I'm trying to print the multiplied fps of a video with ffprobe,
.\ffprobe.exe -v error -select_streams v -of default=noprint_wrappers=1:nokey=1 -show_entries stream=avg_frame_rate -i "$videopath" > rate 2>&1
$script:rate = [IO.File]::ReadAllText(".\rate")
$script:framerate = ($rate/2)
echo "Multiplied = "$framerate
What this is supposed to do is to write the fps into a file, which ffprobe does as a fraction (1/10 in this case), and then multiply it by 2.
Since the variable is read as literal text I can't multiply it by 2, and instead it just gives me this.
Multiplied =
1/10
1/10
Is there any way to make it print Multiplied = 20 instead
Powershell doesn't understand fractions, so you'll need to convert the string into numbers and then do the math.
$rate = "1/10";
# convert the string into numbers
$parts = $rate.Split("/"); # #( "1", "10")
$numerator = [int] $parts[0]; # 1
$denominator = [int] $parts[1]; # 10
# do the math
$framerate = $denominator * 2; # 20
write-host "Multiplied = $framerate"
# Multiplied = 20
I want to calculate the following in Powershell
$x = 100
$y = 25
result = x * y%
Should be really simple, but I can't seem to come up with the right way to calculate the correct result in Powershell
Divide by 100 to get the percentage:
$x = 100
$y = 25
$result = $x * ($y / 100)
Is there an easy way to convert numbers like these
123 -> 123B
1234 -> 1.2K
12345 -> 12.2K
123456 -> 123.4K
1234567 -> 1.2M
12345678 -> 12.3M
123456789 -> 123.4M
...
and ideally also large numbers into days/hours/min. ?
This was discussed on Stackoverflow using Time::Piece. One of the answers comes close to calculating days hours minutes. From what I've read about this question before, I think you can easily code it up like this:
sub dhms {
my $seconds = shift;
my $days = int $seconds / 86400;
$seconds %= 86400;
my $hours = int $seconds / 3600;
$seconds %= 3600;
my $mins = int $seconds / 60;
my $secs = $seconds % 60;
return $days, $hours, $mins, $secs;
}
Update: daxim's answer using DateTime::Format::Duration does this as well
Suffixing is quite simple actually, once we understand the relationship between the letters K, M, G, T and the factors they present:
K = 10^3 = 10^3^1
M = 10^6 = 10^3^2
G = 10^9 = 10^3^3
T = 10^12 = 10^3^4
The next important thing to realize is that 10^0 = 1.
We want to select the largest suffix whose value is smaller than the value we want to transform. To do that, we put the suffixes into an array:
my #suffixes = qw/ B K M G T /;
so that
$suffixes[$i] == 10**3**$i # conceptually
Now it's just a matter of looping over the indices (probably in reverse) and stopping as soon as the $val >= 10**3**$i.
This is a simple number series question, I have numbers in series like
2,4,8,16,32,64,128,256 these numbers are formed by 2,2(square),2(cube) and so on.
Now if I add 2+4+8 = 14. 14 will get only by the addition 2,4 and 8.
so i have 14in my hand now, By some logic i need to get the values which are helped to get 14
Example:
2+4+8 = 14
14(some logic) = 2,4,8.
This is an easy one:
2+4+8=14 ... 14+2=16
2+4+8+16=30 ... 30+2=32
2+4+8+16+32=62 ... 62+2=64
So you just need to add 2 to your sum, then calculate ld (binary logarithm), and then subtract 1. This gives you the number of elements of your sequence you need to add up.
e.g. in PHP:
$target=14;
$count=log($target+2)/log(2)-1;
echo $count;
gives 3, so you have to add the first 3 elements of your sequence to get 14.
Check the following C# code:
x = 14; // In your case
indices = new List<int>();
for (var i = 31; i >= i; i--)
{
var pow = Math.Pow(2, i);
if x - pow >= 0)
{
indices.Add(pow);
x -= pow;
}
}
indices.Reverse();
assuming C:
unsigned int a = 14;
while( a>>=1)
{
printf("%d ", a+1);
}
if this is programming, something like this would suffice:
int myval = 14;
int maxval = 256;
string elements = "";
for (int i = 1; i <= maxval; i*=2)
{
if ((myval & i) != 0)
elements += "," + i.ToString();
}
Use congruency module 2-powers: 14 mod 2 = 0, 14 mod 4 = 2, 14 mod 8 = 6, 14 mod 16 = 14, 14 mod 32 = 14...
The differences of this sequence are the numbers you look for 2 - 0 = 2, 6 - 2 = 4, 14 - 6 = 8, 14 - 14 = 0, ...
It's called the p-adic representation and is formally a bit more difficult to explain, but I hope this gives you an idea for an algorithm.
what i want to do is that i wana show user this
Showing 1 to 10 of 50 entries
Showing 11 to 20 of 50 entries
Showing 21 to 30 of 50 entries
Showing 31 to 40 of 50 entries
Showing 41 to 50 of 50 entries
i have used Zend Paginator in my app lets say
Showing A to B of C entries
I can easily find C which is equal to
$result = $DB->fetchAll($sql);
$total =count($result);
if we see here
$page=$this->_getParam('page',1);
//here we can get the requested page#.
//lets hard code this
$paginator->setItemCountPerPage(10);
$per_page =10;
in my view count($this->paginator) give me total number of pages that is if
if total = 101 = $total
than page = 9 = $page
and paginator = 11 = count($this->paginator)
how can i achieve this but generic mean working with next,previous and so on..
Showing A to B of C entries
Is roughly this:
$page = $paginator->getCurrentPageNumber();
$perPage = $paginator->getItemCountPerPage();
$total = $paginator->getTotalItemCount();
$A = ($page - 1) * $perPage + 1;
$B = min($A + $perPage - 1, $total);
$C = $total;