How to plot all columns of a dataframe?

Question

Good afternoon !

Under R , i developed the following script :

  library("rootSolve")
    
  Sphere_1_pdim = function (x) return(sum(x^2))
    
  RMSprop<-function(objFun ,iter = 50000, alpha = 0.00001,lambda=0.2 ,start_init ){
  
  init = start_init
  
  gradient_1 <- function(init , objFun ) {
    
    p=ceiling(runif(1,min=0,max =length(init) ))
    result <- gradient(objFun, init,pert = 1e-8)    # vector of gradient / partial dérivatives 
    # print("gradient")
    # print(replace(result, sample(length(init),2), 0))
    # return(replace(result,p, 0))
    return(result)
  }
  
  x <- init
  
  # create a vector to contain all xs for all steps
  x.All = matrix(c(paste("X",1:length(start_init)),"objfun"),ncol=length(start_init)+1) 
  tmp<-rep(0,length(init))
  V=(1-lambda)*(gradient_1(x,objFun))^2
  # gradient descent method to find the minimum

  for(i in seq_len(iter)){
    
    V=lambda*V+(1-lambda)*(gradient_1(x,objFun))^2
    tmp = x - alpha*gradient_1(x,objFun)/sqrt(V)
    
    if ( !is.nan(suppressWarnings(objFun(tmp))) ) {
      x <- tmp
      
    }
    x.All = rbind(x.All,c(x,objFun(x)))
    
    
    
    # print(c(i,x,objFun(x)))
  } 
  
  # print result and plot all xs for every iteration
  print(paste("The minimum of f(x) is ", objFun(x), " at position x = ", x, sep = ""))

  # Here i should plot the x.All dataframe columns :

  plot(x.All, type = "l")  
  x.All=as.data.frame(x.All[-1,]) 
  print("x.All")
  print(x.All)
  dput(x.All)
  return(list(x,x.All))
}

Example of output :

       p=RMSprop(objFun=Sphere_1_pdim ,iter = 20, alpha = 0.01,lambda=0.2 ,rep(0.5,4) )

print(p[[2]])
            
[1] "x.All"
                  V1                V2                V3                V4
1  0.489793792738403 0.489793792738403 0.489793792738403 0.489793792738403
2  0.479794218064014 0.479794218064014 0.479794218064014 0.479794218064014
3  0.469836158529361 0.469836158529361 0.469836158529361 0.469836158529361
4  0.459887403385885 0.459887403385885 0.459887403385885 0.459887403385885
5  0.449941475647717 0.449941475647717 0.449941475647717 0.449941475647717
6  0.439997096792552 0.439997096792552 0.439997096792552 0.439997096792552
7  0.430054049875739 0.430054049875739 0.430054049875739 0.430054049875739
8  0.420112337792478 0.420112337792478 0.420112337792478 0.420112337792478
9   0.41017201167119  0.41017201167119  0.41017201167119  0.41017201167119
10 0.400233136231218 0.400233136231218 0.400233136231218 0.400233136231218
11 0.390295782982069 0.390295782982069 0.390295782982069 0.390295782982069
12   0.3803600292637   0.3803600292637   0.3803600292637   0.3803600292637
13 0.370425958630921 0.370425958630921 0.370425958630921 0.370425958630921
14 0.360493661256826 0.360493661256826 0.360493661256826 0.360493661256826
15 0.350563234797495 0.350563234797495 0.350563234797495 0.350563234797495
16 0.340634785238398 0.340634785238398 0.340634785238398 0.340634785238398
17 0.330708427713843 0.330708427713843 0.330708427713843 0.330708427713843
18 0.320784287730144 0.320784287730144 0.320784287730144 0.320784287730144
19 0.310862502424419 0.310862502424419 0.310862502424419 0.310862502424419
20 0.300943221952355 0.300943221952355 0.300943221952355 0.300943221952355
                  V5
1   0.95959183762028
2  0.920809966750635
3  0.882984063446507
4  0.845985695172046
5  0.809789326032179
6  0.774389780743497
7  0.739785943258098
8  0.705977505461846
9  0.672964316633562
10 0.640746253349907
11 0.609323192854345
12 0.578695007445931
13 0.548861563310546
14 0.519822719225405
15 0.491578326366735
16 0.464128227657637
17  0.43747225664385
18 0.411610237018144
19 0.386541981654287
20 0.362267291356257

Output of dput() : 

structure(list(V1 = c("0.489793792738403", "0.479794218064014", 
"0.469836158529361", "0.459887403385885", "0.449941475647717", 
"0.439997096792552", "0.430054049875739", "0.420112337792478", 
"0.41017201167119", "0.400233136231218", "0.390295782982069", 
"0.3803600292637", "0.370425958630921", "0.360493661256826", 
"0.350563234797495", "0.340634785238398", "0.330708427713843", 
"0.320784287730144", "0.310862502424419", "0.300943221952355"
), V2 = c("0.489793792738403", "0.479794218064014", "0.469836158529361", 
"0.459887403385885", "0.449941475647717", "0.439997096792552", 
"0.430054049875739", "0.420112337792478", "0.41017201167119", 
"0.400233136231218", "0.390295782982069", "0.3803600292637", 
"0.370425958630921", "0.360493661256826", "0.350563234797495", 
"0.340634785238398", "0.330708427713843", "0.320784287730144", 
"0.310862502424419", "0.300943221952355"), V3 = c("0.489793792738403", 
"0.479794218064014", "0.469836158529361", "0.459887403385885", 
"0.449941475647717", "0.439997096792552", "0.430054049875739", 
"0.420112337792478", "0.41017201167119", "0.400233136231218", 
"0.390295782982069", "0.3803600292637", "0.370425958630921", 
"0.360493661256826", "0.350563234797495", "0.340634785238398", 
"0.330708427713843", "0.320784287730144", "0.310862502424419", 
"0.300943221952355"), V4 = c("0.489793792738403", "0.479794218064014", 
"0.469836158529361", "0.459887403385885", "0.449941475647717", 
"0.439997096792552", "0.430054049875739", "0.420112337792478", 
"0.41017201167119", "0.400233136231218", "0.390295782982069", 
"0.3803600292637", "0.370425958630921", "0.360493661256826", 
"0.350563234797495", "0.340634785238398", "0.330708427713843", 
"0.320784287730144", "0.310862502424419", "0.300943221952355"
), V5 = c("0.95959183762028", "0.920809966750635", "0.882984063446507", 
"0.845985695172046", "0.809789326032179", "0.774389780743497", 
"0.739785943258098", "0.705977505461846", "0.672964316633562", 
"0.640746253349907", "0.609323192854345", "0.578695007445931", 
"0.548861563310546", "0.519822719225405", "0.491578326366735", 
"0.464128227657637", "0.43747225664385", "0.411610237018144", 
"0.386541981654287", "0.362267291356257")), class = "data.frame", row.names = c(NA, 
-20L))

Here i used a number of 16 iterations (iter=16) . Each row of the dataframe x.All represents one iteration of the 16.

For each row , the row index-1 represents the iteration index. The last entrie is the value of the objective function.

I'm searching a way ( I prefer ggplot2 ) in which i can plot the evolutions of columns ( from column X1 to the objfun column ).

In this case , we will get 5 curves (in the same plot ) where the last curve is the evolution of the objective function Sphere_1_pdim.

I had tried to add ( before return(x) in RMSprop function ) :

  library(ggplot2)
  library(reshape2)

  x.All=as.data.frame(cbind(index=1:nrow(x.All[-1,]),x.All[-1,]))
  colnames(x.All)=c("index",paste("X",1:length(start_init)),"objfun")
  print(x.All)
  
  df <- melt(x.All ,  id.vars = 'index', variable.name = 'series')
  print(df)
  #create line plot for each column in data frame
  ggplot(df, aes(index, value)) + geom_line(aes(colour = series))

However , this had failed to plot the wanted curves .

I hope my question is clear for you !

Thank you a lot for your help !

Leonardo · Accepted Answer · 2021-02-07 21:38:32Z

1

I hope I have understood correctly.

This is your dataframe:

df <- structure(list(X1 = c(0.489793792738403, 0.479794218064014, 0.469836158529361, 
0.459887403385885, 0.449941475647717, 0.439997096792552, 0.430054049875739, 
0.420112337792478, 0.41017201167119, 0.400233136231218, 0.390295782982069, 
0.3803600292637, 0.370425958630921, 0.360493661256826, 0.350563234797495, 
0.340634785238398), X2 = c(0.489793792738403, 0.479794218064014, 
0.469836158529361, 0.459887403385885, 0.449941475647717, 0.439997096792552, 
0.430054049875739, 0.420112337792478, 0.41017201167119, 0.400233136231218, 
0.390295782982069, 0.3803600292637, 0.370425958630921, 0.360493661256826, 
0.350563234797495, 0.340634785238398), X3 = c(0.489793792738403, 
0.479794218064014, 0.469836158529361, 0.459887403385885, 0.449941475647717, 
0.439997096792552, 0.430054049875739, 0.420112337792478, 0.41017201167119, 
0.400233136231218, 0.390295782982069, 0.3803600292637, 0.370425958630921, 
0.360493661256826, 0.350563234797495, 0.340634785238398), X4 = c(0.489793792738403, 
0.479794218064014, 0.469836158529361, 0.459887403385885, 0.449941475647717, 
0.439997096792552, 0.430054049875739, 0.420112337792478, 0.41017201167119, 
0.400233136231218, 0.390295782982069, 0.3803600292637, 0.370425958630921, 
0.360493661256826, 0.350563234797495, 0.340634785238398), objfun = c(0.95959183762028, 
0.920809966750635, 0.882984063446507, 0.845985695172046, 0.809789326032179, 
0.774389780743497, 0.739785943258098, 0.705977505461846, 0.672964316633562, 
0.640746253349907, 0.609323192854345, 0.578695007445931, 0.548861563310546, 
0.519822719225405, 0.491578326366735, 0.464128227657637)), row.names = c(NA, 
-16L), class = c("tbl_df", "tbl", "data.frame"))

First of all you need to reshape df:

library(dplyr)
library(tidyr)

df2 <- df %>% 
  mutate(x = 1:16) %>% 
  pivot_longer(X1:objfun, "variable", "value")

Now you can plot the data:

library(ggplot2)

ggplot(df2) + 
  geom_line(aes(x, value, color = variable))

If your data are character use:

x.All <- as.data.frame(sapply(x.All, as.numeric))

edited Feb 7, 2021 at 21:38

answered Feb 7, 2021 at 18:06

Leonardo

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13 Comments

Leonardo Over a year ago

Sorry, I forgot to add the tidyr library. I edited my answer

Leonardo Over a year ago

Sorry, but without seeing your data I can't know how to fix it. Edit your question with the output of dput(x.All)

Leonardo Over a year ago

it doesn't matter if it changes. The important thing is that you report an output with dput. However, if x.All is a dataframe that has column names V1, V2, V3, V4, V5 and has the form you reported you can use colnames (x.All) <- c("X1", "X2", "X3", "X4", "objfun") and then use my code. If it doesn't work, I can't answer you again if you don't put the output of dput (x.All).

Leonardo Over a year ago

I changed my answer to add character to numeric conversion. If you think it was useful, if you want accept it

Leonardo Over a year ago

No bothering. See you next time!

|

tjebo · Accepted Answer · 2021-02-07 18:18:16Z

1

For this type of exploratory plotting, maybe consider pairs

pairs(mydat, panel = lines)

Thanks to user Leonardo for the data +1

answered Feb 7, 2021 at 18:18

tjebo

24.1k8 gold badges73 silver badges108 bronze badges

Comments

akrun · Accepted Answer · 2021-02-07 18:37:59Z

We can use matplot from base R

matplot(as.matrix(df), type = 'l')

data

df <- structure(list(X1 = c(0.489793792738403, 0.479794218064014, 0.469836158529361, 
0.459887403385885, 0.449941475647717, 0.439997096792552, 0.430054049875739, 
0.420112337792478, 0.41017201167119, 0.400233136231218, 0.390295782982069, 
0.3803600292637, 0.370425958630921, 0.360493661256826, 0.350563234797495, 
0.340634785238398), X2 = c(0.489793792738403, 0.479794218064014, 
0.469836158529361, 0.459887403385885, 0.449941475647717, 0.439997096792552, 
0.430054049875739, 0.420112337792478, 0.41017201167119, 0.400233136231218, 
0.390295782982069, 0.3803600292637, 0.370425958630921, 0.360493661256826, 
0.350563234797495, 0.340634785238398), X3 = c(0.489793792738403, 
0.479794218064014, 0.469836158529361, 0.459887403385885, 0.449941475647717, 
0.439997096792552, 0.430054049875739, 0.420112337792478, 0.41017201167119, 
0.400233136231218, 0.390295782982069, 0.3803600292637, 0.370425958630921, 
0.360493661256826, 0.350563234797495, 0.340634785238398), X4 = c(0.489793792738403, 
0.479794218064014, 0.469836158529361, 0.459887403385885, 0.449941475647717, 
0.439997096792552, 0.430054049875739, 0.420112337792478, 0.41017201167119, 
0.400233136231218, 0.390295782982069, 0.3803600292637, 0.370425958630921, 
0.360493661256826, 0.350563234797495, 0.340634785238398), objfun = c(0.95959183762028, 
0.920809966750635, 0.882984063446507, 0.845985695172046, 0.809789326032179, 
0.774389780743497, 0.739785943258098, 0.705977505461846, 0.672964316633562, 
0.640746253349907, 0.609323192854345, 0.578695007445931, 0.548861563310546, 
0.519822719225405, 0.491578326366735, 0.464128227657637)), row.names = c(NA, 
-16L), class = c("tbl_df", "tbl", "data.frame"))

there is already a variable ` x.All` ! , this data is just one example of what we could obtain after runnig !

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