Why does -103/100 == -2 but 103/100 == 1 in Python? - python

Why does -103/100 == -2 but 103/100 == 1 in Python? I can't seem to understand why.

Integer division always rounds down (towards negative infinity).
Plain or long integer division yields an integer of the same type; the result is that of mathematical division with the floor1 function applied to the result.
http://docs.python.org/2/reference/expressions.html#binary-arithmetic-operations
This allows for the integer division and modulo (remainder, %) operators to connect nicely through the identity x == (x/y)*y + (x%y).
1 floor(x) is the largest integer not greater than x.

Integer division takes (I believe) the floor() of whatever float comes out, more or less.
So that's -2 for the first division and 1 for the second.

Related

What is the C++ equivalent for "//" operator (floor division operator of Python) in C++? [duplicate]

For example,
int result;
result = 125/100;
or
result = 43/100;
Will result always be the floor of the division? What is the defined behavior?
Will result always be the floor of the division? What is the defined behavior?
Not quite. It rounds toward 0, rather than flooring.
6.5.5 Multiplicative operators
6 When integers are divided, the result of the / operator is the algebraic quotient with any
fractional part discarded.88) If the quotient a/b is representable, the expression
(a/b)*b + a%b shall equal a.
and the corresponding footnote:
This is often called ‘‘truncation toward zero’’.
Of course two points to note are:
3 The usual arithmetic conversions are performed on the operands.
and:
5 The result of the / operator is the
quotient from the division of the
first operand by the second; the
result of the % operator is the
remainder. In both operations, if the
value of the second operand is zero,
the behavior is undefined.
[Note: Emphasis mine]
Dirkgently gives an excellent description of integer division in C99, but you should also know that in C89 integer division with a negative operand has an implementation-defined direction.
From the ANSI C draft (3.3.5):
If either operand is negative, whether the result of the / operator is the largest integer less than the algebraic quotient or the smallest integer greater than the algebraic quotient is implementation-defined, as is the sign of the result of the % operator. If the quotient a/b is representable, the expression (a/b)*b + a%b shall equal a.
So watch out with negative numbers when you are stuck with a C89 compiler.
It's a fun fact that C99 chose truncation towards zero because that was how FORTRAN did it. See this message on comp.std.c.
Yes, the result is always truncated towards zero. It will round towards the smallest absolute value.
-5 / 2 = -2
5 / 2 = 2
For unsigned and non-negative signed values, this is the same as floor (rounding towards -Infinity).
Where the result is negative, C truncates towards 0 rather than flooring - I learnt this reading about why Python integer division always floors here: Why Python's Integer Division Floors
Will result always be the floor of the division?
No. The result varies, but variation happens only for negative values.
What is the defined behavior?
To make it clear floor rounds towards negative infinity,while integer division rounds towards zero (truncates)
For positive values they are the same
int integerDivisionResultPositive= 125/100;//= 1
double flooringResultPositive= floor(125.0/100.0);//=1.0
For negative value this is different
int integerDivisionResultNegative= -125/100;//=-1
double flooringResultNegative= floor(-125.0/100.0);//=-2.0
I know people have answered your question but in layman terms:
5 / 2 = 2 //since both 5 and 2 are integers and integers division always truncates decimals
5.0 / 2 or 5 / 2.0 or 5.0 /2.0 = 2.5 //here either 5 or 2 or both has decimal hence the quotient you will get will be in decimal.

Difference between modulus (%) and floor division(//) in NumPy?

Recently, I read a book on Numpy which mentions different types of ufuncs, where I was encountered with two different ufuncs, namely 'modulus', denoted by % symbol and 'floor division' //. Can someone explain the difference between them and why two operators are provided to do the the same thing (display reminder of division, according to me)? Please correct, if I am wrong.
Both are valid mathematical functions with different results.
modulus
The modulus-function computes the remainder of a division, which is the "leftover" of an integral division.
floor
The floor-function provides the lower-bound of an integral division. The upper-bound is computed by the ceil function. (Basically speaking, the floor-function cuts off all decimals).
a=5
b=2
print(a%b) # Prints 1 as leftover
print(a//b) # Prints 2, since 5/2=2.5 and the decimal is cut off
print(a - (a//b)*b) # Prints 1, this is the modulo, calculated by the floor function
Assume a= 10, b = 6
a%b will give you the remainder, that is 4
a//b will give you the quotient, that is 1
The relationship between floor division and modulus may be helpful for understanding this:
denominator = 6 # replace with any integer
numerator = 10 # replace with any integer
floor_division_result = numerator//denominator
modulo_result = numerator % denominator
# this assertion will always be true:
assert denominator * floor_division_result + modulo_result == numerator
One way to think of the difference between modulus and floor division is to use an analogy of a clock face
Imagine a hand on this clock. It moves around and points at a number. Let's say a cycle starts (and ends) when the hand points at 12. In this context
modulus, %, is where the hand is pointing right now
floor division is the number of times the hand has completed a cycle
Example 1
In the first cycle when the hand points at 4 the modulus is 4 but the floor division is 0.
4 % 12 # returns 4
4 // 12 # returns 0
Example 2
In the second cycle when the hand points at 4 the modulus is still 4 but the floor division is now 1
16 % 12 # returns 4
16 // 12 # returns 1

OCaml mod function returns different result compared with %

The modulo function in OCaml mod return results different when compared with the modulo operator in python.
OCaml:
# -1 mod 4
- : int = -1
Python:
>>> -1 % 4
3
Why are the result different?.
Is there any standard module function that operate as % in OCaml?.
Python is a bit different in its usage of the % operator, which really computes the modulo of two values, whereas other programming languages compute the remainder with the same operator. For example, the distinction is clear in Scheme:
(modulo -1 4) ; modulo
=> 3
(remainder -1 4) ; remainder
=> -1
In Python:
-1 % 4 # modulo
=> 3
math.fmod(-1, 4) # remainder
=> -1
But in OCaml, there's only mod (which computes the integer remainder), according to this table and as stated in the documentation:
-1 mod 4 (* remainder *)
=> -1
Of course, you can implement your own modulo operation in terms of remainder, like this:
let modulo x y =
let result = x mod y in
if result >= 0 then result
else result + y
The semantics of modulo are linked with the semantics of integer division (generally, if Q is the result of integer division a / b, and R is the result of a mod b, then a = Q * b + R must always be true), so different methods of rounding the result of integer division to an integer will produce different results for modulo.
The Wikipedia article Modulo operation has a very extensive table about how different languages handle modulo. There are a few common ways:
In languages like C, Java, OCaml, and many others, integer division rounds towards 0, which causes the result of modulo to always have the same sign as the dividend. In this case, the dividend (-1) is negative, so the modulo is also negative (-1).
In languages like Python, Ruby, and many others, integer division always rounds down (towards negative infinity), which causes the result of modulo to always have the same sign as the divisor. In this case, the divisor (4) is positive, so the modulo is also positive (3).

Floor division with negative number

The expression 6 // 4 yields 1, where floor division produces the whole number after dividing a number.
But with a negative number, why does -6 // 4 return -2?
The // operator explicitly floors the result. Quoting the Binary arithmetic operations documentation:
the result is that of mathematical division with the ‘floor’ function applied to the result.
Flooring is not the same thing as rounding to 0; flooring always moves to the lower integer value. See the math.floor() function:
Return the floor of x, the largest integer less than or equal to x.
For -6 // 4, first the result of -6 / 4 is calculated, so -1.5. Flooring then moves to the lower integer value, so -2.
If you want to round towards zero instead, you'll have to do so explicitly; you could do this with the int() function on true division:
>>> int(-6 / 4)
-1
int() removes the decimal portion, so always rounds towards zero instead.
Floor division will also round down to the next lowest number, not the next lowest absolute value.
6 // 4 = 1.5, which rounds down to 1, and up to 2.
-6 // 4 = -1.5, which rounds down to -2, and up to -1.
// in Python is a "floor division" operator. That means that the result of such division is the floor of the result of regular division (performed with / operator).
The floor of the given number is the biggest integer smaller than the this number. For example
7 / 2 = 3.5 so 7 // 2 = floor of 3.5 = 3.
For negative numbers it is less intuitive: -7 / 2 = -3.5, so -7 // 2 = floor of -3.5 = -4. Similarly -1 // 10 = floor of -0.1 = -1.
// is defined to do the same thing as math.floor(): return the largest integer value less than or equal to the floating-point result. Zero is not less than or equal to -0.1.
A useful way to understand why floor division // yields the results it does for negative values is to consider that it complements the modulo, or remainder, % operator, which in Python is defined to always return a non-negative number.
5/3 is equivalent to 1 remainder 2
i.e.
5//3 = 1
5%3 = 2
But
-5/3 = -2
-5%3 = 1
Or
-2 + 1/3rd which is -1.6667 (ish)
It can seem strange, but it ensures results such as
-2,-2,-2,-1,-1,-1,0,0,0,1,1,1,2,2,2,3,3,3 etc. when generating sequences.

What is the result of % in Python?

What does the % in a calculation? I can't seem to work out what it does.
Does it work out a percent of the calculation for example: 4 % 2 is apparently equal to 0. How?
The % (modulo) operator yields the remainder from the division of the first argument by the second. The numeric arguments are first converted to a common type. A zero right argument raises the ZeroDivisionError exception. The arguments may be floating point numbers, e.g., 3.14%0.7 equals 0.34 (since 3.14 equals 4*0.7 + 0.34.) The modulo operator always yields a result with the same sign as its second operand (or zero); the absolute value of the result is strictly smaller than the absolute value of the second operand [2].
Taken from http://docs.python.org/reference/expressions.html
Example 1:
6%2 evaluates to 0 because there's no remainder if 6 is divided by 2 ( 3 times ).
Example 2: 7%2 evaluates to 1 because there's a remainder of 1 when 7 is divided by 2 ( 3 times ).
So to summarise that, it returns the remainder of a division operation, or 0 if there is no remainder. So 6%2 means find the remainder of 6 divided by 2.
Somewhat off topic, the % is also used in string formatting operations like %= to substitute values into a string:
>>> x = 'abc_%(key)s_'
>>> x %= {'key':'value'}
>>> x
'abc_value_'
Again, off topic, but it seems to be a little documented feature which took me awhile to track down, and I thought it was related to Pythons modulo calculation for which this SO page ranks highly.
An expression like x % y evaluates to the remainder of x ÷ y - well, technically it is "modulus" instead of "reminder" so results may be different if you are comparing with other languages where % is the remainder operator. There are some subtle differences (if you are interested in the practical consequences see also "Why Python's Integer Division Floors" bellow).
Precedence is the same as operators / (division) and * (multiplication).
>>> 9 / 2
4
>>> 9 % 2
1
9 divided by 2 is equal to 4.
4 times 2 is 8
9 minus 8 is 1 - the remainder.
Python gotcha: depending on the Python version you are using, % is also the (deprecated) string interpolation operator, so watch out if you are coming from a language with automatic type casting (like PHP or JS) where an expression like '12' % 2 + 3 is legal: in Python it will result in TypeError: not all arguments converted during string formatting which probably will be pretty confusing for you.
[update for Python 3]
User n00p comments:
9/2 is 4.5 in python. You have to do integer division like so: 9//2 if you want python to tell you how many whole objects is left after division(4).
To be precise, integer division used to be the default in Python 2 (mind you, this answer is older than my boy who is already in school and at the time 2.x were mainstream):
$ python2.7
Python 2.7.10 (default, Oct 6 2017, 22:29:07)
[GCC 4.2.1 Compatible Apple LLVM 9.0.0 (clang-900.0.31)] on darwin
Type "help", "copyright", "credits" or "license" for more information.
>>> 9 / 2
4
>>> 9 // 2
4
>>> 9 % 2
1
In modern Python 9 / 2 results 4.5 indeed:
$ python3.6
Python 3.6.1 (default, Apr 27 2017, 00:15:59)
[GCC 4.2.1 Compatible Apple LLVM 8.1.0 (clang-802.0.42)] on darwin
Type "help", "copyright", "credits" or "license" for more information.
>>> 9 / 2
4.5
>>> 9 // 2
4
>>> 9 % 2
1
[update]
User dahiya_boy asked in the comment session:
Q. Can you please explain why -11 % 5 = 4 - dahiya_boy
This is weird, right? If you try this in JavaScript:
> -11 % 5
-1
This is because in JavaScript % is the "remainder" operator while in Python it is the "modulus" (clock math) operator.
You can get the explanation directly from GvR:
Edit - dahiya_boy
In Java and iOS -11 % 5 = -1 whereas in python and ruby -11 % 5 = 4.
Well half of the reason is explained by the Paulo Scardine, and rest of the explanation is below here
In Java and iOS, % gives the remainder that means if you divide 11 % 5 gives Quotient = 2 and remainder = 1 and -11 % 5 gives Quotient = -2 and remainder = -1.
Sample code in swift iOS.
But when we talk about in python its gives clock modulus. And its work with below formula
mod(a,n) = a - {n * Floor(a/n)}
Thats means,
mod(11,5) = 11 - {5 * Floor(11/5)} => 11 - {5 * 2}
So, mod(11,5) = 1
And
mod(-11,5) = -11 - 5 * Floor(-11/5) => -11 - {5 * (-3)}
So, mod(-11,5) = 4
Sample code in python 3.0.
Why Python's Integer Division Floors
I was asked (again) today to explain why integer division in Python returns the floor of the result instead of truncating towards zero like C.
For positive numbers, there's no surprise:
>>> 5//2
2
But if one of the operands is negative, the result is floored, i.e., rounded away from zero (towards negative infinity):
>>> -5//2
-3
>>> 5//-2
-3
This disturbs some people, but there is a good mathematical reason. The integer division operation (//) and its sibling, the modulo operation (%), go together and satisfy a nice mathematical relationship (all variables are integers):
a/b = q with remainder r
such that
b*q + r = a and 0 <= r < b
(assuming a and b are >= 0).
If you want the relationship to extend for negative a (keeping b positive), you have two choices: if you truncate q towards zero, r will become negative, so that the invariant changes to 0 <= abs(r) < otherwise, you can floor q towards negative infinity, and the invariant remains 0 <= r < b. [update: fixed this para]
In mathematical number theory, mathematicians always prefer the latter choice (see e.g. Wikipedia). For Python, I made the same choice because there are some interesting applications of the modulo operation where the sign of a is uninteresting. Consider taking a POSIX timestamp (seconds since the start of 1970) and turning it into the time of day. Since there are 24*3600 = 86400 seconds in a day, this calculation is simply t % 86400. But if we were to express times before 1970 using negative numbers, the "truncate towards zero" rule would give a meaningless result! Using the floor rule it all works out fine.
Other applications I've thought of are computations of pixel positions in computer graphics. I'm sure there are more.
For negative b, by the way, everything just flips, and the invariant becomes:
0 >= r > b.
So why doesn't C do it this way? Probably the hardware didn't do this at the time C was designed. And the hardware probably didn't do it this way because in the oldest hardware, negative numbers were represented as "sign + magnitude" rather than the two's complement representation used these days (at least for integers). My first computer was a Control Data mainframe and it used one's complement for integers as well as floats. A pattern of 60 ones meant negative zero!
Tim Peters, who knows where all Python's floating point skeletons are buried, has expressed some worry about my desire to extend these rules to floating point modulo. He's probably right; the truncate-towards-negative-infinity rule can cause precision loss for x%1.0 when x is a very small negative number. But that's not enough for me to break integer modulo, and // is tightly coupled to that.
PS. Note that I am using // instead of / -- this is Python 3 syntax, and also allowed in Python 2 to emphasize that you know you are invoking integer division. The / operator in Python 2 is ambiguous, since it returns a different result for two integer operands than for an int and a float or two floats. But that's a totally separate story; see PEP 238.
Posted by Guido van Rossum at 9:49 AM
The modulus is a mathematical operation, sometimes described as "clock arithmetic." I find that describing it as simply a remainder is misleading and confusing because it masks the real reason it is used so much in computer science. It really is used to wrap around cycles.
Think of a clock: Suppose you look at a clock in "military" time, where the range of times goes from 0:00 - 23.59. Now if you wanted something to happen every day at midnight, you would want the current time mod 24 to be zero:
if (hour % 24 == 0):
You can think of all hours in history wrapping around a circle of 24 hours over and over and the current hour of the day is that infinitely long number mod 24. It is a much more profound concept than just a remainder, it is a mathematical way to deal with cycles and it is very important in computer science. It is also used to wrap around arrays, allowing you to increase the index and use the modulus to wrap back to the beginning after you reach the end of the array.
Python - Basic Operators
http://www.tutorialspoint.com/python/python_basic_operators.htm
Modulus - Divides left hand operand by right hand operand and returns remainder
a = 10 and b = 20
b % a = 0
In most languages % is used for modulus. Python is no exception.
% Modulo operator can be also used for printing strings (Just like in C) as defined on Google https://developers.google.com/edu/python/strings.
# % operator
text = "%d little pigs come out or I'll %s and %s and %s" % (3, 'huff', 'puff', 'blow down')
This seems to bit off topic but It will certainly help someone.
Also, there is a useful built-in function called divmod:
divmod(a, b)
Take two (non complex) numbers as arguments and return a pair of numbers
consisting of their quotient and
remainder when using long division.
x % y calculates the remainder of the division x divided by y where the quotient is an integer. The remainder has the sign of y.
On Python 3 the calculation yields 6.75; this is because the / does a true division, not integer division like (by default) on Python 2. On Python 2 1 / 4 gives 0, as the result is rounded down.
The integer division can be done on Python 3 too, with // operator, thus to get the 7 as a result, you can execute:
3 + 2 + 1 - 5 + 4 % 2 - 1 // 4 + 6
Also, you can get the Python style division on Python 2, by just adding the line
from __future__ import division
as the first source code line in each source file.
Modulus operator, it is used for remainder division on integers, typically, but in Python can be used for floating point numbers.
http://docs.python.org/reference/expressions.html
The % (modulo) operator yields the remainder from the division of the first argument by the second. The numeric arguments are first converted to a common type. A zero right argument raises the ZeroDivisionError exception. The arguments may be floating point numbers, e.g., 3.14%0.7 equals 0.34 (since 3.14 equals 4*0.7 + 0.34.) The modulo operator always yields a result with the same sign as its second operand (or zero); the absolute value of the result is strictly smaller than the absolute value of the second operand [2].
It's a modulo operation, except when it's an old-fashioned C-style string formatting operator, not a modulo operation. See here for details. You'll see a lot of this in existing code.
It was hard for me to readily find specific use cases for the use of % online ,e.g. why does doing fractional modulus division or negative modulus division result in the answer that it does. Hope this helps clarify questions like this:
Modulus Division In General:
Modulus division returns the remainder of a mathematical division operation. It is does it as follows:
Say we have a dividend of 5 and divisor of 2, the following division operation would be (equated to x):
dividend = 5
divisor = 2
x = 5/2
The first step in the modulus calculation is to conduct integer division:
x_int = 5 // 2 ( integer division in python uses double slash)
x_int = 2
Next, the output of x_int is multiplied by the divisor:
x_mult = x_int * divisor
x_mult = 4
Lastly, the dividend is subtracted from the x_mult
dividend - x_mult = 1
The modulus operation ,therefore, returns 1:
5 % 2 = 1
Application to apply the modulus to a fraction
Example: 2 % 5
The calculation of the modulus when applied to a fraction is the same as above; however, it is important to note that the integer division will result in a value of zero when the divisor is larger than the dividend:
dividend = 2
divisor = 5
The integer division results in 0 whereas the; therefore, when step 3 above is performed, the value of the dividend is carried through (subtracted from zero):
dividend - 0 = 2 —> 2 % 5 = 2
Application to apply the modulus to a negative
Floor division occurs in which the value of the integer division is rounded down to the lowest integer value:
import math
x = -1.1
math.floor(-1.1) = -2
y = 1.1
math.floor = 1
Therefore, when you do integer division you may get a different outcome than you expect!
Applying the steps above on the following dividend and divisor illustrates the modulus concept:
dividend: -5
divisor: 2
Step 1: Apply integer division
x_int = -5 // 2 = -3
Step 2: Multiply the result of the integer division by the divisor
x_mult = x_int * 2 = -6
Step 3: Subtract the dividend from the multiplied variable, notice the double negative.
dividend - x_mult = -5 -(-6) = 1
Therefore:
-5 % 2 = 1
Be aware that
(3 +2 + 1 - 5) + (4 % 2) - (1/4) + 6
even with the brackets results in 6.75 instead of 7 if calculated in Python 3.4.
And the '/' operator is not that easy to understand, too (python2.7): try...
- 1/4
1 - 1/4
This is a bit off-topic here, but should be considered when evaluating the above expression :)
The % (modulo) operator yields the remainder from the division of the first argument by the second. The numeric arguments are first converted to a common type.
3 + 2 + 1 - 5 + 4 % 2 - 1 / 4 + 6 = 7
This is based on operator precedence.
% is modulo. 3 % 2 = 1, 4 % 2 = 0
/ is (an integer in this case) division, so:
3 + 2 + 1 - 5 + 4 % 2 - 1 / 4 + 6
1 + 4%2 - 1/4 + 6
1 + 0 - 0 + 6
7
It's a modulo operation
http://en.wikipedia.org/wiki/Modulo_operation
http://docs.python.org/reference/expressions.html
So with order of operations, that works out to
(3+2+1-5) + (4%2) - (1/4) + 6
(1) + (0) - (0) + 6
7
The 1/4=0 because we're doing integer math here.
It is, as in many C-like languages, the remainder or modulo operation. See the documentation for numeric types — int, float, long, complex.
Modulus - Divides left hand operand by right hand operand and returns remainder.
If it helps:
1:0> 2%6
=> 2
2:0> 8%6
=> 2
3:0> 2%6 == 8%6
=> true
... and so on.
I have found that the easiest way to grasp the modulus operator (%) is through long division. It is the remainder and can be useful in determining a number to be even or odd:
4%2 = 0
2
2|4
-4
0
11%3 = 2
3
3|11
-9
2
def absolute(c):
if c>=0:
return c
else:
return c*-1
x=int(input("Enter the value:"))
a=absolute(x)
print(a)

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