I have a ndarray with shape (x,y,d).
How can I get the number of d dimensional non zero vectors (out of the x*y total d dimensional vectors)?
I tried np.count_nonzero but I don't think it has the option to do what I described.
actually I think that works:
sum(np.any(x) for x in p)
(p is the ndarray)
You need to apply np.count_nonzeo to the last axis(d). It will return a x * y array with the count of non zero elements in the d dimension. If the count is 0 then it is a zero array, so you just need to count the number of elements in the x * y array which are not equal to 0.
np.sum(np.apply_along_axis(np.count_nonzero, 2, your_arr) != 0)
I am declaring a variable with inline declaration 50 * ( 2 / 5 ). The problem is that output result is 0 instead of expected 20.
DATA(exact_result) = 50 * ( 2 / 5 ) .
cl_demo_output=>display( exact_result ).
Can anyone suggest why the result is zero where as 50 * (2/5) = 20.
regards,
Umar Abdullah
The inline declaration assigns a data type depending on the type from the Right-Hand Side (RHS) expression. With an arithmetic expression, the compiler determines a data type based on the overall calculation type.
First, 2 and 5 are considered as type I (4 bytes integer), so the result is also of type I even if the operator is a division (integer division in that precise case).
Then, 50 is also considered as type I, and because it's used with another I-type data object (result of subexpression 2 / 5 which is of type I) the result is also of type I.
So, in your example, EXACT_RESULT is assigned the type I.
At run time, because both LHS and RHS data objects are of type I, then the calculation type is I too. Consequently, 2 / 5 equals 0.4 which is rounded to 0 because it's an integer division and the default ABAP rounding is "half up" (rounding of 0.4 gives 0, but 0.5 gives 1).
The workaround is to define explicitly the data type of EXACT_RESULT as having digits after the decimal point (DECFLOAT16, DECFLOAT34, P type with decimals, F and even C because then the calculation type is P !), because the type of the LHS will have a higher priority than the type of the RHS (I), so the calculation will be deduced from the type of the LHS variable.
DATA(exact_result) = CONV decfloat16( 50 * ( 2 / 5 ) ).
Be careful with this next solution : as I said, C leads to a calculation with type P and many decimals, so we could think this example is a good solution :
DATA(exact_result) = '50' * ( 2 / 5 ). " equals 20
But with inline declarations, a P calculation type leads to a data object of type P but with 0 digits after the decimal point, so the result is truncated with other numbers (8 instead of 50 here) :
DATA(exact_result) = '8' * ( 2 / 5 ). " rounded ! (3 instead of 3.2)
I'm learning NTL and I have a doubt: How will I be able to get any specific element of any finite field?
Here is my code
GF2X P = BuildIrred_GF2X(256);
GF2E::init(P);
GF2E zero = GF2E::zero();
GF2E one;
GF2E r = random_GF2E(); //I want change the function random_GF2E()
I want change the function random_GF2E() by any other to get a specific element.
The Elements of a finite field with 2256 elements are represented as the polynomials f of deg(f) < 256.
If you want a special element, you can declare a polynom p by something like
GF2X p;
p.SetLength(n);
SetCoeff(p,i,1);
There is deg(p) = n. If n < deg(P) (in your case n < 256), then this is a special element of the finite field. If n >= deg(P) you can reduce it modulo P by conv<GF2E>(p).
I hope this is what you were looking for.
One of my friends was asked this question recently:
You have to count how many binary strings are possible of length "K".
Constraint: Every 0 has a 1 in its immediate left.
This question can be reworded:
How many binary sequences of length K are posible if there are no two consecutive 0s, but the first element should be 1 (else the constrains fails). Let us forget about the first element (we can do it bcause it is always fixed).
Then we got a very famous task that sounds like this: "What is the number of binary sequences of length K-1 that have no consecutive 0's." The explanation can be found, for example, here
Then the answer will be F(K+1) where F(K) is the K`th fibonacci number starting from (1 1 2 ...).
∑ From n=0 to ⌊K/2⌋ of (K-n)Cn; n is the number of zeros in the string
The idea is to group every 0 with a 1 and find the number of combinations of the string, for n zeros there will be n ones grouped to them so the string becomes (k-n) elements long. There can be no more than of K/2 zeros as there would not have enough ones to be to the immediate left of each zero.
E.g. 111111[10][10]1[10] for K = 13, n = 3
I know the modulus (%) operator calculates the remainder of a division. How can I identify a situation where I would need to use the modulus operator?
I know I can use the modulus operator to see whether a number is even or odd and prime or composite, but that's about it. I don't often think in terms of remainders. I'm sure the modulus operator is useful, and I would like to learn to take advantage of it.
I just have problems identifying where the modulus operator is applicable. In various programming situations, it is difficult for me to see a problem and realize "Hey! The remainder of division would work here!".
Imagine that you have an elapsed time in seconds and you want to convert this to hours, minutes, and seconds:
h = s / 3600;
m = (s / 60) % 60;
s = s % 60;
0 % 3 = 0;
1 % 3 = 1;
2 % 3 = 2;
3 % 3 = 0;
Did you see what it did? At the last step it went back to zero. This could be used in situations like:
To check if N is divisible by M (for example, odd or even)
or
N is a multiple of M.
To put a cap of a particular value. In this case 3.
To get the last M digits of a number -> N % (10^M).
I use it for progress bars and the like that mark progress through a big loop. The progress is only reported every nth time through the loop, or when count%n == 0.
I've used it when restricting a number to a certain multiple:
temp = x - (x % 10); //Restrict x to being a multiple of 10
Wrapping values (like a clock).
Provide finite fields to symmetric key algorithms.
Bitwise operations.
And so on.
One use case I saw recently was when you need to reverse a number. So that 123456 becomes 654321 for example.
int number = 123456;
int reversed = 0;
while ( number > 0 ) {
# The modulus here retrieves the last digit in the specified number
# In the first iteration of this loop it's going to be 6, then 5, ...
# We are multiplying reversed by 10 first, to move the number one decimal place to the left.
# For example, if we are at the second iteration of this loop,
# reversed gonna be 6, so 6 * 10 + 12345 % 10 => 60 + 5
reversed = reversed * 10 + number % 10;
number = number / 10;
}
Example. You have message of X bytes, but in your protocol maximum size is Y and Y < X. Try to write small app that splits message into packets and you will run into mod :)
There are many instances where it is useful.
If you need to restrict a number to be within a certain range you can use mod. For example, to generate a random number between 0 and 99 you might say:
num = MyRandFunction() % 100;
Any time you have division and want to express the remainder other than in decimal, the mod operator is appropriate. Things that come to mind are generally when you want to do something human-readable with the remainder. Listing how many items you could put into buckets and saying "5 left over" is good.
Also, if you're ever in a situation where you may be accruing rounding errors, modulo division is good. If you're dividing by 3 quite often, for example, you don't want to be passing .33333 around as the remainder. Passing the remainder and divisor (i.e. the fraction) is appropriate.
As #jweyrich says, wrapping values. I've found mod very handy when I have a finite list and I want to iterate over it in a loop - like a fixed list of colors for some UI elements, like chart series, where I want all the series to be different, to the extent possible, but when I've run out of colors, just to start over at the beginning. This can also be used with, say, patterns, so that the second time red comes around, it's dashed; the third time, dotted, etc. - but mod is just used to get red, green, blue, red, green, blue, forever.
Calculation of prime numbers
The modulo can be useful to convert and split total minutes to "hours and minutes":
hours = minutes / 60
minutes_left = minutes % 60
In the hours bit we need to strip the decimal portion and that will depend on the language you are using.
We can then rearrange the output accordingly.
Converting linear data structure to matrix structure:
where a is index of linear data, and b is number of items per row:
row = a/b
column = a mod b
Note above is simplified logic: a must be offset -1 before dividing & the result must be normalized +1.
Example: (3 rows of 4)
1 2 3 4
5 6 7 8
9 10 11 12
(7 - 1)/4 + 1 = 2
7 is in row 2
(7 - 1) mod 4 + 1 = 3
7 is in column 3
Another common use of modulus: hashing a number by place. Suppose you wanted to store year & month in a six digit number 195810. month = 195810 mod 100 all digits 3rd from right are divisible by 100 so the remainder is the 2 rightmost digits in this case the month is 10. To extract the year 195810 / 100 yields 1958.
Modulus is also very useful if for some crazy reason you need to do integer division and get a decimal out, and you can't convert the integer into a number that supports decimal division, or if you need to return a fraction instead of a decimal.
I'll be using % as the modulus operator
For example
2/4 = 0
where doing this
2/4 = 0 and 2 % 4 = 2
So you can be really crazy and let's say that you want to allow the user to input a numerator and a divisor, and then show them the result as a whole number, and then a fractional number.
whole Number = numerator/divisor
fractionNumerator = numerator % divisor
fractionDenominator = divisor
Another case where modulus division is useful is if you are increasing or decreasing a number and you want to contain the number to a certain range of number, but when you get to the top or bottom you don't want to just stop. You want to loop up to the bottom or top of the list respectively.
Imagine a function where you are looping through an array.
Function increase Or Decrease(variable As Integer) As Void
n = (n + variable) % (listString.maxIndex + 1)
Print listString[n]
End Function
The reason that it is n = (n + variable) % (listString.maxIndex + 1) is to allow for the max index to be accounted.
Those are just a few of the things that I have had to use modulus for in my programming of not just desktop applications, but in robotics and simulation environments.
Computing the greatest common divisor
Determining if a number is a palindrome
Determining if a number consists of only ...
Determining how many ... a number consists of...
My favorite use is for iteration.
Say you have a counter you are incrementing and want to then grab from a known list a corresponding items, but you only have n items to choose from and you want to repeat a cycle.
var indexFromB = (counter-1)%n+1;
Results (counter=indexFromB) given n=3:
`1=1`
`2=2`
`3=3`
`4=1`
`5=2`
`6=3`
...
Best use of modulus operator I have seen so for is to check if the Array we have is a rotated version of original array.
A = [1,2,3,4,5,6]
B = [5,6,1,2,3,4]
Now how to check if B is rotated version of A ?
Step 1: If A's length is not same as B's length then for sure its not a rotated version.
Step 2: Check the index of first element of A in B. Here first element of A is 1. And its index in B is 2(assuming your programming language has zero based index).
lets store that index in variable "Key"
Step 3: Now how to check that if B is rotated version of A how ??
This is where modulus function rocks :
for (int i = 0; i< A.length; i++)
{
// here modulus function would check the proper order. Key here is 2 which we recieved from Step 2
int j = [Key+i]%A.length;
if (A[i] != B[j])
{
return false;
}
}
return true;
It's an easy way to tell if a number is even or odd. Just do # mod 2, if it is 0 it is even, 1 it is odd.
Often, in a loop, you want to do something every k'th iteration, where k is 0 < k < n, assuming 0 is the start index and n is the length of the loop.
So, you'd do something like:
int k = 5;
int n = 50;
for(int i = 0;i < n;++i)
{
if(i % k == 0) // true at 0, 5, 10, 15..
{
// do something
}
}
Or, you want to keep something whitin a certain bound. Remember, when you take an arbitrary number mod something, it must produce a value between 0 and that number - 1.