从不同的数据创建图表 [英] Create a chart from different data
问题描述
我需要帮助来创建图表。我解释得更好。
我创建了10个随机图,每个图都有N个节点。
我已经完成了N = 10 ^ 3,10 ^ 4,10 ^ 5。
所以共有30张图。
给他们每个人,我找到了他们拥有的多重链接和selfloops的百分比。
现在我想创建一个显示节点数量百分比函数的图形。
所以像这样:
所以我有3个列表:
- listNets
包含30张图
- listSelf
包含selfloops的百分比
- listMul
包含多重链接的百分比
这就是我所做的:
listN <-c((10 ^ 3),( 10 ^ 4),(10 ^ 5))
#网络列表
listNets< - vector(mode =list,length = 0)
#list of自循环的百分比
listSelf< - vector(mode =list,length = 0)
多链接百分比的列表
listMul< - vector(mode =list,length = 0)
...
(listN中的N){
...
净值< ; - graph_from_adjacency_matrix(adjmatrix = adjacency_matrix,mode =undirected)#它的工作,事实上,如果我绘制它,我看到一个正确的网络
listNets< - c(listNets,net)#我加net到l ($ net
$ -loops e multilinks
netmatr< - as_adjacency_matrix(net,sparse = FALSE)
num_selfloops< - sum(diag(netmatr))
num_multilinks< - sum(netmatr> 1)
#我找到百分比
per_self< - ((num_selfloops / num_vertices)* 100)
per_mul< - ((num_multilinks / num_edges)* 100)
listSelf< - c(listSelf,per_self)
listMul< - c(listMul,per_mul)
}
现在,如果我以这种方式打印 listNets
,我有些奇怪:
> print(listNets)
[[1]]
[1] 9
[[2]]
[1] FALSE
[[3]]
[1] 7 6 3 8 8 8
[[4]]
[1] 0 1 2 4 5 7
[[5]]
[1] 2 1 0 3 4 5
[[6]]
[1] 0 1 2 3 4 5
[[7]]
[1] 0 0 0 0 1 1 1 2 3 6
[[8]]
[1] 0 1 2 3 3 4 5 5 6 6
[[9]]
[[9]] [[1]]
[1] 1 0 1
[[9]] [[2]]
名单列表()
[[9]] [[3]]
list()
$ [$ 9]] [[4]]
list()
[[10]]
< environment:0x000000001a6284a8>
$ b $ [[11]]
[1] 9
[[12]]
[1] FALSE
[[13]]
[1] 2 5 8 8 7 8
[[14]]
[1] 0 1 3 4 6 7
[[15]]
[1] 0 1 4 2 3 5
[[16]]
[1] 0 1 2 3 4 5
[[17]]
[1] 0 0 0 1 1 2 2 3 6
[[18]]
[1] 0 1 2 2 3 4 4 5 6 6
[[19]]
[[19]] [[1]]
[1] 1 0 1
[[19]] [[2]]
名单列表()
[[19]] [[3]]
list()
$ [b] [b]
[b]
...
相反,如果我打印另外两个列表( listSelf
和 listMult
一切正常)。
现在,我如何绘制这些数据?
我阅读了关于数据框的内容,但我不明白如何使用它。
有人可以帮我吗?
我试图通过手工将一个可能的结果表写在一个csv文件中,然后尝试绘制它以查看如果我正朝着正确的方向前进。
这是代码,这就是结果。
注意:我手工创建的表格和我发明的百分比。
> df< - read.csv(./ table.csv,sep =,)#读取csv文件
> df
N perSelf perMul
1 10 ^ 3 2 1
2 10 ^ 3 5 1
3 10 ^ 3 98 15
4 10 ^ 3 50 51
5 10 ^ 3 41 52
6 10 ^ 3 21 100
7 10 ^ 3 36 80
8 10 ^ 3 70 20
9 10 ^ 3 80 55
10 10 ^ 3 100 44
11 10 ^ 4 2 1
12 10 ^ 4 5 18
13 10 ^ 4 100 20
14 10 ^ 4 50 51
15 10 ^ 4 51 52
16 10 ^ 4 21 100
17 10 ^ 4 36 80
18 10 ^ 4 70 20
19 10 ^ 4 73 85
20 10 ^ 4 100 98
21 10 ^ 5 100 10
22 10 ^ 5 5 1
23 10 ^ 5 98 15
24 10 ^ 5 50 51
25 10 ^ 5 41 52
26 10 ^ 5 21 85
27 10 ^ 5 36 80
28 10 ^ 5 65 20
29 10 ^ 5 80 55
30 10 ^ 5 100 44
有s
非常感谢
代码是:
<$ c $从列表(list_all)创建一个矩阵
mat < - matrix(unlist(list_all),
unique(lengths(list_all)),
dimnames = list(NULL,c (N,%selfloops,%multilinks)))
#将矩阵转换为数据帧
df < - as.data.frame(x = mat,row .names = NULL)
df
#plot
dflong < - melt(df,id.vars ='N')
x11( )
ggplot(dflong,aes(x = N,y = value,color = variable))+
geom_point(size = 5,alpha = 0.7,position = position_dodge(width = 0.3))+
scale_x_discrete(labels = parse(text = as.character(unique(dflong $ N))))+
scale_y_continuous('',breaks = seq(0,100,25),labels = paste(seq ( 0',100,25),'%'))+
scale_color_manual('',values = c('red','blue'),
labels = c('Selfloop Percentage of'多重链接的百分比'))+
theme_minimal(base_size = 14)
<$ c $
N%selfloops%multilinks
1 10 11.111111 0.00000
2 10 11.111111 0.00000
3 10 0.000000 0.00000
4 20 0.000000 0.00000
5 20 0.000000 15.38462
6 20 0.000000 0.00000
7 30 3.448276 0.00000
8 30 3.448276 0.00000
9 30 0.000000 0.00000
<以您的 df
数据框为起点,您可以分两步获得所需的结果:
1)使用 reshape2 将数据重塑为长格式:
library( reshape2)
dflong < - melt(df,i d.vars ='N')
2) (ggplot2):
ggplot(dflong,aes(x = N,ggplot2):
,y = value,color = variable))+
geom_point(size = 5,alpha = 0.7,position = position_dodge(width = 0.3))+
scale_x_discrete(labels = parse(text = as.character (unique(dflong $ N))))+
scale_y_continuous('',breaks = seq(0,100,25),labels = paste(seq(0,100,25),'%'))+
scale_color_manual('',values = c('red','blue'),
labels = c('selfloop百分比','多重链接百分比'))+
theme_minimal(base_size = 14)
给出:
我使用透明度( alpha = 0.7
)能够
回应您的评论和问题中的第二个例子:
您必须稍微修改 ggplot2 代码:
x
aes
中的变量作为因子。 没有必要以解析标签的文字,从而删除该部分。
以下代码:
ggplot(dflong,aes(x = factor(N ),y = value,color = variable))+
geom_point(size = 5,alpha = 0.5,position = position_dodge(width = 0.3))+
xlab('N')+ $ b $ (0,20,5),'%'),
limits = c(0,20) ))+
scale_color_manual('',
values = c('red','blue'),
labels = c('自我循环的百分比','多重链接的百分比'))+
theme_minimal(base_size = 14)
会给你:
使用的数据:
df< - structure(list(N =结构(c(1L,1L,1L,1L,1L,1L,1L,1L,1L,1L,2L,2L,2L,2L,2L,2L,2L,2L,2L,2L,3L,3L,3L ,3L,3L,3L,3L,3L,3L,3L),。标签= c(10 ^ 3,10 ^ 4,10 ^ 5),class =factor b perSelf = c(2L,5L,98L,50L,41L,21L,36L,70L,80L,100L,2L,5L,100L,50L,51L,21L,36L,70L,73L,100L,100L,5L,98L (1L,1L,15L,51L,52L,100L,80L,20L,55L,44L,1L,18L,20L,50L,41L,21L,36L,65L,80L,100L) 51L,52L,100L,80L,20L,85L,98L,10L,1L,15L,51L,52L,85L,80L,20L,55L, 44L)),
.Names = c(N,perSelf,perMul),class =data.frame,row.names = c(NA,-30L))
I need help to create a chart. I explain better.
I created 10 random graphs, each with N nodes. I have done that for N = 10^3, 10^4, 10^5. So in total 30 graphs.
To each of them I found the percentage of multilinks and selfloops they have.
Now I would like to create a single graph that shows the percentage in function of the number of nodes. So something like:
So I have a 3 lists:
- listNets
containing 30 graphs
- listSelf
containing the percentage of selfloops
- listMul
containing the percentage of multilinks
This is what I did:
listN <- c((10^3), (10^4), (10^5))
# list of networks
listNets <- vector(mode = "list", length = 0)
# list of percentage of selfloops
listSelf <- vector(mode = "list", length = 0)
#list of percentage of multilinks
listMul <- vector(mode = "list", length = 0)
...
for(N in listN) {
...
net <- graph_from_adjacency_matrix(adjmatrix = adjacency_matrix, mode = "undirected") # it's work, infact if I plot it i saw a correct networks
listNets <- c(listNets, net) # I add net to list of networks
x11()
plot(net, layout = layout.circle(net))
...
# I find self-loops e multilinks
netmatr <- as_adjacency_matrix(net, sparse = FALSE)
num_selfloops <- sum(diag(netmatr))
num_multilinks <- sum(netmatr > 1)
# I find percentage
per_self <- ((num_selfloops/num_vertices)*100)
per_mul <- ((num_multilinks/num_edges)*100)
listSelf <- c(listSelf, per_self)
listMul <- c(listMul, per_mul)
}
Now if I print listNets
in this way I have something strange:
> print(listNets)
[[1]]
[1] 9
[[2]]
[1] FALSE
[[3]]
[1] 7 6 3 8 8 8
[[4]]
[1] 0 1 2 4 5 7
[[5]]
[1] 2 1 0 3 4 5
[[6]]
[1] 0 1 2 3 4 5
[[7]]
[1] 0 0 0 0 1 1 1 2 3 6
[[8]]
[1] 0 1 2 3 3 4 5 5 6 6
[[9]]
[[9]][[1]]
[1] 1 0 1
[[9]][[2]]
named list()
[[9]][[3]]
list()
[[9]][[4]]
list()
[[10]]
<environment: 0x000000001a6284a8>
[[11]]
[1] 9
[[12]]
[1] FALSE
[[13]]
[1] 2 5 8 8 7 8
[[14]]
[1] 0 1 3 4 6 7
[[15]]
[1] 0 1 4 2 3 5
[[16]]
[1] 0 1 2 3 4 5
[[17]]
[1] 0 0 0 1 1 1 2 2 3 6
[[18]]
[1] 0 1 2 2 3 4 4 5 6 6
[[19]]
[[19]][[1]]
[1] 1 0 1
[[19]][[2]]
named list()
[[19]][[3]]
list()
[[19]][[4]]
list()
[[20]]
<environment: 0x000000001a859e28>
...
Instead if I print the other two lists (listSelf
and listMult
everything is ok).
Now, how can I plot this data?
I read about dataframes, but I don't understand how to use it in my case. Can someone help me please?
I tried to bring me back by writing a possible result table on a csv file by hand and tried to plot it to see if I was going in the right direction.
This is the code and this the result. Note: The table I created by hand and I invented the percentages.
> df <- read.csv("./table.csv", sep = ",") # read csv file
> df
N perSelf perMul
1 10^3 2 1
2 10^3 5 1
3 10^3 98 15
4 10^3 50 51
5 10^3 41 52
6 10^3 21 100
7 10^3 36 80
8 10^3 70 20
9 10^3 80 55
10 10^3 100 44
11 10^4 2 1
12 10^4 5 18
13 10^4 100 20
14 10^4 50 51
15 10^4 51 52
16 10^4 21 100
17 10^4 36 80
18 10^4 70 20
19 10^4 73 85
20 10^4 100 98
21 10^5 100 10
22 10^5 5 1
23 10^5 98 15
24 10^5 50 51
25 10^5 41 52
26 10^5 21 85
27 10^5 36 80
28 10^5 65 20
29 10^5 80 55
30 10^5 100 44
There is something wrong.
Thanks a lot
The code is:
# create a matrix from a list (list_all)
mat <- matrix(unlist(list_all),
unique(lengths(list_all)),
dimnames = list(NULL, c("N", "% selfloops", "% multilinks")))
# convert matrix to data frame
df <- as.data.frame(x = mat, row.names = NULL)
df
# plot
dflong <- melt(df, id.vars = 'N')
x11()
ggplot(dflong, aes(x = N, y = value, color = variable)) +
geom_point(size = 5, alpha = 0.7, position = position_dodge(width = 0.3)) +
scale_x_discrete(labels = parse(text = as.character(unique(dflong$N)))) +
scale_y_continuous('', breaks = seq(0, 100, 25), labels = paste(seq(0, 100, 25), '%')) +
scale_color_manual('', values = c('red', 'blue'),
labels = c('Percentage of selfloop','Percentage of multilinks')) +
theme_minimal(base_size = 14)
df
is:
N % selfloops % multilinks
1 10 11.111111 0.00000
2 10 11.111111 0.00000
3 10 0.000000 0.00000
4 20 0.000000 0.00000
5 20 0.000000 15.38462
6 20 0.000000 0.00000
7 30 3.448276 0.00000
8 30 3.448276 0.00000
9 30 0.000000 0.00000
Taking your df
dataframe as a starting point, you can get the desired result in two steps:
1) Reshape your data into long format with reshape2:
library(reshape2)
dflong <- melt(df, id.vars = 'N')
2) Plot the data with ggplot2:
library(ggplot2)
ggplot(dflong, aes(x = N, y = value, color = variable)) +
geom_point(size = 5, alpha = 0.7, position = position_dodge(width = 0.3)) +
scale_x_discrete(labels = parse(text = as.character(unique(dflong$N)))) +
scale_y_continuous('', breaks = seq(0,100,25), labels = paste(seq(0,100,25),'%')) +
scale_color_manual('', values = c('red','blue'),
labels = c('Percentage of selfloop','Percentage of multilinks')) +
theme_minimal(base_size = 14)
which gives:
I used a transparency (alpha = 0.7
) in order to be able to see where points overlap.
In response to your comment and the second example in the question:
You have to alter the ggplot2 code a bit:
- Change the
x
variable in theaes
to a factor. - There is no need to parse the text for the labels anymore, thus that part can be removed.
- Adjust the values and breaks in the y-scale.
The following code:
ggplot(dflong, aes(x = factor(N), y = value, color = variable)) +
geom_point(size = 5, alpha = 0.5, position = position_dodge(width = 0.3)) +
xlab('N') +
scale_y_continuous('', breaks = seq(0, 20, 5),
labels = paste(seq(0, 20, 5), '%'),
limits = c(0,20)) +
scale_color_manual('',
values = c('red', 'blue'),
labels = c('Percentage of selfloop','Percentage of multilinks')) +
theme_minimal(base_size = 14)
will give you:
Used data:
df <- structure(list(N = structure(c(1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 3L, 3L, 3L, 3L, 3L, 3L, 3L, 3L, 3L, 3L), .Label = c("10^3", "10^4", "10^5"), class = "factor"),
perSelf = c(2L, 5L, 98L, 50L, 41L, 21L, 36L, 70L, 80L, 100L, 2L, 5L, 100L, 50L, 51L, 21L, 36L, 70L, 73L, 100L, 100L, 5L, 98L, 50L, 41L, 21L, 36L, 65L, 80L, 100L),
perMul = c(1L, 1L, 15L, 51L, 52L, 100L, 80L, 20L, 55L, 44L, 1L, 18L, 20L, 51L, 52L, 100L, 80L, 20L, 85L, 98L, 10L, 1L, 15L, 51L, 52L, 85L, 80L, 20L, 55L, 44L)),
.Names = c("N", "perSelf", "perMul"), class = "data.frame", row.names = c(NA, -30L))
这篇关于从不同的数据创建图表的文章就介绍到这了,希望我们推荐的答案对大家有所帮助,也希望大家多多支持IT屋!