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Evening all, I am more or less familiar with the order of precedence but this one confuses me and I have an exam tomorrow so yer :d....
Ok so, 6-2/2+5
Is it:
2/2 = 1
1+5 = 6
6-2 = 4
then 4+6 = 10
The part that confused me is once you have in step 3 the value 4 then you just add what ever is left in the equation to the total ?
(2/2) = 1 division is performed first
(6-(1)) = 5 + and - have equal precedence, so we go left to right, hence 6 - (2/2) is done first
(5)+5 = 10 finish it by doing addition
No, it is:
6-2/2+5
= 6 - 2/2 + 5
= (6 - (2/2)) + 5
= (6 - 1 ) + 5
= 5 + 5
= 10
/ and * before + and -. In case of equal precedence it evaluates from left to right.
For more details you might want to consider the documentation (just scroll down to the end).
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i have been coding in python for some time now, and am trying to make a robotic hand in combination with Arduino. I have 2 servo's facing the opposite way (small electric motors) which means that when i rotate them by the same angle, the second one will rotate the opposite way. As the servo can't handle negative integers, i have to come with a solution that basically "reverses" the integer. What i mean by that is that if the number is greater than the middle point (45 in this case) i want it to be smaller, so lets say we have 46 it should be 44 and 47 -> 43 so on. How would i go about creating this? Thanks for reading.
This is just a tricky algorithm:
edge = 45
number = 44
result = edge + (edge - number)
result will be 46
edge = 45
number = 47
result = edge + (edge - number)
result will be 43
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lets say I have 29 apples and put them in boxes where I can fit 10 apples in each box.
when I do print statement I want to see
print(apple)
1 (Full)
2 (Full)
9 (EA)
have any idea how my code should be?
I tried using
full = 10
qty = 29
fullboxes = math.ceil(qty/full)
partials = qty-fullboxes
You can use divmod to get the number of full boxes and the remainder. The rest is just printing.
apples = 29
boxsize = 10
a,b = divmod(apples, boxsize)
for i in range(a):
print(f'{i+1} (Full)', end=' ')
if b:
print(f'{b} (EA)')
output: 1 (Full) 2 (Full) 9 (EA)
output for 30 apples: 1 (Full) 2 (Full) 3 (Full)
Not exactly sure what you are trying to achieve here but
the following code will do what you have described.
d = 29
j = divmod(d, 10)
for i in range(j[0]):
print(f'({i + 1}(full)', end=' ')
else:
print(f'{j[1]}(EA)')
The function divmod(x, y) takes two arguments x and y and gives a tuple of their quotient and remainder.
You can read more about it here
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Why in python the x%y result is x when x is less than y. should the result be zero?
for example when I try 4%5, python give me 4 as a result of and I believe it should be zero
% -> refers to the modulo operator
So you would get the remainder if you use this operator
4%5 will give 4 as 4 is the remainder when you divide 5 by 4
If you wish to get the quotient then you can use // which is used for absolute division
so 4%5 gives 4
and 4//5 gives 0
The % symbol in Python is called the Modulo Operator.
It returns the remainder of dividing the left hand operand by right hand operand. It's used to get the remainder of a division problem.
Example:
5%6
# results in 5
See the long division below
_____0__
6 | 5
| 0
|_____
| 5
Performing 5 / 6 results in reminder of 5, thus 5%6 is 5
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so I'm trying to create as short as possible code for printing an input backwards, and I want to go below 60B. My code takes 79B, and have no idea if it is actually possible to shorten it even more.
tab=[i for i in map(int,input().split())]
print(" ".join(map(str, tab[::-1])))
So when I input:
1 2 3 4 5
I get in output:
5 4 3 2 1
Anybody got idea if it can be even shorter?
print(*input().split()[::-1])
Splits the list by spaces, then reverses and sends to print as a bunch of arguments.
print supplies the separating space automatically.
One liner 41 Bytes
>>> print(''.join(i for i in input()[::-1]))
1 2 3 4 5
5 4 3 2 1
Altough, you could just reverse the input() output w/ subscripting, and print that.
>>> print(input()[::-1])
1 2 3 4 5
5 4 3 2 1
Maybe I don't understand what you're asking, but if it's just to reverse the string as such, then print(input()[::-1]) clocks in at 21B.
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ok, I am studying python in my computing course and i have been challenged with designing a code which validates a GTIN-8 code, I have came to a hurdle that i can not jump. I have looked on how to find the equal or higher multiple of a 10 and i have had no success so far, hope you guys can help me!
Here is a small piece of code, i need to find the equal or higher multiple of 10;
NewNumber = (NewGtin_1 + Gtin_2 + NewGtin_3 + Gtin_4 + NewGtin_5 + Gtin_6 + NewGtin_7)
print (NewNumber)
The easiest way, involving no functions and modules, could be using floor division operator //.
def neareast_higher_multiple_10(number):
return ((number // 10) + 1) * 10
Examples of usage:
>>> neareast_higher_multiple_10(15)
20
>>> neareast_higher_multiple_10(21)
30
>>> neareast_higher_multiple_10(20.1)
30.0
>>> neareast_higher_multiple_10(20)
30
We can also make a generalized version:
def neareast_higher_multiple(number, mult_of):
return ((number // mult_of) + 1) * mult_of
If you need nearest lower multiple, just remove + 1:
def neareast_lower_multiple(number, mult_of):
return (number // mult_of) * mult_of
To find the nearest multiple, you can call both these functions and use that one with a lower difference from the original number.