I can't figure out why a calculation is wrong. I'm trying to calculate the percent of time that the user scores over the median score in my Rails app.
controller
my_scores_over_median_count = 6
total_scores_over_median = 8
#my_over_median_percent = (my_scores_over_median_count / total_scores_over_median) * 100
When I add this to my controller:
puts my_scores_over_median_count
puts total_scores_over_median
puts #my_over_median_percent
I get:
6 (correct)
8 (correct)
0 (incorrect)
Can someone please help me understand why I'm getting 0 instead of 75 ? Thank you.
classic integer division problem. Change it to this and see it work:
my_scores_over_median_count = 6.0
total_scores_over_median = 8.0
#my_over_median_percent = (my_scores_over_median_count / total_scores_over_median) * 100
Ruby interprets 6 as a 'Fixnum' class, essentially an integer. We know this by:
1.9.3-p194 :001 > 6.class
=> Fixnum
Dividing any number by a Fixnum causes Ruby to do integer math which means any remainder gets thrown out. So:
1.9.3-p194 :004 > 4 / 5
=> 0
1.9.3-p194 :005 > 4 / 3
=> 1
If you have a variable that is a Fixnum, and you want to convert it to a float to force floating point division, you use the .to_f method:
1.9.3-p194 :007 > x = 4
=> 4
1.9.3-p194 :008 > y = 5
=> 5
1.9.3-p194 :009 > x / y
=> 0
1.9.3-p194 :010 > x.to_f / y
=> 0.8
Related
My issue is, i don't know how to use the output of a function properly. The output contains multiple lines (j = column , i = testresult)
I want to use the output for some other rules in other functions. (eg. if (i) testresult > 5 then something)
I have a function with two loops. The function goes threw every column and test something. This works fine.
def test():
scope = range(10)
scope2 = range(len(df1.columns))
for (j) in scope2:
for (i) in scope:
if df1.iloc[:,[j]].shift(i).loc[selected_week].item() > df1.iloc[:,[j]].shift(i+1).loc[selected_week].item():
i + 1
else:
print(j,i)
break
Output:
test()
1 0
2 3
3 3
4 1
5 0
6 6
7 0
8 1
9 0
10 1
11 1
12 0
13 0
14 0
15 0
I tried to convert it to list, dataframe etc. However, i miss something here.
What is the best way for that?
Thank you!
A fix of your code would be:
def test():
out = []
scope = range(10)
scope2 = range(len(df1.columns))
for j in scope2:
for i in scope:
if df1.iloc[:,[j]].shift(i).loc[selected_week].item() <= df1.iloc[:,[j]].shift(i+1).loc[selected_week].item():
out.append([i, j])
return pd.DataFrame(out)
out = test()
But you probably don't want to use loops as it's slow, please clarify what is your input with a minimal reproducible example and what you are trying to achieve (expected output and logic), we can probably make it a vectorized solution.
This is the R script that I am attempting to recreate using a CASE WHEN statement in SQL:
dat[ ,X_1_7_Spline := pmax(1,pmin(ifelse(is.na(X),1,X),7))]
It seems that this command is telling the parser to return the parallel maxima of a vector containing a conditional statement as long as the value of variable X lies between 1 and the parallel minima of some value and 7 (as long as the value is not null). It then seems to join the new column containing these values back to the original dataset (dat). I am having some troubles representing the "pmax(1,pmin(ifelse(is.na(X),1,X),7))" portion of the code in my SQL query and would appreciate any ideas on how I might be able to do this effectively.
I have something very remedial right now, which I know does not express this above statement properly:
CASE WHEN MAX(IF(ISNOTNULL(X) AND MIN(X)=1 AND MAX(X)=7) then 1 else X end as X_1_7_Spline
Any thoughts/feedback would be greatly appreciated as I am still trying to understand the R script. Thanks in advance for any insight on this issue.
ifelse(is.na(X),1,X) can be translated into SQL's COALESCE(X, 1); and
pmin and pmax logic can be placed in a CASE WHEN (as you've started)
Perhaps this?
CASE WHEN X < 1 THEN 1
WHEN X > 7 THEN 7
ELSE coalesce(X, 1) END as NewX
We don't need to worry about coalesceing the X < 1 or X > 7 because null < 1 does not resolve as true, so it does not accept that case.
Demo in R using sqldf:
library(data.table)
dat <- data.table(X = c(-1,5,9,NA))
dat[, X_1_7_Spline := pmax(1,pmin(ifelse(is.na(X),1,X),7)) ]
sqldf::sqldf("select *, (CASE WHEN X < 1 THEN 1 WHEN X > 7 THEN 7 ELSE coalesce(X,1) END) as NewX from dat")
# X X_1_7_Spline NewX
# 1 -1 1 1
# 2 5 5 5
# 3 9 7 7
# 4 NA 1 1
I have a KY-022 IR module that I can't get to work on my NodeMCU. I've been searching for some code samples in Lua on the internet with no luck. Can anyone point me in the right direction? Any code samples would be greatly appreciate it.
At the moment I have the following code:
local pin = 4
gpio.mode(pin, gpio.OPENDRAIN, gpio.PULLUP)
gpio.trig(pin, "down", function (level, micro)
print(gpio.read(pin), level, micro)
end)
When I press a button on the remote, I get something like this:
0 0 571940709
0 0 571954086
0 0 571955257
1 0 571958694
1 0 571963275
1 0 571969917
0 0 571974347
0 0 571980989
1 0 571983203
1 0 571987709
0 0 571993359
1 0 572000078
0 0 572004508
0 0 572047513
0 0 572058674
So, how do I get from that to figuring out which key was pressed on the remote?
After a month or so i've reopened this project and played around with it some more. As piglet suggested, I started listening for both high and low signals. The data is still very inconsistent and can't get a stable reading.
(And by the way, thanks for the vote-down piglet, that was greatly appreciated. I wish you could have seen my search history before you decided that i'm ignorant)
I'm going to post my curent code maybe somebody can point out what I'm doing wrong here.
local pin = 4
local prevstate = false
local prevmicro = 0
local prevtime = 0
local count = 0
gpio.mode(pin, gpio.INT)
gpio.trig(pin, "both", function (level, micro)
--local state = gpio.read(pin)
local state = level
if (micro - prevmicro) > 90000 then
prevmicro = 0
prevstate = false
count = 0
print("\n#", "st", "lv", "microtime", "timing")
end
if prevstate ~= state then
time = math.floor((micro - prevmicro)/100)
prevstate = state
prevmicro = micro
if time > 3 and time < 1000 then
if prevtime > 80 and prevtime < 100 then
if time > 17 and time < 25 then
print('Repeat')
elseif time > 40 and time < 50 then
print('Start')
end
else
print(count, gpio.read(pin), level, micro, time)
count = count + 1
end
prevtime = time
end
end
end)
and here are some sample readouts from pushing the same button:
# st lv microtime timing
1 1 1 1504559531 16
2 1 0 1504566995 74
3 0 1 1504567523 5
4 1 0 1504573619 60
5 0 1 1504587422 138
6 1 0 1504588011 5
7 1 1 1504604250 162
8 1 0 1504605908 16
9 1 1 1504659929 540
10 1 0 1504662154 22
# st lv microtime timing
1 1 1 1505483535 16
2 1 0 1505491003 74
3 0 1 1505491558 5
4 1 0 1505497627 60
5 0 1 1505511409 137
6 1 0 1505512023 6
7 1 1 1505518186 61
8 1 0 1505527733 95
9 1 0 1505586167 22
10 1 1 1505586720 5
# st lv microtime timing
1 1 1 1507990937 16
2 1 0 1507998405 74
3 0 1 1507998934 5
4 1 0 1508005029 60
5 0 1 1508018811 137
6 1 0 1508019424 6
7 1 1 1508035641 162
8 1 0 1508037322 16
9 1 1 1508091345 540
10 1 0 1508093570 22
As it turns out, the Lua code required for this is actually quite simple.
Where the code above is falling over is actually the print statements. These are extremely expensive and basically, kill your sampling resolution until it's useless.
You are in essence, writing an interrupt service routine, you have a limited time budget before you have to read the next edge change and if it happens before you are done processing, tough luck! So you need to make the ISR as efficient as you can.
In the example below, we listen to the "both" edge event, when one occurs, we simply record an indication of which edge and what duration.
Periodically (using a timer) we print out the contents of the waveform.
This perfectly matches the waveform on my logic analyzer, you still have the challenge of decoding the signal. Though, there are lots of great protocol docs that explain how to take accurate waveform data and use it to determine the signal being sent. I found that a lot of cheap "brand x" remotes appear to be using the NEC protocol, so this might be a good place to start depending on your project.
IR transmission because of its nature is not completely error-free so you may get a spurious edge signal from time to time but the code below is pretty stable and runs quite well in isolation, I have yet to test it when the Microcontroller is under more load than just listening for IR.
It may turn out that using Lua for this purpose is not the best due to the fact that it is an interpreted language (each command issued is parsed and then executed at runtime, this is not at all efficient.) But I will see how far I can get before I decide to write a c module.
local irpin = 2
local lastTimestamp = 0
local waveform = {}
local i = 1
gpio.mode(irpin,gpio.INT)
gpio.trig(irpin, "both", function(level, ts)
onEdge(level, ts)
end)
function onEdge(level, ts)
waveform[i] = level
waveform[i+1] = ts - lastTimestamp
lastTimestamp = ts
i = i+2
end
-- Print out the waveform
function showWaveform ()
if table.getn(waveform) > 65 then
for k,v in pairs(waveform) do
print(k,v)
end
i = 1;
waveform = {}
end
end
tmr.alarm(0, 1000, 1, showWaveform)
print("Ready")
The following code works for my 17 key remote which came with my cheap KY-022 module. I just finished it and haven't had time to clean it up nor to optimize it, so bear with me.
local IR = 2
local lts, i, wave = 0, 0, {}
local keys = {}
keys['10100010000000100000100010101000'] = '1'
keys['10001010000000100010000010101000'] = '2'
keys['10101010000000100000000010101000'] = '3'
keys['10000010000000100010100010101000'] = '4'
keys['10000000000000100010101010101000'] = '5'
keys['10101000000000100000001010101000'] = '6'
keys['10101010000000000000000010101010'] = '7'
keys['10100010001000000000100010001010'] = '8'
keys['10100000100000000000101000101010'] = '9'
keys['10100000101000000000101000001010'] = '0'
keys['10001010001000000010000010001010'] = '*'
keys['10100010100000000000100000101010'] = '#'
keys['10000000101000000010101000001010'] = 'U'
keys['10000000100000000010101000101010'] = 'L'
keys['10001000101000100010001000001000'] = 'R'
keys['10001000001000100010001010001000'] = 'D'
keys['10000010101000000010100000001010'] = 'OK'
local function getKey()
local data = ''
local len = table.getn(wave)
if len >= 70 then
local pkey = 0
local started = false
for k, v in pairs(wave) do
v = math.floor(v/100)
if (pkey == 87 or pkey == 88 or pkey == 89) and (v > 40 and v < 50) then
started = true
end
pkey = v
if started then
if v > 300 then
started = false
end
--this is just to fix some random skipped edges
if (v > 20 and v < 25) or v == 11 then
if v > 20 and v < 25 then
d = 17
else
d = 6
end
v1 = v - d
data = data .. '' .. math.floor(v1/10)
v2 = v - (v - d)
data = data .. '' .. math.floor(v2/10)
else
if v < 40 then
data = data .. '' .. math.floor(v/10)
end
end
end
end
control = data:sub(0, 32)
if control == '00000000000000000101010101010101' then
data = data:sub(32, 63)
print(len, data, keys[data] or '?')
end
end
lts, i, wave = 0, 0, {}
end
local function onEdge(level, ts)
local time = ts - lts
wave[i] = time
i = i + 1
if time > 75000 then
tmr.alarm(0, 350, 0, getKey)
end
lts = ts
end
gpio.mode(IR,gpio.INT)
gpio.trig(IR, "both", onEdge)
I'm putting this asside and start working on some other parts of my project for the moment, but if anyone has any suggestions on how I could improve it, make it faster, smaller even, leave a comment.
PS: for those that are going to complain about not working for them, you need to adjust the if statement values for the started variable based on your remote timings. In my case it's always 88 or 89 followed by 44.
You have to get the sequence sent by the remote for each button.
Record the IR emitter's on-off sequence by logging the time stamps for high-low and low-high transitions.
Note the various patterns for each button you want to use or emulate.
Here's a in-depth tutorial http://www.instructables.com/id/How-To-Useemulate-remotes-with-Arduino-and-Raspber/
You can find this and similar resources using www.google.com
I have a problem with the round method. When I round a number like 1.876 it give me just 1.8 and I need to have 1.80 I use round(2).
My function is like:
#number.attr = (((#number1.to_f*number2.to_f)/100.0)).round(2).to_s
Round can't do that because 1.8 is a standard float but 1.80 is not a standard float.
irb(main)> 1.80.to_f
=> 1.8
The solution is to concatenate your result string with a zero if there is only 1 number after the '.'
1.9.3-p194 :001 > sprintf("%.2f", 1.8)
# => "1.80"
1.9.3-p194 :002 > sprintf("%.2f", 1.876)
# => "1.88"
I understand the Modulus operator in terms of the following expression:
7 % 5
This would return 2 due to the fact that 5 goes into 7 once and then gives the 2 that is left over, however my confusion comes when you reverse this statement to read:
5 % 7
This gives me the value of 5 which confuses me slightly. Although the whole of 7 doesn't go into 5, part of it does so why is there either no remainder or a remainder of positive or negative 2?
If it is calculating the value of 5 based on the fact that 7 doesn't go into 5 at all why is the remainder then not 7 instead of 5?
I feel like there is something I'm missing here in my understanding of the modulus operator.
(This explanation is only for positive numbers since it depends on the language otherwise)
Definition
The Modulus is the remainder of the euclidean division of one number by another. % is called the modulo operation.
For instance, 9 divided by 4 equals 2 but it remains 1. Here, 9 / 4 = 2 and 9 % 4 = 1.
In your example: 5 divided by 7 gives 0 but it remains 5 (5 % 7 == 5).
Calculation
The modulo operation can be calculated using this equation:
a % b = a - floor(a / b) * b
floor(a / b) represents the number of times you can divide a by b
floor(a / b) * b is the amount that was successfully shared entirely
The total (a) minus what was shared equals the remainder of the division
Applied to the last example, this gives:
5 % 7 = 5 - floor(5 / 7) * 7 = 5
Modular Arithmetic
That said, your intuition was that it could be -2 and not 5. Actually, in modular arithmetic, -2 = 5 (mod 7) because it exists k in Z such that 7k - 2 = 5.
You may not have learned modular arithmetic, but you have probably used angles and know that -90° is the same as 270° because it is modulo 360. It's similar, it wraps! So take a circle, and say that its perimeter is 7. Then you read where is 5. And if you try with 10, it should be at 3 because 10 % 7 is 3.
Two Steps Solution.
Some of the answers here are complicated for me to understand. I will try to add one more answer in an attempt to simplify the way how to look at this.
Short Answer:
Example 1:
7 % 5 = 2
Each person should get one pizza slice.
Divide 7 slices on 5 people and every one of the 5 people will get one pizza slice and we will end up with 2 slices (remaining). 7 % 5 equals 2 is because 7 is larger than 5.
Example 2:
5 % 7 = 5
Each person should get one pizza slice
It gives 5 because 5 is less than 7. So by definition, you cannot divide whole 5items on 7 people. So the division doesn't take place at all and you end up with the same amount you started with which is 5.
Programmatic Answer:
The process is basically to ask two questions:
Example A: (7 % 5)
(Q.1) What number to multiply 5 in order to get 7?
Two Conditions: Multiplier starts from `0`. Output result should not exceed `7`.
Let's try:
Multiplier is zero 0 so, 0 x 5 = 0
Still, we are short so we add one (+1) to multiplier.
1 so, 1 x 5 = 5
We did not get 7 yet, so we add one (+1).
2 so, 2 x 5 = 10
Now we exceeded 7. So 2 is not the correct multiplier.
Let's go back one step (where we used 1) and hold in mind the result which is5. Number 5 is the key here.
(Q.2) How much do we need to add to the 5 (the number we just got from step 1) to get 7?
We deduct the two numbers: 7-5 = 2.
So the answer for: 7 % 5 is 2;
Example B: (5 % 7)
1- What number we use to multiply 7 in order to get 5?
Two Conditions: Multiplier starts from `0`. Output result and should not exceed `5`.
Let's try:
0 so, 0 x 7 = 0
We did not get 5 yet, let's try a higher number.
1 so, 1 x 7 = 7
Oh no, we exceeded 5, let's get back to the previous step where we used 0 and got the result 0.
2- How much we need to add to 0 (the number we just got from step 1) in order to reach the value of the number on the left 5?
It's clear that the number is 5. 5-0 = 5
5 % 7 = 5
Hope that helps.
As others have pointed out modulus is based on remainder system.
I think an easier way to think about modulus is what remains after a dividend (number to be divided) has been fully divided by a divisor. So if we think about 5%7, when you divide 5 by 7, 7 can go into 5 only 0 times and when you subtract 0 (7*0) from 5 (just like we learnt back in elementary school), then the remainder would be 5 ( the mod). See the illustration below.
0
______
7) 5
__-0____
5
With the same logic, -5 mod 7 will be -5 ( only 0 7s can go in -5 and -5-0*7 = -5). With the same token -5 mod -7 will also be -5.
A few more interesting cases:
5 mod (-3) = 2 i.e. 5 - (-3*-1)
(-5) mod (-3) = -2 i.e. -5 - (-3*1) = -5+3
It's just about the remainders. Let me show you how
10 % 5=0
9 % 5=4 (because the remainder of 9 when divided by 5 is 4)
8 % 5=3
7 % 5=2
6 % 5=1
5 % 5=0 (because it is fully divisible by 5)
Now we should remember one thing, mod means remainder so
4 % 5=4
but why 4?
because 5 X 0 = 0
so 0 is the nearest multiple which is less than 4
hence 4-0=4
modulus is remainders system.
So 7 % 5 = 2.
5 % 7 = 5
3 % 7 = 3
2 % 7 = 2
1 % 7 = 1
When used inside a function to determine the array index. Is it safe programming ? That is a different question. I guess.
Step 1 : 5/7 = 0.71
Step 2 : Take the left side of the decimal , so we take 0 from 0.71 and multiply by 7
0*7 = 0;
Step # : 5-0 = 5 ; Therefore , 5%7 =5
Modulus operator gives you the result in 'reduced residue system'. For example for mod 5 there are 5 integers counted: 0,1,2,3,4. In fact 19=12=5=-2=-9 (mod 7). The main difference that the answer is given by programming languages by 'reduced residue system'.
lets put it in this way:
actually Modulus operator does the same division but it does not care about the answer , it DOES CARE ABOUT reminder for example if you divide 7 to 5 ,
so , lets me take you through a simple example:
think 5 is a block, then for example we going to have 3 blocks in 15 (WITH Nothing Left) , but when that loginc comes to this kinda numbers {1,3,5,7,9,11,...} , here is where the Modulus comes out , so take that logic that i said before and apply it for 7 , so the answer gonna be that we have 1 block of 5 in 7 => with 2 reminds in our hand! that is the modulus!!!
but you were asking about 5 % 7 , right ?
so take the logic that i said , how many 7 blocks do we have in 5 ???? 0
so the modulus returns 0...
that's it ...
A novel way to find out the remainder is given below
Statement : Remainder is always constant
ex : 26 divided by 7 gives R : 5
This can be found out easily by finding the number that completely divides 26 which is closer to the
divisor and taking the difference of the both
13 is the next number after 7 that completely divides 26 because after 7 comes 8, 9, 10, 11, 12 where none of them divides 26 completely and give remainder 0.
So 13 is the closest number to 7 which divides to give remainder 0.
Now take the difference (13 ~ 7) = 5 which is the temainder.
Note: for this to work divisor should be reduced to its simplest form ex: if 14 is the divisor, 7 has to be chosen to find the closest number dividing the dividend.
As you say, the % sign is used to take the modulus (division remainder).
In w3schools' JavaScript Arithmetic page we can read in the Remainder section what I think to be a great explanation
In arithmetic, the division of two integers produces a quotient and a
remainder.
In mathematics, the result of a modulo operation is the
remainder of an arithmetic division.
So, in your specific case, when you try to divide 7 bananas into a group of 5 bananas, you're able to create 1 group of 5 (quotient) and you'll be left with 2 bananas (remainder).
If 5 bananas into a group of 7, you won't be able to and so you're left with again the 5 bananas (remainder).