, 0, 0, 0, 0, 0, 0),
"Status_2015" = c(0, 0, 0, 0, 1, 1, 1, 1, 0, 0, 0, 0),
"Status_2016" = c(0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0))
I want to generate a new dummy variable, that takes the value 1, if the rows contains 1 in any of the three columns or otherwise 0 if not. It should end up like this:
df <- data.frame("id" = c(1, 1, 1, 1, 2, 2, 2, 2, 3, 3, 3, 3),
"Status_2014" = c(1, 1, 1, 1, 0, 0, 0, 0, 0, 0, 0, 0),
"Status_2015" = c(0, 0, 0, 0, 1, 1, 1, 1, 0, 0, 0, 0),
"Status_2016" = c(0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0),
"Final_status" = c(1, 1, 1, 1, 1, 1, 1, 1, 0, 0, 0, 0))
Can anyone help me achieve this?
We can use if_any on the columns that starts_with 'Status', to check for any 1 value in a row and it returns TRUE if there is one or else FALSE which is coerced to binary with as.integer/+
library(dplyr)
df %>%
mutate(Final_status = +(if_any(starts_with('Status'), ~ . ==1)))
-outptu
id Status_2014 Status_2015 Status_2016 Final_status
1 1 1 0 0 1
2 1 1 0 0 1
3 1 1 0 0 1
4 1 1 0 0 1
5 2 0 1 0 1
6 2 0 1 0 1
7 2 0 1 0 1
8 2 0 1 0 1
9 3 0 0 0 0
10 3 0 0 0 0
11 3 0 0 0 0
12 3 0 0 0 0
Or using rowSums from base R
df$Final_status <- +(rowSums(df[-1] > 0) > 0)
You write an if condition to define the variable as 1 or 0, and inside this condition the most straight forward ways would be a dplyr pipe.
I don't have the dplyr syntax in my head, to long not used, but dplyr is what you want.
https://www.rstudio.com/wp-content/uploads/2015/02/data-wrangling-cheatsheet.pdf
best greetings
I'm trying to fit neutral model parameter using MLE in R.
I cannot get this error, and how can I solve it?
I've got the following error:
Error in optim(start, f, method = method, hessian = TRUE, ...) :
non-finite finite-difference value [1]
These are the codes that I used.
#Load packages
library(minpack.lm)
library(Hmisc)
library(stats4)
#Calculate the number of individuals per community
N <- mean(apply(otu_f_h, 1, sum))
#Calculate the average relative abundance of each taxa across communities
p.m <- apply(otu_f_h, 2, mean)
p.m <- p.m[p.m != 0]
p <- p.m/N
#Calculate the occurrence frequency of each taxa across communities
spp.bi <- 1*(otu_f_h>0)
freq <- apply(spp.bi, 2, mean)
freq <- freq[freq != 0]
#Combine
C <- merge(p, freq, by=0)
C <- C[order(C[,2]),]
C <- as.data.frame(C)
C.0 <- C[!(apply(C, 1, function(y) any(y == 0))),] #Removes rows with any zero (absent in either source pool or local communities)
p <- C.0[,2]
freq <- C.0[,3]
names(p) <- C.0[,1]
names(freq) <- C.0[,1]
#Calculate the limit of detection
d = 1/N
##Fit model parameter m (or Nm) using Non-linear least squares (NLS)
m.fit <- nlsLM(freq ~ pbeta(d, N*m*p, N*m*(1-p), lower.tail=FALSE), start=list(m=0.1))
m.ci <- confint(m.fit, 'm', level=0.95)
##Fit neutral model parameter m (or Nm) using Maximum likelihood estimation (MLE)
sncm.LL <- function(m, sigma){
R = freq - pbeta(d, N*m*p, N*m*(1-p), lower.tail=FALSE)
R = dnorm(R, 0, sigma)
-sum(log(R))
}
m.mle <- mle(sncm.LL, start=list(m=0.1, sigma=0.1), nobs=length(p))
This is the sample table that I used (otu_f_h)
SampleID
OTU01
OTU02
OTU03
OTU04
OTU05
OTU06
OTU07
OTU08
R01
8
0
8
0
0
0
0
257
R02
0
0
0
0
0
0
0
0
R05
0
0
0
0
51
0
0
0
R06
0
0
0
0
9
0
0
0
R09
0
0
0
0
0
0
0
0
R10
0
0
0
0
0
0
6
0
R39
0
0
0
0
0
0
0
28
R40
0
0
0
0
0
0
0
0
R51
0
0
0
0
0
0
0
0
S01
24
0
4
5
0
0
6
0
S02
0
0
37
0
0
0
0
0
S05
0
0
15
28
0
0
0
0
S06
0
2
0
0
7
0
0
0
S09
0
0
0
0
24
0
0
0
S10
0
0
0
0
0
0
0
0
S39
0
0
0
0
0
0
0
0
S40
0
0
0
0
0
0
0
0
S51
0
0
0
0
0
0
0
0
S52
0
0
46
79
0
0
35
0
structure(list(SampleID = c("R01", "R02", "R05", "R06", "R09",
"R10", "R39", "R40", "R51", "S01", "S02", "S05", "S06", "S09",
"S10", "S39", "S40", "S51", "S52"), OTU01 = c(8, 0, 0, 0, 0,
0, 0, 0, 0, 24, 0, 0, 0, 0, 0, 0, 0, 0, 0), OTU02 = c(0, 0, 0,
0, 0, 0, 0, 0, 0, 0, 0, 0, 2, 0, 0, 0, 0, 0, 0), OTU03 = c(8,
0, 0, 0, 0, 0, 0, 0, 0, 4, 37, 15, 0, 0, 0, 0, 0, 0, 46), OTU04 = c(0,
0, 0, 0, 0, 0, 0, 0, 0, 5, 0, 28, 0, 0, 0, 0, 0, 0, 79), OTU05 = c(0,
0, 51, 9, 0, 0, 0, 0, 0, 0, 0, 0, 7, 24, 0, 0, 0, 0, 0), OTU06 = c(0,
0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0), OTU07 = c(0,
0, 0, 0, 0, 6, 0, 0, 0, 6, 0, 0, 0, 0, 0, 0, 0, 0, 35), OTU08 = c(257,
0, 0, 0, 0, 0, 28, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0)), row.names = 2:20, class = "data.frame")
I am having some issues in realizing a bipartite network in R with the library igraph. Here is my script:
library(igraph)
library(reshape2)
setwd("....")
getwd()
library(readxl)
network=read_excel("network1.xlsx")
print(network)
subjects=as.character(unlist(network[,1]))
agents=colnames(network[-1])
print(network)
network = network[,-1]
g=graph.incidence(network, weighted = T)
V(g)$type
V(g)$name=c(subjects,agents)
V(g)$color = V(g)$type
V(g)$color=gsub("FALSE","red",V(g)$color)
V(g)$color=gsub("TRUE","lightblue",V(g)$color)
plot(g, edge.arrow.width = 0.3,
vertex.size = 5,
edge.arrow.size = 0.5,
vertex.size2 = 5,
vertex.label.cex = 1,
vertex.label.color="black",
asp = 0.35,
margin = 0,
edge.color="grey",
edge.width=(E(g)$weight),
layout=layout_as_bipartite)
The network is properly plotted
as you can see
however I have two issues
(1) I don't understand the order in which the vertexs are showed in the plot. They are not in the same order of the excel file, neither in alphabetical or numerical order. They seem to be in random order. How could I choose the order in which the vertex should be placed?
(2) I don't understand why some vertex are closer toghether, and some are more far apart. I would all vertexes at the same distance. How could I do it?
Thank you a lot for your invaluable help.
Since you do not provide your data, I will illustrate with a made-up example.
Sample graph data
library(igraph)
set.seed(123)
EL = matrix(c(sample(8,18, replace=T),
sample(LETTERS[1:6], 18, replace=T)), ncol=2)
g = simplify(graph_from_edgelist(EL))
V(g)$type = bipartite_mapping(g)$type
VCol = c("#FF000066", "#0000FF66")[as.numeric(V(g)$type)+1]
plot(g, layout=layout_as_bipartite(g), vertex.color=VCol)
As with your graph, this has two problems. The nodes are ordered arbitrarily
and the lower row is oddly spaced. Let's address those problems one at a time.
To do so, we will need to take control of the layout instead of using any of
the automated layout functions. A layout is simply a vcount(g) * 2 matrix
giving the x-y coordinates of the vertices for plotting. Here, I will put one
type of nodes in the top row by specifying the y coordinate as 1 and the other
nodes in a lower row by specifying y=0. We want to specify the order horizontally
by rank (alphabetically) within each group. So
LO = matrix(0, nrow=vcount(g), ncol=2)
LO[!V(g)$type, 2] = 1
LO[V(g)$type, 1] = rank(V(g)$name[V(g)$type])
LO[!V(g)$type, 1] = rank(V(g)$name[!V(g)$type])
plot(g, layout=LO, vertex.color=VCol)
Now both rows are ordered and evenly spaced, but because there are fewer
vertices in the bottom row, there is an unattractive, unbalanced look. We
can fix that by stretching the bottom row. I find it easier to make the right
scale factor if the coordinates go from 0 to (number of nodes) - 1 rather than
1 to (number of nodes) as above. Doing this, we get
LO[V(g)$type, 1] = rank(V(g)$name[V(g)$type]) - 1
LO[!V(g)$type, 1] = (rank(V(g)$name[!V(g)$type]) - 1) *
(sum(V(g)$type) - 1) / (sum(!V(g)$type) - 1)
plot(g, layout=LO, vertex.color=VCol)
thank you a lot. I performed your very very helpful example, and with the step one I did it work properly with my data, keeping the different thickness of the edges and all as in my plot, but with the proper order. This is very important, thank you a lot. However, I have some troubles in understanding how to rescale properly the top and the bottom row with my data, because they always seem to bee too near. probably I did not understand completly the coordinates on which I have to work. Here are my data.
> `> network=read_excel("network1.xlsx",2)
> dput(network)
structure(list(`NA` = c(2333, 2439, 2450, 2451, 2452, 2453, 2454,
2455, 2456, 2457, 2458, 2459, 2460, 2461, 2480, 2490, 2491, 2492,
2493, 2494, 2495), A = c(12, 2, 2, 5, 2, 0, 5, 3, 0, 0, 7, 0,
0, 0, 6, 2, 10, 7, 1, 2, 5), B = c(0, 1, 0, 1, 0, 0, 2, 0, 0,
0, 0, 0, 1, 0, 5, 0, 2, 0, 0, 0, 0), C = c(0, 0, 0, 0, 1, 0,
4, 0, 0, 0, 0, 1, 0, 0, 2, 0, 4, 4, 2, 1, 0), D = c(2, 0, 0,
0, 0, 0, 2, 0, 0, 0, 0, 0, 0, 0, 7, 0, 4, 0, 1, 4, 0), E = c(11,
2, 3, 3, 3, 8, 3, 6, 4, 1, 1, 0, 12, 0, 5, 0, 4, 6, 4, 8, 9),
F = c(2, 0, 0, 3, 1, 0, 10, 1, 0, 0, 0, 1, 0, 0, 9, 0, 0,
1, 1, 3, 3), G = c(0, 3, 1, 1, 0, 0, 0, 0, 0, 3, 2, 0, 0,
0, 1, 0, 0, 2, 0, 1, 0), H = c(0, 0, 2, 0, 0, 0, 1, 0, 0,
0, 0, 0, 0, 0, 2, 0, 0, 0, 0, 0, 1), I = c(0, 0, 0, 0, 0,
0, 3, 0, 6, 3, 0, 0, 1, 0, 7, 0, 0, 4, 1, 2, 0), J = c(0,
0, 0, 0, 0, 1, 0, 0, 0, 0, 0, 0, 0, 0, 1, 0, 0, 0, 0, 0,
0)), class = c("tbl_df", "tbl", "data.frame"), row.names = c(NA,
-21L), .Names = c(NA, "A", "B", "C", "D", "E", "F", "G", "H",
"I", "J"))
> print(network)
NA A B C D E F G H I J
1 2333 12 0 0 2 11 2 0 0 0 0
2 2439 2 1 0 0 2 0 3 0 0 0
3 2450 2 0 0 0 3 0 1 2 0 0
4 2451 5 1 0 0 3 3 1 0 0 0
5 2452 2 0 1 0 3 1 0 0 0 0
6 2453 0 0 0 0 8 0 0 0 0 1
7 2454 5 2 4 2 3 10 0 1 3 0
8 2455 3 0 0 0 6 1 0 0 0 0
9 2456 0 0 0 0 4 0 0 0 6 0
10 2457 0 0 0 0 1 0 3 0 3 0
11 2458 7 0 0 0 1 0 2 0 0 0
12 2459 0 0 1 0 0 1 0 0 0 0
13 2460 0 1 0 0 12 0 0 0 1 0
14 2461 0 0 0 0 0 0 0 0 0 0
15 2480 6 5 2 7 5 9 1 2 7 1
16 2490 2 0 0 0 0 0 0 0 0 0
17 2491 10 2 4 4 4 0 0 0 0 0
18 2492 7 0 4 0 6 1 2 0 4 0
19 2493 1 0 2 1 4 1 0 0 1 0
20 2494 2 0 1 4 8 3 1 0 2 0
21 2495 5 0 0 0 9 3 0 1 0 0
> `