There is a simple trick to convert a number to 1 or -1.
Just raise it to the power of 0.
So:
4^0 = 1
-4^0 = -1
However, in AS3:
Math.pow( 4, 0); // = 1
Math.pow(-4, 0); // = 1
Is there a way to get the right answer without an if else?
This could be done bitwise.
Given the number n (avg time: 0.0065ms):
1 + 2 * (n >> 31);
Or slightly slower (avg time: 0.0095ms):
(n < 0 && -1) || 1;
However, Marty's solution is the fastest (avg time: 0.0055ms)
n < 0 ? -1 : 1;
Not sure if without an if/else includes the ternary operator in your eyes, but if not:
// Where x is your input.
var r:int = x < 0 ? -1 : 1;
Will be more efficient than Math.pow() anyway.
Related
I am trying to list down all the prime numbers up till a specific number e.g. 1000. The code gets slower as the number increase. I am pretty sure it is because of the for loop where (number -1) is checked by all the prime_factors. Need some advise how I can decrease the processing time of the code for larger numbers. Thanks
import time
t0 = time.time()
prime_list = [2]
number = 0
is_not_prime = 0
count = 0
while number < 1000:
print(number)
for i in range (2,number):
count = 0
if (number%i) == 0:
is_not_prime = 1
if is_not_prime == 1:
for j in range (0,len(prime_list)):
if(number-1)%prime_list[j] != 0:
count += 1
if count == len(prime_list):
prime_list.append(number-1)
is_not_prime = 0
count = 0
break
number += 1
print(prime_list)
t1 = time.time()
total = t1-t0
print(total)
Your solution, on top of being confusing, is very inefficient - O(n^3). Please, use the Sieve of Eratosthenes. Also, learn how to use booleans.
Something like this (not optimal, just a mock-up). Essentially, you start with a list of all numbers, 1-1000. Then, you remove ones that are the multiple of something.
amount = 1000
numbers = range(1, amount)
i = 1
while i < len(numbers):
n = i + 1
while n < len(numbers):
if numbers[n] % numbers[i] == 0:
numbers.pop(n)
else:
n += 1
i += 1
print(numbers)
Finally, I was able to answer because your question isn't language-specific, but please tag the question with the language you're using in the example.
I have a series of values that are each being stored as UInt16. Each of these numbers represents a bitmask - these numbers are commands that have been sent to a microprocessor telling it which pins to set high or low. I would like to parse this arrow of commands to find out which pins were being set high each time in such a way that is easier to analyse later.
Consider the example value 0x3c00, which in decimal is 15360 and in binary is 0011110000000000. Currently I have the following function
function read_message(hex_rep)
return findall.(x -> x .== '1',bitstring(hex_rep))
end
Which gets called on every element of the array of UInt16. Is there a better/more efficient way of doing this?
The best approach probably depends on how you want to handle vectors of hex-values. But here's an approach for processing a single hex which is much faster than the one in the OP:
function readmsg(x::UInt16)
N = count_ones(x)
inds = Vector{Int}(undef, N)
if N == 0
return inds
end
k = trailing_zeros(x)
x >>= k + 1
i = N - 1
inds[N] = n = 16 - k
while i >= 1
(x, r) = divrem(x, 0x2)
n -= 1
if r == 1
inds[i] = n
i -= 1
end
end
return inds
end
I can suggest padding your vector into a Vector{UInt64} and use that to manually construct a BitVector. The following should mostly work (even for input element types other than UInt16), but I haven't taken into account specific endianness you might want to respect:
julia> function read_messages(msgs)
bytes = reinterpret(UInt8, msgs)
N = length(bytes)
nchunks, remaining = divrem(N, sizeof(UInt64))
padded_bytes = zeros(UInt8, sizeof(UInt64) * cld(N, sizeof(UInt64)))
copyto!(padded_bytes, bytes)
b = BitVector(undef, N * 8)
b.chunks = reinterpret(UInt64, padded_bytes)
return b
end
read_messages (generic function with 1 method)
julia> msgs
2-element Vector{UInt16}:
0x3c00
0x8000
julia> read_messages(msgs)
32-element BitVector:
0
0
0
0
0
0
0
0
0
⋮
0
0
0
0
0
0
0
1
julia> read_messages(msgs) |> findall
5-element Vector{Int64}:
11
12
13
14
32
julia> bitstring.(msgs)
2-element Vector{String}:
"0011110000000000"
"1000000000000000"
(Getting rid of the unnecessary allocation of the undef bit vector would require some black magic, I belive.)
My function and rounding to nearest even number
function y = rndeven(x)
if x<=1
y=2;
else
y = 2*floor(x);
end
endfunction
When I run it I get:
cc=[0:3]'
both=[cc,rndeven(cc)]
0 0
1 2
2 4
3 6
What I'm trying to get as the Result:
0 2
1 2
2 2
3 4
You can use the modulo 2 to find whether a number is even. If it isn't this will return 1, so just add 1 to this number to find the nearest (larger) even number:
function y = rndeven(x)
x = floor(x);
x(x <= 1) = 2;
y = mod(x,2)+x;
end
This works for any array, order of elements does not matter.
You could also check if it is dividable by 2 if you don't want to use the mod function. The pseudo code would be something like this:
while(x % 2 != 0) x = x + 1
return x
I want to write a function, which will accept parameter and it will print with a condition (It's output will depend on the input). My program is giving key error. I am looking for an output like:
This number is less than 0 and it's spelling is one hundred
thirteen
and my code is:
def word(num):
d1= {0:'Zero',1:'One',2:'Two',3:'Three',4:'Four',5:'Five',6:'Six',7:'Seven',8:'Eight',9:'Nine',10:'Ten',11:'Eleven',12:'Twelve',13:'Thirteen',14:'Fourteen',15:'Fifteen',16:'Sixteen',17:'Seventeen',18:'Eighteen',19:'Ninteen',20:'Twenty',30:'Thirty',40:'Fourty',50:'Fifty',60:'Sixty',70:'Seventy',80:'Eighty',90:'Ninty'}
if (num<20):
return d1[num]
if (num<100):
if num % 10 == 0:
return d1[num]
else:
return d1[num // 10 * 10] + ' ' + d1[num % 10]
if (num < 0):
return "This number is less than 0 and it's spelling is" + word(num)
print (word(- 100))
print (word(13))
You should have your wide condition before narrow condition.
In your code, you have 3 if conditions, num < 20, num < 100, num < 0, which actually is 0 <= num < 20, 20 <= num < 100, num < 0. The last condition is the widest, but you move it incorrectly at the bottom.
Sort your condition train into the order num < 0, num < 20, num < 100 may fix this issue.
Update: You can't use word[num] in your num < 0 block. I can't understand what "is one hundred" in your expected output. Is it a hardcoded text? Then hardcode it, for example:
def word(num):
d1= {0:'Zero',1:'One',2:'Two',3:'Three',4:'Four',5:'Five',6:'Six',7:'Seven',8:'Eight',9:'Nine',10:'Ten',11:'Eleven',12:'Twelve',13:'Thirteen',14:'Fourteen',15:'Fifteen',16:'Sixteen',17:'Seventeen',18:'Eighteen',19:'Ninteen',20:'Twenty',30:'Thirty',40:'Fourty',50:'Fifty',60:'Sixty',70:'Seventy',80:'Eighty',90:'Ninty'}
if num < 0:
return "This number is less than 0 and it's spelling is one hundred"
if num < 20:
return d1[num]
if num < 100:
if num % 10 == 0:
return d1[num]
else:
return d1[num // 10 * 10] + ' ' + d1[num % 10]
print(word(-100))
print(word(13))
I have a function like this:
y=-2 with x<=0
y=-2+3x^2 with 0=1
I need to compute this function on each element of the 1D matrix, without using a loop.
I thought it was possibile defining a function like this one:
function y= foo(x)
if x<=0
y=-2;
elseif x>=1
y=1;
else
y= -2+3*x.^2;
end
end
But this just produces a single result, how to operate on all elements? I know the . operator, but how to access the single element inside an if?
function b = helper(s)
if s<=0
b=-2;
elseif s>=1
b=1;
else
b= -2+3*s^2;
end
end
Then simply call
arrayfun(#helper, x)
to produce the behaviour you want of your function foo.
Another approach which doesn't need arrayfun() would be to multiply by the conditions:
y = -2*(x <= 0) + (-2+3*x.^2).*(x < 1).*(x > 0) + (x >= 1)
which you could also make a function. This will accept vector inputs for x e.g.
x = [1 4 0 -1 0.5];
y = -2*(x <= 0) + (-2+3*x.^2).*(x < 1).*(x > 0) + (x >= 1)
outputs
y =
1.0000 1.0000 -2.0000 -2.0000 -1.2500