Here is my code,
self.play(AnimationGroup(
Write(quadSolution[0]),
Write(quadSolution[1]),
Write(quadSolution[2]),
Write(quadSolution[3]),
Write(quadSolution[4]),
Write(quadSolution[5]),
Write(quadSolution[6]),
Write(quadSolution[7]),
Write(quadSolution[8]),
Write(quadSolution[9]),
lag_ratio = 1
),
AnimationGroup(
Animation(Mobject(), run_time = 4),
Write(text2),
run_time = 4,
lag_ratio = 6
)
)
I don't know any better ways to do that list of animations, I would be thankful if anyone could tell me how to do that, but what I really want to know is why the Write(text2) doesn't have a run time of 4 seconds as it should, but instead has a run time of around a second. I don't know what is wrong with what I am doing.
You might have already found the answer you were looking for. But I think the solution to your problem would be to place the run_time parameter within the Write.
So it should look something like Write(text2, run_time=4). I hope this helps.
Related
Apologies if this is a little unclear, or this question has been asked already. It's a little difficult to explain, but I've bolded my question - it's essentially about shortening formulas.
I'm running a payment plan waterfall model. My code works, but, it's er, you know...
IF($K2=Q$1,Assumptions!$E$56*($N2+$O2),IF(AND($K2<Q$1,$M2>Q$1),Assumptions!$E$56*$P2/$L2,0)) + IF(edate($K2,1)=Q$1,Assumptions!$F$56*($N2+$O2),IF(AND(edate($K2,1)<Q$1,edate($M2,1)>Q$1),Assumptions!$F$56*$P2/$L2,0)) + IF(edate($K2,2)=Q$1,Assumptions!$F$56*($N2+$O2),IF(AND(edate($K2,2)<Q$1,edate($M2,2)>Q$1),Assumptions!$F$56*$P2/$L2,0)) + IF(edate($K2,3)=Q$1,Assumptions!$F$56*($N2+$O2),IF(AND(edate($K2,3)<Q$1,edate($M2,3)>Q$1),Assumptions!$F$56*$P2/$L2,0)) + IF(edate($K2,4)=Q$1,Assumptions!$F$56*($N2+$O2),IF(AND(edate($K2,4)<Q$1,edate($M2,4)>Q$1),Assumptions!$F$56*$P2/$L2,0)) + IF(edate($K2,5)=Q$1,Assumptions!$F$56*($N2+$O2),IF(AND(edate($K2,5)<Q$1,edate($M2,5)>Q$1),Assumptions!$F$56*$P2/$L2,0)) + IF(edate($K2,6)=Q$1,Assumptions!$F$56*($N2+$O2),IF(AND(edate($K2,6)<Q$1,edate($M2,6)>Q$1),Assumptions!$F$56*$P2/$L2,0))
...pretty long.
Essentially what's going on is: the assumption is, when we launch a product, we sell say 80% in the first month, and 2.5% every subsequent month until we each 100%.
I'd like the 80% and the 2.5% to be variables (listed as Assumptions!$E$56 and Assumptions!$E$56) here.
Obviously a little long. But noticed after the first IF clause, the subsequent ones are actually identical, the only difference being the number inside edate(__,2), edate(__,3)...
So my question is - can this code be tidied up into some sort of for loop? Python would make it pretty simple to increment over the variable edate(__,i) and sum over i = 1:6.
Sure, there is. Usually looping is emulated by Sequence(N), which makes an array of numbers from [1,N] vertically, which is somewhat like your Python range. Then you can do stuff to it as an ArrayFormula.
In your case, you end up with two terms: your initial term using $E, and all the looped stuff using $F. I see 6 terms, so I will use Sequence(6):
=IF(
$K2=Q$1,
Assumptions!$E$56*($N2+$O2),
IF(
AND(
$K2<Q$1,
$M2>Q$1
),
Assumptions!$E$56*$P2/$L2,
0
)
) + ArrayFormula(SUM(
IF(
edate($K2,SEQUENCE(6))=Q$1,
Assumptions!$F$56*($N2+$O2),
IF(
(edate($K2,SEQUENCE(6))<Q$1)*
(edate($M2,SEQUENCE(6))>Q$1),
Assumptions!$F$56*$P2/$L2,
0
)
)
))
And if you want, you can give your Assumption values names using named ranges.
As you can see on this photo: https://gyazo.com/9ca355a460c5602ec073bcbae701dede i clicked on every block. My problem is that every block doens't have the same Value. I want every block when i click at it to go down 1/10 everytime i click and still give me 1 in my backpack pr. click. Right now i can see that the value of the blocks if not the same.
Here is the script for the blocks: https://gyazo.com/b627ae8559b9035f0fdfe88297b13364 (Think the problem is here)
As you can see on this photo the value is 0,8. And the other one downstairs is 0,1 and so on https://gyazo.com/ac5359b7f6a7e123ad31507f4384323d
The damage value is 0.2 as u see here: https://gyazo.com/b9de1795c27eb151caf8db2d79e9f58a
Here is the script for the shovel:
https://gyazo.com/14b5e675be5644056bda8b24f828ec6a
I would appreciate if u could help me to fix this problem. Tell me if u need more information about this to fix this problem. Hope u understand my problem.
In this line (26) u r generating a random damage value
dmg.value = math.random(1, 10)/10
Did u try to set it on
dmg.value = 1/10
?
This question already has answers here:
Lua for loop reduce i? Weird behavior [duplicate]
(3 answers)
Closed 7 years ago.
im trying this in lua:
for i = 1, 10,1 do
print(i)
i = i+2
end
I would expect the following output:
1,4,7,10
However, it seems like i is getting not affected, so it gives me:
1,2,3,4,5,6,7,8,9,10
Can someone tell my a bit about the background concept and what is the right way to modify the counter variable?
As Colonel Thirty Two said, there is no way to modify a loop variable in Lua. Or rather more to the point, the loop counter in Lua is hidden from you. The variable i in your case is merely a copy of the counter's current value. So changing it does nothing; it will be overwritten by the actual hidden counter every time the loop cycles.
When you write a for loop in Lua, it always means exactly what it says. This is good, since it makes it abundantly clear when you're doing looping over a fixed sequence (whether a count or a set of data) and when you're doing something more complicated.
for is for fixed loops; if you want dynamic looping, you must use a while loop. That way, the reader of the code is aware that looping is not fixed; that it's under your control.
When using a Numeric for loop, you can change the increment by the third value, in your example you set it to 1.
To see what I mean:
for i = 1,10,3 do
print(i)
end
However this isn't always a practical solution, because often times you'll only want to modify the loop variable under specific conditions. When you wish to do this, you can use a while loop (or if you want your code to run at least once, a repeat loop):
local i = 1
while i < 10 do
print(i)
i = i + 1
end
Using a while loop you have full control over the condition, and any variables (be they global or upvalues).
All answers / comments so far only suggested while loops; here's two more ways of working around this problem:
If you always have the same step size, which just isn't 1, you can explicitly give the step size as in for i =start,end,stepdo … end, e.g. for i = 1, 10, 3 do … or for i = 10, 1, -1 do …. If you need varying step sizes, that won't work.
A "problem" with while-loops is that you always have to manually increment your counter and forgetting this in a sub-branch easily leads to infinite loops. I've seen the following pattern a few times:
local diff = 0
for i = 1, n do
i = i+diff
if i > n then break end
-- code here
-- and to change i for the next round, do something like
if some_condition then
diff = diff + 1 -- skip 1 forward
end
end
This way, you cannot forget incrementing i, and you still have the adjusted i available in your code. The deltas are also kept in a separate variable, so scanning this for bugs is relatively easy. (i autoincrements so must work, any assignment to i below the loop body's first line is an error, check whether you are/n't assigning diff, check branches, …)
This is my code:
def is_prime(i)
j = 2
while j < i do
if i % j == 0
return false
end
j += 1
end
true
end
i = (600851475143 / 2)
while i >= 0 do
if (600851475143 % i == 0) && (is_prime(i) == true)
largest_prime = i
break
end
i -= 1
end
puts largest_prime
Why is it not returning anything? Is it too large of a calculation going through all the numbers? Is there a simple way of doing it without utilizing the Ruby prime library(defeats the purpose)?
All the solutions I found online were too advanced for me, does anyone have a solution that a beginner would be able to understand?
"premature optimization is (the root of all) evil". :)
Here you go right away for the (1) biggest, (2) prime, factor. How about finding all the factors, prime or not, and then taking the last (biggest) of them that is prime. When we solve that, we can start optimizing it.
A factor a of a number n is such that there exists some b (we assume a <= b to avoid duplication) that a * b = n. But that means that for a <= b it will also be a*a <= a*b == n.
So, for each b = n/2, n/2-1, ... the potential corresponding factor is known automatically as a = n / b, there's no need to test a for divisibility at all ... and perhaps you can figure out which of as don't have to be tested for primality as well.
Lastly, if p is the smallest prime factor of n, then the prime factors of n are p and all the prime factors of n / p. Right?
Now you can complete the task.
update: you can find more discussion and a pseudocode of sorts here. Also, search for "600851475143" here on Stack Overflow.
I'll address not so much the answer, but how YOU can pursue the answer.
The most elegant troubleshooting approach is to use a debugger to get insight as to what is actually happening: How do I debug Ruby scripts?
That said, I rarely use a debugger -- I just stick in puts here and there to see what's going on.
Start with adding puts "testing #{i}" as the first line inside the loop. While the screen I/O will be a million times slower than a silent calculation, it will at least give you confidence that it's doing what you think it's doing, and perhaps some insight into how long the whole problem will take. Or it may reveal an error, such as the counter not changing, incrementing in the wrong direction, overshooting the break conditional, etc. Basic sanity check stuff.
If that doesn't set off a lightbulb, go deeper and puts inside the if statement. No revelations yet? Next puts inside is_prime(), then inside is_prime()'s loop. You get the idea.
Also, there's no reason in the world to start with 600851475143 during development! 17, 51, 100 and 1024 will work just as well. (And don't forget edge cases like 0, 1, 2, -1 and such, just for fun.) These will all complete before your finger is off the enter key -- or demonstrate that your algorithm truly never returns and send you back to the drawing board.
Use these two approaches and I'm sure you'll find your answers in a minute or two. Good luck!
Do you know you can solve this with one line of code in Ruby?
Prime.prime_division(600851475143).flatten.max
=> 6857
I have a Time object which contains minutes and seconds like "05:37". I like to know a way to convert this to the following formate : "5m 37s".
If we are talking of of a Time object, this is easy:
t=Time.now
puts "#{t.min}m #{t.sec}s"
=>15m 28s
If you have a string
s="05:37"
t=s.split(':').map{|i| i.to_i}
puts "#{t.first}m #{t.last}s"
=>5m 37s
Or shorter
s.split(':').map{|i| i.to_i}.join("m ")+"s"
While I agree with the comment suggesting strftime, you could also just use gsub:
"05:37".gsub(/(\d+):(\d+)/, '\1m \2s')
#=> "05m 37s"
If you don't want the leading zero, it's easy enough to get rid of.
[your_time_object].strftime('%Mm %Ss').
For further detail try this
Nevermind got it:
[(total_response_time.to_i / counter_for_response_time.to_i) / 60 % 60, (total_response_time.to_i / counter_for_response_time.to_i) % 60].map { |t| t.to_s.rjust(2,'0') }.join('m ')
For anyone who want to convert it like this in future.