BoundsError

Specific Domains Data
1

I have created a function as

function ShortestPath(DAG, K2Score)

for i = 1:length(K2Score)
ind = findall(K2Score .== 0)
#K2Score[1,ind] ~= root,
K2Score[1,ind] = 0.01

end

Finding the root.

G = DAG
M,N = size[DAG]
k = 1
for i in 1:M
for j in 1:N
if G[i,j] ~= 0
a[1,k] = i
a[2,k] = j
k = k +1
end
end
end

n = k-1
k = 1;
for j in 1:N
flag = 0
for i in 1:M
if G[i,j] ~= 0
flag = 1
end
end
if flag .== 0
root[1,k] = j; # Must store all the values
k = k+1
end
end

Applying bellman algorithm

for ii = 1:length(root)

k =1
for i = 1:n
if a[1,i] .== root[1,ii]
b[1,k] = a[1,i]
b[2,k] = a[2,i]
k = k+1
end
end

for i = 1:n
if a[1,i] ~= root[1,ii]
b[1,k] = a[1,i]
b[2,k] = a[2,i]
k = k+1
end
end

for i = 1:M
for ii = 1:length(root)
dist[ii,i] = 20000; # infinity Value
end
end

dist[ii,root[1,ii]] = 0
for i = 1: M
for j = 1:n
k = dist[ii,b[1,j]] + 1
if k .< dist[ii,b[2,j]]
dist[ii,b[2,j]] = k
end
end
end

Finding the minimum of distance

final_dist = minimum(dist, dims = 1)

end

return final_dist

end

And I am us

final_dist_UCII_WOMEN = ShortestPath(UCIV_WOMEN_DAG, UCIV_WOMEN_K2Score)

Throughs an error like

BoundsError: attempt to access 1Ă—112 Array{Float64,2} at index [1, CartesianIndex{2}[CartesianIndex(1, 79), CartesianIndex(1, 82), CartesianIndex(1, 83), CartesianIndex(1, 85), CartesianIndex(1, 88), CartesianIndex(1, 91), CartesianIndex(1, 94), CartesianIndex(1, 97), CartesianIndex(1, 104)]]

Stacktrace:
 [1] throw_boundserror(::Array{Float64,2}, ::Tuple{Int64,Array{CartesianIndex{2},1}}) at .\abstractarray.jl:541
 [2] checkbounds at .\abstractarray.jl:506 [inlined]
 [3] _setindex! at .\multidimensional.jl:784 [inlined]
 [4] setindex! at .\abstractarray.jl:1153 [inlined]
 [5] ShortestPath(::Array{Float64,2}, ::Array{Float64,2}) at .\In[33]:6
 [6] top-level scope at In[34]:1
 [7] include_string(::Function, ::Module, ::String, ::String) at .\loading.jl:1091

The data in UCIV_WOMEN_K2Score is like

Tables.MatrixTable{Array{Float64,2}}:
 :Column1    [-282.75936567503123]
 :Column2    [-305.58336610157846]
 :Column3    [-292.41728385977933]
 :Column4    [-305.18462549235426]
 :Column5    [-274.38053518866104]
 :Column6    [-264.9541448871945]
 :Column7    [-220.70659792617909]
 :Column8    [-134.11407429858826]
 :Column9    [-238.25801615743833]
 :Column10   [-223.2858117418622]
 :Column11   [-297.1694634952932]
 :Column12   [-296.66690599815234]
 :Column13   [-280.3308279437817]
 :Column14   [-232.9550950909364]
 :Column15   [-232.13675627716506]
 :Column16   [-236.38528000947178]
 :Column17   [-261.930952318306]
 :Column18   [-260.73925958765756]
 :Column19   [-289.29985505843]
 :Column20   [-248.3846110642878]
 :Column21   [-265.17573796201276]
 :Column22   [-304.0344529707333]
 :Column23   [-289.8263398616875]
 :Column24   [-257.0691832080685]
 :Column25   [-291.3757440508586]
 :Column26   [-264.0948838305343]
 :Column27   [-292.5600596311132]
 :Column28   [-249.35945573391615]
 :Column29   [-81.92327266391918]
 :Column30   [-119.47535227825449]
 :Column31   [-290.87671876102786]
 :Column32   [-259.1160299476976]
 :Column33   [-274.5405597958887]
 :Column34   [-282.42420745421214]
 :Column35   [-250.5227823754762]
 :Column36   [-237.45806324419493]
 :Column37   [-311.6203721601196]
 :Column38   [-284.93117212740185]
 :Column39   [-273.39518795019046]
 :Column40   [-217.78300320183075]
 :Column41   [-251.84915951112367]
 :Column42   [-272.9138143613708]
 :Column43   [-287.4156101302241]
 :Column44   [-114.37471605094095]
 :Column45   [-161.41934549656662]
 :Column46   [-261.84687097011204]
 :Column47   [-274.9514937520016]
 :Column48   [-158.65000549478196]
 :Column49   [-109.86505012401156]
 :Column50   [-213.1359440025262]
 :Column51   [-75.29153514406694]
 :Column52   [-195.70559894473286]
 :Column53   [-213.0431581230352]
 :Column54   [-281.5760210196869]
 :Column55   [-295.204849892014]
 :Column56   [-8.409162447202561]
 :Column57   [-129.96265278868083]
 :Column58   [-199.24228981544687]
 :Column59   [-279.023929653169]
 :Column60   [-170.92260998595782]
 :Column61   [-96.28484854268876]
 :Column62   [-84.50709805496444]
 :Column63   [-28.22050948456115]
 :Column64   [-5.105945473900647]
 :Column65   [-46.82916759238583]
 :Column66   [-8.409162447202561]
 :Column67   [-8.409162447202561]
 :Column68   [-8.409162447202561]
 :Column69   [-8.409162447202561]
 :Column70   [-8.409162447202561]
 :Column71   [-15.468565275793821]
 :Column72   [-8.409162447202561]
 :Column73   [-8.409162447202561]
 :Column74   [-8.409162447202561]
 :Column75   [-8.409162447202561]
 :Column76   [-13.299511575424276]
 :Column77   [-8.409162447202561]
 :Column78   [-8.409162447202561]
 :Column79   [0.0]
 :Column80   [-8.409162447202561]
 :Column81   [-8.409162447202561]
 :Column82   [0.0]
 :Column83   [0.0]
 :Column84   [-8.409162447202561]
 :Column85   [0.0]
 :Column86   [-8.409162447202561]
 :Column87   [-8.409162447202561]
 :Column88   [0.0]
 :Column89   [-8.409162447202561]
 :Column90   [-8.409162447202561]
 :Column91   [0.0]
 :Column92   [-15.468565275793821]
 :Column93   [-15.468565275793821]
 :Column94   [0.0]
 :Column95   [-8.409162447202561]
 :Column96   [-8.409162447202561]
 :Column97   [0.0]
 :Column98   [-5.105945473900647]
 :Column99   [-5.105945473900647]
 :Column100  [-5.105945473900647]
 :Column101  [-5.105945473900647]
 :Column102  [-15.468565275793821]
 :Column103  [-5.105945473900647]
 :Column104  [0.0]
 :Column105  [-8.409162447202561]
 :Column106  [-15.468565275793821]
 :Column107  [-5.105945473900647]
 :Column108  [-5.105945473900647]
 :Column109  [-5.105945473900647]
 :Column110  [-5.105945473900647]
 :Column111  [-8.409162447202561]
 :Column112  [-15.468565275793821]

Please help me to figure out the problem.

Thanks

2

Hi @ashwanimalviya, welcome to our community! :confetti_ball:

I recommend you look at our list of good practices. I think you used the blockquote instead of triple backticks to post your code and it lost its identation, making it harder to read. If you use triple backticks instead the spacing is preserved (and a monospace font is used).

If you break your code in smaller functions the error message will probably be more precise. For now it is hard to understand which exact line of your problem that is throwing the error.

Giving your code a look, I found strange that you have M, N = size[DAG], did you mean M, N = size(DAG)? Do you have some data to test you function with that throws the same error?

2 Likes
3

Hi @Henrique_Becker Thanks for your response. I follow your suggestion regarding use of blockquote. I recently started using Julia I follow your suggestion for small functions too.
I am reinserting the original code

function ShortestPath(DAG, K2Score)
    
for i = 1:length(K2Score)
  ind = findall(K2Score .== 0.0)
   #K2Score[1,ind] ~= root,
      K2Score[1,ind] = 0.01
    
end


#  Finding the root.
G = DAG
M,N = size(DAG)
k = 1
for i in 1:M
    for j in 1:N
        if G[i,j] ~= 0
            a[1,k] = i
            a[2,k] = j
            k = k +1
        end
    end
end


##
n = k-1
k = 1;     
for j in 1:N
    flag = 0
    for i in 1:M
        if G[i,j] ~= 0
            flag = 1
        end
    end
    if flag .== 0
        root[1,k] = j; # Must store all the values
        k = k+1
    end
end

## Applying bellman algorithm


for ii = 1:length(root)

k =1
for i = 1:n
    if a[1,i] .== root[1,ii]
        b[1,k] = a[1,i]
        b[2,k] = a[2,i]
        k = k+1
    end
end


for i = 1:n
    if a[1,i] ~= root[1,ii]
        b[1,k] = a[1,i]
        b[2,k] = a[2,i]
        k = k+1
    end
end



for i = 1:M
    for ii = 1:length(root)
    dist[ii,i] = 20000; # infinity Value 
    end
end

dist[ii,root[1,ii]] = 0
 for i = 1: M
     for j = 1:n
         k = dist[ii,b[1,j]] + 1
         if k .< dist[ii,b[2,j]]
             dist[ii,b[2,j]] = k
         end
     end
 end
## Finding the minimum of distance
final_dist = minimum(dist, dims = 1)

end
    
return final_dist
    
end

can you please suggest me the way to attached the input data .csv file as per your request to test the function.
Thank you very much for taking your time.

2 Likes
4

I think you can use https://pastebin.com/ for the CSV data.

5

How do I send you the data on pastebin?

6

Paste the code here, press “Create new paste”, and copy here the link of the page it will redirect you.

7

Hi @Henrique_Becker , I have mailed you the data. Please give your suggestion here as you feel right.
Thanks

8

@ashwanimalviya, I did not have time to dive into your problem until now.

I downloaded the CSV files, and executed:

using CSV, DataFrames

dag = CSV.read("UCIV_WOMEN_DAG.csv", DataFrame)
k2s = CSV.read("UCIV_WOMEN_K2Score.csv", DataFrame)
ShortestPath(dag, k2s)

This already fails at the start, with:

ERROR: MethodError: no method matching length(::DataFrame)

What data structures are your code expecting? How do you read them from the CSV?

9

Hi @Henrique_Becker

Thanks for your reply and Sorry for my late reply. I was travelling this week.
The function “ShortestPath” is expecting the inputs(dag, k2s) in matrix form.
Since these inputs are also an output from different function and needs to be feed as input for ShortestPath function.

I hope you get my point.
Please feel free to write if you need to know anything about it further.

Thanks
Ashwani

10

Ashwani,

I read the CSVs in matrix form, and started trying to put it to work, but the problems kept appearing:

  1. The problem you report was easy to fix. It is just that the ind returned by findall already gives you a list of CartesianIndex{2}, so when you use it to index K2Score, you should not do K2Score[1,ind] but instead just K2Score[ind].
  2. After this, I found the problem that ~= is not defined. I suppose you meant isapprox. However, isapprox needs an absolute tolerance to compare with zero (what seems to be your intent), so I guessed 0.01 was a good enough tolerance, but you may need to change this.
  3. Finally, you refer to some a variables in your code, but your code does not define them. I had nothing to do here, as I do not know anything about them. At this point I stopped.

The code I changed is here.

2 Likes
11

Hi @Henrique_Becker

Thanks for your time taken to look into the matter.
Following your suggestion to break the code into smaller functions, I have recreated the code to solve the whole task into smaller tasks. The first and second point you have mentioned is solved in this way. Thanks.

I would like to mention that ~= is for comparing not equal to in MATLAB. I have used != in Julia and it is working fine.

Now, the next thing is, I want to create a matrix a of 2 rows and variable columns out of DAG

# creating vector a out of DAG
function Creat_Vector_a(DAG)
G = DAG
M,N = size(DAG)
a1 = reshape(Int64[],0,1)
a2 = reshape(Int64[],0,1)   
k = 1
for i in 1:M
    for j in 1:N
        if G[i,j]!= 0'
           a1[k] = i
           a2[k] = j
            k = k +1
                if k > N
                    break
                end
                a = [a1; a2]   

        end
    end
end
    
return a
    
end

I am getting the following error

BoundsError: attempt to access 0Ă—1 Array{Int64,2} at index [1]

Stacktrace:
 [1] setindex! at .\array.jl:847 [inlined]
 [2] Creat_Vector_a(::Array{Float64,2}) at .\In[87]:13
 [3] top-level scope at In[88]:1
 [4] include_string(::Function, ::Module, ::String, ::String) at .\loading.jl:1091

Please help me to understand it.

Thanks

12

The problems is that both a1 and a2 have no valid positions (as they have zero rows), so you cannot ever save anything inside them.

Did you read the documentation for reshape? What exactly you intend to do with a matrix that has zero rows (and one column)? Also, why would you reshape a just-created Int64[], you do reshape to make already allocated data take another shape. If you want to allocate a matrix do it directly, i.e., Matrix{Int64}(undef, number_of_rows, number_of_columns) if you want uninitialized data, or zeros(Int64, number_of_rows, number_of_columns) if you want zeroed data.

13

Hi @Henrique_Becker

Thank for the correction. The code worked.
But I am not getting the desired result.

form the previous code, I want to search in every column of DAG starting from 1 to N and if 0 is not found, I want to store the index i in a1 and j in a2 and finally a matrix a.

for example,
If there are two non zero values in column 1 of DAG at rows 3 and 84, then it must give me a1[1,1] = 1, a2[1,1] = 3 and a1[1,2] = 1, a2[1,2] = 84

in this way our matrix a must look like

a[1,1] = 1, a[1,2] = 1
a[2,1] = 3, a[2,2] = 84 
and so on

Instead of getting this result, I am getting

 :Column1    [1.0, 1.0]
 :Column2    [1.0, 2.0]
 :Column3    [1.0, 3.0]
 :Column4    [1.0, 5.0]
 :Column5    [1.0, 6.0]
 :Column6    [1.0, 7.0]
 :Column7    [1.0, 8.0]
 :Column8    [1.0, 9.0]
 :Column9    [1.0, 10.0]
 :Column10   [1.0, 11.0]
 :Column11   [1.0, 12.0]
 :Column12   [1.0, 13.0]
 :Column13   [1.0, 14.0]
 :Column14   [1.0, 15.0]
 :Column15   [1.0, 16.0]
 :Column16   [1.0, 17.0]
 :Column17   [1.0, 18.0]
 :Column18   [1.0, 19.0]
 :Column19   [1.0, 20.0]
 :Column20   [1.0, 21.0]
 :Column21   [1.0, 22.0]
 :Column22   [1.0, 23.0]
 :Column23   [1.0, 24.0]
 :Column24   [1.0, 25.0]
 :Column25   [1.0, 26.0]
 :Column26   [1.0, 27.0]
 :Column27   [1.0, 28.0]
 :Column28   [1.0, 29.0]
 :Column29   [1.0, 30.0]
 :Column30   [1.0, 31.0]
 :Column31   [1.0, 32.0]
 :Column32   [1.0, 33.0]
 :Column33   [1.0, 34.0]
 :Column34   [1.0, 35.0]
 :Column35   [1.0, 36.0]
 :Column36   [1.0, 37.0]
 :Column37   [1.0, 38.0]
 :Column38   [1.0, 39.0]
 :Column39   [1.0, 40.0]
 :Column40   [1.0, 41.0]
 :Column41   [1.0, 42.0]
 :Column42   [1.0, 43.0]
 :Column43   [1.0, 44.0]
 :Column44   [1.0, 45.0]
 :Column45   [1.0, 46.0]
 :Column46   [1.0, 47.0]
 :Column47   [1.0, 48.0]
 :Column48   [1.0, 49.0]
 :Column49   [1.0, 50.0]
 :Column50   [1.0, 51.0]
 :Column51   [1.0, 52.0]
 :Column52   [1.0, 53.0]
 :Column53   [1.0, 54.0]
 :Column54   [1.0, 55.0]
 :Column55   [1.0, 56.0]
 :Column56   [1.0, 57.0]
 :Column57   [1.0, 58.0]
 :Column58   [1.0, 59.0]
 :Column59   [1.0, 60.0]
 :Column60   [1.0, 61.0]
 :Column61   [1.0, 62.0]
 :Column62   [1.0, 63.0]
 :Column63   [1.0, 64.0]
 :Column64   [1.0, 65.0]
 :Column65   [1.0, 66.0]
 :Column66   [1.0, 67.0]
 :Column67   [1.0, 68.0]
 :Column68   [1.0, 69.0]
 :Column69   [1.0, 70.0]
 :Column70   [1.0, 71.0]
 :Column71   [1.0, 72.0]
 :Column72   [1.0, 73.0]
 :Column73   [1.0, 74.0]
 :Column74   [1.0, 75.0]
 :Column75   [1.0, 76.0]
 :Column76   [1.0, 77.0]
 :Column77   [1.0, 78.0]
 :Column78   [1.0, 79.0]
 :Column79   [1.0, 80.0]
 :Column80   [1.0, 81.0]
 :Column81   [1.0, 82.0]
 :Column82   [1.0, 83.0]
 :Column83   [1.0, 84.0]
 :Column84   [1.0, 85.0]
 :Column85   [1.0, 86.0]
 :Column86   [1.0, 87.0]
 :Column87   [1.0, 88.0]
 :Column88   [1.0, 89.0]
 :Column89   [1.0, 90.0]
 :Column90   [1.0, 91.0]
 :Column91   [1.0, 92.0]
 :Column92   [1.0, 93.0]
 :Column93   [1.0, 94.0]
 :Column94   [1.0, 95.0]
 :Column95   [1.0, 96.0]
 :Column96   [1.0, 97.0]
 :Column97   [1.0, 98.0]
 :Column98   [1.0, 99.0]
 :Column99   [1.0, 100.0]
 :Column100  [1.0, 101.0]
 :Column101  [1.0, 102.0]
 :Column102  [1.0, 103.0]
 :Column103  [1.0, 104.0]
 :Column104  [1.0, 105.0]
 :Column105  [1.0, 106.0]
 :Column106  [1.0, 107.0]
 :Column107  [1.0, 108.0]
 :Column108  [1.0, 109.0]
 :Column109  [1.0, 110.0]
 :Column110  [1.0, 111.0]
 :Column111  [1.0, 112.0]
 :Column112  [2.0, 2.0]

Please let me know where am I making mistake in the code.

Thanks

14

I have updated the code like below

# creating vector a out of DAG
function Creat_Vector_a(DAG)
G = DAG
M,N = size(DAG)
a1 = Matrix{Int64}(undef, 1, 2000)
a2 = Matrix{Int64}(undef, 1, 2000)
#b = zeros(2,N)    
k = 1
for i in 1:M
    for j in 1:N
        if G[i,j]!== 0
           a1[1,k] = i
                #push!(a1[k], i)
           a2[1,k] = j
                #push!(a2[k], j)
            k = k +1
                if k > N
                    break
                end
                a = [a1; a2]   

        end
    end
end
    
return a
    
end
15

Why a1 and a2 are matrices, or better, why a1 and a2 have both rows and columns, in the example you only change the second index, the first index will change in some situation? They are not better as Vectors? In fact, in your code you save them by using:

           a1[k] = i
           a2[k] = j

In other words, you only use one index, it seems to me that they are Vectors that for some reason you want to treat as Matrices. As you do not know the size of the vectors before starting you could define them as:

a1 = Int[]
a2 = Int[]

and then when you fill them you could instead

push!(a1, i)
push!(a2, j)

There is some reason to why

                if k > N
                    break
                end

? You do want to limit the number of nonzero entries obtained?

Also, you are making a with a = [a1; a2] every time you finda a non-zero, why? Do you not want to do this only at the end, and as they would be vectors it would be written this way:

a = [a1'; a2']
16

Hi @Henrique_Becker

Thanks for your clear explanation.

I followed your suggestion and updated the code as

function Creat_Vector_a(DAG)
G = DAG
M,N = size(DAG)
a1 = Int[]
a2 = Int[]   
k = 1
for i in 1:M
    for j in 1:N
        if G[i,j]!== 0               
           a1[k] = i           
           a2[k] = j               
           push!(a1, i)
           push!(a2, j)
           k = k +1
        end
            a = [a1'; a2']
    end
end
    
return a
    
end

I am getting the following error now

BoundsError: attempt to access 0-element Array{Int64,1} at index [1]

Stacktrace:
 [1] setindex! at .\array.jl:847 [inlined]
 [2] Creat_Vector_a(::Array{Float64,2}) at .\In[167]:12
 [3] top-level scope at In[168]:1
 [4] include_string(::Function, ::Module, ::String, ::String) at .\loading.jl:1091
17

I know I do not explain every detail but, unfortunately, I do not have the time to do so. I ask you that you try to pick up the pieces and try to make sense of what I tell you.

The error happens because you are trying to save an element to position 1 when the Vector has zero positions (it is empty). Maybe in other languages the position would be created for you but this is not the case in Julia. My suggestion was that you should use push! instead of the attribution (i.e., a1[k] = i and a2[k] = j), so keep only the two last lines of these four. push! will create a new position for you (after all current positions), and save the value you passed in such position. Also, you do not need k anymore, because push! will always add the new element after the last one so you do not need to control the current position.

Finally, did you change the position of a = [a1'; a2']? It keeps making a lot of unnecessary effort where it is in your last code. My suggestion would be creating only in the line just before the return a (i.e., after closing the two loops).