I tried to implement sha256 without using library, so I took this code : https://github.com/dmarman/sha256algorithm/blob/main/src/classes/sha.js
It's working well, taking in input string and returning the good hash (also a string)
The problem I have is with CryptoJS library. Let's see the three following lines :
x = "00000000";
console.log(CryptoJS.SHA256(CryptoJS.enc.Hex.parse(x))+"");
console.log(sha256(x));
It gives to me two distincts results :
df3f619804a92fdb4057192dc43dd748ea778adc52bc498ce80524c014b81119
7e071fd9b023ed8f18458a73613a0834f6220bd5cc50357ba3493c6040a9ea8c
I understood that the CryptoJS.enc.Hex.parse(x) changes the result with CryptoJS, If I use x directly, I don't have difference
Another point, if you try x == CryptoJS.enc.Hex.parse(x) it gives you true, but CryptoJS.SHA256 seems to give different result following the type of the input
The problem, is that, in my exercice, I want to get the result of CryptoJS.SHA256(CryptoJS.enc.Hex.parse(x)) with my sha256 function
So I'm looking for function f such as sha256(f(x)) = CryptoJS.SHA256(CryptoJS.enc.Hex.parse(x))+""
I know that the condition x == CryptoJS.enc.Hex.parse(x) is not always true, it's here because I took 00000000, but we case suppose the condition always satisfied for the exercise.
Do you have any explanation please ?
Related
This code is originally written in Python 2 and I need to translate it in python 3!
I'm sorry for not sharing enough information:
Also, here's the part where self.D was first assigned:
def __init__(self,instance,transformed,describe1,describe2):
self.D=[]
self.instance=instance
self.transformed=transformed
self.describe1,self.describe2=describe1,describe2
self.describe=self.describe1+', '+self.describe2 if self.describe2 else self.describe1
self.column_num=self.tuple_num=self.view_num=0
self.names=[]
self.types=[]
self.origins=[]
self.features=[]
self.views=[]
self.classify_id=-1
self.classify_num = 1
self.classes=[]
def generateViews(self):
T=map(list,zip(*self.D))
if self.transformed==0:
s= int( self.column_num)
for column_id in range(s):
f = Features(self.names[column_id],self.types[column_id],self.origins[column_id])
#calculate min,max for numerical,temporal
if f.type==Type.numerical or f.type==Type.temporal:
f.min,f.max=min(T[column_id]),max(T[column_id])
if f.min==f.max:
self.types[column_id]=f.type=Type.none
self.features.append(f)
continue
d={}
#calculate distinct,ratio for categorical,temporal
if f.type == Type.categorical or f.type == Type.temporal:
for i in range(self.tuple_num):
print([type(self.D[i]) for i in range(self.tuple_num)])
if self.D[i][column_id] in d:
d[self.D[i][column_id]]+=1
else:
d[self.D[i][column_id]]=1
f.distinct = len(d)
f.ratio = 1.0 * f.distinct / self.tuple_num
f.distinct_values=[(k,d[k]) for k in sorted(d)]
if f.type==Type.temporal:
self.getIntervalBins(f)
self.features.append(f)
TypeError: 'map' object is not subscriptable
The snippet you have given is not enough to solve the problem. The problem lies in self.D which you are trying to subscript using self.D[i]. Please look into your code where self.D is instantiated and make sure that its an array-like variable so that you can subscript it.
Edit
based on your edit, please confirm that whether self.D[i] is also array-like for all i in the range mentioned in the code. you can do that by simply
print([type(self.D[i]) for i in range(self.tuple_num))
share the response of this code, so that I may help further.
Edit-2
As per your comments and the edited code snippet, it seems that self.D is the output of some map function. In python 2, map is a function that returns a list. However, in python3 map is a class that when invoked, creates a map object.
The simplest way to resolve this is the find out the line where self.D was first assigned, and whatever code is in the RHS, wrap it with a list(...) function.
Alternately, just after this line
T=map(list,zip(*self.D))
add the following
self.D = list(self.D)
Hope this will resolve the issue
We don't have quite enough information to answer the question, but in Python 3, generator and map objects are not subscriptable. I think it may be in your
self.D[i]
variable, because you claim that self.D is a list, but it is possible that self.D[i] is a map object.
In your case, to access the indexes, you can convert it to a list:
list(self.D)[i]
Or use unpacking to implicitly convert to a list (this may be more condensed, but remember that explicit is better than implicit):
[*self.D[i]]
I'm trying to make Maple solve a complex equation, but it produces an incorrect result.
The following images tells it all :
At (3) I would expect to get something close to 1 (as (2) shows), yet it gives me something that doesn't make any sense. Is it that the || (to express the complex number modulus) operator has another significance in the solve() function?
The more appropriate function here is fsolve.
Example 1
restart:
G:=(w,L)->(5+I*L*2*Pi*w)/(150+I*L*2*Pi*w);
evalf(5*abs(G(10,1)));
fsolve(5*abs(G(10,L))=%,L=0..10)
Example 2
As above, you need to specify the interval L=0..1 where the solution might be.
G:=(f,L)->(256.4+I*L*2*Pi*f)/(256.4+9845+I*L*2*Pi*f);
evalf(5*abs(G(20000,0.03602197444)));
fsolve(5*abs(G(20000,L))=%,L=0..1);
If you are facing difficulties to specify the interval then you should plot it first, it will give you an idea about it?
plot(5*abs(G(20000,L)),L=0..1)
Restrict the values of L in the solve command with the assuming command.
Example 1:
G:= (w,L) -> (50+I*L*2*Pi*w)/(150+I*L*2*Pi*w);
result := evalf(5*abs(G(10,1)));
solve({5*abs(G(10,L)) = result},L) assuming L::real;
{L = 1.000000000}, {L = -1.000000000}
Example 2:
G:=(f,L) -> (256.4+I*2*Pi*L*f)/(256.4+9845+I*2*Pi*L*f);
result := 5*abs(G(20000,0.03602197444));
solve({5*abs(G(20000,L)) = result},L) assuming L::real;
{L = 0.03602197445}, {L = -0.03602197445}
I'm attempting to make a program (for homework) which reads in a file, and then counts the number of times each word is used. For tackling the issue efficiently, I've decided to map all unique words to keys, and then increment the key value each time the word comes up.
function [] = problem2
file_open = fopen('austen.txt');
complete_string = textscan(file_open, '%s');
numel(complete_string{1,1})
unique_words = unique(complete_string{1,1});
length(unique_words);
frequency = zeros(numel(unique_words), 1);
found_frequency = containers.Map(unique_words, frequency);
for i=1:numel(complete_string{1,1})
found_frequency(complete_string{1,1}(i)) = found_frequency(complete_string{1,1}(i))+1;
end
fclose(file_open)
Sadly, this code does not work. When the line comes up to increment, I receive an error stating that "specified key type does not match the type expected for this container", which makes no sense to me - I'm using strings as the keys. Any ideas as to why I'm receiving this error?
The issue was in the use of Cell type - complete_String{1,1}(i) would actually return a Cell rather than a String (per spec, though). Wrapped it in char(*) and it worked fine.
I have been using MATLAB fminunc function to solve my optimization problem. I want to try the minFunc package :
http://www.di.ens.fr/~mschmidt/Software/minFunc.html
When using fminunc, I defined a function funObj.m which gives me the objective value and the gradient at any point 'x'. It also takes in several external inputs say, {a,b,c} which are matrices. So the function prototype looks like :
function [objVal,G] = funObj(x,a,b,c)
I want to use the same setup in the minFunc package. From the examples, I figured this should work :
options.Method='lbfgs';
f = #(x)funObj(x,a,b,c);
x = minFunc(f,x_init,options);
But when I call this way, I get an error as:
Error using funObj
Too many output arguments.
What is the correct way to call minFunc for my case?
**EDIT : Alright, here is a sample function that I want to use with minFunc. Lets say I want to find the minimum of a*(b-x)^2, where a,b are scalar parameters and x being a scalar too. The MATLAB objective function will then look like :
function obj = testFunc(x,a,b)
obj = a*(b-x)^2;
The function call to minimize this using fminunc (in MATLAB ) is simply:
f = #(x)testFunc(x,a,b);
x = fminunc(f,x_init);
This gives me the minimum of x = 10. Now, How do I do the same using minFunc ?
"Note that by default minFunc assumes that the gradient is supplied, unless the 'numDiff' option is set to 1 (for forward-differencing) or 2 (for central-differencing)."
The error is because only one argument is returned by the function. You can either return the gradient as a second argument or turn on numerical differencing.
Agree with Mark. I think the simplest way to solve it is
minFunc(#testFunc, x_init, a, b, c)
In MATLAB temporary function can only have one return value. So f = #(x)testFunc(x,a,b); let your method drop gradient part every time. Because minFunc can accept extra paramters, you can pass a, b and c after x_init. I think this would work.
I would like to use a dataset filename "AUDUSD" in several functions. It would be easier for me, just to change the filename "AUDUSD" to a more general name like "FX" and then using the abbreviation "FX" in other_matlab functions, e.g. double(). But matlab does not know the name "FX" (that should be assigned to the dataset "AUDUSD") in the code below... Any suggestions?
CODE:
FX = 'AUDUSD';
load(FX); %OKAY !!! FX works as input to open file AUDUSD!
Svars = {'S_bid','S_offer'};
Fvars = {'F_bid','F_offer'};
vS = double(FX,Svars); % FX does NOT work as input for the file AUDUSD
There is no double() function that accepts multiple cell arrays as arguments (this is what happens when you call double(FX,Svars)).
If you call double(FX), then each character in FX is interpreted for its ASCII value and then cast to double. So you get [ 65.0 85.0 68.0 85.0 83.0 68.0 ]. This is the behavior for the double() function if you provide a vector: each individual value in the vector is cast to double.
You'd have to provide more details on what you're trying to accomplish to give any more suggestions.
I have a different example, maybe you will better understand my point. The key work I would like to process is as follows:
I have got a folder with "dataset" files. I would like to loop through this folder, entering in any datasetfile, extracting the 2nd and 3rd column of each dataset file, and constructing only ONE new datasetfile with all 2nd and 3rd columns of the datasetfiles.
One problem is that the size of the datasetfiles are not the same, so I tried to translate a datasetfile into a double-matrix and then consolidate all double matrices into ONE double matrx.
Here my code:
folder_string = 'Diss_Data/Raw';
FolderContent = dir(folder_string);
No_ds = numel(FolderContent);
for i = 1:No_ds
if isdir(FolderContent(i).name)==0
file_string = FolderContent(i).name;
file_path = strcat(folder_string,'/',file_string)
dataset_filename = file_string(1:6);
load(file_path); %loads the suggested datasetfile; OKAY
M = double(dataset_filename);% returns an ASCII code number; WRONG; should transfer the datasetfile into a matrix M
vS = M(:,2:3);
%... to be continued
end
end